(a)Gas laws
- Matter is made up of small particle in accordance to Kinetic Theory of matter:
Naturally, there are basically three states of matter: Solid, Liquid and gas:
(i)A solid is made up of particles which are very closely packed with a definite/fixed shape and fixed/definite volume /occupies definite space. It has a very high density.
(ii) A liquid is made up of particles which have some degree of freedom. It thus has no definite/fixed shape. It takes the shape of the container it is put. A liquid has fixed/definite volume/occupies definite space.
(iii)A gas is made up of particles free from each other. It thus has no definite /fixed shape. It takes the shape of the container it is put. It has no fixed/definite volume/occupies every space in a container.
2.Gases are affected by physical conditions. There are two physical conditions:
(i)Temperature
(ii)Pressure
- The SI unit of temperature is Kelvin(K).
Degrees Celsius/Centigrade(oC) are also used.
The two units can be interconverted from the relationship:
oC + 273= K
K -273 = oC
Practice examples
- Convert the following into Kelvin.
(i) O oC
oC + 273 = K substituting : O oC + 273 = 273 K
(ii) -273 oC
oC + 273 = K substituting : -273oC + 273 = 0 K
(iii) 25 oC
oC + 273 = K substituting : 25 oC + 273 = 298 K
(iv) 100 oC
oC + 273 = K substituting : 100 oC + 273 = 373 K
- Convert the following into degrees Celsius/Centigrade(oC).
(i) 10 K
K -273 = oC substituting: 10 – 273 = -263 oC
(ii) (i) 1 K
K -273 = oC substituting: 1 – 273 = -272 oC
(iii) 110 K
K -273 = oC substituting: 110 – 273 = -163 oC
(iv) -24 K
K -273 = oC substituting: -24 – 273 = -297 oC
The standard temperature is 273K = 0 oC.
The room temperature is assumed to be 298K = 25oC
- The SI unit of pressure is Pascal(Pa) / Newton per metre squared (Nm-2) . Millimeters’ of mercury(mmHg) ,centimeters of mercury(cmHg) and atmospheres are also commonly used.
The units are not interconvertible but Pascals(Pa) are equal to Newton per metre squared(Nm-2).
The standard pressure is the atmospheric pressure.
Atmospheric pressure is equal to about:
(i)101325 Pa
(ii)101325 Nm-2
(iii)760 mmHg
(iv)76 cmHg
(v)one atmosphere.
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CHEMISTRY NOTES F1-4: LATEST NOTES
- Molecules of gases are always in continuous random motion at high speed. This motion is affected by the physical conditions of temperature and pressure.
Physical conditions change the volume occupied by gases in a closed system.
The effect of physical conditions of temperature and pressure was investigated and expressed in both Boyles and Charles laws.
- Boyles law states that
“the volume of a fixed mass of a gas is inversely proportional to the pressure at constant/fixed temperature ”
Mathematically:
Volume α 1 (Fixed /constant Temperature)
Pressure
V α 1 (Fixed /constant T) ie PV = Constant(k)
P
From Boyles law , an increase in pressure of a gas cause a decrease in volume. i.e doubling the pressure cause the volume to be halved.
Graphically therefore a plot of volume(V) against pressure (P) produces a curve.
V
P
Graphically a plot of volume(V) against inverse/reciprocal of pressure (1/p) produces a straight line
V
1/P
For two gases then P1 V1 = P2 V2
P1 = Pressure of gas 1
V1 = Volume of gas 1
P2 = Pressure of gas 2
V2 = Volume of gas 2
Practice examples:
- A fixed mass of gas at 102300Pa pressure has a volume of 25cm3.Calculate its volume if the pressure is doubled.
Working
P1 V1 = P2 V2 Substituting :102300 x 25 = (102300 x 2) x V2
V2 = 102300 x 25 = 12.5cm3
(102300 x 2)
- Calculate the pressure which must be applied to a fixed mass of 100cm3 of Oxygen for its volume to triple at 100000Nm-2.
P1 V1 = P2 V2 Substituting :100000 x 100 = P2 x (100 x 3)
V2 = 100000 x 100 = 33333.3333 Nm-2
(100 x 3)
3.A 60cm3 weather ballon full of Hydrogen at atmospheric pressure of 101325Pa was released into the atmosphere. Will the ballon reach stratosphere where the pressure is 90000Pa?
P1 V1 = P2 V2 Substituting :101325 x 60 = 90000 x V2
V2 = 101325 x 60 = 67.55 cm3
90000
The new volume at 67.55 cm3 exceed ballon capacity of 60.00 cm3.It will burst before reaching destination.
7.Charles law states that“the volume of a fixed mass of a gas is directly proportional to the absolute temperature at constant/fixed pressure ”
Mathematically:
Volume α Pressure (Fixed /constant pressure)
V α T (Fixed /constant P) ie V = Constant(k)
T
From Charles law , an increase in temperature of a gas cause an increase in volume. i.e doubling the temperature cause the volume to be doubled.
Gases expand/increase by 1/273 by volume on heating.Gases contact/decrease by 1/273 by volume on cooling at constant/fixed pressure.
The volume of a gas continue decreasing with decrease in temperature until at -273oC /0 K the volume is zero. i.e. there is no gas.
This temperature is called absolute zero. It is the lowest temperature at which a gas can exist.
Graphically therefore a plot of volume(V) against Temperature(T) in:
(i)oC produces a straight line that is extrapolated to the absolute zero of -273oC .
V
-273oC 0oC
T(oC)
(ii)Kelvin/K produces a straight line from absolute zero of O Kelvin
V
0 T(Kelvin)
For two gases then V1 = V2
T1 T2
T1 = Temperature in Kelvin of gas 1
V1 = Volume of gas 1
T2 = Temperature in Kelvin of gas 2
V2 = Volume of gas 2
Practice examples:
- 500cm3 of carbon(IV)oxide at 0oC was transfered into a cylinder at -4oC. If the capacity of the cylinder is 450 cm3,explain what happened.
V1 = V2 substituting 500 = V2
T1 T2 (0 +273) (-4 +273)
= 500 x (-4 x 273) = 492.674cm3
(0 + 273)
The capacity of cylinder (500cm3) is less than new volume(492.674cm3).
7.326cm3(500-492.674cm3)of carbon(IV)oxide gas did not fit into the cylinder.
- A mechanic was filling a deflated tyre with air in his closed garage using a hand pump. The capacity of the tyre was 40,000cm3 at room temperature. He rolled the tyre into the car outside. The temperature outside was 30oC.Explain what happens.
V1 = V2 substituting 40000 = V2
T1 T2 (25 +273) (30 +273)
= 40000 x (30 x 273) = 40671.1409cm3
(25 + 273)
The capacity of a tyre (40000cm3) is less than new volume(40671.1409cm3).
The tyre thus bursts.
- A hydrogen gas balloon with 80cm3 was released from a research station at room temperature. If the temperature of the highest point it rose is -30oC , explain what happened.
V1 = V2 substituting 80 = V2
T1 T2 (25 +273) (-30 +273)
= 80 x (-30 x 273) = 65.2349cm3
(25 + 273)
The capacity of balloon (80cm3) is more than new volume (65.2349cm3).
The balloon thus remained intact.
- The continuous random motion of gases differ from gas to the other.The movement of molecules (of a gas) from region of high concentration to a region of low concentration is called diffusion.
The rate of diffusion of a gas depends on its density. i.e. The higher the rate of diffusion, the less dense the gas.
The density of a gas depends on its molar mass/relative molecular mass. i.e. The higher the density the higher the molar mass/relative atomic mass and thus the lower the rate of diffusion.
Examples
1.Carbon (IV)oxide(CO2) has a molar mass of 44g.Nitrogen(N2)has a molar mass of 28g. (N2)is thus lighter/less dense than Carbon (IV)oxide(CO2). N2 diffuses faster than CO2.
2.Ammonia(NH3) has a molar mass of 17g.Nitrogen(N2)has a molar mass of 28g. (N2)is thus about twice lighter/less dense than Ammonia(NH3). Ammonia(NH3) diffuses twice faster than N2.
- Ammonia(NH3) has a molar mass of 17g.Hydrogen chloride gas has a molar mass of 36.5g.Both gases on contact react to form white fumes of ammonium chloride .When a glass/cotton wool dipped in ammonia and another glass/cotton wool dipped in hydrochloric acid are placed at opposite ends of a glass tube, both gases diffuse towards each other. A white disk appears near to glass/cotton wool dipped in hydrochloric acid. This is because hydrogen chloride is heavier/denser than Ammonia and thus its rate of diffusion is lower .
The rate of diffusion of a gas is in accordance to Grahams law of diffusion. Grahams law states that:
“the rate of diffusion of a gas is inversely proportional to the square root of its density, at the same/constant/fixed temperature and pressure”
Mathematically
R α 1 and since density is proportional to mass then R α 1
√ p √ m
For two gases then:
R1 = R2 where: R1 and R2 is the rate of diffusion of 1st and 2nd gas.
√M2 √M1 M1 and M2 is the molar mass of 1st and 2nd gas.
Since rate is inverse of time. i.e. the higher the rate the less the time:
For two gases then:
T1 = T2 where: T1 and T2 is the time taken for 1st and 2nd gas to diffuse.
√M1 √M2 M1 and M2 is the molar mass of 1st and 2nd gas.
Practice examples:
- It takes 30 seconds for 100cm3 of carbon(IV)oxide to diffuse across a porous plate. How long will it take 150cm3 of nitrogen(IV)oxide to diffuse across the same plate under the same conditions of temperature and pressure. (C=12.0,N=14.0=16.0)
Molar mass CO2=44.0 Molar mass NO2=46.0
Method 1
100cm3 CO2 takes 30seconds
150cm3 takes 150 x30 = 45seconds
100
T CO2 = √ molar mass CO2 => 45seconds = √ 44.0
T NO2 √ molar mass NO2 T NO2 √ 46.0
T NO2 =45seconds x √ 46.0 = 46.0114 seconds
√ 44.0
Method 2
100cm3 CO2 takes 30seconds
1cm3 takes 100 x1 = 3.3333cm3sec-1
30
R CO2 = √ molar mass NO2 => 3.3333cm3sec-1 = √ 46.0
R NO2 √ molar mass CO2 R NO2 √ 44.0
R NO2 = 3.3333cm3sec-1 x √ 44.0 = 3.2601cm3sec-1
√ 46.0
3.2601cm3 takes 1seconds
150cm3 take 150cm3 = 46.0109seconds
3.2601cm3
- How long would 200cm3 of Hydrogen chloride take to diffuse through a porous plug if carbon(IV)oxide takes 200seconds to diffuse through.
Molar mass CO2 = 44g Molar mass HCl = 36.5g
T CO2 = √ molar mass CO2 => 200 seconds = √ 44.0
T HCl √ molar mass HCl T HCl √ 36.5
T HCl = 200seconds x √ 36.5 = 182.1588 seconds
√ 44.0
- Oxygen gas takes 250 seconds to diffuse through a porous diaphragm. Calculate the molar mass of gas Z which takes 227 second to diffuse.
Molar mass O2 = 32g Molar mass Z = x g
T O2 = √ molar mass O2 => 250 seconds = √ 32.0
T Z √ molar mass Z 227seconds √ x
√ x = 227seconds x √ 32 = 26.3828 grams
250
- 25cm3 of carbon(II)oxide diffuses across a porous plate in 25seconds. How long will it take 75cm3 of Carbon(IV)oxide to diffuse across the same plate under the same conditions of temperature and pressure. (C=12.0,0=16.0)
Molar mass CO2 = 44.0 Molar mass CO = 28.0
Method 1
25cm3 CO takes 25seconds
75cm3 takes 75 x25 = 75seconds
25
T CO2 = √ molar mass CO2 => T CO2seconds = √ 44.0
T CO √ molar mass CO 75 √ 28.0
T CO2 =75seconds x √ 44.0 = 94.0175 seconds
√ 28.0
Method 2
25cm3 CO2 takes 25seconds
1cm3 takes 25 x1 = 1.0cm3sec-1
25
R CO2 = √ molar mass CO => x cm3sec-1 = √ 28.0
R CO √ molar mass CO2 1.0cm3sec-1 √ 44.0
R CO2 = 1.0cm3sec-1 x √ 28.0 = 0.7977cm3sec-1
√ 44.0
0.7977cm3 takes 1 seconds
75cm3 takes 75cm3 = 94.0203seconds
0.7977cm3
(b)Introduction to the mole, molar masses and Relative atomic masses
- The mole is the SI unit of the amount of substance.
- The number of particles e.g. atoms, ions, molecules, electrons, cows, cars are all measured in terms of moles.
- The number of particles in one mole is called the Avogadros Constant. It is denoted “L”.
The Avogadros Constant contain 6.023 x10 23 particles. i.e.
1mole = 6.023 x10 23 particles = 6.023 x10 23
2 moles = 2 x 6.023 x10 23 particles = 1.205 x10 24
0.2 moles = 0.2 x 6.023 x10 23 particles = 1.205 x10 22
0.0065 moles = 0.0065 x 6.023 x10 23 particles = 3.914 x10 21
- The mass of one mole of a substance is called molar mass. The molar mass of:
(i)an element has mass equal to relative atomic mass /RAM(in grams)of the element e.g.
Molar mass of carbon(C)= relative atomic mass = 12.0g
6.023 x10 23 particles of carbon = 1 mole =12.0 g
Molar mass of sodium(Na) = relative atomic mass = 23.0g
6.023 x10 23 particles of sodium = 1 mole =23.0 g
Molar mass of Iron (Fe) = relative atomic mass = 56.0g
6.023 x10 23 particles of iron = 1 mole =56.0 g
(ii)a molecule has mass equal to relative molecular mass /RMM (in grams)of the molecule. Relative molecular mass is the sum of the relative atomic masses of the elements making the molecule.
The number of atoms making a molecule is called atomicity. Most gaseous molecules are diatomic (e.g. O2, H2, N2, F2, Cl2, Br2, I2)noble gases are monoatomic(e.g. He, Ar, Ne, Xe),Ozone gas(O3) is triatomic e.g.
Molar mass Oxygen molecule(O2) =relative molecular mass =(16.0x 2)g =32.0g
6.023 x10 23 particles of Oxygen molecule = 1 mole = 32.0 g
Molar mass chlorine molecule(Cl2) =relative molecular mass =(35.5x 2)g =71.0g
6.023 x10 23 particles of chlorine molecule = 1 mole = 71.0 g
Molar mass Nitrogen molecule(N2) =relative molecular mass =(14.0x 2)g =28.0g
6.023 x10 23 particles of Nitrogen molecule = 1 mole = 28.0 g
(ii)a compound has mass equal to relative formular mass /RFM (in grams)of the molecule. Relative formular mass is the sum of the relative atomic masses of the elements making the compound. e.g.
(i)Molar mass Water(H2O) = relative formular mass =[(1.0 x 2 ) + 16.0]g =18.0g
6.023 x10 23 particles of Water molecule = 1 mole = 18.0 g
6.023 x10 23 particles of Water molecule has:
– 2 x 6.023 x10 23 particles of Hydrogen atoms
-1 x 6.023 x10 23 particles of Oxygen atoms
(ii)Molar mass sulphuric(VI)acid(H2SO4) = relative formular mass
=[(1.0 x 2 ) + 32.0 + (16.0 x 4)]g =98.0g
6.023 x10 23 particles of sulphuric(VI)acid(H2SO4) = 1 mole = 98.0g
6.023 x10 23 particles of sulphuric(VI)acid(H2SO4) has:
– 2 x 6.023 x10 23 particles of Hydrogen atoms
–1 x 6.023 x10 23 particles of Sulphur atoms
-4 x 6.023 x10 23 particles of Oxygen atoms
(iii)Molar mass sodium carbonate(IV)(Na2CO3) = relative formular mass
=[(23.0 x 2 ) + 12.0 + (16.0 x 3)]g =106.0g
6.023 x10 23 particles of sodium carbonate(IV)(Na2CO3) = 1 mole = 106.0g
6.023 x10 23 particles of sodium carbonate(IV)(Na2CO3) has:
– 2 x 6.023 x10 23 particles of Sodium atoms
–1 x 6.023 x10 23 particles of Carbon atoms
-3 x 6.023 x10 23 particles of Oxygen atoms
(iv)Molar mass Calcium carbonate(IV)(CaCO3) = relative formular mass
=[(40.0+ 12.0 + (16.0 x 3)]g =100.0g.
6.023 x10 23 particles of Calcium carbonate(IV)(CaCO3) = 1 mole = 100.0g
6.023 x10 23 particles of Calcium carbonate(IV)(CaCO3) has:
– 1 x 6.023 x10 23 particles of Calcium atoms
–1 x 6.023 x10 23 particles of Carbon atoms
-3 x 6.023 x10 23 particles of Oxygen atoms
(v)Molar mass Water(H2O) = relative formular mass
=[(2 x 1.0 )+ 16.0 ]g =18.0g
6.023 x10 23 particles of Water(H2O) = 1 mole = 18.0g
6.023 x10 23 particles of Water(H2O) has:
– 2 x 6.023 x10 23 particles of Hydrogen atoms
-2 x 6.023 x10 23 particles of Oxygen atoms
Practice
- Calculate the number of moles present in:
(i)0.23 g of Sodium atoms
Molar mass of Sodium atoms = 23g
Moles = mass in grams = > 0.23g = 0.01moles
Molar mass 23
(ii) 0.23 g of Chlorine atoms
Molar mass of Chlorine atoms = 35.5 g
Moles = mass in grams = > 0.23g = 0.0065moles /6.5 x 10-3 moles
Molar mass 35.5
(iii) 0.23 g of Chlorine molecules
Molar mass of Chlorine molecules =( 35.5 x 2) = 71.0 g
Moles = mass in grams = > 0.23g = 0.0032moles /3.2 x 10-3 moles
Molar mass 71
(iv) 0.23 g of dilute sulphuric(VI)acid
Molar mass of H2SO4 = [(2 x 1) + 32 + (4 x14)] = 98.0g
Moles = mass in grams = > 0.23g = 0.0023moles /2.3 x 10-3 moles
Molar mass 98
- Calculate the number of atoms present in:(Avogadros constant L = 6.0 x 10 23)
(i) 0.23 g of dilute sulphuric (VI)acid
Method I
Molar mass of H2SO4 = [(2 x 1) + 32 + (4 x14)] = 98.0g
Moles = mass in grams = > 0.23g = 0.0023moles /2.3 x 10-3 moles
Molar mass 98
1 mole has 6.0 x 10 23 atoms
2.3 x 10-3 moles has (2.3 x 10-3 x 6.0 x 10 23) = 1.38 x 10 21 atoms
1
Method II
Molar mass of H2SO4 = [(2 x 1) + 32 + (4 x14)] = 98.0g
98.0g = 1 mole has 6.0 x 10 23 atoms
0.23 g therefore has (0.23 g x 6.0 x 10 23 ) = 1.38 x 10 21 atoms
98
(ii)0.23 g of sodium carbonate(IV)decahydrate
Molar mass of Na2CO3.10H2 O=
[(2 x 23) + 12 + (3 x16) + (10 x 1.0) + (10 x 16)] = 276.0g
Method I
Moles = mass in grams = > 0.23g = 0.00083moles /
Molar mass 276 8.3 x 10-4 moles
1 mole has 6.0 x 10 23 atoms
8.3 x 10-4 moles has (8.3 x 10-4 moles x 6.0 x 10 23) = 4.98 x 10 20 atoms
1
Method II
276.0g = 1 mole has 6.0 x 10 23 atoms
0.23 g therefore has (0.23 g x 6.0 x 10 23 ) = 4.98 x 10 20 atoms
276.0
(iii)0.23 g of Oxygen gas
Molar mass of O2 = (2 x16) = 32.0 g
Method I
Moles = mass in grams = > 0.23g = 0.00718moles /
Molar mass 32 7.18 x 10-3 moles
1 mole has 2 x 6.0 x 10 23 atoms in O2
7.18 x 10-3moles has (7.18 x 10-3moles x 2 x 6.0 x 10 23) =8.616 x 10 21atoms
1
Method II
32.0g = 1 mole has 2 x 6.0 x 10 23 atoms in O2
0.23 g therefore has (0.23 g x 2 x 6.0 x 10 23 ) = 8.616 x 10 21atoms
32.0
(iv)0.23 g of Carbon(IV)oxide gas
Molar mass of CO2 = [12 + (2 x16)] = 44.0 g
Method I
Moles = mass in grams = > 0.23g = 0.00522moles /
Molar mass 44 5.22 x 10-3 moles
1 mole has 3 x 6.0 x 10 23 atoms in CO2
7.18 x 10-3moles has (5.22 x 10-3moles x 3 x 6.0 x 10 23) =9.396 x 10 21atoms
1
Method II
44.0g = 1 mole has 3 x 6.0 x 10 23 atoms in CO2
0.23 g therefore has (0.23 g x 3 x 6.0 x 10 23 ) = 9.409 x 10 21atoms
44.0
(c)Empirical and molecular formula
1.The empirical formula of a compound is its simplest formula. It is the simplest whole number ratios in which atoms of elements combine to form the compound. 2.It is mathematically the lowest common multiple (LCM) of the atoms of the elements in the compound
3.Practically the empirical formula of a compound can be determined as in the following examples.
To determine the empirical formula of copper oxide
(a)Method 1:From copper to copper(II)oxide
Procedure.
Weigh a clean dry covered crucible(M1).Put two spatula full of copper powder into the crucible. Weigh again (M2).Heat the crucible on a strong Bunsen flame for five minutes. Lift the lid, and swirl the crucible carefully using a pair of tong. Cover the crucible and continue heating for another five minutes. Remove the lid and stop heating. Allow the crucible to cool. When cool replace the lid and weigh the contents again (M3).
Sample results
| Mass of crucible(M1) | 15.6g |
| Mass of crucible + copper before heating(M2) | 18.4 |
| Mass of crucible + copper after heating(M3) | 19.1 |
Sample questions
- Calculate the mass of copper powder used.
Mass of crucible + copper before heating(M2) = 18.4
Less Mass of crucible(M1) = – 15.6g
Mass of copper 2.8 g
- Calculate the mass of Oxygen used to react with copper.
Method I
Mass of crucible + copper after heating(M3) = 19.1g
Mass of crucible + copper before heating(M2) = – 18.4g
Mass of Oxygen = 0.7 g
Method II
Mass of crucible + copper after heating(M3) = 19.1g
Mass of crucible = – 15.6g
Mass of copper(II)Oxide = 3.5 g
Mass of copper(II)Oxide = 3.5 g
Mass of copper = – 2.8 g
Mass of Oxygen = 0.7 g
- Calculate the number of moles of:
(i) copper used (Cu = 63.5)
number of moles of copper = mass used => 2.8 = 0.0441moles
Molar mass 63.5
(ii) Oxygen used (O = 16.0)
number of moles of oxygen = mass used => 0.7 = 0.0441moles
Molar mass 16.0
- Determine the mole ratio of the reactants
Moles of copper = 0.0441moles = 1 => Mole ratio Cu: O = 1:1
Moles of oxygen 0.0441moles 1
5.What is the empirical, formula of copper oxide formed.
CuO (copper(II)oxide
- State and explain the observations made during the experiment.
Observation
Colour change from brown to black
Explanation
Copper powder is brown. On heating it reacts with oxygen from the air to form black copper(II)oxide
- Explain why magnesium ribbon/shavings would be unsuitable in a similar experiment as the one above.
Hot magnesium generates enough heat energy to react with both Oxygen and Nitrogen in the air forming a white solid mixture of Magnesium oxide and magnesium nitride. This causes experimental mass errors.
(b)Method 2:From copper(II)oxide to copper
Procedure.
Weigh a clean dry porcelain boat (M1). Put two spatula full of copper(II)oxide powder into the crucible. Reweigh the porcelain boat (M2).Put the porcelain boat in a glass tube and set up the apparatus as below;
Pass slowly(to prevent copper(II)oxide from being blown away)a stream of either dry Hydrogen /ammonia/laboratory gas/ carbon(II)oxide gas for about two minutes from a suitable generator.
When all the in the apparatus set up is driven out ,heat the copper(II)oxide strongly for about five minutes until there is no further change. Stop heating.
Continue passing the gases until the glass tube is cool.
Turn off the gas generator.
Carefully remove the porcelain boat form the combustion tube.
Reweigh (M3).
Sample results
| Mass of boat(M1) | 15.6g |
| Mass of boat before heating(M2) | 19.1 |
| Mass of boat after heating(M3) | 18.4 |
Sample questions
- Calculate the mass of copper(II)oxide used.
Mass of boat before heating(M2) = 19.1
Mass of empty boat(M1) = – 15.6g
Mass of copper(II)Oxide 3.5 g
- Calculate the mass of
(i) Oxygen.
Mass of boat before heating(M2) = 19.1
Mass of boat after heating (M3) = – 18.4g
Mass of oxygen = 0.7 g
(ii)Copper
Mass of copper(II)Oxide = 3.5 g
Mass of oxygen = 0.7 g
Mass of oxygen = 2.8 g
- Calculate the number of moles of:
(i) Copper used (Cu = 63.5)
number of moles of copper = mass used => 2.8 = 0.0441moles
Molar mass 63.5
(ii) Oxygen used (O = 16.0)
number of moles of oxygen = mass used => 0.7 = 0.0441moles
Molar mass 16.0
- Determine the mole ratio of the reactants
Moles of copper = 0.0441moles = 1 => Mole ratio Cu: O = 1:1
Moles of oxygen 0.0441moles 1
5.What is the empirical, formula of copper oxide formed.
CuO (copper(II)oxide
- State and explain the observations made during the experiment.
Observation
Colour change from black to brown
Explanation
Copper(II)oxide powder is black. On heating it is reduced by a suitable reducing agent to brown copper metal.
- Explain why magnesium oxide would be unsuitable in a similar experiment as the one above.
Magnesium is high in the reactivity series. None of the above reducing agents is strong enough to reduce the oxide to the metal.
- Write the equation for the reaction that would take place when the reducing agent is:
(i) Hydrogen
CuO(s) + H2(g) -> Cu(s) + H2O(l)
(Black) (brown) (colourless liquid form
on cooler parts )
(ii)Carbon(II)oxide
CuO(s) + CO (g) -> Cu(s) + CO2(g)
(Black) (brown) (colourless gas, form
white ppt with lime water )
(iii)Ammonia
3CuO(s) + 2NH3(g) -> 3Cu(s) + N2 (g) + 3H2O(l)
(Black) (brown) (colourless liquid form
on cooler parts )
- Explain why the following is necessary during the above experiment;
(i)A stream of dry hydrogen gas should be passed before heating copper (II) Oxide.
Air combine with hydrogen in presence of heat causing an explosion
(ii)A stream of dry hydrogen gas should be passed after heating copper (II) Oxide has been stopped.
Hot metallic copper can be re-oxidized back to copper(II)oxide
(iii) A stream of excess carbon (II)oxide gas should be ignited to burn
Carbon (II)oxide is highly poisonous/toxic. On ignition it burns to form less toxic carbon (IV)oxide gas.
- State two sources of error in this experiment.
(i)All copper(II)oxide may not be reduced to copper.
(ii)Some copper(II)oxide may be blown out the boat by the reducing agent.
4.Theoreticaly the empirical formula of a compound can be determined as in the following examples.
(a)A oxide of copper contain 80% by mass of copper. Determine its empirical formula. (Cu = 63.5, 16.0)
% of Oxygen = 100% – % of Copper => 100- 80 = 20% of Oxygen
| Element | Copper | Oxygen |
| Symbol | Cu | O |
| Moles present = % composition
Molar mass |
80
63.5 |
20
16 |
| Divide by the smallest value | 1.25
1.25 |
1.25
1.25 |
| Mole ratios | 1 | 1 |
Empirical formula is CuO
(b)1.60g of an oxide of Magnesium contain 0.84g by mass of Magnesium. Determine its empirical formula(Mg = 24.0, 16.0)
Mass of Oxygen = 1.60 – 0.84 => 0.56 g of Oxygen
| Element | Magnesium | Oxygen |
| Symbol | Mg | O |
| Moles present = % composition
Molar mass |
0.84
24 |
0.56
16 |
| Divide by the smallest value | 0.35
0.35 |
0.35
0.35 |
| Mole ratios | 1 | 1 |
Empirical formula is MgO
(c)An oxide of Silicon contain 47% by mass of Silicon. What is its empirical formula(Si = 28.0, 16.0)
Mass of Oxygen = 100 – 47 => 53% of Oxygen
| Element | Silicon | Oxygen |
| Symbol | Si | O |
| Moles present = % composition
Molar mass |
47
28 |
53
16 |
| Divide by the smallest value | 1.68
1.68 |
3.31
1.68 |
| Mole ratios | 1 | 1.94 = 2 |
Empirical formula is SiO2
(d)A compound contain 70% by mass of Iron and 30% Oxygen. What is its empirical formula(Fe = 56.0, 16.0)
Mass of Oxygen = 100 – 47 => 53% of Oxygen
| Element | Silicon | Oxygen |
| Symbol | Si | O |
| Moles present = % composition
Molar mass |
47
28 |
53
16 |
| Divide by the smallest value | 1.68
1.68 |
3.31
1.68 |
| Mole ratios | 1 | 1.94 = 2 |
Empirical formula is SiO2
2.During heating of a hydrated copper (II)sulphate(VI) crystals, the following readings were obtained:
Mass of evaporating dish =300.0g
Mass of evaporating dish + hydrated salt = 305.0g
Mass of evaporating dish + anhydrous salt = 303.2g
Calculate the number of water of crystallization molecules in hydrated copper (II)sulphate(VI)
(Cu =64.5, S = 32.0,O=16.0, H = 1.0)
Working
Mass of Hydrated salt = 305.0g -300.0g = 5.0g
Mass of anhydrous salt = 303.2 g -300.0g = 3.2 g
Mass of water in hydrated salt = 5.0g -3.2 g = 1.8g
Molar mass of water(H2O) = 18.0g
Molar mass of anhydrous copper (II)sulphate(VI) (CuSO4) = 160.5g
| Element/compound | anhydrous copper (II)
sulphate(VI) |
Oxygen |
| Symbol | CuSO4 | O |
| Moles present = composition by mass
Molar mass |
3,2
160.5 |
1.8
18 |
| Divide by the smallest value | 0.0199
0.0199 |
0.1
18 |
| Mole ratios | 1 | 5 |
The empirical formula of hydrated salt = CuSO4.5H2O
Hydrated salt has five/5 molecules of water of crystallizations
- The molecular formula is the actual number of each kind of atoms present in a molecule of a compound.
The empirical formula of an ionic compound is the same as the chemical formula but for simple molecular structured compounds, the empirical formula may not be the same as the chemical formula.
The molecular formula is a multiple of empirical formula .It is determined from the relationship:
(i) n = Relative formular mass
Relative empirical formula
where n is a whole number.
(ii) Relative empirical formula x n = Relative formular mass
where n is a whole number.
Practice sample examples
- A hydrocarbon was found to contain 92.3% carbon and the remaining Hydrogen.
If the molecular mass of the compound is 78, determine the molecular formula(C=12.0, H =1.0)
Mass of Hydrogen = 100 – 92.3 => 7.7% of Oxygen
| Element | Carbon | Hydrogen |
| Symbol | C | H |
| Moles present = % composition
Molar mass |
92.3
12 |
7.7
1 |
| Divide by the smallest value | 7.7
7.7 |
7.7
7.7 |
| Mole ratios | 1 | 1 |
Empirical formula is CH
The molecular formular is thus determined :
n = Relative formular mass = 78 = 6
Relative empirical formula 13
The molecular formula is (C H ) x 6 = C6H6
- A compound of carbon, hydrogen and oxygen contain 54.55% carbon, 9.09% and remaining 36.36% oxygen.
If its relative molecular mass is 88, determine its molecular formula(C=12.0, H =1.0, O= 16.0)
| Element | Carbon | Hydrogen | Oxygen |
| Symbol | C | H | O |
| Moles present = % composition
Molar mass |
54.55
12 |
9.09
1 |
36.36
16 |
| Divide by the smallest value | 4.5458
2.2725 |
9.09
2.2725 |
2.2725
2.2725 |
| Mole ratios | 2 | 4 | 1 |
Empirical formula is C2H4O
The molecular formula is thus determined :
n = Relative formular mass = 88 = 2
Relative empirical formula 44
The molecular formula is (C2H4O ) x 2 = C4H8O2.
4.A hydrocarbon burns completely in excess air to form 5.28 g of carbon (IV) oxide and 2,16g of water.
If the molecular mass of the hydrocarbon is 84, draw and name its molecular structure.
Since a hydrocarbon is a compound containing Carbon and Hydrogen only. Then:
Mass of carbon in CO2 = Mass of C in CO2 x mass of CO2 =>
Molar mass of CO2
12 x 5.28 = 1.44g√
44
Mass of Hydrogen in H2O = Mass of C in H2O x mass of H2O =>
Molar mass of H2O
2 x 2.16 = 0.24g√
18
| Element | Carbon | Hydrogen |
| Symbol | C | H |
| Moles present = mass
Molar mass |
1.44g
12 |
0.24g√
1 |
| Divide by the smallest value | 0.12
0.12 |
0.24
0.12
|
| Mole ratios | 1 | 2√ |
Empirical formula is CH2√
The molecular formular is thus determined :
n = Relative formular mass = 84 = 6√
Relative empirical formula 14
The molecular formula is (CH2 ) x 6 = C6H12. √
molecular name Hexene√/Hex-1-ene (or any position isomer of Hexene)
Molecular structure
H H H H H H
H C C C C C C H√
H H H H
- Compound A contain 5.2% by mass of Nitrogen .The other elements present are Carbon, hydrogen and Oxygen. On combustion of 0.085g of A in excess Oxygen,0.224g of carbon(IV)oxide and 0.0372g of water was formed. Determine the empirical formula of A (N=14.0, O=16.0 , C=12.0 , H=1.0)
Mass of N in A = 5.2% x 0.085 = 0.00442 g
Mass of C in A = 12 x 0.224 = 0.0611g
44
Mass of H in A = 2 x 0.0372 = 0.0041g
18
Mass of O in A = 0.085g – 0.004442g = 0.0806g (Mass of C,H,O)
=> 0.0611g + 0.0041g = 0.0652g (Mass of C,H)
0.0806g (Mass of C,H,O)- 0.0652g (Mass of C,H) = 0.0154 g
| Element | Nitrogen | Carbon | Hydrogen | Oxygen |
| Symbol | N | C | H | O |
| Moles present = mass
Molar mass |
0.00442 g
14 |
0.0611g
12 |
0.0041g
1 |
0.0154 g 16 |
| Divide by the smallest value | 0.00032
0.00032 |
0.00509
0.00032 |
0.0041g
0.00032 |
0.00096
0.00032 |
| Mole ratios | 1 | 16 | 13 | 3 |
Empirical formula = C16H13NO3
(d)Molar gas volume
The volume occupied by one mole of all gases at the same temperature and pressure is a constant.It is:
(i) 24dm3/24litres/24000cm3 at room temperature(25oC/298K)and pressure(r.t.p).
i.e. 1mole of all gases =24dm3/24litres/24000cm3 at r.t.p
Examples
1mole of O2 = 32g =6.0 x1023 particles= 24dm3/24litres/24000cm3 at r.t.p
1mole of H2 = 2g =6.0 x1023 particles =24dm3/24litres/24000cm3 at r.t.p
1mole of CO2 = 44g = 6.0 x1023 particles =24dm3/24litres/24000cm3 at r.t.p
1mole of NH3 = 17g =6.0 x1023 particles = 24dm3/24litres/24000cm3 at r.t.p
1mole of CH4 = 16g =6.0 x1023 particles =24dm3/24litres/24000cm3 at r.t.p
(ii)22.4dm3/22.4litres/22400cm3 at standard temperature(0oC/273K) and pressure(s.t.p)
i.e. 1mole of all gases =22.4dm3/22.4litres/22400cm3 at s.t.p
Examples
1mole of O2 = 32g =6.0 x1023 particles= 22.4dm3/22.4litres/22400cm3 at s.t.p
1mole of H2 = 2g =6.0 x1023 particles = 22.4dm3/22.4litres/22400cm3 at s.t.p
1mole of CO2 = 44g = 6.0 x1023 particles = 22.4dm3/22.4litres/22400cm3 at s.t.p
1mole of NH3 = 17g =6.0 x1023 particles= 22.4dm3/22.4litres/22400cm3 at s.t.p
1mole of CH4 = 16g =6.0 x1023 particles = 22.4dm3/22.4litres/22400cm3 at s.t.p
The volume occupied by one mole of a gas at r.t.p or s.t.p is commonly called the molar gas volume. Whether the molar gas volume is at r.t.p or s.t.p must always be specified.
From the above therefore a less or more volume can be determined as in the examples below.
Practice examples
- Calculate the number of particles present in:
(Avogadros constant =6.0 x1023mole-1 )
(i) 2.24dm3 of Oxygen.
22.4dm3 -> 6.0 x1023
2.24dm3 -> 2.24 x 6.0 x1023
22.4
=6.0 x1022 molecules = 2 x 6.0 x1022. = 1.2 x1023 atoms
(ii) 2.24dm3 of Carbon(IV)oxide.
22.4dm3 -> 6.0 x1023
2.24dm3 -> 2.24 x 6.0 x1023
22.4
=6.0 x1022 molecules = (CO2) = 3 x 6.0 x1022. = 1.8 x1023 atoms
- 0.135 g of a gaseous hydrocarbon X on complete combustion produces 0.41g of carbon(IV)oxide and 0.209g of water.0.29g of X occupy 120cm3 at room temperature and 1 atmosphere pressure .Name X and draw its molecular structure.(C=12.0,O= 16.O,H=1.0,1 mole of gas occupies 24dm3 at r.t.p)
Molar mass CO2= 44 gmole-1√ Molar mass H2O = 18 gmole-1√
Molar mass X = 0.29 x (24 x 1000)cm3 = 58 gmole-1√
120cm3
Since a hydrocarbon is a compound containing Carbon and Hydrogen only. Then:
Mass of carbon in CO2 = Mass of C in CO2 x mass of CO2 =>
Molar mass of CO2
12 x 0.41 = 0.1118g√
44
Mass of Hydrogen in H2O = Mass of C in H2O x mass of H2O =>
Molar mass of H2O
2 x 0.209 = 0.0232g√
18
| Element | Carbon | Hydrogen |
| Symbol | C | H |
| Moles present = % composition
Molar mass |
0.g118
12 |
0.0232g√
1 |
| Divide by the smallest value | 0.0093
0.0093 |
0.0232
0.0093√
|
| Mole ratios | 1 x2 | 2.5×2 |
| 2 | 5√ |
Empirical formula is C2H5√
The molecular formular is thus determined :
n = Relative formular mass = 58 = 2√
Relative empirical formula 29
The molecular formula is (C2H5 ) x 2 = C4H10.√
Molecule name Butane
Molecula structure
H H H H
H C C C C H√
H H H H
(e)Gravimetric analysis
Gravimetric analysis is the relationship between reacting masses and the volumes and /or masses of products. All reactants are in mole ratios to their products in accordance to their stoichiometric equation. Using the mole ration of reactants and products any volume and/or mass can be determined as in the examples:
- Calculate the volume of carbon(IV)oxide at r.t.p produced when 5.0 g of calcium carbonate is strongly heated.(Ca=40.0, C= 12.0,O = 16.0,1 mole of gas =22.4 at r.t.p)
Chemical equation
CaCO3(s) -> CaO(s) + CO2(g)
Mole ratios 1: 1: 1
Molar Mass CaCO3 =100g
Method 1
100g CaCO3(s) -> 24dm3 CO2(g) at r.t.p
5.0 g CaCO3(s) -> 5.0 g x 24dm3 = 1.2dm3/1200cm3
100g
Method 2
Moles of 5.0 g CaCO3(s) = 5.0 g = 0.05 moles
100 g
Mole ratio 1:1
Moles of CO2(g) = 0.05moles
Volume of CO2(g) = 0.05 x 24000cm3 =1200cm3 /1.2dm3
- 1.0g of an alloy of aluminium and copper were reacted with excess hydrochloric acid. If 840cm3 of hydrogen at s.t.p was produced, calculate the % of copper in the alloy.(Al =27.0,one mole of a gas at s.t.p =22.4dm3 )
Chemical equation
Copper does not react with hydrochloric acid
2Al(s) + 6HCl(aq) -> 2AlCl3(aq) + 3H2(g)
Method 1
3H2(g) = 3 moles x (22.4 x 1000)cm3 => 2 x 27 g Al
840cm3 => 840cm3 x 2 x 27 = 0.675g of Aluminium
3 x 22.4 x 1000
Total mass of alloy – mass of aluminium = mass of copper
=> 1.0g – 0.675g =0.325g of copper
% copper = mass of copper x100% = 32.5%
Mass of alloy
Method 2
Mole ratio 2Al: 3H2 = 2:3
Moles of Hydrogen gas = volume of gas => 840cm3 = 0.0375moles
Molar gas volume 22400cm3
Moles of Al = 2/3 moles of H2 => 2/3x 0.0375moles = 0.025moles
Mass of Al = moles x molar mass =>0.025moles x 27 = 0.675g
Total mass of alloy – mass of aluminium = mass of copper
=> 1.0g – 0.675g = 0.325 g of copper
% copper = mass of copper x100% = 32.5%
Mass of alloy
(f)Gay Lussac’s law
Gay Lussacs law states that “when gases combine/react they do so in simple volume ratios to each other and to their gaseous products at constant/same temperature and pressure”
Gay Lussacs law thus only apply to gases
Given the volume of one gas reactant, the other gaseous reactants can be deduced thus:
Examples
- Calculate the volume of Oxygen required to completely react with 50cm3 of Hydrogen.
Chemical equation: 2H2 (g) + O2 (g) -> 2H2O(l)
Volume ratios 2 : 1 : 0
Reacting volumes 50cm3 : 25cm3
50cm3 of Oxygen is used
- Calculate the volume of air required to completely reacts with 50cm3 of Hydrogen.(assume Oxygen is 21% by volume of air)
Chemical equation: 2H2 (g) + O2 (g) -> 2H2O(l)
Volume ratios 2 : 1 : 0
Reacting volumes 50cm3 : 25cm3
50cm3 of Oxygen is used
21% = 25cm3
100% = 100 x 25 =
21
3.If 5cm3 of a hydrocarbon CxHy burn in 15cm3 of Oxygen to form 10cm3 of Carbon(IV)oxide and 10cm3 of water vapour/steam, obtain the equation for the reaction and hence find the value of x and y in CxHy.
Chemical equation: CxHy (g) + O2 (g) -> H2O(g) + CO2(g)
Volumes 5cm3 : 15cm3 : 10cm3 : 10cm3
Volume ratios 5cm3 : 15cm3 : 10cm3 : 10cm3 (divide by lowest volume) 5 5 5 5
Reacting volume ratios 1volume 3 volume 2 volume 2 volume
Balanced chemical equation: CxHy (g) + 3O2 (g) -> 2H2O(g) + 2CO2(g)
If “4H” are in 2H2O(g) the y=4
If “2C” are in 2CO2 (g) the x=2
Thus(i) chemical formula of hydrocarbon = C2H4
(ii) chemical name of hydrocarbon = Ethene
4.100cm3 of nitrogen (II)oxide NO combine with 50cm3 of Oxygen to form 100cm3 of a single gaseous compound of nitrogen. All volumes measured at the same temperature and pressure. Obtain the equation for the reaction and name the gaseous product.
Chemical equation: NO (g) + O2 (g) -> NOx
Volumes 100cm3 : 50cm3 : 100
Volume ratios 100cm3 : 50cm3 : 100cm3 (divide by lowest volume) 50 50 50
Reacting volume ratios 2volume 1 volume 2 volume
Balanced chemical equation: 2 NO (g) + O2 (g) -> 2NO x(g)
Thus(i) chemical formula of the nitrogen compound = 2 NO2
(ii) chemical name of compound = Nitrogen(IV)oxide
5.When 15cm3 of a gaseous hydrocarbon was burnt in 100cm3 of Oxygen ,the resulting gaseous mixture occupied70cm3 at room temperature and pressure. When the gaseous mixture was passed through, potassium hydroxide its volume decreased to 25cm3.
(a)What volume of Oxygen was used during the reaction.(1mk)
Volume of Oxygen used =100-25 =75cm3√
(P was completely burnt)
(b)Determine the molecular formula of the hydrocarbon(2mk)
CxHy + O2 -> xCO2 + yH2O
15cm3 : 75cm3
15 15
1 : 3√
=> 1 atom of C react with 6 (3×2)atoms of Oxygen
Thus x = 1 and y = 2 => P has molecula formula CH4√
(g) Ionic equations
An ionic equation is a chemical statement showing the movement of ions (cations and anions ) from reactants to products.
Solids, gases and liquids do not ionize/dissociate into free ions. Only ionic compounds in aqueous/solution or molten state ionize/dissociate into free cations and anions (ions)
An ionic equation is usually derived from a stoichiometric equation by using the following guidelines
Guidelines for writing ionic equations
1.Write the balanced stoichiometric equation
2.Indicate the state symbols of the reactants and products
3.Split into cations and anions all the reactants and products that exist in aqueous state.
4.Cancel out any cation and anion that appear on both the product and reactant side.
- Rewrite the chemical equation. It is an ionic equation.
Practice
(a)Precipitation of an insoluble salt
All insoluble salts are prepared in the laboratory from double decomposition /precipitation. This involves mixing two soluble salts to form one soluble and one insoluble salt
- When silver nitrate(V) solution is added to sodium chloride solution,sodium nitrate(V) solution and a white precipitate of silver chloride are formed.
Balanced stoichiometric equation
AgNO3(aq) + NaCl(aq) -> AgCl (s) + NaNO3 (aq)
Split reactants product existing in aqueous state as cation/anion
Ag+(aq) + NO3– (aq) + Na+(aq) + Cl–(aq) -> AgCl(s) + Na+(aq)+ NO3– (aq)
Cancel out ions appearing on reactant and product side
Ag+(aq) + NO3– (aq) + Na+(aq) + Cl–(aq) -> AgCl(s) + Na+(aq)+ NO3– (aq)
Rewrite the equation
Ag+(aq) + Cl–(aq) -> AgCl(s) (ionic equation)
- When barium nitrate(V) solution is added to copper(II)sulphate(VI) solution, copper(II) nitrate(V) solution and a white precipitate of barium sulphate(VI) are formed.
Balanced stoichiometric equation
Ba(NO3)2(aq) + CuSO4(aq) -> BaSO4 (s) + Cu(NO3) 2 (aq)
Split reactants product existing in aqueous state as cation/anion
Ba2+(aq) + 2NO3– (aq) + Cu2+(aq) + SO42-(aq) -> BaSO4 (s) + 2NO3– (aq)+ Cu2+(aq)
Cancel out ions appearing on reactant and product side
Ba2+(aq) + 2NO3– (aq) +Cu2+ (aq) + SO42-(aq)-> BaSO4(s) + 2NO3– (aq) + Cu2+(aq)
Rewrite the equation
Ba2+(aq) + SO42-(aq) -> BaSO4(s) (ionic equation)
3.A yellow precipitate of Potassium Iodide is formed from the reaction of Lead(II)nitrate(v) and potassium iodide.
Balanced stoichiometric equation
Pb(NO3)2(aq) + 2KI (aq) -> PbI2 (s) + 2KNO3 (aq)
Split reactants product existing in aqueous state as cation/anion
Pb2+(aq) + 2NO3– (aq) + 2K +(aq) + 2I – (aq) -> PbI2 (s) + 2NO3– (aq)+ 2K +(aq)
Cancel out ions appearing on reactant and product side
Pb2+(aq) + 2NO3– (aq) + 2K +(aq) + 2I – (aq) -> PbI2 (s) + 2NO3– (aq)+ 2K +(aq)
Rewrite the equation
Pb2+(aq) + 2I – (aq) -> PbI2 (s) (ionic equation)
(b)Neutralization
Neutralization is the reaction of an acid with a soluble base/alkali or insoluble base.
(i)Reaction of alkalis with acids
1.Reaction of nitric(V)acid with potassium hydroxide
Balanced stoichiometric equation
HNO3(aq) + KOH (aq) -> H2O (l) + KNO3 (aq)
Split reactants product existing in aqueous state as cation/anion
H+(aq) + NO3– (aq) + K +(aq) + OH – (aq) -> H2O (l) + NO3– (aq)+ K +(aq)
Cancel out ions appearing on reactant and product side
H+(aq) + NO3– (aq) + K +(aq) + OH – (aq) -> H2O (l) + NO3– (aq)+ K +(aq)
Rewrite the equation
H+ (aq) + OH – (aq) -> H2O (l) (ionic equation)
2.Reaction of sulphuric(VI)acid with ammonia solution
Balanced stoichiometric equation
H2SO4(aq) + 2NH4OH (aq) -> H2O (l) + (NH4) 2SO4 (aq)
Split reactants product existing in aqueous state as cation/anion
2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH – (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq)
Cancel out ions appearing on reactant and product side
2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH – (aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq)
Rewrite the equation
2H+ (aq) + 2OH – (aq) -> 2H2O (l)
H+ (aq) + OH – (aq) -> H2O (l) (ionic equation)
3.Reaction of hydrochloric acid with Zinc hydroxide
Balanced stoichiometric equation
2HCl(aq) + Zn(OH)2 (s) -> 2H2O (l) + ZnCl 2 (aq)
Split reactants product existing in aqueous state as cation/anion
2H+(aq) + 2Cl– (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl– (aq)+ Zn 2+ (aq)
Cancel out ions appearing on reactant and product side
2H+(aq) + 2Cl– (aq) + Zn(OH)2 (s) ->2H2O (l) + 2Cl– (aq)+ Zn 2+ (aq)
Rewrite the equation
2H+(aq) + Zn(OH)2 (s) ->2H2O (l) + Zn 2+ (aq) (ionic equation)
(h)Molar solutions
A molar solution is one whose concentration is known. The SI unit of concentration is Molarity denoted M.
Molarity may be defined as the number of moles of solute present in one cubic decimeter of solution.
One cubic decimeter is equal to one litre and also equal to 1000cm3.
The higher the molarity the higher the concentration and the higher/more solute has been dissolved in the solvent to make one cubic decimeter/ litre/1000cm3 solution.
Examples
2M sodium hydroxide means 2 moles of sodium hydroxide solute is dissolved in enough water to make one cubic decimeter/ litre/1000cm3 uniform solution mixture of sodium hydroxide and water.
0.02M sodium hydroxide means 0.02 moles of sodium hydroxide solute is dissolved in enough water to make one cubic decimeter/ litre/1000cm3 uniform solution mixture of sodium hydroxide and water.
“2M” is more concentrated than“0.02M”.
Preparation of molar solution
Procedure
Weigh accurately 4.0 g of sodium hydroxide pellets into a 250cm3 volumetric flask.
Using a wash bottle add about 200cm3 of distilled water.
Stopper the flask.
Shake vigorously for three minutes.
Remove the stopper for a second then continue to shake for about another two minutes until all the solid has dissolved.
Add more water slowly upto exactly the 250 cm3 mark.
Sample questions
1.Calculate the number of moles of sodium hydroxide pellets present in:
(i) 4.0 g.
Molar mass of NaOH = (23 + 16 + 1) = 40g
Moles = Mass => 4.0 = 0.1 / 1.0 x 10 -1 moles
Molar mass 40
(ii) 250 cm3 solution in the volumetric flask.
Moles in 250 cm3 = 0.1 / 1.0 x 10 -1 moles
(iii) one decimeter of solution
Method 1
Moles in decimeters = Molarity = Moles x 1000cm3/1dm3
Volume of solution
=> 1.0 x 10 -1 moles x 1000cm3 =
250cm3
= 0.4 M / 0.4 molesdm-3
Method 2
250cm3 solution contain 1.0 x 10 -1 moles
1000cm3 solution = Molarity contain 1000 x 1.0 x 10 -1 moles
250 cm3
= 0.4 M / 0.4 molesdm-3
Theoretical sample practice
- Calculate the molarity of a solution containing:
(i) 4.0 g sodium hydroxide dissolved in 500cm3 solution
Molar mass of NaOH = (23 + 16 + 1) = 40g
Moles = Mass => 4.0 = 0.1 / 1.0 x 10 -1 moles
Molar mass 40
Method 1
Moles in decimeters = Molarity = Moles x 1000cm3/1dm3
Volume of solution
=> 1.0 x 10 -1 moles x 1000cm3
500cm3
= 0.2 M / 0.2 molesdm-3
Method 2
500 cm3 solution contain 1.0 x 10 -1 moles
1000cm3 solution = Molarity contain 1000 x 1.0 x 10 -1 moles
500 cm3
= 0.2 M / 0.2 molesdm-3
(ii) 5.3 g anhydrous sodium carbonate dissolved in 50cm3 solution
Molar mass of Na2CO3 = (23 x 2 + 12 + 16 x 3) = 106 g
Moles = Mass => 5.3 = 0.05 / 5. 0 x 10-2 moles
Molar mass 106
Method 1
Moles in decimeters = Molarity = Moles x 1000cm3/1dm3
Volume of solution
=> 1.0 moles x 1000cm3 =
50cm3
=1.0 M
Method 2
50 cm3 solution contain 5.0 x 10 -2 moles
1000cm3 solution = Molarity contain 1000 x 5.0 x 10 -2 moles
50 cm3
= 1.0M / 1.0 molesdm-3
(iii) 5.3 g hydrated sodium carbonate decahydrate dissolved in 50cm3 solution
Molar mass of Na2CO3.10H2O = (23 x 2 + 12 + 16 x 3 + 20 x 1 + 10 x 16) =286g
Moles = Mass => 5.3 = 0.0185 / 1.85 x 10 -2 moles
Molar mass 286
Method 1
Moles in decimeters = Molarity = Moles x 1000cm3/1dm3
Volume of solution
=> 1.85 x 10 -2 moles x 1000cm3 =
50cm3
= 0.37 M/0.37 molesdm-3
Method 2
50 cm3 solution contain 1.85 x 10 -2 moles
1000cm3 solution = Molarity contain 1000 x 1.85 x 10 -2 moles
50 cm3
= 3.7 x 10-1 M / 3.7 x 10-1 molesdm-3
(iv) 7.1 g of anhydrous sodium sulphate(VI)was dissolved in 20.0 cm3 solution. Calculate the molarity of the solution.
Method 1
20.0cm3 solution ->7.1 g
1000cm3 solution -> 1000 x 71 = 3550 g dm-3
20
Molar mass Na2SO4 = 142 g
Moles dm-3 = Molarity = Mass 3550 = 2.5 M/ molesdm-3
Molar mass 142
Method 2
Molar mass Na2SO4 = 142 g
Moles = Mass => 7.1 = 0.05 / 5.0 x 10 -2 moles
Molar mass 142
Method 2(a)
Moles in decimeters = Molarity = Moles x 1000cm3/1dm3
Volume of solution
=> 5.0 x 10 -2 moles x 1000cm3
20cm3
= 2.5 M/2.5 molesdm-3
Method 2(b)
20 cm3 solution contain 5.0 x 10 -2 moles
1000cm3 solution = Molarity contain 1000 x 5.0 x 10 -2 moles
20 cm3
= 2.5 M/2.5 molesdm-3
(iv) The density of sulphuric(VI) is 1.84gcm-3 Calculate the molarity of the acid.
Method 1
1.0cm3 solution ->1.84 g
1000cm3 solution -> 1000 x 1.84 = 1840 g dm-3
1
Molar mass H2SO4 = 98 g
Moles dm-3 = Molarity = Mass = 1840
Molar mass 98
= 18.7755 M/ molesdm-3
Method 2
Molar mass H2SO4 = 98 g
Moles = Mass => 1.84 = 0.0188 / 1.88 x 10 -2 moles
Molar mass 98
Method 2(a)
Moles in decimeters = Molarity = Moles x 1000cm3/1dm3
Volume of solution
=> 1.88 x 10 -2 moles x 1000cm3
1.0cm3
= 18.8M/18.8 molesdm-3
Method 2(b)
20 cm3 solution contain 1.88 x 10 -2 moles
1000cm3 solution = Molarity contain 1000 x 1.88 x 10 -2 moles
1.0 cm3
= 18.8M/18.8 molesdm-3
- Calculate the mass of :
(i) 25 cm3 of 0.2M sodium hydroxide solution(Na =23.0.O =16.0, H=1.0)
Molar mass NaOH = 40g
Moles in 25 cm3 = Molarity x volume => 0.2 x 25 = 0.005/5.0 x 10-3moles
1000 1000
Mass of NaOH =Moles x molar mass = 5.0 x 10-3 x 40 = 0.2 g
(ii) 20 cm3 of 0.625 M sulphuric(VI)acid (S =32.0.O =16.0, H=1.0)
Molar mass H2SO4 = 98g
Moles in 20 cm3 = Molarity x volume=> 0.625 x 20 = 0.0125/1.25.0 x 10-3moles
1000 1000
Mass of H2SO4 =Moles x molar mass => 5.0 x 10-3 x 40 = 0.2 g
(iii) 1.0 cm3 of 2.5 M Nitric(V)acid (N =14.0.O =16.0, H=1.0)
Molar mass HNO3 = 63 g
Moles in 1 cm3 = Molarity x volume => 2.5 x 1 = 0.0025 / 2.5. x 10-3moles
1000 1000
Mass of HNO3 =Moles x molar mass => 2.5 x 10-3 x 40 = 0.1 g
- Calculate the volume required to dissolve :
(a)(i) 0.25moles of sodium hydroxide solution to form a 0.8M solution
Volume (in cm3) = moles x 1000 => 0.25 x 1000 = 312.5cm3
Molarity 0.8
(ii) 100cm3 was added to the sodium hydroxide solution above. Calculate the concentration of the solution.
C1 x V1 = C2 x V2 where:
C1 = molarity/concentration before diluting/adding water
C2 = molarity/concentration after diluting/adding water
V1 = volume before diluting/adding water
V2 = volume after diluting/adding water
=> 0.8M x 312.5cm3 = C2 x (312.5 + 100)
C2 = 0.8M x 312.5cm3 = 0.6061M
412.5
(b)(ii) 0.01M solution containing 0.01moles of sodium hydroxide solution .
Volume (in cm3) = moles x 1000 => 0.01 x 1000 = 1000 cm3
Molarity 0.01
(ii) Determine the quantity of water which must be added to the sodium hydroxide solution above to form a 0.008M solution.
C1 x V1 = C2 x V2 where:
C1 = molarity/concentration before diluting/adding water
C2 = molarity/concentration after diluting/adding water
V1 = volume before diluting/adding water
V2 = volume after diluting/adding water
=> 0.01M x 1000 cm3 = 0.008 x V2
V2 = 0.01M x 1000cm3 = 1250cm3
0.008
Volume added = 1250 – 1000 = 250cm3
(c)Volumetric analysis/Titration
Volumetric analysis/Titration is the process of determining unknown concentration of one reactant from a known concentration and volume of another.
Reactions take place in simple mole ratio of reactants and products.
Knowing the concentration/ volume of one reactant, the other can be determined from the relationship:
M1V1 = M2V2 where:
n1 n2
M1 = Molarity of 1st reactant
M2 = Molarity of 2nd reactant
V1 = Volume of 1st reactant
V1 = Volume of 2nd reactant
n1 = number of moles of 1st reactant from stoichiometric equation
n2 = number of moles of 2nd reactant from stoichiometric equation
Examples
1.Calculate the molarity of MCO3 if 5.0cm3 of MCO3 react with 25.0cm3 of 0.5M hydrochloric acid.(C=12.0 ,O =16.0)
Stoichiometric equation: MCO3(s) + 2HCl(aq) -> MCl2(aq) + CO2(g) + H2O(l)
Method 1
M1V1 = M2V2 -> M1 x 5.0cm3 = 0.5M x 25.0cm3
n1 n2 1 2
=> M1 = 0.5 x 25.0 x1 = 1.25M / 1.25 moledm-3
5.0 x 2
Method 2
Moles of HCl used = molarity x volume
1000
=> 0.5 x 25.0 = 0.0125 /1.25 x 10-2moles
1000
Mole ratio MCO3 : HCl = 1:2
Moles MCO3 = 0.0125 /1.25 x 10-2moles = 0.00625 / 6.25 x 10-3 moles
2
Molarity MCO3 = moles x 1000 => 0.00625 / 6.25 x 10-3 x 1000
Volume 5
= 1.25M / 1.25 moledm-3
- 2.0cm3 of 0.5M hydrochloric acid react with 0.1M of M2CO3. Calculate the volume of 0.1M M2CO3 used.
Stoichiometric equation: M2CO3 (aq) + 2HCl(aq) -> 2MCl (aq) + CO2(g) + H2O(l)
Method 1
M1V1 = M2V2 -> 0.5 x 2.0cm3 = 0.1M x V2 cm3
n1 n2 2 1
=> V2 = 0.5 x 2.0 x1 = 1.25M / 1.25 moledm-3
0.1 x 2
Method 2
Moles of HCl used = molarity x volume
1000
=> 0.5 x 2.0 = 0.0125 /1.25 x 10-2moles
1000
Mole ratio M2CO3 : HCl = 1:2
Moles M2CO3 = 0.0125 /1.25 x 10-2moles = 0.00625 / 6.25 x 10-3 moles
2
Molarity M2CO3 = moles x 1000 => 0.00625 / 6.25 x 10-3 x 1000
Volume 5
= 1.25M / 1.25 moledm-3
- 5.0cm3 of 0.1M sodium iodide react with 0.1M of Lead(II)nitrate(V). Calculate(i) the volume of Lead(II)nitrate(V) used.
(ii)the mass of Lead(II)Iodide formed
(Pb=207.0, I =127.0)
Stoichiometric equation: 2NaI(aq) + Pb(NO3)2(aq) -> 2NaNO3(aq) + PbI2(s)
(i)Volume of Lead(II)nitrate(V) used
Method 1
M1V1 = M2V2 -> 5 x 0.1cm3 = 0.1M x V2 cm3
n1 n2 2 1
=> V2 = 0.1 x 5.0 x 1 = 1.25M / 1.25 moledm-3
0.1 x 2
Method 2
Moles of HCl used = molarity x volume
1000
=> 0.1 x 5.0 = 0.0125 /1.25 x 10-2moles
1000
Mole ratio M2CO3 : HCl = 1:2
Moles M2CO3 = 0.0125 /1.25 x 10-2moles = 0.00625 / 6.25 x 10-3 moles
2
Molarity M2CO3 = moles x 1000 => 0.00625 / 6.25 x 10-3 x 1000
Volume 5
= 1.25M / 1.25 moledm-3
- 0.388g of a monobasic organic acid B required 46.5 cm3 of 0.095M sodium hydroxide for complete neutralization. Name and draw the structural formula of B
Moles of NaOH used = molarity x volume
1000
=> 0.095 x 46.5 = 0.0044175 /4.4175 x 10-3moles
1000
Mole ratio B : NaOH = 1:1
Moles B = 0.0044175 /4.4175 x 10-3moles
Molar mass B = mass => 0.388
moles 0.0044175 /4.4175 x 10-3moles
= 87.8324 gmole-1
X-COOH = 87.8324 where X is an alkyl group
X =87.8324- 42 = 42.8324=43
By elimination: CH3 = 15 CH3CH2 = 29 CH3CH2 CH2 = 43
Molecula formula : CH3CH2 CH2COOH
Molecule name : Butan-1-oic acid
Molecular structure
H H H O
H C C C C O H
H H H H
- 10.5 g of an impure sample containing ammonium sulphate (VI) fertilizer was warmed with 250cm3 of o.8M sodium hydroxide solution.The excess of the alkali was neutralized by 85cm3 of 0.5M hydrochloric acid. Calculate the % of impurities in the ammonium sulphate (VI)fertilizer. (N=14.0,S=32.0,O=16.0, H=1.0)
Equation for neutralization
NaOH(aq) + HCl(aq) -> NaOH(aq) + H2O(l)
Mole ratio NaOH(aq):HCl(aq)= 1:1
Moles of HCl = Molarity x volume => 0.5 x 85 = 0.0425 moles
1000 1000
Excess moles of NaOH(aq)= 0.0425 moles
Equation for reaction with ammonium salt
2NaOH(aq) + (NH4) 2SO4(aq) -> Na 2SO4(aq) + 2NH3 (g)+ 2H2O(l)
Mole ratio NaOH(aq): (NH4) 2SO4(aq)= 2:1
Total moles of NaOH = Molarity x volume => 0.8 x 250 = 0.2 moles
1000 1000
Moles of NaOH that reacted with(NH4) 2SO4 = 0.2 – 0.0425 = 0.1575moles
Moles (NH4) 2SO4 = ½ x 0.1575moles = 0. 07875moles
Molar mass (NH4) 2SO4= 132 gmole-1
Mass of in impure sample = moles x molar mass =>0. 07875 x 132 = 10.395 g
Mass of impurities = 10.5 -10.395 = 0.105 g
% impurities = 0.105 x 100 = 1.0 %
10.5
Practically volumetric analysis involves titration.
Titration generally involves filling a burette with known/unknown concentration of a solution then adding the solution to unknown/known concentration of another solution in a conical flask until there is complete reaction. If the solutions used are both colourless, an indicator is added to the conical flask. When the reaction is over, a slight/little excess of burette contents change the colour of the indicator. This is called the end point.
Set up of titration apparatus
The titration process involve involves determination of titre. The titre is the volume of burette contents/reading before and after the end point. Burette contents/reading before titration is usually called the Initial burette reading. Burette contents/reading after titration is usually called the Final burette reading. The titre value is thus a sum of the Final less Initial burette readings.
To reduce errors, titration process should be repeated at least once more.
The results of titration are recorded in a titration table as below
Sample titration table
| Titration number | 1 | 2 | 3 |
| Final burette reading (cm3) | 20.0 | 20.0 | 20.0 |
| Initial burette reading (cm3) | 0.0 | 0.0 | 0.0 |
| Volume of solution used(cm3) | 20.0 | 20.0 | 20.0 |
As evidence of a titration actually done examining body requires the candidate to record their burette readings before and after the titration.
For KCSE candidates burette readings must be recorded in a titration table in the format provided by the Kenya National Examination Council.
As evidence of all titration actually done Kenya National Examination Council require the candidate to record their burette readings before and after the titration to complete the titration table in the format provided.
Calculate the average volume of solution used
24.0 + 24.0 + 24.0 = 24.0 cm3
3
As evidence of understanding the degree of accuracy of burettes , all readings must be recorded to a decimal point.
As evidence of accuracy in carrying the out the titration , candidates value should be within 0.2 of the school value .
The school value is the teachers readings presented to the examining body/council based on the concentrations of the solutions s/he presented to her/his candidates.
Bonus mark is awarded for averaged reading within 0.1 school value as Final answer.
Calculations involved after the titration require candidates thorough practical and theoretical practice mastery on the:
(i)relationship among the mole, molar mass, mole ratios, concentration, molarity.
(ii) mathematical application of 1st principles.
Very useful information which candidates forget appears usually in the beginning of the question paper as:
“You are provided with…”
All calculation must be to the 4th decimal point unless they divide fully to a lesser decimal point.
Candidates are expected to use a non programmable scientific calculator.
(a)Sample Titration Practice 1 (Simple Titration)
You are provided with:
0.1M sodium hydroxide solution A
Hydrochloric acid solution B
You are required to determine the concentration of solution B in moles per litre.
Procedure
Fill the burette with solution B. Pipette 25.0cm3 of solution A into a conical flask. Titrate solution A with solution B using phenolphthalein indicator to complete the titration table 1
Sample results Titration table 1
| Titration number | 1 | 2 | 3 |
| Final burette reading (cm3) | 20.0 | 20.0 | 20.0 |
| Initial burette reading (cm3) | 0.0 | 0.0 | 0.0 |
| Volume of solution B used(cm3) | 20.0 | 20.0 | 20.0 |
Sample worked questions
- Calculate the average volume of solution B used
Average titre = Titre 1 + Titre 2 +Titre 3 => ( 20.0 +20.0 +20.0 ) = 20.0cm3
3 3
- How many moles of:
(i)solution A were present in 25cm3 solution.
Moles of solution A = Molarity x volume = 0.1 x 25 = 2.5 x 10-3 moles
1000 1000
(ii)solution B were present in the average volume.
Chemical equation: NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l)
Mole ratio 1:1 => Moles of A = Moles of B = 2.5 x 10-3 moles
(iii) solution B in moles per litre.
Moles of B per litre = moles x 1000 = 2.5 x 10-3 x 1000 = 0.1M
Volume 20
(b)Sample Titration Practice 2 (Redox Titration)
You are provided with:
Acidified Potassium manganate(VII) solution A
0.1M of an iron (II)salt solution B
8.5g of ammonium iron(II)sulphate(VI) crystals(NH4)2 SO4FeSO4.xH2O solid C
You are required to
(i)standardize acidified potassium manganate(VII)
(ii)determine the value of x in the formula (NH4)2 SO4FeSO4.xH2O.
Procedure 1
Fill the burette with solution A. Pipette 25.0cm3 of solution B into a conical flask. Titrate solution A with solution B until a pink colour just appears.
Record your results to complete table 1.
Table 1:Sample results
| Titration number | 1 | 2 | 3 |
| Final burette reading (cm3) | 20.0 | 20.0 | 20.0 |
| Initial burette reading (cm3) | 0.0 | 0.0 | 0.0 |
| Volume of solution A used(cm3) | 20.0 | 20.0 | 20.0 |
Sample worked questions
- Calculate the average volume of solution A used
Average titre = Titre 1 + Titre 2 +Titre 3 => ( 20.0 +20.0 +20.0 ) = 20.0cm3
3 3
- How many moles of:
(i)solution B were present in 25cm3 solution.
Moles of solution A = Molarity x volume = 0.1 x 25 = 2.5 x 10-3 moles
1000 1000
(ii)solution A were present in the average volume. Assume one mole of B react with five moles of B
Mole ratio A : B = 1:5
=> Moles of A = Moles of B = 2.5 x 10-3 moles = 5.0 x 10 -4 moles
5 5
(iii) solution B in moles per litre.
Moles of B per litre = moles x 1000 = 2.5 x 10-3 x 1000
Volume 20
= 0.025 M /moles per litre /moles l-1
Procedure 2
Place all the solid C into the 250cm3 volumetric flask carefully. Add about 200cm3 of distilled water. Shake to dissolve. Make up to the 250cm3 of solution by adding more distilled water. Label this solution C. Pipette 25cm3 of solution C into a conical flask, Titrate solution C with solution A until a permanent pink colour just appears. Complete table 2.
Table 2:Sample results
| Titration number | 1 | 2 | 3 |
| Final burette reading (cm3) | 20.0 | 20.0 | 20.0 |
| Initial burette reading (cm3) | 0.0 | 0.0 | 0.0 |
| Volume of solution A used(cm3) | 20.0 | 20.0 | 20.0 |
Sample worked questions
- Calculate the average volume of solution A used
Average titre = Titre 1 + Titre 2 +Titre 3 => ( 20.0 +20.0 +20.0 ) = 20.0cm3
3 3
- How many moles of:
(i)solution A were present inin the average titre.
Moles of solution A = Molarity x volume = 0.025 x 20 = 5.0 x 10-4 moles
1000 1000
(ii)solution C in 25cm3 solution given the equation for the reaction:
MnO4– (aq) + 8H+(aq) + 5Fe2+ (aq) -> Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
Mole ratio MnO4– (aq): 5Fe2+ (aq) = 1:5 => Moles of 5Fe2+ (aq) = Moles of MnO4– (aq) = 5.0 x 10-4 moles = 1.0 x 10 -4 moles
5 5
(iii) solution B in 250cm3.
Moles of B per litre = moles x 250 = 1.0 x 10 -4 x 250 = 1.0 x 10 -3 moles Volume 25
- Calculate the molar mass of solid C and hence the value of x in the chemical formula (NH4)2SO4FeSO4.xH2O.
(N=14.0, S=32.0, Fe=56.0, H=1.0 O=16.0)
Molar mass = mass perlitre = 8.5 = 8500 g
Moles per litre 1.0 x 10 -3 moles
NH4)2SO4FeSO4.xH2O = 8500
284 + 18x =8500
8500 – 284 = 8216 = 18x = 454.4444
18 18
x = 454 (whole number)
(c)Sample Titration Practice 3 (Back titration)
You are provided with:
(i)an impure calcium carbonate labeled M
(ii)Hydrochloric acid labeled solution N
(iii)solution L containing 20g per litre sodium hydroxide.
You are required to determine the concentration of N in moles per litre and the % of calcium carbonate in mixture M.
Procedure 1
Pipette 25.0cm3 of solution L into a conical flask. Add 2-3 drops of phenolphthalein indicator. Titrate with dilute hydrochloric acid solution N and record your results in table 1(4mark)
Sample Table 1
| 1 | 2 | 3 | |
| Final burette reading (cm3) | 6.5 | 6.5 | 6.5 |
| Initial burette reading (cm3) | 0.0 | 0.0 | 0.0 |
| Volume of N used (cm3) | 6.5 | 6.5 | 6.5 |
Sample questions
(a) Calculate the average volume of solution N used
6.5 + 6.5 + 6.5 = 6.5 cm3
3
(b) How many moles of sodium hydroxide are contained in 25cm3of solution L
Molar mass NaOH =40g
Molarity of L = mass per litre => 20 = 0.5M
Molar mass NaOH 40
Moles NaOH in 25cm3 = molarity x volume => 0.5M x 25cm3 = 0.0125 moles
1000 1000
(c)Calculate:
(i)the number of moles of hydrochloric acidthat react with sodium hydroxide in (b)above.
Mole ratio NaOH : HCl from stoichiometric equation= 1:1
Moles HCl =Moles NaOH => 0.0125 moles
(ii)the molarity of hydrochloric acid solution N.
Molarity = moles x 1000 => 0.0125 moles x 1000 =1.9231M/moledm-3
6.5 6.5
Procedure 2
Place the 4.0 g of M provided into a conical flask and add 25.0cm3 of the dilute hydrochloric acid to it using a clean pipette. Swirl the contents of the flask vigorously until effervescence stop.Using a 100ml measuring cylinder add 175cm3 distilled waterto make up the solution up to 200cm3.Label this solution K.Using a clean pipettetransfer 25.0cm3 of the solution into a clean conical flask and titrate with solution L from the burette using 2-3 drops of methyl orange indicator.Record your observations in table 2.
Sample Table 2
| 1 | 2 | 3 | |
| Final burette reading (cm3) | 24.5 | 24.5 | 24.5 |
| Initial burette reading (cm3) | 0.0 | 0.0 | 0.0 |
| Volume of N used (cm3) | 24.5 | 24.5 | 24.5 |
Sample calculations
(a)Calculate the average volume of solution L used(1mk)
24.5 + 24.5 + 24.5 = 24.5cm3
3
(b)How many moles of sodium hydroxide are present in the average volume of solution L used?
Moles = molarity x average burette volume => 0.5 x 24.5
1000 1000
= 0.01225 /1.225 x 10-2 moles
(c) How many moles of hydrochloric acid are present in the original 200cm3 of solution K?
Mole ratio NaOH: HCl = 1:1 => moles of HCl = 0.01225 /1.225 x 10-2 moles
Moles in 200cm3 = 200cm3 x 0.01225 /1.225 x 10-2moles
25cm3(volume pipetted)
=0.49 /4.9 x 10-1moles
(d)How many moles of hydrochloric acid were contained in original 25 cm3 solution N used
Original moles = Original molarity x pipetted volume =>
1000cm3
1.9231M/moledm-3 x 25 = 0.04807/4.807 x 10-2 moles
1000
(e)How many moles of hydrochloric acid were used to react with calcium carbonate present?
Moles that reacted = original moles –moles in average titre =>
0.04807/4.807 x 10-2moles – 0.01225 /1.225 x 10-2moles
= 0.03582/3.582 x 10 -2 moles
(f)Write the equation for the reaction between calcium carbonate and hydrochloric acid.
CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l)
(g)Calculate the number of moles of calcium carbonate that reacted with hydrochloric acid.
From the equation CaCO3(s):2HCl(aq) = 1:2
=> Moles CaCO3(s) = 1/2moles HCl
= 1/2 x 0.03582/3.582 x 10 -2 moles
= 0.01791 /1.791 x 10-2moles
(h)Calculate the mass of calcium carbonate in 4.0g of mixture M (Ca=40.0,O = 16.0,C=12.0)
Molar mass CaCO3 = 100g
Mass CaCO3 = moles x molar mass => 0.01791 /1.791 x 10-2moles x 100g
= 1.791g
(i)Determine the % of calcium carbonate present in the mixture
% CaCO3 = mass of pure x 100% => 1.791g x 100% = 44.775%
Mass of impure 4.0
(d)Sample titration practice 4 (Multiple titration)
You are provided with:
(i)sodium L containing 5.0g per litre of a dibasic organic acid H2X.2H2O.
(ii)solution M which is acidified potassium manganate(VII)
(iii)solution N a mixture of sodium ethanedioate and ethanedioic acid
(iv)0.1M sodium hydroxide solution P
(v)1.0M sulphuric(VI)
You are required to:
(i)standardize solution M using solution L
(ii)use standardized solution M and solution P to determine the % of sodium ethanedioate in the mixture.
Procedure 1
Fill the burette with solution M. Pipette 25.0cm3 of solution L into a conical flask. Heat this solution to about 70oC(but not to boil).Titrate the hot solution L with solution M until a permanent pink colour just appears .Shake thoroughly during the titration. Repeat this procedure to complete table 1.
Sample Table 1
| 1 | 2 | 3 | |
| Final burette reading (cm3) | 24.0 | 24.0 | 24.0 |
| Initial burette reading (cm3) | 0.0 | 0.0 | 0.0 |
| Volume of N used (cm3) | 24.0 | 24.0 | 24.0 |
Sample calculations
(a)Calculate the average volume of solution L used (1mk)
24.0 + 24.0 + 24.0 = 24.0cm3
3
(b)Given that the concentration of the dibasic acid is 0.05molesdm-3.determine the value of x in the formula H2X.2H2O (H=1.0,O=16.0)
Molar mass H2X.2H2O= mass per litre => 5.0g/litre = 100g
Moles/litre 0.05molesdm-3
H2X.2H2O =100
X = 100 – ((2 x1) + 2 x (2 x1) + (2 x 16) => 100 – 34 = 66
(c) Calculate the number of moles of the dibasic acid H2X.2H2O.
Moles = molarity x pipette volume => 0.5 x 25 = 0.0125/1.25 x10 -2 moles
1000 1000
(d)Given the mole ratio manganate(VII)(MnO4–): acid H2X is 2:5, calculate the number of moles of manganate(VII) (MnO4–) in the average titre.
Moles H2X = 2/5 moles of MnO4–
=> 2/5 x 0.0125/1.25 x10 -2 moles
= 0.005/5.0 x 10 -3moles
(e)Calculate the concentration of the manganate(VII)(MnO4–) in moles per litre.
Moles per litre/molarity = moles x 1000
average burette volume
=>0.005/5.0 x 10 -3moles x 1000 = 0.2083 molesl-1/M
24.0
Procedure 2
With solution M still in the burette ,pipette 25.0cm3 of solution N into a conical flask. Heat the conical flask containing solution N to about 70oC.Titrate while hot with solution M.Repeat the experiment to complete table 2.
Sample Table 2
| 1 | 2 | 3 | |
| Final burette reading (cm3) | 12.5 | 12.5 | 12.5 |
| Initial burette reading (cm3) | 0.0 | 0.0 | 0.0 |
| Volume of N used (cm3) | 12.5 | 12.5 | 12.5 |
Sample calculations
(a)Calculate the average volume of solution L used (1mk)
12.5 + 12.5 + 12.5 =12.5cm3
3
(b)Calculations:
(i)How many moles of manganate(VII)ions are contained in the average volume of solution M used?
Moles = molarity of solution M x average burette volume
1000
=> 0.2083 molesl-1/ M x 12.5 = 0.0026 / 2.5 x 10-3 moles
1000
(ii)The reaction between manganate(VII)ions and ethanedioate ions that reacted with is as in the equation:
2MnO4– (aq) + 5C2O42- (aq) + 16H+ (aq) -> 2Mn2+(aq) + 10CO2(g) + 8H2O(l)
Calculate the number of moles of ethanedioate ions that reacted with manganate (VII) ions in the average volume of solution M.
From the stoichiometric equation,mole ratio MnO4– (aq): C2O42- (aq) = 2:5
=> moles C2O42- = 5/2 moles MnO4– => 5/2 x 0.0026 / 2.5 x 10-3 moles
= 0.0065 /6.5 x10-3 moles
(iii)Calculate the number of moles of ethanedioate ions contained in 250cm3 solution N.
25cm3 pipette volume -> 0.0065 /6.5 x10-3 moles
250cm3 ->
0.0065 /6.5 x10-3 moles x 250 = 0.065 / 6.5 x10-2 moles
25
Procedure 3
Remove solution M from the burette and rinse it with distilled water. Fill the burette with sodium hydroxide solution P. Pipette 25cm3 of solution N into a conical flask and add 2-3 drops of phenolphthalein indicator. Titrate this solution N with solution P from the burette. Repeat the procedure to complete table 3.
Sample Table 2
| 1 | 2 | 3 | |
| Final burette reading (cm3) | 24.9 | 24.9 | 24.9 |
| Initial burette reading (cm3) | 0.0 | 0.0 | 0.0 |
| Volume of N used (cm3) | 24.9 | 24.9 | 24.9 |
Sample calculations
(a)Calculate the average volume of solution L used (1mk)
24.9 + 24.9 + 24.9 = 24.9 cm3
3
(b)Calculations:
(i)How many moles of sodium hydroxide solution P were contained in the average volume?
Moles = molarity of solution P x average burette volume
1000
=> 0.1 molesl-1 x 24.9 = 0.00249 / 2.49 x 10-3 moles
1000
(ii)Given that NaOH solution P reacted with the ethanedioate ions from the acid only and the equation for the reaction is:
2NaOH (aq) + H2C2O4 (aq) -> Na2C2O4(g) + 2H2O(l)
Calculate the number of moles of ethanedioic acid that were used in the reaction
From the stoichiometric equation,mole ratio NaOH(aq): H2C2O4 (aq) = 2:1
=> moles H2C2O4 = 1/2 moles NaOH => 1/2 x 0.00249 / 2.49 x 10-3 moles
= 0.001245/1.245 x10-3 moles.
(iii)How many moles of ethanedioic acid were contained in 250cm3 of solution N?
25cm3 pipette volume -> 0.001245/1.245 x10-3 moles
250cm3 ->
0.001245/1.245 x10-3 moles x 250 = 0.01245/1.245 x10-2 moles
25
(iii)Determine the % by mass of sodium ethanedioate in the micture (H= 1.0,O=16.0,C=12.0 and total mass of mixture =2.0 g in 250cm3 solution)
Molar mass H2C2O4 = 90.0g
Mass of H2C2O4 in 250cm3 = moles in 250cm3 x molar mass H2C2O4
=>0.01245/1.245 x10-2 moles x 90.0
= 1.1205g
% by mass of sodium ethanedioate
=(Mass of mixture – mass of H2C2O4) x 100%
Mass of mixture
=> 2.0 – 1.1205 g = 43.975%
2.0
Note
(i) L is 0.05M Oxalic acid
(ii) M is 0.01M KMnO4
(iii) N is 0.03M oxalic acid(without sodium oxalate)
Practice example 5.(Determining equation for a reaction)
You are provided with
-0.1M hydrochloric acid solution A
-0.5M sodium hydroxide solution B
You are to determine the equation for thereaction between solution A and B
Procedure
Fill the burette with solution A.Using a pipette and pipette filler transfer 25.0cm3 of solution B into a conical flask.Add 2-3 drops of phenolphthalein indicator.Run solution A into solution B until a permanent pink colour just appears.Record your results in Table 1.Repeat the experiment to obtain three concordant results to complete Table 1
Table 1(Sample results)
| Titration | 1 | 2 | 3 |
| Final volume(cm3) | 12.5 | 25.0 | 37.5 |
| Initial volume(cm3) | 0.0 | 12.5 | 25.0 |
| Volume of solution A used(cm3) | 12.5 | 12.5 | 12.5 |
Sample questions
Calculate the average volume of solution A used.
12.5+12.5+12.5 = 12.5cm3
3
Theoretical Practice examples
- 1.0g of dibasic acid HOOC(CH2)xCOOH was dissolved in 250cm3 solution. 25.0 cm3 of this solution reacted with 30.0cm3 of 0.06M sodium hydroxide solution. Calculate the value of x in HOOC(CH2)xCOOH. (C=12.0,H=1.0,O=16.)
Chemical equation
2NaOH(aq) + H2X(aq) -> Na2X (aq) + 2H2O(aq)
Mole ratio NaOH(aq) :H2X(aq) = 2:1
Method 1
Ma Va = na => Ma x 25.0 = 1 => Ma =0.06 x 30.0 x1
Mb Vb = nb 0.06 x 30.0 2 25.0 x 2
Molarity of acid = 0.036M/Mole l-1
Mass of acid per lite = 1.0 x1000 = 4.0 g/l
250
0.036M/ Mole l-1 -> 4.0 g /l
1 mole= molar mass of HOOC(CH2)xCOOH = 4.0 x 1 = 111.1111 g
0.036
Molar mass (CH2)x = 111.1111 – (HOOCCOOH = 90.0) = 21.1111
(CH2)x = 14x = 21.1111 = 1.5 = 1 (whole number)
14
Method 2
Moles of sodium hydroxide = Molarity x volume = 0.06 x 30 = 1.8 x 10 -3moles
1000
Moles of Hydrochloric acid = 1/2 x 1.8 x 10 -3moles = 9.0 x10 -4moles
Molarity of Hydrochloric acid = moles x 1000 = 9.0 x10 -4moles x1000
Volume 25
Molarity of acid = 0.036M/Mole l-1
Mass of acid per lite = 1.0 x1000 = 4.0 g/l
250
0.036M/ Mole l-1 -> 4.0 g /l
1 mole= molar mass of HOOC(CH2)xCOOH = 4.0 x 1 = 111.1111 g
0.036
Molar mass (CH2)x = 111.1111 – (HOOCCOOH = 90.0) = 21.1111
(CH2)x = 14x = 21.1111 = 1.5 = 1 (whole number)
14
- 20.0cm3 of 0.05 M acidified potassium manganate(VII)solution oxidized 25.0cm3 of Fe2+(aq) ions in 40.0g/l of impure Iron (II)sulphate(VI) to Fe3+(aq) ions. Calculate the percentage impurities in the Iron (II)sulphate(VI).
MnO4– (aq) + 8H+(aq)+ 5Fe2+(aq)-> 5Fe3+(aq) + Mn2+(aq) + 4H2O(aq)
Fe=56.0,S= 32.0, O=16.0).
Moles of MnO4– (aq) = Molarity x volume = 0.05 x 20.0 = 0.001 Moles
1000 1000
Mole ratio MnO4– (aq): 5Fe2+(aq)= 1:5
Moles 5Fe2+(aq) = 5 x0.001 = 0.005 Moles
Moles of 5Fe2+(aq) per litre/molarity = Moles x 1000 = 0005 x 1000
Volume 25.0
= 0.2 M/ Moles/litre
Molar mass =FeSO4=152 g
Mass of in the mixture = Moles x molar mass => 0.2 x 152 = 30.4 g
Mass of impurity = 40.0 – 30.4 =9.6 g
% impurity = 9.6 g x100 = 24.0 % impurity
40.0
3.9.7 g of a mixture of Potassium hydroxide and Potassium chloride was dissolved to make one litre solution.20.0cm3 of this solution required 25.0cm3 of 0.12M hydrochloric acid for completed neutralization. Calculate the percentage by mass of Potassium chloride.(K=39.0,Cl= 35.5)
Chemical equation
KOH(aq) + HCl(aq) -> KCl(aq) + H2O(l)
Moles of HCl = Molarity x volume => 0.12 x 25.0 = 0.003/3.0 x 10 -3 moles
1000 1000
Mole ratio KOH(aq) : HCl(aq) -= 1:1
Moles KOH =0.003/3.0 x 10 -3 moles
Method 1
Molar mass KOH =56.0g
Mass KOH in 25cm3 =0.003/3.0 x 10 -3 moles x56.0 = 0.168g
Mass KOH in 1000cm3/1 litre = 0.168 x1000= 8.4 g/l
20
Mass of KCl = 9.7g – 8.4g = 1.3 g
% of KCl = 1.3 x 100 = 13.4021%
9.7
Method 2
Moles KOH in 1000cm3 /1 litre = Moles in 20cm3 x 1000 =>0.003 x 1000
20 20
=0.15M/Moles /litre
Molar mass KOH =56.0g
Mass KOH in 1000/1 litre = 0.15M/Moles /litre x 56.0 = 8.4g/l
Mass of KCl = 9.7g – 8.4g = 1.3 g
% of KCl = 1.3 x 100 = 13.4021%
9.7
4.A certain carbonate, GCO3, reacts with dilute hydrochloric acid according to the equation given below:
GCO3(s) + 2HCl(aq) -> GCl2 (aq) + CO2 (g) + H2O(l)
If 1 g of the carbonate reacts completely with 20 cm3 of 1 M hydrochloric acid ,calculate the relative atomic mass of G (C = 12.0 = 16.0)
Moles of HCl = Molarity x volume=> 1 x20 = 0.02 moles
1000 1000
Mole ratio HCl; GCO3 = 2:1
Moles of GCO3= 0.02 moles = 0.01moles
2
Molar mass of GCO3 = mass => 1 = 100 g
moles 0.01moles
G= GCO3 – CO3 =>100g – (12+ 16 x3 = 60) = 40(no units)
- 46.0g of a metal carbonate MCO3 was dissolved 160cm3 of 0.1M excess hydrochloric acid and the resultant solution diluted to one litre.25.0cm3 of this solution required 20.0cm3 of 0.1M sodium hydroxide solution for complete neutralization. Calculate the atomic mass of ‘M’
Equation
Chemical equation
NaOH(aq) + HCl(aq) -> KCl(aq) + H2O(l)
Moles of NaOH = Molarity x volume=> 0.1 x20 = 0.002 moles
1000 1000
Mole ratio HCl; NaOH = 1:1
Excess moles of HCl = 0.002 moles
25cm3 -> 0.002 moles
1000cm3 -> 1000 x 0.002 = 0.08moles
25cm3
Original moles of HCl = Molarity x volume => 1M x 1litre = 1.0 moles
Moles of HCl reacted with MCO3 = 1.0 – 0.08 moles = 0.92moles
Chemical equation
MCO3(s) + 2HCl(aq) -> MCl2 (aq) + CO2 (g) + H2O(l)
Mole ratio MCO3(s) : HCl(aq) =1:2
Moles of MCO3 = 0.92moles => 0.46moles
2
Molar mass of MCO3= mass => 46g = 100 g
moles 0.46moles
M= MCO3 – CO3 =>100g – (12+ 16 x3 = 60) = 40
- 25.0cm3 of a mixture of Fe2+ and Fe3+ ions in an aqueous salt was acidified with sulphuric(VI)acid then titrated against potassium manganate(VI).The salt required 15cm3 ofe0.02M potassium manganate(VI) for complete reaction.
A second 25cm3 portion of the Fe2+ and Fe3+ ion salt was reduced by Zinc then titrated against the same concentration of potassium manganate(VI).19.0cm3 of potassium manganate(VI)solution was used for complete reaction. Calculate the concentration of Fe2+ and Fe3+ ion in the solution on moles per litre.
Mole ratio Fe2+ :Mn04– = 5:1
Moles Mn04– used = 0.02 x 15 = 3.0 x 10-4 moles
1000
Moles Fe2+ = 3.0 x 10-4 moles = 6.0 x 10-5 moles
5
Molarity of Fe2+ = 6.0 x 10-4 moles x 1000 = 2.4 x 10-3 moles l-1
25
Since Zinc reduces Fe3+ to Fe2+ in the mixture:
Moles Mn04– that reacted with all Fe2+= 0.02 x 19 = 3.8 x 10-4 moles
1000
Moles of all Fe2+ = 3.8 x 10-4 moles = 7.6 x 10-5 moles
5
Moles of Fe3+ = 3.8 x 10-4 – 6.0 x 10-5 = 1.6 x 10-5 moles
Molarity of Fe3+ = 1.6 x 10-5 moles x 1000 = 4.0 x 10-4 moles l-1
25
|
UPGRADE
CHEMISTRY
FORM 3
ORGANIC Chemistry 1&2
Comprehensive tutorial notes
MUTHOMI S.G 0720096206
|
Introduction to Organic chemistry
Organic chemistry is the branch of chemistry that studies carbon compounds present in living things, once living things or synthetic/man-made.
Compounds that makes up living things whether alive or dead mainly contain carbon. Carbon is tetravalent.
It is able to form stable covalent bonds with itself and many non-metals like hydrogen, nitrogen ,oxygen and halogens to form a variety of compounds. This is because:
(i) carbon uses all the four valence electrons to form four strong covalent bond.
(ii)carbon can covalently bond to form a single, double or triple covalent bond with itself.
(iii)carbon atoms can covalently bond to form a very long chain or ring.
When carbon covalently bond with Hydrogen, it forms a group of organic compounds called Hydrocarbons
A.HYDROCARBONS (HCs)
Hydrocarbons are a group of organic compounds containing /made up of hydrogen and carbon atoms only.
Depending on the type of bond that exist between the individual carbon atoms, hydrocarbon are classified as:
(i) Alkanes
(ii) Alkenes
(iii) Alkynes
(i) Alkanes
(a)Nomenclature/Naming
These are hydrocarbons with a general formula CnH2n+2 where n is the number of Carbon atoms in a molecule.
The carbon atoms are linked by single bond to each other and to hydrogen atoms.
They include:
| n | General/
Molecular formula |
Structural formula | Name
|
| 1 | CH4 | H
H C H
H
|
Methane |
| 2 | C2H6 | H H
H C C H
H H
|
Ethane
|
| 3 | C3H8 | H H H
H C C C H
H H H
|
Propane |
| 4 | C4H10 | H H H H
H C C C C H
H H H H
|
Butane |
| 5 | C5H12 | H H H H H
H C C C C C H CH3 (CH2) 6CH3
H H H H H
|
Pentane |
| 6 | C6H14 | H H H H H H
H C C C C C C H CH3 (CH2) 6CH3
H H H H H H
|
Hexane |
| 7 | C7H16 | H H H H H H H
H C C C C C C C H
H H H H H H H
|
Heptane |
| 8 | C8H18 | H H H H H H H H
H C C C C C C C C H
H H H H H H H H
|
Octane |
| 9 | C9H20 | H H H H H H H H H
H C C C C C C C C C H
H H H H H H H H H
|
Nonane |
| 10 | C10H22 | H H H H H H H H H H
H C C C C C C C C C C H
H H H H H H H H H H
|
decane |
Note
1.The general formula/molecular formular of a compound shows the number of each atoms of elements making the compound e.g.
Decane has a general/molecular formula C10H22 ;this means there are 10 carbon atoms and 22 hydrogen atoms in a molecule of decane.
2.The structural formula shows the arrangement/bonding of atoms of each element making the compound e.g
Decane has the structural formula as in the table above ;this means the 1st carbon from left to right is bonded to three hydrogen atoms and one carbon atom.
The 2nd carbon atom is joined/bonded to two other carbon atoms and two Hydrogen atoms.
3.Since carbon is tetravalent ,each atom of carbon in the alkane MUST always be bonded using four covalent bond /four shared pairs of electrons.
4.Since Hydrogen is monovalent ,each atom of hydrogen in the alkane MUST always be bonded using one covalent bond/one shared pair of electrons.
5.One member of the alkane differ from the next/previous by a CH2 group.
e.g
Propane differ from ethane by one carbon and two Hydrogen atoms form ethane. Ethane differ from methane also by one carbon and two Hydrogen atoms
6.A group of compounds that differ by a CH2 group from the next /previous consecutively is called a homologous series.
7.A homologous series:
(i) differ by a CH2 group from the next /previous consecutively
(ii)have similar chemical properties
(iii)have similar chemical formula that can be represented by a general formula e.g alkanes have the general formula CnH2n+2.
(iv)the physical properties (e.g.melting/boiling points)show steady gradual change)
8.The 1st four alkanes have the prefix meth_,eth_,prop_ and but_ to represent 1,2,3 and 4 carbons in the compound. All other use the numeral prefix pent_,Hex_,hept_ , etc to show also the number of carbon atoms.
9.If one hydrogen atom in an alkane is removed, an alkyl group is formed.e.g
| Alkane name | molecular structure
CnH2n+2 |
Alkyl name | Molecula structure
CnH2n+1 |
| methane | CH4 | methyl | CH3 |
| ethane | CH3CH3 | ethyl | CH3 CH2 |
| propane | CH3 CH2 CH3 | propyl | CH3 CH2 CH2 |
| butane | CH3 CH2 CH2 CH3 | butyl | CH3 CH2 CH2 CH2 |
(b)Isomers of alkanes
Isomers are compounds with the same molecular general formula but different molecular structural formula.
Isomerism is the existence of a compounds having the same general/molecular formula but different structural formula.
The 1st three alkanes do not form isomers.Isomers are named by using the IUPAC(International Union of Pure and Applied Chemistry) system of nomenclature/naming.
The IUPAC system of nomenclature uses the following basic rules/guidelines:
1.Identify the longest continuous carbon chain to get/determine the parent alkane.
2.Number the longest chain form the end of the chain that is near the branches so as the branch get the lowest number possible
- Determine the position, number and type of branches. Name them as methyl, ethyl, propyl e.tc. according to the number of carbon chains attached to the parent alkane. Name them fluoro-,chloro-,bromo-,iodo- if they are halogens
4.Use prefix di-,tri-,tetra-,penta-,hexa- to show the number of branches attached to the parent alkane.
Practice on IUPAC nomenclature of alkanes
(a)Draw the structure of:
(i)2-methylpentane
Procedure
- Identify the longest continuous carbon chain to get/determine the parent alkane.
Butane is the parent name CH3 CH2 CH2 CH3
- Number the longest chain form the end of the chain that is near the branches so as the branch get the lowest number possible
The methyl group is attached to Carbon “2”
- Determine the position, number and type of branches. Name them as methyl, ethyl, propyl e.tc. according to the number of carbon chains attached to the parent alkane i.e
Position of the branch at carbon “2”
Number of branches at carbon “1”
Type of the branch “methyl” hence
Molecular formula
CH3
CH3 CH CH2 CH3 // CH3 CH (CH3 ) CH2CH3
Structural formula
H H H H
H C C C C H
H H H
H C H
H
(ii)2,2-dimethylpentane
Procedure
- Identify the longest continuous carbon chain to get/determine the parent alkane.
Butane is the parent name CH3 CH2 CH2 CH3
- Number the longest chain form the end of the chain that is near the branches so as the branch get the lowest number possible
The methyl group is attached to Carbon “2”
- Determine the position, number and type of branches. Name them as methyl, ethyl, propyl e.tc. according to the number of carbon chains attached to the parent alkane i.e
Position of the branch at carbon “2”
Number of branches at carbon “2”
Type of the branch two“methyl” hence
Molecular formular
CH3
CH3 C CH2 CH3 // CH3 C (CH3 )2 CH2CH3
CH3
Structural formula
H
H C H
H H H
H C C C C H
H H H
H C H
H
(iii) 2,2,3-trimethylbutane
Procedure
- Identify the longest continuous carbon chain to get/determine the parent alkane.
Butane is the parent name CH3 CH2 CH2 CH3
- Number the longest chain form the end of the chain that is near the branches so as the branch get the lowest number possible
The methyl group is attached to Carbon “2 and 3”
- Determine the position, number and type of branches. Name them as methyl, ethyl, propyl e.tc. according to the number of carbon chains attached to the parent alkane i.e
Position of the branch at carbon “2 and 3”
Number of branches at carbon “3”
Type of the branch three “methyl” hence
Molecular formular
CH3
CH3 C CH CH3 // CH3 C (CH3 )3 CH2CH3
CH3 CH3
Structural formula
H
H C H
H H
H C C C H
H H
H
H C C H
H
H C H
H
(iv) 1,1,1,2,2,2-hexabromoethane
Molecular formula
CBr3 CBr3
Structural formula
Br Br
Br C C Br
Br Br
(v) 1,1,1-tetrachloro-2,2-dimethylbutane
CH3
CCl 3 C CH3 // C Cl 3 C (CH3 )2 CH3
CH3
Structural formula
Cl
Cl C Cl
H H
H C C C H
H H
H C H
H
(c)Occurrence and extraction
Crude oil ,natural gas and biogas are the main sources of alkanes:
(i)Natural gas is found on top of crude oil deposits and consists mainly of methane.
(ii)Biogas is formed from the decay of waste organic products like animal dung and cellulose. When the decay takes place in absence of oxygen , 60-75% by volume of the gaseous mixture of methane gas is produced.
(iii)Crude oil is a mixture of many flammable hydrocarbons/substances. Using fractional distillation, each hydrocarbon fraction can be separated from the other. The hydrocarbon with lower /smaller number of carbon atoms in the chain have lower boiling point and thus collected first.
As the carbon chain increase, the boiling point, viscosity (ease of flow) and colour intensity increase as flammability decrease. Hydrocarbons in crude oil are not pure. They thus have no sharp fixed boiling point.
Uses of different crude oil fractions
| Carbon atoms in a molecule | Common name of fraction | Uses of fraction |
| 1-4 | Gas | L.P.G gas for domestic use |
| 5-12 | Petrol | Fuel for petrol engines |
| 9-16 | Kerosene/Paraffin | Jet fuel and domestic lighting/cooking |
| 15-18 | Light diesel | Heavy diesel engine fuel |
| 18-25 | Diesel oil | Light diesel engine fuel |
| 20-70 | Lubricating oil | Lubricating oil to reduce friction. |
| Over 70 | Bitumen/Asphalt | Tarmacking roads |
(d)School laboratory preparation of alkanes
In a school laboratory, alkanes may be prepared from the reaction of a sodium alkanoate with solid sodium hydroxide/soda lime.
Chemical equation:
Sodium alkanoate + soda lime -> alkane + Sodium carbonate
CnH2n+1COONa(s) + NaOH(s) -> C n H2n+2 + Na2CO3(s)
The “H” in NaOH is transferred/moves to the CnH2n+1 in CnH2n+1COONa(s) to form C n H2n+2.
Examples
- Methane is prepared from the heating of a mixture of sodium ethanoate and soda lime and collecting over water
Sodium ethanoate + soda lime -> methane + Sodium carbonate
CH3COONa(s) + NaOH(s) -> C H4 + Na2CO3(s)
The “H” in NaOH is transferred/moves to the CH3 in CH3COONa(s) to form CH4.
- Ethane is prepared from the heating of a mixture of sodium propanoate and soda lime and collecting over water
Sodium propanoate + soda lime -> ethane + Sodium carbonate
CH3 CH2COONa(s) + NaOH(s) -> CH3 CH3 + Na2CO3(s)
The “H” in NaOH is transferred/moves to the CH3 CH2 in CH3 CH2COONa (s) to form CH3 CH3
- Propane is prepared from the heating of a mixture of sodium butanoate and soda lime and collecting over water
Sodium butanoate + soda lime -> propane + Sodium carbonate
CH3 CH2CH2COONa(s) + NaOH(s) -> CH3 CH2CH3 + Na2CO3(s)
The “H” in NaOH is transferred/moves to the CH3 CH2 CH2 in CH3 CH2CH2COONa (s) to form CH3 CH2CH3
- Butane is prepared from the heating of a mixture of sodium pentanoate and soda lime and collecting over water
Sodium pentanoate + soda lime -> butane + Sodium carbonate
CH3 CH2 CH2CH2COONa(s)+NaOH(s) -> CH3 CH2CH2CH3 + Na2CO3(s)
The “H” in NaOH is transferred/moves to the CH3CH2 CH2 CH2 in CH3 CH2CH2 CH2COONa (s) to form CH3 CH2 CH2CH3
Laboratory set up for the preparation of alkanes
(d)Properties of alkanes
- Physical properties
Alkanes are colourless gases, solids and liquids that are not poisonous.
They are slightly soluble in water.
The solubility decrease as the carbon chain and thus the molar mass increase
The melting and boiling point increase as the carbon chain increase.
This is because of the increase in van-der-waals /intermolecular forces as the carbon chain increase.
The 1st four straight chain alkanes (methane,ethane,propane and butane)are therefore gases ,the nect six(pentane ,hexane, heptane,octane,nonane, and decane) are liquids while the rest from unidecane(11 carbon atoms) are solids .
The density of straight chain alkanes increase with increasing carbon chain as the intermolecular forces increases.
This reduces the volume occupied by a given mass of the compound.
Summary of physical properties of alkanes
| Alkane | General formula | Melting point(K) | Boiling point(K) | Density gcm-3 | State at room(298K) temperature and pressure atmosphere (101300Pa) |
| Methane | CH4 | 90 | 112 | 0.424 | gas |
| Ethane | CH3CH3 | 91 | 184 | 0.546 | gas |
| Propane | CH3CH2CH3 | 105 | 231 | 0.501 | gas |
| Butane | CH3(CH2)2CH3 | 138 | 275 | 0.579 | gas |
| Pentane | CH3(CH2)3CH3 | 143 | 309 | 0.626 | liquid |
| Hexane | CH3(CH2)4CH3 | 178 | 342 | 0.657 | liquid |
| Heptane | CH3(CH2)5CH3 | 182 | 372 | 0.684 | liquid |
| Octane | CH3(CH2)6CH3 | 216 | 399 | 0.703 | liquid |
| Nonane | CH3(CH2)7CH3 | 219 | 424 | 0.708 | liquid |
| Octane | CH3(CH2)8CH3 | 243 | 447 | 0.730 | liquid |
II.Chemical properties
(i)Burning.
Alkanes burn with a blue/non-luminous non-sooty/non-smoky flame in excess air to form carbon(IV) oxide and water.
Alkane + Air -> carbon(IV) oxide + water (excess air/oxygen)
Alkanes burn with a blue/non-luminous no-sooty/non-smoky flame in limited air to form carbon(II) oxide and water.
Alkane + Air -> carbon(II) oxide + water (limited air)
Examples
1.(a) Methane when ignited burns with a blue non sooty flame in excess air to form carbon(IV) oxide and water.
Methane + Air -> carbon(IV) oxide + water (excess air/oxygen)
CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l/g)
(b) Methane when ignited burns with a blue non sooty flame in limited air to form carbon(II) oxide and water.
Methane + Air -> carbon(II) oxide + water (excess air/oxygen)
2CH4(g) + 3O2(g) -> 2CO(g) + 4H2O(l/g)
2.(a) Ethane when ignited burns with a blue non sooty flame in excess air to form carbon(IV) oxide and water.
Ethane + Air -> carbon(IV) oxide + water (excess air/oxygen)
2C2H6(g) + 7O2(g) -> 4CO2(g) + 6H2O(l/g)
(b) Ethane when ignited burns with a blue non sooty flame in limited air to form carbon(II) oxide and water.
Ethane + Air -> carbon(II) oxide + water (excess air/oxygen)
2C2H6(g) + 5O2(g) -> 4CO(g) + 6H2O(l/g)
3.(a) Propane when ignited burns with a blue non sooty flame in excess air to form carbon(IV) oxide and water.
Propane + Air -> carbon(IV) oxide + water (excess air/oxygen)
C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l/g)
(b) Ethane when ignited burns with a blue non sooty flame in limited air to form carbon(II) oxide and water.
Ethane + Air -> carbon(II) oxide + water (excess air/oxygen)
2C3H8(g) + 7O2(g) -> 6CO(g) + 8H2O(l/g)
ii)Substitution
Substitution reaction is one in which a hydrogen atom is replaced by a halogen in presence of ultraviolet light.
Alkanes react with halogens in presence of ultraviolet light to form halogenoalkanes.
During substitution:
(i)the halogen molecule is split into free atom/radicals.
(ii)one free halogen radical/atoms knock /remove one hydrogen from the alkane leaving an alkyl radical.
(iii) the alkyl radical combine with the other free halogen atom/radical to form halogenoalkane.
(iv)the chlorine atoms substitute repeatedly in the alkane. Each substitution removes a hydrogen atom from the alkane and form hydrogen halide.
(v)substitution stops when all the hydrogen in alkanes are replaced with halogens.
Substitution reaction is a highly explosive reaction in presence of sunlight / ultraviolet light that act as catalyst.
Examples of substitution reactions
Methane has no effect on bromine or chlorine in diffused light/dark. In sunlight , a mixture of chlorine and methane explode to form colourless mixture of chloromethane and hydrogen chloride gas. The pale green colour of chlorine gas fades.
Chemical equation
1.(a)Methane + chlorine -> Chloromethane + Hydrogen chloride
CH4(g) + Cl2(g) -> CH3Cl (g) + HCl (g)
H H
H C H + Cl Cl -> H C Cl + H Cl
H H
(b) Chloromethane + chlorine -> dichloromethane + Hydrogen chloride
CH3Cl (g) + Cl2(g) -> CH2Cl2 (g) + HCl (g)
H H
H C Cl + Cl Cl -> H C Cl + H Cl
H Cl
(c) dichloromethane + chlorine -> trichloromethane + Hydrogen chloride
CH2Cl2 (g) + Cl2(g) -> CHCl3 (g) + HCl (g)
Cl H
H C Cl + Cl Cl -> Cl C Cl + H Cl
H Cl
(c) trichloromethane + chlorine -> tetrachloromethane + Hydrogen chloride
CHCl3 (g) + Cl2(g) -> CCl4 (g) + HCl (g)
H Cl
Cl C Cl + Cl Cl -> Cl C Cl + H Cl
Cl Cl
Ethane has no effect on bromine or chlorine in diffused light/dark. In sunlight , a mixture of bromine and ethane explode to form colourless mixture of bromoethane and hydrogen chloride gas. The red/brown colour of bromine gas fades.
Chemical equation
(a)Ethane + chlorine -> Chloroethane + Hydrogen chloride
CH3CH3(g) + Br2(g) -> CH3CH2Br (g) + HBr (g)
H H H H
H C C H + Br Br -> H C C H + H Br
H H H Br
Bromoethane
H H H Br
H C C H + Br Br -> H C C H + H Br
H Br H Br
1,1-dibromoethane
H Br H Br
H C C H + Br Br -> H C C Br + H Br
H Br H Br
1,1,1-tribromoethane
H Br H Br
H C C Br + Br Br -> H C C Br + H Br
H Br Br Br
1,1,1,2-tetrabromoethane
H Br H Br
H C C Br + Br Br -> Br C C Br + H Br
Br Br Br Br
1,1,1,2,2-pentabromoethane
H Br Br Br
Br C C Br + Br Br -> Br C C Br + H Br
Br Br Br Br
1,1,1,2,2,2-hexabromoethane
Uses of alkanes
1.Most alkanes are used as fuel e.g. Methane is used as biogas in homes.Butane is used as the Laboratory gas.
2.On cracking ,alkanes are a major source of Hydrogen for the manufacture of ammonia/Haber process.
3.In manufacture of Carbon black which is a component in printers ink.
4.In manufacture of useful industrial chemicals like methanol, methanol, and chloromethane.
(ii) Alkenes
(a)Nomenclature/Naming
These are hydrocarbons with a general formula CnH2n and C C double bond as the functional group . n is the number of Carbon atoms in the molecule.
The carbon atoms are linked by at least one double bond to each other and single bonds to hydrogen atoms.
They include:
| n | General/
Molecular formula |
Structural formula | Name
|
| 1 | Does not exist
|
||
| 2 | C2H6
|
H H
H C C H
CH2 CH2
|
Ethene |
| 3 | C3H8 | H H H
H C C C H
H
CH2 CH CH3
|
Propene |
| 4 | C4H10 | H H H H
H C C C C H
H H
CH2 CH CH2CH3
|
Butene |
| 5 | C5H12 | H H H H H
H C C C C C H
H H H CH2 CH (CH2)2CH3
|
Pentene |
| 6 | C6H14 | H H H H H H
H C C C C C C H
H H H H
CH2 CH (CH2)3CH3
|
Hexene |
| 7 | C7H16 | H H H H H H H
H C C C C C C C H
H H H H H H H
CH2 CH (CH2)4CH3
|
Heptene |
| 8 | C8H18 | H H H H H H H H
H C C C C C C C C H
H H H H H H
CH2 CH (CH2)5CH3
|
Octene |
| 9 | C9H20 | H H H H H H H H H
H C C C C C C C C C H
H H H H H H H
CH2 CH (CH2)6CH3
|
Nonene |
| 10 | C10H22 | H H H H H H H H H H
H C C C C C C C C C C H
H H H H H H H H
CH2 CH (CH2)7CH3 |
decene |
Note
1.Since carbon is tetravalent ,each atom of carbon in the alkene MUST always be bonded using four covalent bond /four shared pairs of electrons including at the double bond.
2.Since Hydrogen is monovalent ,each atom of hydrogen in the alkene MUST always be bonded using one covalent bond/one shared pair of electrons.
3.One member of the alkene ,like alkanes,differ from the next/previous by a CH2 group.They also form a homologous series.
e.g
Propene differ from ethene by one carbon and two Hydrogen atoms from ethene. 4.A homologous series of alkenes like that of alkanes:
(i) differ by a CH2 group from the next /previous consecutively
(ii)have similar chemical properties
(iii)have similar chemical formula represented by the general formula CnH2n
(iv)the physical properties also show steady gradual change
5.The = C= C = double bond in alkene is the functional group. A functional group is the reacting site of a molecule/compound.
- The = C= C = double bond in alkene can easily be broken to accommodate more two more monovalent atoms. The = C= C = double bond in alkenes make it thus unsaturated.
- An unsaturated hydrocarbon is one with a double =C=C= or triple – C C – carbon bonds in their molecular structure. Unsaturated hydrocarbon easily reacts to be saturated.
8.A saturated hydrocarbon is one without a double =C=C= or triple – C C – carbon bonds in their molecular structure.
Most of the reactions of alkenes take place at the = C = C =bond.
(b)Isomers of alkenes
Isomers are alkenes lie alkanes have the same molecular general formula but different molecular structural formula.
Ethene and propene do not form isomers. Isomers of alkenes are also named by using the IUPAC(International Union of Pure and Applied Chemistry) system of nomenclature/naming.
The IUPAC system of nomenclature of naming alkenes uses the following basic rules/guidelines:
1.Identify the longest continuous/straight carbon chain which contains the =C = C= double bond get/determine the parent alkene.
2.Number the longest chain form the end of the chain which contains the =C = C= double bond so he =C = C= double bond lowest number possible.
3 Indicate the positions by splitting “alk-positions-ene” e.g. but-2-ene, pent-1,3-diene.
4.The position indicated must be for the carbon atom at the lower position in the =C = C= double bond.i.e
But-2-ene means the double =C = C= is between Carbon “2”and “3”
Pent-1,3-diene means there are two double bond one between carbon “1” and “2”and another between carbon “3” and “4”
- Determine the position, number and type of branches. Name them as methyl, ethyl, propyl e.tc. according to the number of alkyl carbon chains attached to the alkene. Name them fluoro-,chloro-,bromo-,iodo- if they are halogens
6.Use prefix di-,tri-,tetra-,penta-,hexa- to show the number of double C = C bonds and branches attached to the alkene.
7.Position isomers can be formed when the=C = C= double bond is shifted between carbon atoms e.g.
But-2-ene means the double =C = C= is between Carbon “2”and “3”
But-1-ene means the double =C = C= is between Carbon “1”and “2”
Both But-1-ene and But-2-ene are position isomers of Butene
8.Position isomers are molecules/compounds having the same general formular but different position of the functional group.i.e.
Butene has the molecular/general formular C4H8 position but can form both But-1-ene and But-2-ene as position isomers.
- Like alkanes ,an alkyl group can be attached to the alkene. Chain/branch isomers are thus formed.
10.Chain/branch isomers are molecules/compounds having the same general formula but different structural formula e.g
Butene and 2-methyl propene both have the same general formualr but different branching chain.
Practice on IUPAC nomenclature of alkenes
Name the following isomers of alkene
H H H H
H C C C C H But-1-ene
H H
H H H H
H C C C C H But-2-ene
H H
H H H H H H
H C C C C C C H 4-methylhex-1-ene
H H H
H C H
H
H
H C H
H H H H H
H C C C C C C H 4,4-dimethylhex-1-ene
H H H
H C H
H
- H
H C H
H H H H
H C C C C C H 4,4-dimethylpent -1- ene
H H
H C H
H
- H
H C H
H H H H
H C C C C C H 5,5-dimethylhex-2- ene
H C H H H
H C H
H
H
- H
H C H
H H H
H C C C C H 2,2-dimethylbut -2- ene
H H
H C H
H
8.H2C CHCH2 CH2 CH3 pent -1- ene
9.H2C C(CH3)CH2 CH2 CH3 2-methylpent -1- ene
10.H2C C(CH3)C(CH3)2 CH2 CH3 2,3,3-trimethylpent -1- ene
11.H2C C(CH3)C(CH3)2 C(CH3)2 CH3 2,3,3,4,4-pentamethylpent -1- ene
12.H3C C(CH3)C(CH3) C(CH3)2 CH3 2,3,4,4-tetramethylpent -2- ene
- H2C C(CH3)C(CH3) C(CH3) CH3 2,3,4-trimethylpent -1,3- diene
- H2C CBrCBr CBr CH3 2,3,4-tribromopent -1,3- diene
- H2C CHCH CH2 But -1,3- diene
- Br2C CBrCBr CBr2 1,1,2,3,4,4-hexabromobut -1,3- diene
- I2C CICI CI2 1,1,2,3,4,4-hexaiodobut -1,3- diene
- H2C C(CH3)C(CH3) CH2 2,3-dimethylbut -1,3- diene
(c)Occurrence and extraction
At indusrial level,alkenes are obtained from the cracking of alkanes.Cracking is the process of breaking long chain alkanes to smaller/shorter alkanes, an alkene and hydrogen gas at high temperatures.
Cracking is a major source of useful hydrogen gas for manufacture of ammonia/nitric(V)acid/HCl i.e.
Long chain alkane -> smaller/shorter alkane + Alkene + Hydrogen gas
Examples
1.When irradiated with high energy radiation,Propane undergo cracking to form methane gas, ethene and hydrogen gas.
Chemical equation
CH3CH2CH3 (g) -> CH4(g) + CH2=CH2(g) + H2(g)
2.Octane undergo cracking to form hydrogen gas, butene and butane gases
Chemical equation
CH3(CH2) 6 CH3 (g) -> CH3CH2CH2CH3(g) + CH3 CH2CH=CH2(g) + H2(g)
(d)School laboratory preparation of alkenes
In a school laboratory, alkenes may be prepared from dehydration of alkanols using:
(i) concentrated sulphuric(VI)acid(H2SO4).
(a) aluminium(III)oxide(Al2O3) i.e
Alkanol –Conc. H2SO4 –> Alkene + Water
Alkanol –Al2O3 –> Alkene + Water e.g.
1.(a)At about 180oC,concentrated sulphuric(VI)acid dehydrates/removes water from ethanol to form ethene.
The gas produced contain traces of carbon(IV)oxide and sulphur(IV)oxide gas as impurities.
It is thus passed through concentrated sodium/potassium hydroxide solution to remove the impurities.
Chemical equation
CH3CH2OH (l) –conc H2SO4/180oC–> CH2=CH2(g) + H2O(l)
(b)On heating strongly aluminium(III)oxide(Al2O3),it dehydrates/removes water from ethanol to form ethene.
Ethanol vapour passes through the hot aluminium (III) oxide which catalyses the dehydration.
Activated aluminium(III)oxide has a very high affinity for water molecules/elements of water and thus dehydrates/ removes water from ethanol to form ethene.
Chemical equation
CH3CH2OH (l) –(Al2O3/strong heat–> CH2=CH2(g) + H2O(l)
2(a) Propan-1-ol and Propan-2-ol(position isomers of propanol) are dehydrated by conc H2SO4 at about 180oC to propene(propene has no position isomers).
Chemical equation
CH3CH2 CH2OH (l) — conc H2SO4/180oC –> CH3CH2=CH2(g) + H2O(l)
Propan-1-ol Prop-1-ene
CH3CHOH CH3 (l) — conc H2SO4/180oC –> CH3CH2=CH2(g) + H2O(l)
Propan-2-ol Prop-1-ene
(b) Propan-1-ol and Propan-2-ol(position isomers of propanol) are dehydrated by heating strongly aluminium(III)oxide(Al2O3) form propene
Chemical equation
CH3CH2 CH2OH (l) — Heat/Al2O3 –> CH3CH2=CH2(g) + H2O(l)
Propan-1-ol Prop-1-ene
CH3CHOH CH3 (l) — Heat/Al2O3 –> CH3CH2=CH2(g) + H2O(l)
Propan-2-ol Prop-1-ene
3(a) Butan-1-ol and Butan-2-ol(position isomers of butanol) are dehydrated by conc H2SO4 at about 180oC to But-1-ene and But-2-ene respectively
Chemical equation
CH3CH2 CH2 CH2OH (l) — conc H2SO4/180oC –>CH3 CH2CH2=CH2(g) + H2O(l)
Butan-1-ol But-1-ene
CH3CHOH CH2CH3 (l)– conc H2SO4/180oC –>CH3CH=CH CH2(g) + H2O(l)
Butan-2-ol But-2-ene
(b) Butan-1-ol and Butan-2-ol are dehydrated by heating strongly aluminium (III) oxide (Al2O3) form But-1-ene and But-2-ene respectively.
Chemical equation
CH3CH2 CH2 CH2OH (l) — Heat/Al2O3 –> CH3 CH2CH2=CH2(g) + H2O(l)
Butan-1-ol But-1-ene
CH3CHOH CH2CH3 (l) — Heat/Al2O3 –> CH3CH=CH CH2(g) + H2O(l)
Butan-2-ol But-2-ene
Laboratory set up for the preparation of alkenes/ethene
Caution
(i)Ethanol is highly inflammable
(ii)Conc H2SO4 is highly corrosive on skin contact.
(iii)Common school thermometer has maximum calibration of 110oC and thus cannot be used. It breaks/cracks.
(i)Using conentrated sulphuric(VI)acid
Some broken porcelain or sand should be put in the flask when heating to:
(i)prevent bumping which may break the flask.
(ii)ensure uniform and smooth boiling of the mixture
The temperatures should be maintained at above160oC.
At lower temperatures another compound –ether is predominantly formed instead of ethene gas.
(ii)Using aluminium(III)oxide
(e)Properties of alkenes
- Physical properties
Like alkanes, alkenes are colourles gases, solids and liquids that are not poisonous.
They are slightly soluble in water.
The solubility in water decrease as the carbon chain and as the molar mass increase but very soluble in organic solvents like tetrachloromethane and methylbenzene.
The melting and boiling point increase as the carbon chain increase.
This is because of the increase in van-der-waals /intermolecular forces as the carbon chain increase.
The 1st four straight chain alkenes (ethene,propane,but-1-ene and pent-1-ene)are gases at room temperature and pressure.
The density of straight chain alkenes,like alkanes, increase with increasing carbon chain as the intermolecular forces increases reducing the volume occupied by a given mass of the alkene.
Summary of physical properties of the 1st five alkenes
| Alkene | General formula | Melting point(oC) | Boiling point(K) | State at room(298K) temperature and pressure atmosphere (101300Pa) |
| Ethene | CH2CH2 | -169 | -104 | gas |
| Propene | CH3 CHCH2 | -145 | -47 | gas |
| Butene | CH3CH2 CHCH2 | -141 | -26 | gas |
| Pent-1-ene | CH3(CH2 CHCH2 | -138 | 30 | liquid |
| Hex-1-ene | CH3(CH2) CHCH2 | -98 | 64 | liquid |
- Chemical properties
(a)Burning/combustion
Alkenes burn with a yellow/ luminous sooty/ smoky flame in excess air to form carbon(IV) oxide and water.
Alkene + Air -> carbon(IV) oxide + water (excess air/oxygen)
Alkenes burn with a yellow/ luminous sooty/ smoky flame in limited air to form carbon(II) oxide and water.
Alkene + Air -> carbon(II) oxide + water (limited air)
Burning of alkenes with a yellow/ luminous sooty/ smoky flame is a confirmatory test for the presence of the =C=C= double bond because they have higher C:H ratio.
A homologous series with C = C double or C C triple bond is said to be unsaturated.
A homologous series with C C single bond is said to be saturated.Most of the reactions of the unsaturated compound involve trying to be saturated to form a
C C single bond .
Examples of burning alkenes
1.(a) Ethene when ignited burns with a yellow sooty flame in excess air to form carbon(IV) oxide and water.
Ethene + Air -> carbon(IV) oxide + water (excess air/oxygen)
C2H4(g) + 3O2(g) -> 2CO2(g) + 2H2O(l/g)
(b) Ethene when ignited burns with a yellow sooty flame in limited air to form carbon(II) oxide and water.
Ethene + Air -> carbon(II) oxide + water (limited air )
C2H4(g) + 3O2(g) -> 2CO2(g) + 2H2O(l/g)
2.(a) Propene when ignited burns with a yellow sooty flame in excess air to form carbon(IV) oxide and water.
Propene + Air -> carbon(IV) oxide + water (excess air/oxygen)
2C3H6(g) + 9O2(g) -> 6CO2(g) + 6H2O(l/g)
(a) Propene when ignited burns with a yellow sooty flame in limited air to form carbon(II) oxide and water.
Propene + Air -> carbon(IV) oxide + water (excess air/oxygen)
C3H6(g) + 3O2(g) -> 3CO(g) + 3H2O(l/g)
(b)Addition reactions
An addition reaction is one which an unsaturated compound reacts to form a saturated compound.Addition reactions of alkenes are named from the reagent used to cause the addtion/convert the double =C=C= to single C-C bond.
(i)Hydrogenation
Hydrogenation is an addition reaction in which hydrogen in presence of Palladium/Nickel catalyst at high temperatures react with alkenes to form alkanes.
Examples
1.When Hydrogen gas is passed through liquid vegetable and animal oil at about 180oC in presence of Nickel catalyst,solid fat is formed.
Hydrogenation is thus used to harden oils to solid fat especially margarine.
During hydrogenation, one hydrogen atom in the hydrogen molecule attach itself to one carbon and the other hydrogen to the second carbon breaking the double bond to single bond.
Chemical equation
H2C=CH2 + H2 -Ni/Pa-> H3C – CH3
H H H H
C = C + H – H – Ni/Pa -> H – C – C – H
H H H H
2.Propene undergo hydrogenation to form Propane
Chemical equation
H3C CH=CH2 + H2 -Ni/Pa-> H3C CH – CH3
H H H H H H
H C C = C + H – H – Ni/Pa-> H – C – C – C- H
H H H H H
3.Both But-1-ene and But-2-ene undergo hydrogenation to form Butane
Chemical equation
But-1-ene + Hydrogen –Ni/Pa-> Butane
H3C CH2 CH=CH2 + H2 -Ni/Pa-> H3C CH2CH – CH3
H H H H H H H H
H C C – C = C + H – H – Ni/Pa-> H – C- C – C – C- H
H H H H H H H
But-2-ene + Hydrogen –Ni/Pa-> Butane
H3C CH2 =CH CH2 + H2 -Ni/Pa-> H3C CH2CH – CH3
H H H H H H H H
H C C = C – C -H + H – H – Ni/Pa-> H – C- C – C – C- H
H H H H H H
- But-1,3-diene should undergo hydrogenation to form Butane. The reaction uses two moles of hydrogen molecules/four hydrogen atoms to break the two double bonds.
But-1,3-diene + Hydrogen –Ni/Pa-> Butane
H2C CH CH=CH2 + 2H2 -Ni/Pa-> H3C CH2CH – CH3
H H H H H H H H
H C C – C = C -H + 2(H – H) – Ni/Pa-> H – C- C – C – C- H
H H H H
(ii) Halogenation.
Halogenation is an addition reaction in which a halogen (Fluorine, chlorine, bromine, iodine) reacts with an alkene to form an alkane.
The double bond in the alkene break and form a single bond.
The colour of the halogen fades as the number of moles of the halogens remaining unreacted decreases/reduces.
One bromine atom bond at the 1st carbon in the double bond while the other goes to the 2nd carbon.
Examples
1Ethene reacts with bromine to form 1,2-dibromoethane.
Chemical equation
H2C=CH2 + Br2 H2 Br C – CH2 Br
H H H H
C = C + Br – Br Br – C – C – Br
H H H H
Ethene + Bromine 1,2-dibromoethane
2.Propene reacts with chlorine to form 1,2-dichloropropane.
Chemical equation
H3C CH=CH2 + Cl2 H3C CHCl – CH2Cl
Propene + Chlorine 1,2-dichloropropane
H H H H H H
H C C = C + Cl – Cl H – C – C – C- Cl
H H H Cl H
H H H H H H H H
H C C – C = C + I – I H – C- C – C – C- I
H H H H H H H H
3.Both But-1-ene and But-2-ene undergo halogenation with iodine to form 1,2-diiodobutane and 2,3-diiodobutane
Chemical equation
But-1-ene + iodine 1,2 diiodobutane
H3C CH2 CH=CH2 + I2 H3C CH2CH I – CH2I
But-2-ene + Iodine 2,3-diiodobutane
H3C CH= CH-CH2 + F2 H3C CHICHI – CH3
H H H H H H H H
H C C = C – C -H + I – I H – C- C – C – C- H
H H H I I H
- But-1,3-diene should undergo halogenation to form Butane. The reaction uses two moles of iodine molecules/four iodine atoms to break the two double bonds.
But-1,3-diene + iodine 1,2,3,4-tetraiodobutane
H2C= CH CH=CH2 + 2I2 H2CI CHICHI – CHI
H H H H H H H H
H C C – C = C -H + 2(I – I) H – C- C – C – C- H
I I I I
(iii) Reaction with hydrogen halides.
Hydrogen halides reacts with alkene to form a halogenoalkane. The double bond in the alkene break and form a single bond.
The main compound is one which the hydrogen atom bond at the carbon with more hydrogen .
Examples
- Ethene reacts with hydrogen bromide to form bromoethane.
Chemical equation
H2C=CH2 + HBr H3 C – CH2 Br
H H H H
C = C + H – Br H – C – C – Br
H H H H
Ethene + Bromine bromoethane
- Propene reacts with hydrogen iodide to form 2-iodopropane.
Chemical equation
H3C CH=CH2 + HI H3C CHI – CH3
| Carbon atom with more Hydrogen atoms gets extra hydrogen |
Propene + Chlorine 2-chloropropane
H H H H H H
H C C = C + H – Cl H – C – C – C– H
H H H Cl H
- Both But-1-ene and But-2-ene reacts with hydrogen bromide to form 2- bromobutane
Chemical equation
But-1-ene + hydrogen bromide 2-bromobutane
H3C CH2 CH=CH2 + HBr H3C CH2CHBr -CH3
H H H H H H H H
H C C – C = C + H – Br H – C- C – C – C- H
H H H H H Br H
But-2-ene + Hydrogen bromide 2-bromobutane
H3C CH= CH-CH2 + HBr H3C CHBrCH2 – CH3
H H H H H H H H
H C C = C – C -H + Br – H H – C- C – C – C- H
H H H Br H H
- But-1,3-diene react with hydrogen iodide to form 2,3- diiodobutane. The reaction uses two moles of hydrogen iodide molecules/two iodine atoms and two hydrogen atoms to break the two double bonds.
But-1,3-diene + iodine 2,3-diiodobutane
H2C= CH CH=CH2 + 2HI2 H3CCHICHI – CH3
H H H H H H H H
H C C – C = C -H + 2(H – I) H – C- C – C – C- H
H I I H
(iv) Reaction with bromine/chlorine water.
Chlorine and bromine water is formed when the halogen is dissolved in distilled water.Chlorine water has the formular HOCl(hypochlorous/chloric(I)acid) .Bromine water has the formular HOBr(hydrobromic(I)acid).
During the addition reaction .the halogen move to one carbon and the OH to the other carbon in the alkene at the =C=C= double bond to form a halogenoalkanol.
Bromine water + Alkene -> bromoalkanol
Chlorine water + Alkene -> bromoalkanol
Examples
1Ethene reacts with bromine water to form bromoethanol.
Chemical equation
H2C=CH2 + HOBr H2 Br C – CH2 OH
H H H H
C = C + Br – OH Br – C – C – OH
H H H H
Ethene + Bromine water bromoethanol
2.Propene reacts with chlorine water to form chloropropan-2-ol / 2-chloropropan-1-ol.
Chemical equation
I.H3C CH=CH2 + HOCl H3C CHCl – CH2OH
Propene + Chlorine water 2-chloropropane
H H H H H H
H C C = C + HO – Cl H – C – C – C- OH
H H H Cl H
II.H3C CH=CH2 + HOCl H3C CHOH – CH2Cl
Propene + Chlorine chloropropan-2-ol
H H H H H H
H C C = C + HO – Cl H – C – C – C- Cl
H H H OH H
3.Both But-1-ene and But-2-ene react with bromine water to form 2-bromobutan-1-ol /3-bromobutan-2-ol respectively
Chemical equation
I.But-1-ene + bromine water 2-bromobutan-1-ol
H3C CH2 CH=CH2 + HOBr H3C CH2CH Br – CH2OH
H H H H H H H H
H C C – C = C + HO– Br H – C- C – C – C- OH
H H H H H Br H
II.But-2-ene + bromine water 3-bromobutan-2-ol
H3C CH= CHCH3 + HOBr H3C CH2OHCH Br CH3
H H H H H H H H
H C C – C = C + HO– Br H – C- C – C – C- OH
H H H H H Br H
- But-1,3-diene reacts with bromine water to form Butan-1,3-diol.
The reaction uses two moles of bromine water molecules to break the two double bonds.
But-1,3-diene + bromine water 2,4-dibromobutan-1,3-diol
H2C= CH CH=CH2 + 2HOBr H2COH CHBrCHOH CHBr
H H H H H H H H
H C C – C = C -H + 2(HO – Br) H – C- C – C – C- H
HO Br HO Br
(v) Oxidation.
Alkenes are oxidized to alkanols with duo/double functional groups by oxidizing agents.
When an alkene is bubbled into orange acidified potassium/sodium dichromate (VI) solution,the colour of the oxidizing agent changes to green.
When an alkene is bubbled into purple acidified potassium/sodium manganate(VII) solution, the oxidizing agent is decolorized.
Examples
1Ethene is oxidized to ethan-1,2-diol by acidified potassium/sodium manganate(VII) solution/ acidified potassium/sodium dichromate(VI) solution.
The purple acidified potassium/sodium manganate(VII) solution is decolorized.
The orange acidified potassium/sodium dichromate(VI) solution turns to green.
Chemical equation
H2C=CH2 [O] in H+/K2Cr2O7 HO CH2 – CH2 OH
H H H H
C = C+ [O] in H+/KMnO4 H – C – C – H
H H OH OH
Ethene + [O] in H+/KMnO4 ethan-1,2-diol
- Propene is oxidized to propan-1,2-diol by acidified potassium/sodium manganate(VII) solution/ acidified potassium/sodium dichromate(VI) solution.
The purple acidified potassium/sodium manganate(VII) solution is decolorized.
The orange acidified potassium/sodium dichromate(VI) solution turns to green.
Chemical equation
H3C CH=CH2 [O] in H+/KMnO4 H3C CHOH – CH2OH
Propene [O] in H+/KMnO4 propan-1,2-diol
H H H H H H
H C C = C [O] in H+/KMnO4 H – C – C – C- OH
H H H OH H
3.Both But-1-ene and But-2-ene react with bromine water to form butan-1,2-diol and butan-2,3-diol
Chemical equation
I.But-1-ene + [O] in H+/KMnO4 butan-1,2-diol
H3C CH2 CH=CH2 + [O] H3C CH2CHOH – CH2OH
H H H H H H H H
H C C – C = C + [O] H – C- C – C – C- OH
H H H H H OH H
(v) Hydrolysis.
Hydrolysis is the reaction of a compound with water/addition of H-OH to a compound.
Alkenes undergo hydrolysis to form alkanols .
This takes place in two steps:
(i)Alkenes react with concentrated sulphuric(VI)acid at room temperature and pressure to form alkylhydrogen sulphate(VI).
Alkenes + concentrated sulphuric(VI)acid -> alkylhydrogen sulphate(VI)
(ii)On adding water to alkylhydrogen sulphate(VI) then warming, an alkanol is formed.
alkylhydrogen sulphate(VI) + water -warm-> Alkanol.
Examples
(i)Ethene reacts with cold concentrated sulphuric(VI)acid to form ethyl hydrogen sulphate(VII)
Chemical equation
H2C=CH2 + H2SO4 CH3 – CH2OSO3H
H H H O-SO3H
C = C + H2SO4 H – C – C – H
H H H H
Ethene + H2SO4 ethylhydrogen sulphate(VI)
(ii) Ethylhydrogen sulphate(VI) is hydrolysed by water to ethanol
Chemical equation
CH3 – CH2OSO3H + H2O CH3 – CH2OH + H2SO4
H OSO3H H OH
H – C – C – H + H2O H – C – C – H + H2SO4
H H H H
ethylhydrogen sulphate(VI) + H2O Ethanol
- Propene reacts with cold concentrated sulphuric(VI)acid to form propyl hydrogen sulphate(VII)
Chemical equation
CH3H2C=CH2 + H2SO4 CH3CH2 – CH2OSO3H
H H H H H O-SO3H
C = C – C – H + H2SO4 H – C – C – C – H
H H H H H H
Propene + H2SO4 propylhydrogen sulphate(VI)
(ii) Propylhydrogen sulphate(VI) is hydrolysed by water to propanol
Chemical equation
CH3 – CH2OSO3H + H2O CH3 – CH2OH + H2SO4
H H OSO3H H H OH
H – C – C – C – H + H2O H – C – C – C – H + H2SO4
H H H H H H
propylhydrogen sulphate(VI) + H2O propanol
(vi) Polymerization/self addition
Addition polymerization is the process where a small unsaturated monomer (alkene ) molecule join together to form a large saturated molecule.
Only alkenes undergo addition polymerization.
Addition polymers are named from the alkene/monomer making the polymer and adding the prefix “poly” before the name of monomer to form a polyalkene
During addition polymerization
(i)the double bond in alkenes break
(ii)free radicals are formed
(iii)the free radicals collide with each other and join to form a larger molecule. The more collisions the larger the molecule.
Examples of addition polymerization
1.Formation of Polyethene
Polyethene is an addition polymer formed when ethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.
During polymerization:
(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting paticles)
H H H H H H H H
C = C + C = C + C = C + C = C + …
H H H H H H H H
Ethene + Ethene + Ethene + Ethene + …
(ii)the double bond joining the ethane molecule break to free readicals
H H H H H H H H
- C – C• + •C – C• + •C – C• + •C – C• + …
H H H H H H H H
Ethene radical + Ethene radical + Ethene radical + Ethene radical + …
(iii)the free radicals collide with each other and join to form a larger molecule
H H H H H H H H lone pair of electrons
- C – C – C – C – C – C – C – C• + …
H H H H H H H H
Lone pair of electrons can be used to join more monomers to form longer polyethene.
Polyethene molecule can be represented as:
H H H H H H H H extension of
molecule/polymer
– C – C – C – C – C – C – C – C-
H H H H H H H H
Since the molecule is a repetition of one monomer, then the polymer is:
H H
( C – C )n
H H
Where n is the number of monomers in the polymer. The number of monomers in the polymer can be determined from the molar mass of the polymer and monomer from the relationship:
Number of monomers/repeating units in monomer = Molar mass polymer
Molar mass monomer
Examples
Polythene has a molar mass of 4760.Calculate the number of ethene molecules in the polymer(C=12.0, H=1.0 )
Number of monomers/repeating units in polyomer = Molar mass polymer
Molar mass monomer
=> Molar mass ethene (C2H4 )= 28 Molar mass polyethene = 4760
Substituting 4760 = 170 ethene molecules
28
The commercial name of polyethene is polythene.
It is an elastic, tough, transparent and durable plastic.
Polythene is used:
(i)in making plastic bag
(ii)bowls and plastic bags
(iii)packaging materials
2.Formation of Polychlorethene
Polychloroethene is an addition polymer formed when chloroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.
During polymerization:
(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles)
H H H H H H H H
C = C + C = C + C = C + C = C + …
H Cl H Cl H Cl H Cl
chloroethene + chloroethene + chloroethene + chloroethene + …
(ii)the double bond joining the chloroethene molecule break to free radicals
H H H H H H H H
- C – C• + •C – C• + •C – C• + •C – C• + …
H Cl H Cl H Cl H Cl
(iii)the free radicals collide with each other and join to form a larger molecule
H H H H H H H H lone pair of electrons
- C – C – C – C – C – C – C – C• + …
H Cl H Cl H Cl H Cl
Lone pair of electrons can be used to join more monomers to form longer polychloroethene.
Polychloroethene molecule can be represented as:
H H H H H H H H extension of
molecule/polymer
– C – C – C – C – C – C – C – C- + …
H Cl H Cl H Cl H Cl
Since the molecule is a repetition of one monomer, then the polymer is:
H H
( C – C )n
H Cl
Examples
Polychlorothene has a molar mass of 4760.Calculate the number of chlorethene molecules in the polymer(C=12.0, H=1.0,Cl=35.5 )
Number of monomers/repeating units in monomer = Molar mass polymer
Molar mass monomer
=> Molar mass ethene (C2H3Cl )= 62.5 Molar mass polyethene = 4760
Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number)
62.5
The commercial name of polychloroethene is polyvinylchloride(PVC). It is a tough, non-transparent and durable plastic. PVC is used:
(i)in making plastic rope
(ii)water pipes
(iii)crates and boxes
3.Formation of Polyphenylethene
Polyphenylethene is an addition polymer formed when phenylethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.
During polymerization:
(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles)
H H H H H H H H
C = C + C = C + C = C + C = C + …
H C6H5 H C6H5 H C6H5 H C6H5
phenylethene + phenylethene + phenylethene + phenylethene + …
(ii)the double bond joining the phenylethene molecule break to free radicals
H H H H H H H H
- C – C• + •C – C• + •C – C• + •C – C• + …
H C6H5 H C6H5 H C6H5 H C6H5
(iii)the free radicals collide with each other and join to form a larger molecule
H H H H H H H H lone pair of electrons
- C – C – C – C – C – C – C – C • + …
H C6H5 H C6H5 H C6H5 H C6H5
Lone pair of electrons can be used to join more monomers to form longer polyphenylethene.
Polyphenylethene molecule can be represented as:
H H H H H H H H
– C – C – C – C – C – C – C – C –
H C6H5 H C6H5 H C6H5 H C6H5
Since the molecule is a repetition of one monomer, then the polymer is:
H H
( C – C )n
H C6H5
Examples
Polyphenylthene has a molar mass of 4760.Calculate the number of phenylethene molecules in the polymer(C=12.0, H=1.0, )
Number of monomers/repeating units in monomer = Molar mass polymer
Molar mass monomer
=> Molar mass ethene (C8H8 )= 104 Molar mass polyethene = 4760
Substituting 4760 = 45.7692 =>45 polyphenylethene molecules(whole number)
104
The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic. Polystyrene is used:
(i)in making packaging material for carrying delicate items like computers, radion,calculators.
(ii)ceiling tiles
(iii)clothe linings
4.Formation of Polypropene
Polypropene is an addition polymer formed when propene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.
During polymerization:
(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles)
H H H H H H H H
C = C + C = C + C = C + C = C + …
H CH3 H CH3 H CH3 H CH3
propene + propene + propene + propene + …
(ii)the double bond joining the phenylethene molecule break to free radicals
H H H H H H H H
- C – C• + •C – C• + •C – C• + •C – C• + …
H CH3 H CH3 H CH3 H CH3
(iii)the free radicals collide with each other and join to form a larger molecule
H H H H H H H H lone pair of electrons
- C – C – C – C – C – C – C – C • + …
H CH3 H CH3 H CH3 H CH3
Lone pair of electrons can be used to join more monomers to form longer propene.
propene molecule can be represented as:
H H H H H H H H
– C – C – C – C – C – C – C – C –
H CH3 H CH3 H CH3 H CH3
Since the molecule is a repetition of one monomer, then the polymer is:
H H
( C – C )n
H CH3
Examples
Polypropene has a molar mass of 4760.Calculate the number of propene molecules in the polymer(C=12.0, H=1.0, )
Number of monomers/repeating units in monomer = Molar mass polymer
Molar mass monomer
=> Molar mass propene (C3H8 )= 44 Molar mass polyethene = 4760
Substituting 4760 = 108.1818 =>108 propene molecules(whole number)
44
The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic. Polystyrene is used:
(i)in making packaging material for carrying delicate items like computers, radion,calculators.
(ii)ceiling tiles
(iii)clothe linings
5.Formation of Polytetrafluorothene
Polytetrafluorothene is an addition polymer formed when tetrafluoroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.
During polymerization:
(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles)
F F F F F F F F
C = C + C = C + C = C + C = C + …
F F F F F F F F
tetrafluoroethene + tetrafluoroethene+ tetrafluoroethene+ tetrafluoroethene + …
(ii)the double bond joining the tetrafluoroethene molecule break to free radicals
F F F F F F F F
- C – C• + •C – C• + •C – C• + •C – C• + …
F F F F F F F F
(iii)the free radicals collide with each other and join to form a larger molecule
F F F F F F F F lone pair of electrons
- C – C – C – C – C – C – C – C• + …
F F F F F F F F
Lone pair of electrons can be used to join more monomers to form longer polytetrafluoroethene.
polytetrafluoroethene molecule can be represented as:
F F F F F F F F extension of
molecule/polymer
– C – C – C – C – C – C – C – C- + …
F F F F F F F F
Since the molecule is a repetition of one monomer, then the polymer is:
F F
( C – C )n
F F
Examples
Polytetrafluorothene has a molar mass of 4760.Calculate the number of tetrafluoroethene molecules in the polymer(C=12.0, ,F=19 )
Number of monomers/repeating units in monomer = Molar mass polymer
Molar mass monomer
=> Molar mass ethene (C2F4 )= 62.5 Molar mass polyethene = 4760
Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number)
62.5
The commercial name of polytetrafluorethene(P.T.F.E) is Teflon(P.T.F.E). It is a tough, non-transparent and durable plastic. PVC is used:
(i)in making plastic rope
(ii)water pipes
(iii)crates and boxes
6.Formation of rubber from Latex
Natural rubber is obtained from rubber trees.
During harvesting an incision is made on the rubber tree to produce a milky white substance called latex.
Latex is a mixture of rubber and lots of water.
The latex is then added an acid to coagulate the rubber.
Natural rubber is a polymer of 2-methylbut-1,3-diene ;
H CH3 H H
CH2=C (CH3) CH = CH2 H – C = C – C = C – H
During natural polymerization to rubber, one double C=C bond break to self add to another molecule.The double bond remaining move to carbon “2” thus;
H CH3 H H H CH3 H H
– C – C = C – C – C – C = C – C –
H H H H
Generally the structure of rubber is thus;
H CH3 H H
-(- C – C = C – C -)n–
H H
Pure rubber is soft and sticky.It is used to make erasers, car tyres. Most of it is vulcanized.Vulcanization is the process of heating rubber with sulphur to make it harder/tougher.
During vulcanization the sulphur atoms form a cross link between chains of rubber molecules/polymers. This decreases the number of C=C double bonds in the polymer.
| Sulphur atoms make cross link between polymers |
H CH3 H H H CH3 H H
– C – C – C – C – C – C – C – C –
H S H H S H
H CH3 S H H CH3 S H
– C – C – C – C – C – C – C – C –
H H H H H H
Vulcanized rubber is used to make tyres, shoes and valves.
7.Formation of synthetic rubber
Synthetic rubber is able to resist action of oil,abrasion and organic solvents which rubber cannot.
Common synthetic rubber is a polymer of 2-chlorobut-1,3-diene ;
H Cl H H
CH2=C (Cl CH = CH2 H – C = C – C = C – H
During polymerization to synthetic rubber, one double C=C bond is broken to self add to another molecule. The double bond remaining move to carbon “2” thus;
H Cl H H H Cl H H
– C – C = C – C – C – C = C – C –
H H H H
Generally the structure of rubber is thus;
H Cl H H
-(- C – C = C – C -)n–
H H
Rubber is thus strengthened through vulcanization and manufacture of synthetic rubber.
(c)Test for the presence of – C = C – double bond.
(i)Burning/combustion
All unsaturated hydrocarbons with a – C = C – or – C = C – bond burn with a yellow sooty flame.
Experiment
Scoop a sample of the substance provided in a clean metallic spatula. Introduce it on a Bunsen burner.
| Observation | Inference |
| Solid melt then burns with a yellow sooty flame
|
– C = C –,
– C = C – bond |
(ii)Oxidation by acidified KMnO4/K2Cr2O7
Bromine water ,Chlorine water and Oxidizing agents acidified KMnO4/K2Cr2O7 change to unique colour in presence of – C = C –
or – C = C – bond.
Experiment
Scoop a sample of the substance provided into a clean test tube. Add 10cm3 of distilled water. Shake. Take a portion of the solution mixture. Add three drops of acidified KMnO4/K2Cr2O7 .
| Observation | Inference |
| Acidified KMnO4 decolorized
Orange colour of acidified K2Cr2O7 turns green
Bromine water is decolorized
Chlorine water is decolorized |
– C = C –
– C = C – bond
|
(d)Some uses of Alkenes
- In the manufacture of plastic
- Hydrolysis of ethene is used in industrial manufacture of ethanol.
- In ripening of fruits.
- In the manufacture of detergents.
(iii) Alkynes
(a)Nomenclature/Naming
These are hydrocarbons with a general formula CnH2n–2 and C C double bond as the functional group . n is the number of Carbon atoms in the molecule.
The carbon atoms are linked by at least one triple bond to each other and single bonds to hydrogen atoms.
They include:
| n | General/
Molecular formula |
Structural formula | Name |
| 1 | Does not exist
|
– | |
| 2 | C2H2
|
H C C H CH CH
|
Ethyne |
| 3 | C3H4 | H
H C C C H
H CH C CH3
|
Propyne |
| 4 | C4H6 | H H
H C C C C H
H H CH C CH2CH3
|
Butyne |
| 5 | C5H8 | H H H
H C C C C C H
H H H CH C (CH2)2CH3
|
Pentyne |
| 6 | C6H10 | H H H H
H C C C C C C H
H H H H CH C (CH2)3CH3
|
Hexyne |
| 7 | C7H12 | H H H H H
H C C C C C C C H
H H H H H H H CH C (CH2)4CH3 |
Heptyne |
| 8 | C8H14 | H H H H H H
H C C C C C C C C H
H H H H H H CH C (CH2)5CH3 |
Octyne |
| 9 | C9H16 | H H H H H H H
H C C C C C C C C C H
H H H H H H H CH C (CH2)6CH3 |
Nonyne |
| 10 | C10H18 | H H H H H H H H
H C C C C C C C C C C H
H H H H H H H H CH C (CH2)7CH3 |
Decyne |
Note
- Since carbon is tetravalent ,each atom of carbon in the alkyne MUST always be bonded using four covalent bond /four shared pairs of electrons including at the triple bond.
- Since Hydrogen is monovalent ,each atom of hydrogen in the alkyne MUST always be bonded using one covalent bond/one shared pair of electrons.
- One member of the alkyne ,like alkenes and alkanes, differ from the next/previous by a CH2 group(molar mass of 14 atomic mass units).They thus form a homologous series.
e.g
Propyne differ from ethyne by (14 a.m.u) one carbon and two Hydrogen atoms from ethyne.
4.A homologous series of alkenes like that of alkanes:
(i) differ by a CH2 group from the next /previous consecutively
(ii) have similar chemical properties
(iii)have similar chemical formula with general formula CnH2n-2
(iv)the physical properties also show steady gradual change
5.The – C = C – triple bond in alkyne is the functional group. The functional group is the reacting site of the alkynes.
- The – C = C – triple bond in alkyne can easily be broken to accommodate more /four more monovalent atoms. The – C = C – triple bond in alkynes make it thus unsaturated like alkenes.
- Most of the reactions of alkynes like alkenes take place at the – C = C- triple bond.
(b)Isomers of alkynes
Isomers of alkynes have the same molecular general formula but different molecular structural formula.
Isomers of alkynes are also named by using the IUPAC(International Union of Pure and Applied Chemistry) system of nomenclature/naming.
The IUPAC system of nomenclature of naming alkynes uses the following basic rules/guidelines:
1.Identify the longest continuous/straight carbon chain which contains the – C = C- triple bond to get/determine the parent alkene.
- Number the longest chain form the end of the chain which contains the -C = C- triple bond so as – C = C- triple bond get lowest number possible.
3 Indicate the positions by splitting “alk-positions-yne” e.g. but-2-yne, pent-1,3-diyne.
4.The position indicated must be for the carbon atom at the lower position in the
-C = C- triple bond. i.e
But-2-yne means the triple -C = C- is between Carbon “2”and “3”
Pent-1,3-diyne means there are two triple bonds; one between carbon “1” and “2”and another between carbon “3” and “4”
- Determine the position, number and type of branches. Name them as methyl, ethyl, propyl e.tc. according to the number of alkyl carbon chains attached to the alkyne. Name them fluoro-,chloro-,bromo-,iodo- if they are halogens
6.Use prefix di-,tri-,tetra-,penta-,hexa- to show the number of triple – C = C- bonds and branches attached to the alkyne.
7.Position isomers can be formed when the – C = C- triple bond is shifted between carbon atoms e.g.
But-2-yne means the double – C = C- is between Carbon “2”and “3”
But-1-yne means the double – C = C- is between Carbon “1”and “2”
Both But-1-yne and But-2-yne are position isomers of Butyne.
- Like alkanes and alkynes , an alkyl group can be attached to the alkyne. Chain/branch isomers are thus formed.
Butyne and 2-methyl propyne both have the same general formular but different branching chain.
(More on powerpoint)
(c)Preparation of Alkynes.
Ethyne is prepared from the reaction of water on calcium carbide. The reaction is highly exothermic and thus a layer of sand should be put above the calcium carbide to absorb excess heat to prevent the reaction flask from breaking. Copper(II)sulphate(VI) is used to catalyze the reaction
Chemical equation
CaC2(s) + 2 H2O(l) -> Ca(OH) 2 (aq) + C2H2 (g)
(d)Properties of alkynes
- Physical properties
Like alkanes and alkenes,alkynes are colourles gases, solids and liquids that are not poisonous.
They are slightly soluble in water. The solubility in water decrease as the carbon chain and as the molar mass increase but very soluble in organic solvents like tetrachloromethane and methylbenzene. Ethyne has a pleasant taste when pure.
The melting and boiling point increase as the carbon chain increase.
This is because of the increase in van-der-waals /intermolecular forces as the carbon chain increase. The 1st three straight chain alkynes (ethyne,propyne and but-1-yne)are gases at room temperature and pressure.
The density of straight chain alkynes increase with increasing carbon chain as the intermolecular forces increases reducing the volume occupied by a given mass of the alkyne.
Summary of physical properties of the 1st five alkenes
| Alkyne | General formula | Melting point(oC) | Boiling point(oC) | State at room(298K) temperature and pressure atmosphere (101300Pa) |
| Ethyne | CH CH | -82 | -84 | gas |
| Propyne | CH3 C CH | -103 | -23 | gas |
| Butyne | CH3CH2 CCH | -122 | 8 | gas |
| Pent-1-yne | CH3(CH2) 2 CCH | -119 | 39 | liquid |
| Hex-1-yne | CH3(CH2) 3C CH | -132 | 71 | liquid |
- Chemical properties
(a)Burning/combustion
Alkynes burn with a yellow/ luminous very sooty/ smoky flame in excess air to form carbon(IV) oxide and water.
Alkyne + Air -> carbon(IV) oxide + water (excess air/oxygen)
Alkenes burn with a yellow/ luminous verysooty/ smoky flame in limited air to form carbon(II) oxide/carbon and water.
Alkyne + Air -> carbon(II) oxide /carbon + water (limited air)
Burning of alkynes with a yellow/ luminous sooty/ smoky flame is a confirmatory test for the presence of the – C = C – triple bond because they have very high C:H ratio.
Examples of burning alkynes
1.(a) Ethyne when ignited burns with a yellow very sooty flame in excess air to form carbon(IV) oxide and water.
Ethyne + Air -> carbon(IV) oxide + water (excess air/oxygen)
2C2H2(g) + 5O2(g) -> 4CO2(g) + 2H2O(l/g)
(b) Ethyne when ignited burns with a yellow sooty flame in limited air to form a mixture of unburnt carbon and carbon(II) oxide and water.
Ethyne + Air -> carbon(II) oxide + water (limited air )
C2H2(g) + O2(g) -> 2CO2(g) + C + 2H2O(l/g)
2.(a) Propyne when ignited burns with a yellow sooty flame in excess air to form carbon(IV) oxide and water.
Propyne + Air -> carbon(IV) oxide + water (excess air/oxygen)
C3H4(g) + 4O2(g) -> 3CO2(g) + 2H2O(l/g)
(a) Propyne when ignited burns with a yellow sooty flame in limited air to form carbon(II) oxide and water.
Propene + Air -> carbon(IV) oxide + water (excess air/oxygen)
2C3H4(g) + 5O2(g) -> 6CO(g) + 4H2O(l/g)
(b)Addition reactions
An addition reaction is one which an unsaturated compound reacts to form a saturated compound. Addition reactions of alkynes are also named from the reagent used to cause the addition/convert the triple – C = C- to single C- C bond.
(i)Hydrogenation
Hydrogenation is an addition reaction in which hydrogen in presence of Palladium/Nickel catalyst at 150oC temperatures react with alkynes to form alkenes then alkanes.
Examples
1.During hydrogenation, two hydrogen atom in the hydrogen molecule attach itself to one carbon and the other two hydrogen to the second carbon breaking the triple bond to double the single.
Chemical equation
HC = CH + H2 -Ni/Pa -> H2C = CH2 + H2 -Ni/Pa -> H2C – CH2
H H H H H H
C = C + H – H – Ni/Pa -> H – C = C – H + H – H – Ni/Pa -> H – C – C – H
H H H H H H
2.Propyne undergo hydrogenation to form Propane
Chemical equation
H3C CH = CH2 + 2H2 -Ni/Pa-> H3C CH – CH3
H H H H H H
H C C = C + 2H – H – Ni/Pa-> H – C – C – C- H
H H H H H
3(a) But-1-yne undergo hydrogenation to form Butane
Chemical equation
But-1-yne + Hydrogen –Ni/Pa-> Butane
H3C CH2 C = CH + 2H2 -Ni/Pa-> H3C CH2CH – CH3
H H H H H H H
H C C – C = C + 2H – H – Ni/Pa-> H – C- C – C – C- H
H H H H H H
(b) But-2-yne undergo hydrogenation to form Butane
Chemical equation
But-2-yne + Hydrogen –Ni/Pa-> Butane
H3C C = C CH2 + 2H2 -Ni/Pa-> H3C CH2CH – CH3
H H H H H H
H C C = C – C H + 2H – H- Ni/Pa-> H – C- C – C – C- H
H H H H H H
(ii) Halogenation.
Halogenation is an addition reaction in which a halogen (Fluorine, chlorine, bromine, iodine) reacts with an alkyne to form an alkene then alkane.
The reaction of alkynes with halogens with alkynes is faster than with alkenes. The triple bond in the alkyne break and form a double then single bond.
The colour of the halogen fades as the number of moles of the halogens remaining unreacted decreases.
Two bromine atoms bond at the 1st carbon in the triple bond while the other two goes to the 2nd carbon.
Examples
1Ethyne reacts with brown bromine vapour to form 1,1,2,2-tetrabromoethane.
Chemical equation
HC = CH + 2Br2 H Br2 C – CH Br2
H H H H
C = C + 2Br – Br Br – C – C – Br
Br Br
Ethyne + Bromine 1,1,2,1-tetrabromoethane
2.Propyne reacts with chlorine to form 1,1,2,2-tetrachloropropane.
Chemical equation
H3C C = CH + 2Cl2 H3C CHCl2 – CHCl2
Propyne + Chlorine 1,1,2,2-tetrachloropropane
H H Cl H
H C C = C + 2Cl – Cl H – C – C – C- Cl
H H H Cl Cl
Propyne + Iodine 1,1,2,2-tetraiodopropane
H3C C = CH + 2I2 H3C CHI2 – CHI2
H H H H H I H
H C C – C = C + 2I – I H – C- C – C – C- I
H H H H I I
3(a)But-1-yne undergo halogenation to form 1,1,2,2-tetraiodobutane with iodine
Chemical equation
But-1-yne + iodine 1,1,2,2-tetrabromobutane
H3C CH2 C = CH + 2I2 H3C CH2C I2 – CHI2
H H H H I I
H C C – C = C -H + 2I – I H – C- C – C – C- H
H H H H H I I
(b) But-2-yne undergo halogenation to form 2,2,3,3-tetrafluorobutane with fluorine
But-2-yne + Fluorine 2,2,3,3-tetrafluorobutane
H3C C = C -CH2 + 2F2 H3C CF2CF2 – CH3
H H H H H H H H
H C C = C – C -H + F – F H – C- C – C – C- H
H H H H H H
- But-1,3-diyne should undergo halogenation to form 1,1,2,3,3,4,4 octaiodobutane. The reaction uses four moles of iodine molecules/eight iodine atoms to break the two(2) triple double bonds at carbon “1” and “2”.
But-1,3-diene + iodine 1,2,3,4-tetraiodobutane
H C = C C = C H + 4I2 H C I2 C I2 C I2 C H I2
I I I I
H C C – C = C -H + 4(I – I) H – C- C – C – C- H
I I I I
(iii) Reaction with hydrogen halides.
Hydrogen halides reacts with alkyne to form a halogenoalkene then halogenoalkane. The triple bond in the alkyne break and form a double then single bond.
The main compound is one which the hydrogen atom bond at the carbon with more hydrogen .
Examples
- Ethyne reacts with hydrogen bromide to form bromoethane.
Chemical equation
H C = C H + 2HBr H3 C – CH Br2
H H H H
C = C + 2H – Br H – C – C – Br
H Br
Ethyne + Bromine 1,1-dibromoethane
- Propyne reacts with hydrogen iodide to form 2,2-diiodopropane (as the main product )
Chemical equation
H3C C = CH + 2HI H3C CHI2 – CH3
| Carbon atom with more Hydrogen atoms gets extra hydrogen |
Propene + Chlorine 2,2-dichloropropane
H H I H
H C C = C + 2H – I H – C – C – C– H
H H H I H
- Both But-1-yne and But-2-yne reacts with hydrogen bromide to form 2,2- dibromobutane
Chemical equation
But-1-ene + hydrogen bromide 2,2-dibromobutane
H3C CH2 C = CH + 2HBr H3C CH2CHBr -CH3
H H H H Br H
H C C – C = C + 2H – Br H – C- C – C – C- H
H H H H H Br H
But-2-yne + Hydrogen bromide 2,2-dibromobutane
H3C C = C -CH3 + 2HBr H3C CBr2CH2 – CH3
H H H Br H H
H C C = C – C -H + 2Br – H H – C- C – C – C- H
H H H Br H H
- But-1,3-diene react with hydrogen iodide to form 2,3- diiodobutane. The reaction uses four moles of hydrogen iodide molecules/four iodine atoms and two hydrogen atoms to break the two double bonds.
But-1,3-diyne + iodine 2,2,3,3-tetraiodobutane
H C = C C = C H + 4HI H3C C I2 C I2 CH3
H H H I I H
H C C – C = C -H + 4(H – I) H – C- C – C – C- H
H I I H
B.ALKANOLS(Alcohols)
(A) INTRODUCTION.
Alkanols belong to a homologous series of organic compounds with a general formula CnH2n +1 OH and thus -OH as the functional group .The 1st ten alkanols include
| n | General / molecular formular | Structural formula | IUPAC name |
| 1 | CH3OH |
H – C –O – H │ H
|
Methanol |
| 2 | CH3 CH2OH
C2H5 OH |
H H
H C – C –O – H │ H H
|
Ethanol |
| 3 | CH3 (CH2)2OH
C3H7 OH |
H H H
H C – C – C –O – H │ H H H
|
Propanol |
| 4 | CH3 (CH2)3OH
C4H9 OH |
H H H H
H C – C – C – C –O – H │ H H H H |
Butanol |
| 5 | CH3(CH2)4OH
C5H11 OH |
H H H H H
H C – C – C- C- C –O – H │ H H H H H
|
Pentanol |
| 6 | CH3(CH2)5OH
C6H13 OH |
H H H H H H
H C – C – C- C- C– C – O – H │ H H H H H H
|
Hexanol |
| 7 | CH3(CH2)6OH
C7H15 OH |
H H H H H H H
H C – C – C- C- C– C –C- O – H │ H H H H H H H
|
Heptanol |
| 8 | CH3(CH2)7OH
C8H17 OH |
H H H H H H H H
H C – C – C- C- C– C –C- C -O – H │ H H H H H H H H
|
Octanol |
| 9 | CH3(CH2)8OH
C9H19 OH |
H H H H H H H H H
H C – C – C- C- C– C –C- C –C- O – H │ H H H H H H H H H
|
Nonanol |
| 10 | CH3(CH2)9OH
C10H21 OH |
H H H H H H H H H H
H C – C – C- C- C– C –C- C –C- C-O – H │ H H H H H H H H H H
|
Decanol |
Alkanols like Hydrocarbons( alkanes/alkenes/alkynes) form a homologous series where:
(i)general name is derived from the alkane name then ending with “-ol”
(ii)the members have –OH as the fuctional group
(iii)they have the same general formula represented by R-OH where R is an alkyl group.
(iv) each member differ by –CH2 group from the next/previous.
(v)they show a similar and gradual change in their physical properties e.g. boiling and melting points.
(vi)they show similar and gradual change in their chemical properties.
- ISOMERS OF ALKANOLS.
Alkanols exhibit both structural and position isomerism. The isomers are named by using the following basic guidelines:
(i)Like alkanes , identify the longest carbon chain to be the parent name.
(ii)Identify the position of the -OH functional group to give it the smallest /lowest position.
(iii) Identify the type and position of the side branches.
Practice examples of isomers of alkanols
(i)Isomers of propanol C3H7OH
CH3CH2CH2OH – Propan-1-ol
OH
CH3CHCH3 – Propan-2-ol
Propan-2-ol and Propan-1-ol are position isomers because only the position of the –OH functional group changes.
(ii)Isomers of Butanol C4H9OH
CH3 CH2 CH3 CH2 OH Butan-1-ol
CH3 CH2 CH CH3
OH Butan-2-ol
CH3
CH3 CH3 CH3
OH 2-methylpropan-2-ol
Butan-2-ol and Butan-1-ol are position isomers because only the position of the -OH functional group changes.
2-methylpropan-2-ol is both a structural and position isomers because both the position of the functional group and the arrangement of the atoms in the molecule changes.
(iii)Isomers of Pentanol C5H11OH
CH3 CH2 CH2CH2CH2 OH Pentan-1-ol (Position isomer)
CH3 CH2 CH CH3
OH Pentan-2-ol (Position isomer)
CH3 CH2 CH CH2 CH3
OH Pentan-3-ol (Position isomer)
CH3
CH3 CH2 CH2 C CH3
OH 2-methylbutan-2-ol (Position /structural isomer)
CH3
CH3 CH2 CH2 C CHOH
CH3 2,2-dimethylbutan-1-ol (Position /structural isomer)
CH3
CH3 CH2 CH C CH3
CH3 OH 2,3-dimethylbutan-1-ol (Position /structural isomer)
(iv)1,2-dichloropropan-2-ol
CClH2 CCl CH3
OH
(v)1,2-dichloropropan-1-ol
CClH2 CHCl CH2
OH
(vi) Ethan1,2-diol
H H
HOCH2CH2OH H-O – C – C – O-H
H H
(vii) Propan1,2,3-triol H OH H
HOCH2CHOHCH2OH H-O – C- C – C – O-H
H H H
- LABORATORY PREPARATION OF ALKANOLS.
For decades the world over, people have been fermenting grapes juice, sugar, carbohydrates and starch to produce ethanol as a social drug for relaxation.
In large amount, drinking of ethanol by mammals /human beings causes mental and physical lack of coordination.
Prolonged intake of ethanol causes permanent mental and physical lack of coordination because it damages vital organs like the liver.
Fermentation is the reaction where sugar is converted to alcohol/alkanol using biological catalyst/enzymes in yeast.
It involves three processes:
(i)Conversion of starch to maltose using the enzyme diastase.
(C6H10O5)n (s) + H2O(l) –diastase enzyme –> C12H22O11(aq)
(Starch) (Maltose)
(ii)Hydrolysis of Maltose to glucose using the enzyme maltase.
C12H22O11(aq)+ H2O(l) — maltase enzyme –>2 C6H12O6(aq)
(Maltose) (glucose)
(iii)Conversion of glucose to ethanol and carbon(IV)oxide gas using the enzyme zymase.
C6H12O6(aq) — zymase enzyme –> 2 C2H5OH(aq) + 2CO2(g)
(glucose) (Ethanol)
At concentration greater than 15% by volume, the ethanol produced kills the yeast enzyme stopping the reaction.
To increases the concentration, fractional distillation is done to produce spirits (e.g. Brandy=40% ethanol).
Methanol is much more poisonous /toxic than ethanol.
Taken large quantity in small quantity it causes instant blindness and liver, killing the consumer victim within hours.
School laboratory preparation of ethanol from fermentation of glucose
Measure 100cm3 of pure water into a conical flask.
Add about five spatula end full of glucose.
Stir the mixture to dissolve.
Add about one spatula end full of yeast.
Set up the apparatus as below.
Preserve the mixture for about three days.
D.PHYSICAL AND CHEMICAL PROPERTIES OF ALKANOLS
Use the prepared sample above for the following experiments that shows the characteristic properties of alkanols
- Role of yeast
Yeast is a single cell fungus which contains the enzyme maltase and zymase that catalyse the fermentation process.
- Observations in lime water.
A white precipitate is formed that dissolve to a colourless solution later. Lime water/Calcium hydroxide reacts with carbon(IV)0xide produced during the fermentation to form insoluble calcium carbonate and water.
More carbon (IV)0xide produced during fermentation react with the insoluble calcium carbonate and water to form soluble calcium hydrogen carbonate.
Ca(OH)2(aq) + CO2 (g) -> CaCO3(s)
H2O(l) + CO2 (g) + CaCO3(s) -> Ca(HCO3) 2 (aq)
(c)Effects on litmus paper
Experiment
Take the prepared sample and test with both blue and red litmus papers.
Repeat the same with pure ethanol and methylated spirit.
Sample Observation table
| Substance/alkanol | Effect on litmus paper |
| Prepared sample | Blue litmus paper remain blue
Red litmus paper remain red |
| Absolute ethanol | Blue litmus paper remain blue
Red litmus paper remain red |
| Methylated spirit | Blue litmus paper remain blue
Red litmus paper remain red |
Explanation
Alkanols are neutral compounds/solution that have characteristic sweet smell and taste.
They have no effect on both blue and red litmus papers.
(d)Solubility in water.
Experiment
Place about 5cm3 of prepared sample into a clean test tube Add equal amount of distilled water.
Repeat the same with pure ethanol and methylated spirit.
Observation
No layers formed between the two liquids.
Explanation
Ethanol is miscible in water.Both ethanol and water are polar compounds .
The solubility of alkanols decrease with increase in the alkyl chain/molecular mass.
The alkyl group is insoluble in water while –OH functional group is soluble in water.
As the molecular chain becomes longer ,the effect of the alkyl group increases as the effect of the functional group decreases.
e)Melting/boiling point.
Experiment
Place pure ethanol in a long boiling tube .Determine its boiling point.
Observation
Pure ethanol has a boiling point of 78oC at sea level/one atmosphere pressure.
Explanation
The melting and boiling point of alkanols increase with increase in molecular chain/mass .
This is because the intermolecular/van-der-waals forces of attraction between the molecules increase.
More heat energy is thus required to weaken the longer chain during melting and break during boiling.
f)Density
Density of alkanols increase with increase in the intermolecular/van-der-waals forces of attraction between the molecule, making it very close to each other.
This reduces the volume occupied by the molecule and thus increase the their mass per unit volume (density).
Summary table showing the trend in physical properties of alkanols
| Alkanol | Melting point
(oC) |
Boiling point
(oC) |
Density
gcm-3 |
Solubility in water |
| Methanol | -98 | 65 | 0.791 | soluble |
| Ethanol | -117 | 78 | 0.789 | soluble |
| Propanol | -103 | 97 | 0.803 | soluble |
| Butanol | -89 | 117 | 0.810 | Slightly soluble |
| Pentanol | -78 | 138 | 0.814 | Slightly soluble |
| Hexanol | -52 | 157 | 0.815 | Slightly soluble |
| Heptanol | -34 | 176 | 0.822 | Slightly soluble |
| Octanol | -15 | 195 | 0.824 | Slightly soluble |
| Nonanol | -7 | 212 | 0.827 | Slightly soluble |
| Decanol | 6 | 228 | 0.827 | Slightly soluble |
g)Burning
Experiment
Place the prepared sample in a watch glass. Ignite. Repeat with pure ethanol and methylated spirit.
Observation/Explanation
Fermentation produce ethanol with a lot of water(about a ratio of 1:3)which prevent the alcohol from igniting.
Pure ethanol and methylated spirit easily catch fire / highly flammable.
They burn with an almost colourless non-sooty/non-smoky blue flame to form carbon(IV) oxide (in excess air/oxygen)or carbon(II) oxide (limited air) and water.
Ethanol is thus a saturated compound like alkanes.
Chemica equation
C2 H5OH(l) + 3O2 (g) -> 3H2O(l) + 2CO2 (g) ( excess air)
C2 H5OH(l) + 2O2 (g) -> 3H2O(l) + 2CO (g) ( limited air)
2CH3OH(l) + 3O2 (g) -> 4H2O(l) + 2CO2 (g) ( excess air)
2 CH3OH(l) + 2O2 (g) -> 4H2O(l) + 2CO (g) ( limited air)
2C3 H7OH(l) + 9O2 (g) -> 8H2O(l) + 6CO2 (g) ( excess air)
C3 H7OH(l) + 3O2 (g) -> 4H2O(l) + 3CO (g) ( limited air)
2C4 H9OH(l) + 13O2 (g) -> 20H2O(l) + 8CO2 (g) ( excess air)
C4 H9OH(l) + 3O2 (g) -> 4H2O(l) + 3CO (g) ( limited air)
Due to its flammability, ethanol is used;
- as a fuel in spirit lamps
- as gasohol when blended with gasoline
(h)Formation of alkoxides
Experiment
Cut a very small piece of sodium. Put it in a beaker containing about 20cm3 of the prepared sample in a beaker.
Test the products with litmus papers. Repeat with absolute ethanol and methylated spirit.
Sample observations
| Substance/alkanol | Effect of adding sodium |
| Fermentation prepared sample | (i)effervescence/fizzing/bubbles
(ii)colourless gas produced that extinguish burning splint with explosion/ “Pop” sound (iii)colourless solution formed (iv)blue litmus papers remain blue (v)red litmus papers turn blue |
| Pure/absolute ethanol/methylated spirit | (i)slow effervescence/fizzing/bubbles
(ii)colourless gas slowly produced that extinguish burning splint with explosion/ “Pop” sound (iii)colourless solution formed (iv)blue litmus papers remain blue (v)red litmus papers turn blue |
Explanations
Sodium/potassium reacts slowly with alkanols to form basic solution called alkoxides and producing hydrogen gas.
If the alkanol has some water the metals react faster with the water to form soluble hydroxides/alkalis i.e.
Sodium + Alkanol -> Sodium alkoxides + Hydrogen gas
Potassium + Alkanol -> Potassium alkoxides + Hydrogen gas
Sodium + Water -> Sodium hydroxides + Hydrogen gas
Potassium + Water -> Potassium hydroxides + Hydrogen gas
Examples
1.Sodium metal reacts with ethanol to form sodium ethoxide
Sodium metal reacts with water to form sodium Hydroxide
2CH3CH2OH(l) + 2Na(s) -> 2CH3CH2ONa (aq) + H2 (s)
2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s)
2.Potassium metal reacts with ethanol to form Potassium ethoxide
Potassium metal reacts with water to form Potassium Hydroxide
2CH3CH2OH(l) + 2K(s) -> 2CH3CH2OK (aq) + H2 (s)
2H2O(l) + 2K(s) -> 2KOH (aq) + H2 (s)
3.Sodium metal reacts with propanol to form sodium propoxide
Sodium metal reacts with water to form sodium Hydroxide
2CH3CH2 CH2OH(l) + 2Na(s) -> 2CH3CH2 CH2ONa (aq) + H2 (s)
2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s)
4.Potassium metal reacts with propanol to form Potassium propoxide
Potassium metal reacts with water to form Potassium Hydroxide
2CH3CH2 CH2OH(l) + 2K(s) -> 2CH3CH2 CH2OK (aq) + H2 (s)
2H2O(l) + 2K(s) -> 2KOH (aq) + H2 (s)
5.Sodium metal reacts with butanol to form sodium butoxide
Sodium metal reacts with water to form sodium Hydroxide
2CH3CH2 CH2 CH2OH(l) + 2Na(s) -> 2CH3CH2 CH2 CH2ONa (aq) + H2 (s)
2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s)
6.Sodium metal reacts with pentanol to form sodium pentoxide
Sodium metal reacts with water to form sodium Hydroxide
2CH3CH2 CH2 CH2 CH2OH(l)+2Na(s) -> 2CH3CH2 CH2 CH2 CH2ONa (aq) + H2 (s)
2H2O(l) + 2Na(s) -> 2NaOH (aq) + H2 (s)
(i)Formation of Esters/Esterification
Experiment
Place 2cm3 of ethanol in a boiling tube.
Add equal amount of ethanoic acid.To the mixture add carefully 2drops of concentrated sulphuric(VI)acid.
Warm/Heat gently.
Pour the mixture into a beaker containing about 50cm3 of cold water.
Smell the products.
Repeat with methanol
Sample observations
| Substance/alkanol | Effect on adding equal amount of ethanol/concentrated sulphuric(VI)acid |
| Absolute ethanol | Sweet fruity smell |
| Methanol | Sweet fruity smell |
Explanation
Alkanols react with alkanoic acids to form a group of homologous series of sweet smelling compounds called esters and water. This reaction is catalyzed by concentrated sulphuric(VI)acid in the laboratory.
Alkanol + Alkanoic acid –Conc. H2SO4-> Ester + water
Naturally esterification is catalyzed by sunlight. Each ester has a characteristic smell derived from the many possible combinations of alkanols and alkanoic acids that create a variety of known natural(mostly in fruits) and synthetic(mostly in juices) esters .
Esters derive their names from the alkanol first then alkanoic acids. The alkanol “becomes” an alkyl group and the alkanoic acid “becomes” alkanoate hence alkylalkanoate. e.g.
Ethanol + Ethanoic acid -> Ethylethanoate + Water
Ethanol + Propanoic acid -> Ethylpropanoate + Water
Ethanol + Methanoic acid -> Ethylmethanoate + Water
Ethanol + butanoic acid -> Ethylbutanoate + Water
Propanol + Ethanoic acid -> Propylethanoate + Water
Methanol + Ethanoic acid -> Methyethanoate + Water
Methanol + Decanoic acid -> Methyldecanoate + Water
Decanol + Methanoic acid -> Decylmethanoate + Water
During the formation of the ester, the “O” joining the alkanol and alkanoic acid comes from the alkanol.
R1 -COOH + R2 –OH -> R1 -COO –R2 + H2O
e.g.
- Ethanol reacts with ethanoic acid to form the ester ethyl ethanoate and water.
Ethanol + Ethanoic acid –Conc. H2SO4 –>Ethylethanoate + Water
C2H5OH (l) + CH3COOH(l) –Conc. H2SO4 –> CH3COO C2H5(aq) +H2O(l)
CH3CH2OH (l)+ CH3COOH(l) –Conc. H2SO4 –> CH3COOCH2CH3(aq) +H2O(l)
- Ethanol reacts with propanoic acid to form the ester ethylpropanoate and water.
Ethanol + Propanoic acid –Conc. H2SO4 –>Ethylethanoate + Water
C2H5OH (l)+ CH3 CH2COOH(l) –Conc. H2SO4 –>CH3CH2COO C2H5(aq) +H2O(l)
CH3CH2OH (l)+ CH3 CH2COOH(l) –Conc. H2SO4 –>
CH3 CH2COOCH2CH3(aq) +H2O(l)
- Methanol reacts with ethanoic acid to form the ester methyl ethanoate and water.
Methanol + Ethanoic acid –Conc. H2SO4 –>Methylethanoate + Water
CH3OH (l) + CH3COOH(l) –Conc. H2SO4 –> CH3COO CH3(aq) +H2O(l)
- Methanol reacts with propanoic acid to form the ester methyl propanoate and water.
Methanol + propanoic acid –Conc. H2SO4 –>Methylpropanoate + Water
CH3OH (l)+ CH3 CH2COOH(l) –Conc. H2SO4 –> CH3 CH2COO CH3(aq) +H2O(l)
- Propanol reacts with propanoic acid to form the ester propylpropanoate and water.
Propanol + Propanoic acid –Conc. H2SO4 –>Ethylethanoate + Water
C3H7OH (l)+ CH3 CH2COOH(l) –Conc. H2SO4 –>CH3CH2COO C3H7(aq) +H2O(l)
CH3CH2 CH2OH (l)+ CH3 CH2COOH(l) –Conc. H2SO4 –>
CH3 CH2COOCH2 CH2CH3(aq) +H2O(l)
(j)Oxidation
Experiment
Place 5cm3 of absolute ethanol in a test tube.Add three drops of acidified potassium manganate(VII).Shake thoroughly for one minute/warm.Test the solution mixture using pH paper. Repeat by adding acidified potassium dichromate(VII).
Sample observation table
| Substance/alkanol | Adding acidified KMnO4/K2Cr2O7 | pH of resulting solution/mixture | Nature of resulting solution/mixture |
| Pure ethanol | (i)Purple colour of KMnO4decolorized
(ii) Orange colour of K2Cr2O7turns green. |
pH= 4/5/6
pH = 4/5/6 |
Weakly acidic
Weakly acidic
|
Explanation
Both acidified KMnO4 and K2Cr2O7 are oxidizing agents(add oxygen to other compounds. They oxidize alkanols to a group of homologous series called alkanals then further oxidize them to alkanoic acids.The oxidizing agents are themselves reduced hence changing their colour:
(i) Purple KMnO4 is reduced to colourless Mn2+
(ii)Orange K2Cr2O7is reduced to green Cr3+
The pH of alkanoic acids show they have few H+ because they are weak acids i.e
Alkanol + [O] -> Alkanal + [O] -> alkanoic acid
NB The [O] comes from the oxidizing agents acidified KMnO4 or K2Cr2O7
Examples
1.When ethanol is warmed with three drops of acidified KMnO4 there is decolorization of KMnO4
Ethanol + [O] -> Ethanal + [O] -> Ethanoic acid
CH3CH2OH + [O] -> CH3CH2O + [O] -> CH3COOH
2.When methanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green.
methanol + [O] -> methanal + [O] -> methanoic acid
CH3OH + [O] -> CH3O + [O] -> HCOOH
3.When propanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green.
Propanol + [O] -> Propanal + [O] -> Propanoic acid
CH3CH2 CH2OH + [O] -> CH3CH2 CH2O + [O] -> CH3 CH2COOH
4.When butanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green.
Butanol + [O] -> Butanal + [O] -> Butanoic acid
CH3CH2 CH2 CH2OH + [O] ->CH3CH2 CH2CH2O +[O] -> CH3 CH2COOH
Air slowly oxidizes ethanol to dilute ethanoic acid commonly called vinegar. If beer is not tightly corked, a lot of carbon(IV)oxide escapes and there is slow oxidation of the beer making it “flat”.
(k)Hydrolysis /Hydration and Dehydration
- Hydrolysis/Hydration is the reaction of a compound/substance with water.
Alkenes react with water vapour/steam at high temperatures and high pressures in presence of phosphoric acid catalyst to form alkanols.i.e.
Alkenes + Water – H3PO4 catalyst-> Alkanol
Examples
(i)Ethene is mixed with steam over a phosphoric acid catalyst at 300oC temperature and 60 atmosphere pressure to form ethanol
Ethene + water —60 atm/300oC/ H3PO4 –> Ethanol
H2C =CH2 (g) + H2O(l) –60 atm/300oC/ H3PO4 –> CH3 CH2OH(l)
This is the main method of producing large quantities of ethanol instead of fermentation
(ii) Propene + water —60 atm/300oC/ H3PO4 –> Propanol
CH3C =CH2 (g) + H2O(l) –60 atm/300oC/ H3PO4 –> CH3 CH2 CH2OH(l)
(iii) Butene + water —60 atm/300oC/ H3PO4 –> Butanol
CH3 CH2 C=CH2 (g) + H2O(l) –60 atm/300oC/ H3PO4 –> CH3 CH2 CH2 CH2OH(l)
- Dehydration is the process which concentrated sulphuric(VI)acid (dehydrating agent) removes water from a compound/substances.
Concentrated sulphuric(VI)acid dehydrates alkanols to the corresponding alkenes at about 180oC. i.e
Alkanol –Conc. H2 SO4/180oC–> Alkene + Water
Examples
- At 180oC and in presence of Concentrated sulphuric(VI)acid, ethanol undergoes dehydration to form ethene.
Ethanol —180oC/ H2SO4 –> Ethene + Water
CH3 CH2OH(l) –180oC/ H2SO4 –> H2C =CH2 (g) + H2O(l)
- Propanol undergoes dehydration to form propene.
Propanol —180oC/ H2SO4 –> Propene + Water
CH3 CH2 CH2OH(l) –180oC/ H2SO4 –> CH3CH =CH2 (g) + H2O(l)
- Butanol undergoes dehydration to form Butene.
Butanol —180oC/ H2SO4 –> Butene + Water
CH3 CH2 CH2CH2OH(l) –180oC/ H2SO4 –> CH3 CH2C =CH2 (g) + H2O(l)
- Pentanol undergoes dehydration to form Pentene.
Pentanol —180oC/ H2SO4 –> Pentene + Water
CH3 CH2 CH2 CH2 CH2OH(l)–180oC/ H2SO4–>CH3 CH2 CH2C =CH2 (g)+H2O(l)
(l)Similarities of alkanols with Hydrocarbons
- Similarity with alkanes
Both alkanols and alkanes burn with a blue non-sooty flame to form carbon(IV)oxide(in excess air/oxygen)/carbon(II)oxide(in limited air) and water. This shows they are saturated with high C:H ratio. e.g.
Both ethanol and ethane ignite and burns in air with a blue non-sooty flame to form carbon(IV)oxide(in excess air/oxygen)/carbon(II)oxide(in limited air) and water.
CH2 CH2OH(l) + 3O2(g) -Excess air-> 2CO2 (g) + 3H2 O(l)
CH2 CH2OH(l) + 2O2(g) -Limited air-> 2CO (g) + 3H2 O(l)
CH3 CH3(g) + 3O2(g) -Excess air-> 2CO2 (g) + 3H2 O(l)
2CH3 CH3(g) + 5O2(g) -Limited air-> 4CO (g) + 6H2 O(l)
- Similarity with alkenes/alkynes
Both alkanols(R-OH) and alkenes/alkynes(with = C = C = double and – C = C- triple ) bond:
(i)decolorize acidified KMnO4
(ii)turns Orange acidified K2Cr2O7 to green.
Alkanols(R-OH) are oxidized to alkanals(R-O) ant then alkanoic acids(R-OOH).
Alkenes are oxidized to alkanols with duo/double functional groups.
Examples
1.When ethanol is warmed with three drops of acidified K2Cr2O7 the orange of acidified K2Cr2O7 turns to green. Ethanol is oxidized to ethanol and then to ethanoic acid.
Ethanol + [O] -> Ethanal + [O] -> Ethanoic acid
CH3CH2OH + [O] -> CH3CH2O + [O] -> CH3COOH
2.When ethene is bubbled in a test tube containing acidified K2Cr2O7 ,the orange of acidified K2Cr2O7 turns to green. Ethene is oxidized to ethan-1,2-diol.
Ethene + [O] -> Ethan-1,2-diol.
H2C=CH2 + [O] -> HOCH2 -CH2OH
III. Differences with alkenes/alkynes
Alkanols do not decolorize bromine and chlorine water.
Alkenes decolorizes bromine and chlorine water to form halogenoalkanols
Example
When ethene is bubbled in a test tube containing bromine water,the bromine water is decolorized. Ethene is oxidized to bromoethanol.
Ethene + Bromine water -> Bromoethanol.
H2C=CH2 + HOBr -> BrCH2 -CH2OH
- Differences in melting and boiling point with Hydrocarbons
Alkanos have higher melting point than the corresponding hydrocarbon (alkane/alkene/alkyne)
This is because most alkanols exist as dimer.A dimer is a molecule made up of two other molecules joined usually by van-der-waals forces/hydrogen bond or dative bonding.
Two alkanol molecules form a dimer joined by hydrogen bonding.
Example
In Ethanol the oxygen atom attracts/pulls the shared electrons in the covalent bond more to itself than Hydrogen.
This creates a partial negative charge (δ-) on oxygen and partial positive charge(δ+) on hydrogen.
Two ethanol molecules attract each other at the partial charges through Hydrogen bonding forming a dimmer.
| Hydrogen bonds |
H H H
| Covalent bonds |
H C C O H H
H H H O C C H
H H
Dimerization of alkanols means more energy is needed to break/weaken the Hydrogen bonds before breaking/weakening the intermolecular forces joining the molecules of all organic compounds during boiling/melting.
E.USES OF SOME ALKANOLS
(a)Methanol is used as industrial alcohol and making methylated spirit
(b)Ethanol is used:
- as alcohol in alcoholic drinks e.g Beer, wines and spirits.
2.as antiseptic to wash woulds
3.in manufacture of vanishes, ink ,glue and paint because it is volatile and thus easily evaporate
4.as a fuel when blended with petrol to make gasohol.
B.ALKANOIC ACIDS (Carboxylic acids)
(A) INTRODUCTION.
Alkanoic acids belong to a homologous series of organic compounds with a general formula CnH2n +1 COOH and thus -COOH as the functional group .The 1st ten alkanoic acids include:
| n | General /molecular
formular |
Structural formula | IUPAC name |
| 0 | HCOOH |
H – C –O – H │ O |
Methanoic acid
|
| 1 | CH3 COOH
|
H
H – C – C – O – H │ H O |
Ethanoic acid |
| 2 | CH3 CH2 COOH
C2 H5 COOH |
H H
H-C – C – C – O – H
H H O |
Propanoic acid |
| 3 | CH3 CH2 CH2 COOH
C3 H7 COOH |
H H H
H- C – C – C – C – O – H
H H H O |
Butanoic acid |
| 4 | CH3CH2CH2CH2 COOH
C4 H9 COOH |
H H H H
H – C – C – C – C – C – O – H
H H H H O |
Pentanoic acid |
| 5 | CH3CH2 CH2CH2CH2 COOH
C5 H11 COOH |
H H H H H
H C – C – C – C – C – C – O – H
H H H H H O
|
Hexanoic acid |
| 6 | CH3CH2 CH2 CH2CH2CH2 COOH
C6 H13 COOH |
H H H H H H
H C C – C – C – C – C – C – O – H
H H H H H H O
|
Pentanoic acid
|
Alkanoic acids like alkanols /alkanes/alkenes/alkynes form a homologous series where:
(i)the general name of an alkanoic acids is derived from the alkane name then ending with “–oic” acid as the table above shows.
(ii) the members have R-COOH/R C-O-H as the functional group.
O
(iii)they have the same general formula represented by R-COOH where R is an alkyl group.
(iv)each member differ by –CH2– group from the next/previous.
(v)they show a similar and gradual change in their physical properties e.g. boiling and melting point.
(vi)they show similar and gradual change in their chemical properties.
(vii) since they are acids they show similar properties with mineral acids.
(B) ISOMERS OF ALKANOIC ACIDS.
Alkanoic acids exhibit both structural and position isomerism. The isomers are named by using the following basic guidelines
(i)Like alkanes. identify the longest carbon chain to be the parent name.
(ii)Identify the position of the -C-O-H functional group to give it the smallest
O
/lowest position.
(iii)Identify the type and position of the side group branches.
Practice examples on isomers of alkanoic acids
1.Isomers of butanoic acid C3H7COOH
CH3 CH2 CH2 COOH
Butan-1-oic acid
CH3
H2C C COOH 2-methylpropan-1-oic acid
2-methylpropan-1-oic acid and Butan-1-oic acid are structural isomers because the position of the functional group does not change but the arrangement of the atoms in the molecule does.
2.Isomers of pentanoic acid C4H9COOH
CH3CH2CH2CH2 COOH pentan-1-oic acid
CH3
CH3CH2CH COOH 2-methylbutan-1-oic acid
CH3
H3C C COOH 2,2-dimethylpropan-1-oic acid
CH3
3.Ethan-1,2-dioic acid
O O
HOOC- COOH // H – O – C – C – O – H
4.Propan-1,3-dioic acid
O H O
HOOC- CH2COOH // H – O – C – C – C – O – H
H
5.Butan-1,4-dioic acid
O H H O
HOOC CH2 CH2 COOH H- O – C – C – C – C –O – H
H H
6.2,2-dichloroethan-1,2-dioic acid
HOOCCHCl2 Cl
H – O – C – C – Cl
O H
(C) LABORATORY AND INDUSTRIAL PREPARATIONOF ALKANOIC ACIDS.
In a school laboratory, alkanoic acids can be prepared by adding an oxidizing agent (H+/KMnO4 or H+/K2Cr2O7)to the corresponding alkanol then warming.
The oxidation converts the alkanol first to an alkanal the alkanoic acid.
NB Acidified KMnO4 is a stronger oxidizing agent than acidified K2Cr2O7
General equation:
R- CH2 – OH + [O] –H+/KMnO4–> R- CH –O + H2O(l)
(alkanol) (alkanal)
R- CH – O + [O] –H+/KMnO4–> R- C –OOH
(alkanal) (alkanoic acid)
Examples
1.Ethanol on warming in acidified KMnO4 is oxidized to ethanal then ethanoic acid .
CH3– CH2 – OH + [O] –H+/KMnO4–> CH3– CH –O + H2O(l)
(ethanol) (ethanal)
CH3– CH – O + [O] –H+/KMnO4–> CH3– C –OOH
(ethanal) (ethanoic acid)
2Propanol on warming in acidified KMnO4 is oxidized to propanal then propanoic acid
CH3– CH2 CH2 – OH + [O] –H+/KMnO4–> CH3– CH2 CH –O + H2O(l)
(propanol) (propanal)
CH3– CH – O + [O] –H+/KMnO4–> CH3– C –OOH
(propanal) (propanoic acid)
Industrially,large scale manufacture of alkanoic acid like ethanoic acid is obtained from:
(a)Alkenes reacting with steam at high temperatures and pressure in presence of phosphoric(V)acid catalyst and undergo hydrolysis to form alkanols. i.e.
Alkenes + Steam/water — H2PO4 Catalyst–> Alkanol
The alkanol is then oxidized by air at 5 atmosphere pressure with Manganese (II)sulphate(VI) catalyst to form the alkanoic acid.
Alkanol + Air — MnSO4 Catalyst/5 atm pressure–> Alkanoic acid
Example
Ethene is mixed with steam over a phosphoric(V)acid catalyst,300oC temperature and 60 atmosphere pressure to form ethanol.
CH2=CH2 + H2O -> CH3 CH2OH
(Ethene) (Ethanol)
This is the industrial large scale method of manufacturing ethanol
Ethanol is then oxidized by air at 5 atmosphere pressure with Manganese (II)sulphate(VI) catalyst to form the ethanoic acid.
CH3 CH2OH + [O] — MnSO4 Catalyst/5 atm pressure–> CH3 COOH
(Ethanol) (Ethanoic acid)
(b)Alkynes react with liquid water at high temperatures and pressure in presence of Mercury(II)sulphate(VI)catalyst and 30% concentrated sulphuric(VI)acid to form alkanals.
Alkyne + Water — Mercury(II)sulphate(VI)catalyst–> Alkanal
The alkanal is then oxidized by air at 5 atmosphere pressure with Manganese (II) sulphate(VI) catalyst to form the alkanoic acid.
Alkanal + air/oxygen — Manganese(II)sulphate(VI)catalyst–> Alkanoic acid
Example
Ethyne react with liquid water at high temperature and pressure with Mercury (II) sulphate (VI)catalyst and 30% concentrated sulphuric(VI)acid to form ethanal.
CH = CH + H2O –HgSO4–> CH3 CH2O
(Ethyne) (Ethanal)
This is another industrial large scale method of manufacturing ethanol from large quantities of ethyne found in natural gas.
Ethanal is then oxidized by air at 5 atmosphere pressure with Manganese (II)sulphate(VI) catalyst to form the ethanoic acid.
CH3 CH2O + [O] — MnSO4 Catalyst/5 atm pressure–> CH3 COOH
(Ethanal) (Oxygen from air) (Ethanoic acid)
(D) PHYSICAL AND CHEMICAL PROPERTIES OF ALKANOIC ACIDS.
I.Physical properties of alkanoic acids
The table below shows some physical properties of alkanoic acids
| Alkanol | Melting point(oC) | Boiling point(oC) | Density(gcm-3) | Solubility in water |
| Methanoic acid | 18.4 | 101 | 1.22 | soluble |
| Ethanoic acid | 16.6 | 118 | 1.05 | soluble |
| Propanoic acid | -2.8 | 141 | 0.992 | soluble |
| Butanoic acid | -8.0 | 164 | 0.964 | soluble |
| Pentanoic acid | -9.0 | 187 | 0.939 | Slightly soluble |
| Hexanoic acid | -11 | 205 | 0.927 | Slightly soluble |
| Heptanoic acid | -3 | 223 | 0.920 | Slightly soluble |
| Octanoic acid | 11 | 239 | 0.910 | Slightly soluble |
| Nonanoic acid | 16 | 253 | 0.907 | Slightly soluble |
| Decanoic acid | 31 | 269 | 0.905 | Slightly soluble |
From the table note the following:
- Melting and boiling point decrease as the carbon chain increases due to increase in intermolecular forces of attraction between the molecules requiring more energy to separate the molecules.
- The density decreases as the carbon chain increases as the intermolecular forces of attraction increases between the molecules making the molecule very close reducing their volume in unit mass.
- Solubility decreases as the carbon chain increases as the soluble –COOH end is shielded by increasing insoluble alkyl/hydrocarbon chain.
- Like alkanols ,alkanoic acids exist as dimmers due to the hydrogen bonds within the molecule. i.e..
II Chemical properties of alkanoic acids
The following experiments shows the main chemical properties of ethanoic (alkanoic) acid.
(a)Effect on litmus papers
Experiment
Dip both blue and red litmus papers in ethanoic acid. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute nitric(V)acid.
Sample observations
| Solution/acid | Observations/effect on litmus papers | Inference |
| Ethanoic acid | Blue litmus paper turn red
Red litmus paper remain red |
H3O+/H+(aq)ion |
| Succinic acid | Blue litmus paper turn red
Red litmus paper remain red |
H3O+/H+(aq)ion |
| Citric acid | Blue litmus paper turn red
Red litmus paper remain red |
H3O+/H+(aq)ion |
| Oxalic acid | Blue litmus paper turn red
Red litmus paper remain red |
H3O+/H+(aq)ion |
| Tartaric acid | Blue litmus paper turn red
Red litmus paper remain red |
H3O+/H+(aq)ion |
| Nitric(V)acid | Blue litmus paper turn red
Red litmus paper remain red |
H3O+/H+(aq)ion |
Explanation
All acidic solutions contains H+/H3O+(aq) ions. The H+ /H3O+ (aq) ions is responsible for turning blue litmus paper/solution to red
(b)pH
Experiment
Place 2cm3 of ethaoic acid in a test tube. Add 2 drops of universal indicator solution and determine its pH. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute sulphuric (VI)acid.
Sample observations
| Solution/acid | pH | Inference |
| Ethanoic acid | 4/5/6 | Weakly acidic |
| Succinic acid | 4/5/6 | Weakly acidic |
| Citric acid | 4/5/6 | Weakly acidic |
| Oxalic acid | 4/5/6 | Weakly acidic |
| Tartaric acid | 4/5/6 | Weakly acidic |
| Sulphuric(VI)acid | 1/2/3 | Strongly acidic |
Explanations
Alkanoic acids are weak acids that partially/partly dissociate to release few H+ ions in solution. The pH of their solution is thus 4/5/6 showing they form weakly acidic solutions when dissolved in water.
All alkanoic acid dissociate to releases the “H” at the functional group in -COOH to form the alkanoate ion; –COO–
Mineral acids(Sulphuric(VI)acid, Nitric(V)acid and Hydrochloric acid) are strong acids that wholly/fully dissociate to release many H+ ions in solution. The pH of their solution is thus 1/2/3 showing they form strongly acidic solutions when dissolved in water.i.e
Examples
- CH3COOH(aq) CH3COO–(aq) + H+(aq)
(ethanoic acid) (ethanoate ion) (few H+ ion)
- CH3 CH2COOH(aq) CH3 CH2COO–(aq) + H+(aq)
(propanoic acid) (propanoate ion) (few H+ ion)
- CH3 CH2 CH2COOH(aq) CH3 CH2 CH2COO–(aq) + H+(aq)
(Butanoic acid) (butanoate ion) (few H+ ion)
- HOOH(aq) HOO–(aq) + H+(aq)
(methanoic acid) (methanoate ion) (few H+ ion)
- H2 SO4 (aq) SO42- (aq) + 2H+(aq)
(sulphuric(VI) acid) (sulphate(VI) ion) (many H+ ion)
- HNO3 (aq) NO3– (aq) + H+(aq)
(nitric(V) acid) (nitrate(V) ion) (many H+ ion)
(c)Reaction with metals
Experiment
Place about 4cm3 of ethanoic acid in a test tube. Put about 1cm length of polished magnesium ribbon. Test any gas produced using a burning splint. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute sulphuric (VI) acid.
Sample observations
| Solution/acid | Observations | Inference |
| Ethanoic acid | (i)effervescence, fizzing, bubbles
(ii)colourless gas produced that burn with “pop” sound/explosion |
H3O+/H+(aq)ion |
| Succinic acid | (i)effervescence, fizzing, bubbles
(ii)colourless gas produced that burn with “pop” sound/explosion |
H3O+/H+(aq)ion |
| Citric acid | (i)effervescence, fizzing, bubbles
(ii)colourless gas produced that burn with “pop” sound/explosion |
H3O+/H+(aq)ion |
| Oxalic acid | (i)effervescence, fizzing, bubbles
(ii)colourless gas produced that burn with “pop” sound/explosion |
H3O+/H+(aq)ion |
| Tartaric acid | (i)effervescence, fizzing, bubbles
(ii)colourless gas produced that burn with “pop” sound/explosion |
H3O+/H+(aq)ion |
| Nitric(V)acid | (i)effervescence, fizzing, bubbles
(ii)colourless gas produced that burn with “pop” sound/explosion |
H3O+/H+(aq)ion |
Explanation
Metals higher in the reactivity series displace the hydrogen in all acids to evolve/produce hydrogen gas and form a salt. Alkanoic acids react with metals with metals to form alkanoates salt and produce/evolve hydrogen gas .Hydrogen extinguishes a burning splint with a pop sound/explosion. Only the “H”in the functional group -COOH is /are displaced and not in the alkyl hydrocarbon chain.
Alkanoic acid + Metal -> Alkanoate + Hydrogen gas. i.e.
Examples
- For a monovalent metal with monobasic acid
2R – COOH + 2M -> 2R- COOM + 2H2(g)
2.For a divalent metal with monobasic acid
2R – COOH + M -> (R- COO) 2M + H2(g)
3.For a divalent metal with dibasic acid
HOOC-R-COOH+ M -> MOOC-R-COOM + H2(g)
4.For a monovalent metal with dibasic acid
HOOC-R-COOH+ 2M -> MOOC-R-COOM + H2(g)
5 For mineral acids
(i)Sulphuric(VI)acid is a dibasic acid
H2 SO4 (aq) + 2M -> M2 SO4 (aq) + H2(g)
H2 SO4 (aq) + M -> MSO4 (aq) + H2(g)
(ii)Nitric(V) and hydrochloric acid are monobasic acid
HNO3 (aq) + 2M -> 2MNO3 (aq) + H2(g)
HNO3 (aq) + M -> M(NO3 ) 2 (aq) + H2(g)
Examples
1.Sodium reacts with ethanoic acid to form sodium ethanoate and produce. hydrogen gas.
Caution: This reaction is explosive.
CH3COOH (aq) + Na(s) -> CH3COONa (aq) + H2(g)
(Ethanoic acid) (Sodium ethanoate)
2.Calcium reacts with ethanoic acid to form calcium ethanoate and produce. hydrogen gas.
2CH3COOH (aq) + Ca(s) -> (CH3COO) 2Ca (aq) + H2(g)
(Ethanoic acid) (Calcium ethanoate)
3.Sodium reacts with ethan-1,2-dioic acid to form sodium ethan-1,2-dioate and produce. hydrogen gas.
HOOC-COOH+ 2Na -> NaOOC – COONa + H2(g)
(ethan-1,2-dioic acid) (sodium ethan-1,2-dioate)
Commercial name of ethan-1,2-dioic acid is oxalic acid. The salt is sodium oxalate.
4.Magnesium reacts with ethan-1,2-dioic acid to form magnesium ethan-1,2-dioate and produce. hydrogen gas.
HOOC-R-COOH+ Mg -> ( OOC – COO) Mg + H2(g)
(ethan-1,2-dioic acid) (magnesium ethan-1,2-dioate)
5.Magnesium reacts with
(i)Sulphuric(VI)acid to form Magnesium sulphate(VI)
H2 SO4 (aq) + Mg -> MgSO4 (aq) + H2(g)
(ii)Nitric(V) and hydrochloric acid are monobasic acid
2HNO3 (aq) + Mg -> M(NO3 ) 2 (aq) + H2(g)
(d)Reaction with hydrogen carbonates and carbonates
Experiment
Place about 3cm3 of ethanoic acid in a test tube. Add about 0.5g/ ½ spatula end full of sodium hydrogen carbonate/sodium carbonate. Test the gas produced using lime water. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute sulphuric (VI) acid.
Sample observations
| Solution/acid | Observations | Inference |
| Ethanoic acid | (i)effervescence, fizzing, bubbles
(ii)colourless gas produced that forms a white precipitate with lime water |
H3O+/H+(aq)ion |
| Succinic acid | (i)effervescence, fizzing, bubbles
(ii)colourless gas produced that forms a white precipitate with lime water |
H3O+/H+(aq)ion |
| Citric acid | (i)effervescence, fizzing, bubbles
(ii)colourless gas produced that forms a white precipitate with lime water |
H3O+/H+(aq)ion |
| Oxalic acid | (i)effervescence, fizzing, bubbles
(ii)colourless gas produced that forms a white precipitate with lime water |
H3O+/H+(aq)ion |
| Tartaric acid | (i)effervescence, fizzing, bubbles
(ii)colourless gas produced that forms a white precipitate with lime water |
H3O+/H+(aq)ion |
| Nitric(V)acid | (i)effervescence, fizzing, bubbles
(ii)colourless gas produced that forms a white precipitate with lime water |
H3O+/H+(aq)ion |
All acids react with hydrogen carbonate/carbonate to form salt ,water and evolve/produce bubbles of carbon(IV)oxide and water.
Carbon(IV)oxide forms a white precipitate when bubbled in lime water/extinguishes a burning splint.
Alkanoic acids react with hydrogen carbonate/carbonate to form alkanoates ,water and evolve/produce bubbles of carbon(IV)oxide and water.
Alkanoic acid + hydrogen carbonate -> alkanoate + water + carbon(IV)oxide
Alkanoic acid + carbonate -> alkanoate + water + carbon(IV)oxide
Examples
- Sodium hydrogen carbonate reacts with ethanoic acid to form sodium ethanoate ,water and carbon(IV)oxide gas.
CH3COOH (aq) + NaHCO3 (s) -> CH3COONa (aq) + H2O(l) + CO2 (g)
(Ethanoic acid) (Sodium ethanoate)
2.Sodium carbonate reacts with ethanoic acid to form sodium ethanoate ,water and carbon(IV)oxide gas.
2CH3COOH (aq) + Na2CO3 (s) -> 2CH3COONa (aq) + H2O(l) + CO2 (g)
(Ethanoic acid) (Sodium ethanoate)
3.Sodium carbonate reacts with ethan-1,2-dioic acid to form sodium ethanoate ,water and carbon(IV)oxide gas.
HOOC-COOH+ Na2CO3 (s) -> NaOOC – COONa + H2O(l) + CO2 (g)
(ethan-1,2-dioic acid) (sodium ethan-1,2-dioate)
4.Sodium hydrogen carbonate reacts with ethan-1,2-dioic acid to form sodium ethanoate ,water and carbon(IV)oxide gas.
HOOC-COOH+ 2NaHCO3 (s) -> NaOOC – COONa + H2O(l) + 2CO2 (g)
(ethan-1,2-dioic acid) (sodium ethan-1,2-dioate)
(e)Esterification
Experiment
Place 4cm3 of ethanol acid in a boiling tube.
Add equal volume of ethanoic acid. To the mixture, add 2 drops of concentrated sulphuric(VI)acid carefully. Warm/heat gently on Bunsen flame.
Pour the mixture into a beaker containing 50cm3 of water. Smell the products. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute sulphuric (VI) acid.
Sample observations
| Solution/acid | Observations |
| Ethanoic acid | Sweet fruity smell |
| Succinic acid | Sweet fruity smell |
| Citric acid | Sweet fruity smell |
| Oxalic acid | Sweet fruity smell |
| Tartaric acid | Sweet fruity smell |
| Dilute sulphuric(VI)acid | No sweet fruity smell |
Explanation
Alkanols react with alkanoic acid to form the sweet smelling homologous series of esters and water.The reaction is catalysed by concentrated sulphuric(VI)acid in the laboratory but naturally by sunlight /heat.Each ester has a characteristic smell derived from the many possible combinations of alkanols and alkanoic acids.
Alkanol + Alkanoic acids -> Ester + water
Esters derive their names from the alkanol first then alkanoic acids. The alkanol “becomes” an alkyl group and the alkanoic acid “becomes” alkanoate hence alkylalkanoate. e.g.
Ethanol + Ethanoic acid -> Ethylethanoate + Water
Ethanol + Propanoic acid -> Ethylpropanoate + Water
Ethanol + Methanoic acid -> Ethylmethanoate + Water
Ethanol + butanoic acid -> Ethylbutanoate + Water
Propanol + Ethanoic acid -> Propylethanoate + Water
Methanol + Ethanoic acid -> Methyethanoate + Water
Methanol + Decanoic acid -> Methyldecanoate + Water
Decanol + Methanoic acid -> Decylmethanoate + Water
During the formation of the ester, the “O” joining the alkanol and alkanoic acid comes from the alkanol.
R1 -COOH + R2 –OH -> R1 -COO –R2 + H2O
Examples
- Ethanol reacts with ethanoic acid to form the ester ethyl ethanoate and water.
Ethanol + Ethanoic acid –Conc. H2SO4 –>Ethylethanoate + Water
C2H5OH (l) + CH3COOH(l) –Conc. H2SO4 –> CH3COO C2H5(aq) +H2O(l)
CH3CH2OH (l)+ CH3COOH(l) –Conc. H2SO4 –> CH3COOCH2CH3(aq) +H2O(l)
- Ethanol reacts with propanoic acid to form the ester ethylpropanoate and water.
Ethanol + Propanoic acid –Conc. H2SO4 –>Ethylethanoate + Water
C2H5OH (l)+ CH3 CH2COOH(l) –Conc. H2SO4 –>CH3CH2COO C2H5(aq) +H2O(l)
CH3CH2OH (l)+ CH3 CH2COOH(l) –Conc. H2SO4 –>
CH3 CH2COOCH2CH3(aq) +H2O(l)
- Methanol reacts with ethanoic acid to form the ester methyl ethanoate and water.
Methanol + Ethanoic acid –Conc. H2SO4 –>Methylethanoate + Water
CH3OH (l) + CH3COOH(l) –Conc. H2SO4 –> CH3COO CH3(aq) +H2O(l)
- Methanol reacts with propanoic acid to form the ester methyl propanoate and water.
Methanol + propanoic acid –Conc. H2SO4 –>Methylpropanoate + Water
CH3OH (l)+ CH3 CH2COOH(l) –Conc. H2SO4 –> CH3 CH2COO CH3(aq) +H2O(l)
- Propanol reacts with propanoic acid to form the ester propylpropanoate and water.
Propanol + Propanoic acid –Conc. H2SO4 –>Ethylethanoate + Water
C3H7OH (l)+ CH3 CH2COOH(l) –Conc. H2SO4 –>CH3CH2COO C3H7(aq) +H2O(l)
CH3CH2 CH2OH (l)+ CH3 CH2COOH(l) –Conc. H2SO4 –>
CH3 CH2COOCH2 CH2CH3(aq) +H2O(l)
- DETERGENTS
Detergents are cleaning agents that improve the cleaning power /properties of water.A detergent therefore should be able to:
(i)dissolve substances which water can not e.g grease ,oil, fat
(ii)be washed away after cleaning.
There are two types of detergents:
(a)Soapy detergents
(b)Soapless detergents
- SOAPY DETERGENTS
Soapy detergents usually called soap is long chain salt of organic alkanoic acids.Common soap is sodium octadecanoate .It is derived from reacting concentrated sodium hydroxide solution with octadecanoic acid(18 carbon alkanoic acid) i.e.
Sodium hydroxide + octadecanoic acid -> Sodium octadecanoate + water
NaOH(aq) + CH3 (CH2) 16 COOH(aq) -> CH3 (CH2) 16 COO – Na+ (aq) +H2 O(l)
Commonly ,soap can thus be represented ;
R- COO – Na+ where;
R is a long chain alkyl group and -COO – Na+ is the alkanoate ion.
In a school laboratory and at industrial and domestic level,soap is made by reacting concentrated sodium hydroxide solution with esters from (animal) fat and oil. The process of making soap is called saponification. During saponification ,the ester is hydrolyzed by the alkali to form sodium salt /soap and glycerol/propan-1,2,3-triol is produced.
Fat/oil(ester)+sodium/potassium hydroxide->sodium/potassium salt(soap)+ glycerol
Fats/Oils are esters with fatty acids and glycerol parts in their structure;
C17H35COOCH2
C17H35COOCH
C17H35COOCH2
When boiled with concentrated sodium hydroxide solution NaOH;
(i)NaOH ionizes/dissociates into Na+ and OH– ions
(ii)fat/oil split into three C17H35COO– and one CH2 CH CH2
(iii) the three Na+ combine with the three C17H35COO– to form the salt C17H35COO– Na+
(iv)the three OH–ions combine with the CH2 CH CH2 to form an alkanol with three functional groups CH2 OH CH OH CH2 OH(propan-1,2,3-triol)
C17H35COOCH2 CH2OH
C17H35COOCH +NaOH -> 3 C17H35COO– Na+ + CHOH
C17H35COOCH2 CH2OH
Ester Alkali Soap glycerol
Generally:
CnH2n+1COOCH2 CH2OH
CnH2n+1COOCH +NaOH -> 3 CnH2n+1COO– Na+ + CHOH
CnH2n+1COOCH2 CH2OH
Ester Alkali Soap glycerol
R – COOCH2 CH2OH
R – COOCH +NaOH -> 3R-COO– Na+ + CHOH
R- COOCH2 CH2OH
Ester Alkali Soap glycerol
During this process a little sodium chloride is added to precipitate the soap by reducing its solubility. This is called salting out.
The soap is then added colouring agents ,perfumes and herbs of choice.
School laboratory preparation of soap
Place about 40 g of fatty (animal fat)beef/meat in 100cm3 beaker .Add about 15cm3 of 4.0M sodium hydroxide solution. Boil the mixture for about 15minutes.Stir the mixture .Add about 5.0cm3 of distilled water as you boil to make up for evaporation. Boil for about another 15minutes.Add about four spatula end full of pure sodium chloride crystals. Continue stirring for another five minutes. Allow to cool. Filter of /decant and wash off the residue with distilled water .Transfer the clean residue into a dry beaker. Preserve.
The action of soap
Soapy detergents:
(i)act by reducing the surface tension of water by forming a thin layer on top of the water.
(ii)is made of a non-polar alkyl /hydrocarbon tail and a polar -COO–Na+ head. The non-polar alkyl /hydrocarbon tail is hydrophobic (water hating) and thus does not dissolve in water .It dissolves in non-polar solvent like grease, oil and fat. The polar -COO–Na+ head is hydrophilic (water loving)and thus dissolve in water. When washing with soapy detergent, the non-polar tail of the soapy detergent surround/dissolve in the dirt on the garment /grease/oil while the polar head dissolve in water.
Through mechanical agitation/stirring/sqeezing/rubbing/beating/kneading, some grease is dislodged/lifted of the surface of the garment. It is immediately surrounded by more soap molecules It float and spread in the water as tiny droplets that scatter light in form of emulsion making the water cloudy and shinny. It is removed from the garment by rinsing with fresh water.The repulsion of the soap head prevent /ensure the droplets do not mix.Once removed, the dirt molecules cannot be redeposited back because it is surrounded by soap molecules.
Advantages and disadvantages of using soapy detergents
Soapy detergents are biodegradable. They are acted upon by bacteria and rot.They thus do not cause environmental pollution.
Soapy detergents have the diadvatage in that:
(i)they are made from fat and oils which are better eaten as food than make soap.
(ii)forms an insoluble precipitate with hard water called scum. Scum is insoluble calcium octadecanoate and Magnesium octadecanoate formed when soap reacts with Ca2+ and Mg2+ present in hard water.
Chemical equation
2C17H35COO– Na+ (aq) + Ca2+(aq) -> (C17H35COO– )Ca2+ (s) + 2Na+(aq)
(insoluble Calcium octadecanote/scum)
2C17H35COO– Na+ (aq) + Mg2+(aq) -> (C17H35COO– )Mg2+ (s) + 2Na+(aq)
(insoluble Magnesium octadecanote/scum)
This causes wastage of soap.
Potassium soaps are better than Sodium soap. Potassium is more expensive than sodium and thus its soap is also more expensive.
(b)SOAPLESS DETERGENTS
Soapless detergent usually called detergent is a long chain salt fromed from by-products of fractional distillation of crude oil.Commonly used soaps include:
(i)washing agents
(ii)toothpaste
(iii)emulsifiers/wetting agents/shampoo
Soapless detergents are derived from reacting:
(i)concentrated sulphuric(VI)acid with a long chain alkanol e.g. Octadecanol(18 carbon alkanol) to form alkyl hydrogen sulphate(VI)
Alkanol + Conc sulphuric(VI)acid -> alkyl hydrogen sulphate(VI) + Water
R –OH + H2SO4 -> R –O-SO3H + H2O
(ii)the alkyl hydrogen sulphate(VI) is then neutralized with sodium/potassium hydroxide to form sodium/potassium alkyl hydrogen sulphate(VI)
Sodium/potassium alkyl hydrogen sulphate(VI) is the soapless detergent.
alkyl hydrogen + Potassium/sodium -> Sodium/potassium + Water
sulphate(VI) hydroxide alkyl hydrogen sulphate(VI)
R –O-SO3H + NaOH -> R –O-SO3– Na+ + H2O
Example
Step I : Reaction of Octadecanol with Conc.H2SO4
C17H35CH2OH (aq) + H2SO4 -> C17H35CH2–O- SO3– H+ (aq) + H2O (l)
octadecanol + sulphuric(VI)acid -> Octadecyl hydrogen sulphate(VI) + water
Step II: Neutralization by an alkali
C17H35CH2–O- SO3– H+ (aq) + NaOH -> C17H35CH2–O- SO3– Na+ (aq) + H2O (l)
Octadecyl hydrogen + sodium/potassium -> sodium/potassium octadecyl+Water
sulphate(VI) hydroxide hydrogen sulphate(VI)
School laboratory preparation of soapless detergent
Place about 20g of olive oil in a 100cm3 beaker. Put it in a trough containing ice cold water.
Add dropwise carefully 18M concentrated sulphuric(VI)acid stirring continuously into the olive oil until the oil turns brown.Add 30cm3 of 6M sodium hydroxide solution.Stir.This is a soapless detergent.
The action of soapless detergents
The action of soapless detergents is similar to that of soapy detergents.The soapless detergents contain the hydrophilic head and a long hydrophobic tail. i.e.
vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv-COO–Na+
vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv-O-SO3– Na+
(long hydrophobic /non-polar alkyl tail) (hydrophilic/polar/ionic head)
The tail dissolves in fat/grease/oil while the ionic/polar/ionic head dissolves in water.
The tail stick to the dirt which is removed by the attraction of water molecules and the polar/ionic/hydrophilic head by mechanical agitation /squeezing/kneading/ beating/rubbing/scrubbing/scatching.
The suspended dirt is then surrounded by detergent molecules and repulsion of the anion head preventing the dirt from sticking on the material garment.
The tiny droplets of dirt emulsion makes the water cloudy. On rinsing the cloudy emulsion is washed away.
Advantages and disadvantages of using soapless detergents
Soapless detergents are non-biodegradable unlike soapy detergents.
They persist in water during sewage treatment by causing foaming in rivers ,lakes and streams leading to marine /aquatic death.
Soapless detergents have the advantage in that they:
(i)do not form scum with hard water.
(ii)are cheap to manufacture/buying
(iii)are made from petroleum products but soapis made from fats/oil for human consumption.
Sample revision questions
- Study the scheme below
| KOH |
| Fat/oil |
| Residue X |
| Filtrate Y |
| Filtration |
| Sodium Chloride |
| Boiling |
(a)Identify the process
Saponification
(b)Fats and oils are esters. Write the formula of the a common structure of ester
C17H35COOCH2
C17H35COOCH
C17H35COOCH2
(c)Write a balanced equation for the reaction taking place during boiling
C17H35COOCH2 CH2OH
C17H35COOCH +3NaOH -> 3 C17H35COO– Na+ + CHOH
C17H35COOCH2 CH2OH
Ester Alkali Soap glycerol
(d)Give the IUPAC name of:
(i)Residue X
Potassium octadecanoate
(ii)Filtrate Y
Propan-1,2,3-triol
(e)Give one use of fitrate Y
Making paint
(f)What is the function of sodium chloride
To reduce the solubility of the soap hence helping in precipitating it out
(g)Explain how residue X helps in washing.
Has a non-polar hydrophobic tail that dissolves in dirt/grease /oil/fat
Has a polar /ionic hydrophilic head that dissolves in water.
From mechanical agitation,the dirt is plucked out of the garment and surrounded by the tail end preventing it from being deposited back on the garment.
(h)State one:
(i)advantage of continued use of residue X on the environment
Is biodegradable and thus do not pollute the environment
(ii)disadvantage of using residue X
Uses fat/oil during preparation/manufacture which are better used for human consumption.
(i)Residue X was added dropwise to some water.The number of drops used before lather forms is as in the table below.
| Water sample
|
|||
|
A |
B |
C |
|
| Drops of residue X | 15 | 2 | 15 |
| Drops of residue X in boiled water | 2 | 2 | 15 |
(i)State and explain which sample of water is:
- Soft
Sample B .Very little soap is used and no effect on amount of soap even on boiling/heating.
- Permanent hard
Sample C . A lot of soap is used and no effect on amount of soap even on boiling/heating. Boiling does not remove permanent hardness of water.
III. Temporary hard
Sample A . A lot of soap is used before boiling. Very little soap is used on boiling/heating. Boiling remove temporary hardness of water.
(ii)Write the equation for the reaction at water sample C.
Chemical equation
2C17H35COO– K+ (aq) + CaSO4(aq) -> (C17H35COO– )Ca2+ (s) + K2SO4(aq)
(insoluble Calcium octadecanote/scum)
Ionic equation
2C17H35COO– K+ (aq) + Ca2+(aq) -> (C17H35COO– )Ca2+ (s) + 2K+(aq)
(insoluble Calcium octadecanote/scum)
Chemical equation
2C17H35COO– K+ (aq) + MgSO4(aq) -> (C17H35COO– )Mg2+ (s) + K2SO4(aq)
(insoluble Calcium octadecanote/scum)
Ionic equation
2C17H35COO– K+ (aq) + Mg2+(aq) -> (C17H35COO– )Mg2+ (s) + 2K+(aq)
(insoluble Magnesium octadecanote/scum)
(iii)Write the equation for the reaction at water sample A before boiling.
Chemical equation
2C17H35COO– K+ (aq) + Ca(HCO3)(aq) ->(C17H35COO– )Ca2+ (s) + 2KHCO3 (aq)
(insoluble Calcium octadecanote/scum)
Ionic equation
2C17H35COO– K+ (aq) + Ca2+(aq) -> (C17H35COO– )Ca2+ (s) + 2K+(aq)
(insoluble Calcium octadecanote/scum)
Chemical equation
2C17H35COO– K+ (aq) + Mg(HCO3)(aq) ->(C17H35COO– )Mg2+ (s) + 2KHCO3 (aq)
(insoluble Calcium octadecanote/scum)
Ionic equation
2C17H35COO– K+ (aq) + Mg2+(aq) -> (C17H35COO– )Mg2+ (s) + 2K+(aq)
(insoluble Magnesium octadecanote/scum)
(iv)Explain how water becomes hard
Natural or rain water flowing /passing through rocks containing calcium (chalk, gypsum, limestone)and magnesium compounds (dolomite)dissolve them to form soluble Ca2+ and Mg2+ ions that causes water hardness.
(v)State two useful benefits of hard water
-Used in bone and teeth formation
-Coral polyps use hard water to form coral reefs
-Snails use hard water to make their shells
2.Study the scheme below and use it to answer the questions that follow.
| Conc. H2SO4 |
| Substance B |
| 6M sodium hydroxide |
| Brown solid A |
| Ice cold water |
| Olive oil |
(a)Identify :
(i)brown solid A
Alkyl hydrogen sulphate(VI)
(ii)substance B
Sodium alkyl hydrogen sulphate(VI)
(b)Write a general formula of:
(i)Substance A.
O
R-O-S O3 H // R- O – S – O – H
O
(ii)Substance B O
R-O-S O3 – Na+ R- O – S – O – Na+
O
(c)State one
(i) advantage of continued use of substance B
–Does not form scum with hard water
-Is cheap to make
-Does not use food for human as a raw material.
(ii)disadvantage of continued use of substance B.
Is non-biodegradable therefore do not pollute the environment
(d)Explain the action of B during washing.
Has a non-polar hydrocarbon long tail that dissolves in dirt/grease/oil/fat.
Has a polar/ionic hydrophilic head that dissolves in water
Through mechanical agitation the dirt is plucked /removed from the garment and surrounded by the tail end preventing it from being deposited back on the garment.
(e) Ethene was substituted for olive oil in the above process. Write the equation and name of the new products A and B.
Product A
Ethene + Sulphuric(VI)acid -> Ethyl hydrogen sulphate(VI)
H2C=CH2 + H2SO4 –> H3C – CH2 –O-SO3H
Product B
Ethyl hydrogen sulphate(VI) + sodium hydroxide -> sodium Ethyl + Water
hydrogen sulphate(VI)
H3C – CH2 –O-SO3H + NaOH -> H3C – CH2 –O-SO3–Na+ + H2O
(f)Ethanol can also undergo similar reactions forming new products A and B.Show this using a chemical equation.
Product A
Ethanol + Sulphuric(VI)acid ->Ethyl hydrogen sulphate(VI) + water
H3C-CH2OH + H2SO4 –> H3C – CH2 –O-SO3H + H2O
Product B
Ethyl hydrogen sulphate(VI) + sodium hydroxide -> sodium Ethyl + Water
hydrogen sulphate(VI)
H3C – CH2 –O-SO3H + NaOH -> H3C – CH2 –O-SO3–Na+ + H2O
3.Below is part of a detergent
H3C – (CH2 )16 – O – SO3 – K +
(a)Write the formular of the polar and non-polar end
Polar end
H3C – (CH2 )16 –
Non-polar end
– O – SO3 – K +
(b)Is the molecule a soapy or saopless detergent?
Soapless detergent
(c)State one advantage of using the above detergent
-does not form scum with hard water
-is cheap to manufacture
4.The structure of a detergent is
H H H H H H H H H H H H H
H- C- C- C-C- C- C- C- C- C- C -C- C- -C- COO–Na+
H H H H H H H H H H H H H
- a) Write the molecular formula of the detergent. (1mk)
CH3(CH2)12COO–Na+
- b) What type of detergent is represented by the formula? (1mk)
Soapy detergent
- c) When this type of detergent is used to wash linen in hard water, spots (marks) are left on the linen. Write the formula of the substance responsible for the spots (CH3(CH2)12COO–)2Ca2+ / CH3(CH2)12COO–)2Mg2+
- POLYMERS AND FIBRES
Polymers and fibres are giant molecules of organic compounds. Polymers and fibres are formed when small molecules called monomers join together to form large molecules called polymers at high temperatures and pressures. This process is called polymerization.
Polymers and fibres are either:
(a)Natural polymers and fibres
(b)Synthetic polymers and fibres
Natural polymers and fibres are found in living things(plants and animals) Natural polymers/fibres include:
-proteins/polypeptides making amino acids in animals
-cellulose that make cotton,wool,paper and silk
-Starch that come from glucose
-Fats and oils
-Rubber from latex in rubber trees.
Synthetic polymers and fibres are man-made. They include:
-polyethene
-polychloroethene
-polyphenylethene(polystyrene)
-Terylene(Dacron)
-Nylon-6,6
-Perspex(artificial glass)
Synthetic polymers and fibres have the following characteristic advantages over natural polymers
- They are light and portable
- They are easy to manufacture.
- They can easily be molded into shape of choice.
- They are resistant to corrosion, water, air , acids, bases and salts.
- They are comparatively cheap, affordable, colourful and aesthetic
Synthetic polymers and fibres however have the following disadvantages over natural polymers
- They are non-biodegradable and hence cause environmental pollution during disposal
- They give out highly poisonous gases when burnt like chlorine/carbon(II)oxide
- Some on burning produce Carbon(IV)oxide. Carbon(IV)oxide is a green house gas that cause global warming.
- Compared to some metals, they are poor conductors of heat,electricity and have lower tensile strength.
To reduce environmental pollution from synthetic polymers and fibres, the followitn methods of disposal should be used:
1.Recycling: Once produced all synthetic polymers and fibres should be recycled to a new product. This prevents accumulation of the synthetic polymers and fibres in the environment.
2.Production of biodegradable synthetic polymers and fibres that rot away.
There are two types of polymerization:
(a)addition polymerization
(b)condensation polymerization
(a)addition polymerization
Addition polymerization is the process where a small unsaturated monomer (alkene ) molecule join together to form a large saturated molecule. Only alkenes undergo addition polymerization.
Addition polymers are named from the alkene/monomer making the polymer and adding the prefix “poly” before the name of monomer to form a polyalkene
During addition polymerization
(i)the double bond in alkenes break
(ii)free radicals are formed
(iii)the free radicals collide with each other and join to form a larger molecule. The more collisions the larger the molecule.
Examples of addition polymerization
1.Formation of Polyethene
Polyethene is an addition polymer formed when ethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.
During polymerization:
(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting paticles)
H H H H H H H H
C = C + C = C + C = C + C = C + …
H H H H H H H H
Ethene + Ethene + Ethene + Ethene + …
(ii)the double bond joining the ethane molecule break to free readicals
H H H H H H H H
- C – C• + •C – C• + •C – C• + •C – C• + …
H H H H H H H H
Ethene radical + Ethene radical + Ethene radical + Ethene radical + …
(iii)the free radicals collide with each other and join to form a larger molecule
H H H H H H H H lone pair of electrons
- C – C – C – C – C – C – C – C• + …
H H H H H H H H
Lone pair of electrons can be used to join more monomers to form longer polyethene.
Polyethene molecule can be represented as:
H H H H H H H H extension of
molecule/polymer
– C – C – C – C – C – C – C – C- + …
H H H H H H H H
Since the molecule is a repetition of one monomer, then the polymer is:
H H
( C – C )n
H H
Where n is the number of monomers in the polymer. The number of monomers in the polymer can be determined from the molar mass of the polymer and monomer from the relationship:
Number of monomers/repeating units in monomer = Molar mass polymer
Molar mass monomer
Examples
Polythene has a molar mass of 4760.Calculate the number of ethene molecules in the polymer(C=12.0, H=1.0 )
Number of monomers/repeating units in polyomer = Molar mass polymer
Molar mass monomer
=> Molar mass ethene (C2H4 )= 28 Molar mass polyethene = 4760
Substituting 4760 = 170 ethene molecules
28
The commercial name of polyethene is polythene. It is an elastic, tough, transparent and durable plastic. Polythene is used:
(i)in making plastic bag
(ii)bowls and plastic bags
(iii)packaging materials
2.Formation of Polychlorethene
Polychloroethene is an addition polymer formed when chloroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.
During polymerization:
(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles)
H H H H H H H H
C = C + C = C + C = C + C = C + …
H Cl H Cl H Cl H Cl
chloroethene + chloroethene + chloroethene + chloroethene + …
(ii)the double bond joining the chloroethene molecule break to free radicals
H H H H H H H H
- C – C• + •C – C• + •C – C• + •C – C• + …
H Cl H Cl H Cl H Cl
(iii)the free radicals collide with each other and join to form a larger molecule
H H H H H H H H lone pair of electrons
- C – C – C – C – C – C – C – C• + …
H Cl H Cl H Cl H Cl
Lone pair of electrons can be used to join more monomers to form longer polychloroethene.
Polychloroethene molecule can be represented as:
H H H H H H H H extension of
molecule/polymer
– C – C – C – C – C – C – C – C- + …
H Cl H Cl H Cl H Cl
Since the molecule is a repetition of one monomer, then the polymer is:
H H
( C – C )n
H Cl
Examples
Polychlorothene has a molar mass of 4760.Calculate the number of chlorethene molecules in the polymer(C=12.0, H=1.0,Cl=35.5 )
Number of monomers/repeating units in monomer = Molar mass polymer
Molar mass monomer
=> Molar mass ethene (C2H3Cl )= 62.5 Molar mass polyethene = 4760
Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number)
62.5
The commercial name of polychloroethene is polyvinylchloride(PVC). It is a tough, non-transparent and durable plastic. PVC is used:
(i)in making plastic rope
(ii)water pipes
(iii)crates and boxes
3.Formation of Polyphenylethene
Polyphenylethene is an addition polymer formed when phenylethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.
During polymerization:
(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles)
H H H H H H H H
C = C + C = C + C = C + C = C + …
H C6H5 H C6H5 H C6H5 H C6H5
phenylethene + phenylethene + phenylethene + phenylethene + …
(ii)the double bond joining the phenylethene molecule break to free radicals
H H H H H H H H
- C – C• + •C – C• + •C – C• + •C – C• + …
H C6H5 H C6H5 H C6H5 H C6H5
(iii)the free radicals collide with each other and join to form a larger molecule
H H H H H H H H lone pair of electrons
- C – C – C – C – C – C – C – C • + …
H C6H5 H C6H5 H C6H5 H C6H5
Lone pair of electrons can be used to join more monomers to form longer polyphenylethene.
Polyphenylethene molecule can be represented as:
H H H H H H H H
– C – C – C – C – C – C – C – C –
H C6H5 H C6H5 H C6H5 H C6H5
Since the molecule is a repetition of one monomer, then the polymer is:
H H
( C – C )n
H C6H5
Examples
Polyphenylthene has a molar mass of 4760.Calculate the number of phenylethene molecules in the polymer(C=12.0, H=1.0, )
Number of monomers/repeating units in monomer = Molar mass polymer
Molar mass monomer
=> Molar mass ethene (C8H8 )= 104 Molar mass polyethene = 4760
Substituting 4760 = 45.7692 =>45 polyphenylethene molecules(whole number)
104
The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic. Polystyrene is used:
(i)in making packaging material for carrying delicate items like computers, radion,calculators.
(ii)ceiling tiles
(iii)clothe linings
4.Formation of Polypropene
Polypropene is an addition polymer formed when propene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.
During polymerization:
(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles)
H H H H H H H H
C = C + C = C + C = C + C = C + …
H CH3 H CH3 H CH3 H CH3
propene + propene + propene + propene + …
(ii)the double bond joining the phenylethene molecule break to free radicals
H H H H H H H H
- C – C• + •C – C• + •C – C• + •C – C• + …
H CH3 H CH3 H CH3 H CH3
(iii)the free radicals collide with each other and join to form a larger molecule
H H H H H H H H lone pair of electrons
- C – C – C – C – C – C – C – C • + …
H CH3 H CH3 H CH3 H CH3
Lone pair of electrons can be used to join more monomers to form longer propene.
propene molecule can be represented as:
H H H H H H H H
– C – C – C – C – C – C – C – C –
H CH3 H CH3 H CH3 H CH3
Since the molecule is a repetition of one monomer, then the polymer is:
H H
( C – C )n
H CH3
Examples
Polypropene has a molar mass of 4760.Calculate the number of propene molecules in the polymer(C=12.0, H=1.0, )
Number of monomers/repeating units in monomer = Molar mass polymer
Molar mass monomer
=> Molar mass propene (C3H8 )= 44 Molar mass polyethene = 4760
Substituting 4760 = 108.1818 =>108 propene molecules(whole number)
44
The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic. Polystyrene is used:
(i)in making packaging material for carrying delicate items like computers, radion,calculators.
(ii)ceiling tiles
(iii)clothe linings
5.Formation of Polytetrafluorothene
Polytetrafluorothene is an addition polymer formed when tetrafluoroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.
During polymerization:
(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles)
F F F F F F F F
C = C + C = C + C = C + C = C + …
F F F F F F F F
tetrafluoroethene + tetrafluoroethene+ tetrafluoroethene+ tetrafluoroethene + …
(ii)the double bond joining the tetrafluoroethene molecule break to free radicals
F F F F F F F F
- C – C• + •C – C• + •C – C• + •C – C• + …
F F F F F F F F
(iii)the free radicals collide with each other and join to form a larger molecule
F F F F F F F F lone pair of electrons
- C – C – C – C – C – C – C – C• + …
F F F F F F F F
Lone pair of electrons can be used to join more monomers to form longer polytetrafluoroethene.
polytetrafluoroethene molecule can be represented as:
F F F F F F F F extension of
molecule/polymer
– C – C – C – C – C – C – C – C- + …
F F F F F F F F
Since the molecule is a repetition of one monomer, then the polymer is:
F F
( C – C )n
F F
Examples
Polytetrafluorothene has a molar mass of 4760.Calculate the number of tetrafluoroethene molecules in the polymer(C=12.0, ,F=19 )
Number of monomers/repeating units in monomer = Molar mass polymer
Molar mass monomer
=> Molar mass ethene (C2F4 )= 62.5 Molar mass polyethene = 4760
Substituting 4760 = 77.16 => 77 polychloroethene molecules(whole number)
62.5
The commercial name of polytetrafluorethene(P.T.F.E) is Teflon(P.T.F.E). It is a tough, non-transparent and durable plastic. PVC is used:
(i)in making plastic rope
(ii)water pipes
(iii)crates and boxes
5.Formation of rubber from Latex
Natural rubber is obtained from rubber trees.
During harvesting an incision is made on the rubber tree to produce a milky white substance called latex.
Latex is a mixture of rubber and lots of water.
The latex is then added an acid to coagulate the rubber.
Natural rubber is a polymer of 2-methylbut-1,3-diene ;
H CH3 H H
CH2=C (CH3) CH = CH2 H – C = C – C = C – H
During natural polymerization to rubber, one double C=C bond break to self add to another molecule.The double bond remaining move to carbon “2” thus;
H CH3 H H H CH3 H H
– C – C = C – C – C – C = C – C –
H H H H
Generally the structure of rubber is thus;
H CH3 H H
-(- C – C = C – C -)n–
H H
Pure rubber is soft and sticky.It is used to make erasers, car tyres. Most of it is vulcanized.Vulcanization is the process of heating rubber with sulphur to make it harder/tougher.
During vulcanization the sulphur atoms form a cross link between chains of rubber molecules/polymers. This decreases the number of C=C double bonds in the polymer.
| Sulphur atoms make cross link between polymers |
H CH3 H H H CH3 H H
– C – C – C – C – C – C – C – C –
H S H H S H
H CH3 S H H CH3 S H
– C – C – C – C – C – C – C – C –
H H H H H H
Vulcanized rubber is used to make tyres, shoes and valves.
6.Formation of synthetic rubber
Synthetic rubber is able to resist action of oil,abrasion and organic solvents which rubber cannot.
Common synthetic rubber is a polymer of 2-chlorobut-1,3-diene ;
H Cl H H
CH2=C (Cl CH = CH2 H – C = C – C = C – H
During polymerization to synthetic rubber, one double C=C bond is broken to self add to another molecule. The double bond remaining move to carbon “2” thus;
H Cl H H H Cl H H
– C – C = C – C – C – C = C – C –
H H H H
Generally the structure of rubber is thus;
H Cl H H
-(- C – C = C – C -)n–
H H
Rubber is thus strengthened through vulcanization and manufacture of synthetic rubber.
(b)Condensation polymerization
Condensation polymerization is the process where two or more small monomers join together to form a larger molecule by elimination/removal of a simple molecule. (usually water).
Condensation polymers acquire a different name from the monomers because the two monomers are two different compounds
During condensation polymerization:
(i)the two monomers are brought together by high pressure to reduce distance between them.
(ii)monomers realign themselves at the functional group.
(iii)from each functional group an element is removed so as to form simple molecule (of usually H2O/HCl)
(iv)the two monomers join without the simple molecule of H2O/HCl
Examples of condensation polymerization
1.Formation of Nylon-6,6
Method 1: Nylon-6,6 can be made from the condensation polymerization of hexan-1,6-dioic acid with hexan-1,6-diamine.Amines are a group of homologous series with a general formula R-NH2 and thus -NH2 as the functional group.
During the formation of Nylon-6,6:
(i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups.
O O H H
H- O – C – (CH2 ) 4 – C – O – H + H –N – (CH2) 6 – N – H
(iii)from each functional group an element is removed so as to form a molecule of H2O and the two monomers join at the linkage .
O O H H
H- O – C – (CH2 ) 4 – C – N – (CH2) 6 – N – H + H 2O
.
Polymer bond linkage
Nylon-6,6 derive its name from the two monomers each with six carbon chain
Method 2: Nylon-6,6 can be made from the condensation polymerization of hexan-1,6-dioyl dichloride with hexan-1,6-diamine.
Hexan-1,6-dioyl dichloride belong to a group of homologous series with a general formula R-OCl and thus -OCl as the functional group.
The R-OCl is formed when the “OH” in R-OOH/alkanoic acid is replaced by Cl/chlorine/Halogen
During the formation of Nylon-6,6:
(i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups.
O O H H
Cl – C – (CH2 ) 4 – C – Cl + H –N – (CH2) 6 – N – H
(iii)from each functional group an element is removed so as to form a molecule of HCl and the two monomers join at the linkage .
O O H H
Cl – C – (CH2 ) 4 – C – N – (CH2) 6 – N – H + HCl
.
Polymer bond linkage
The two monomers each has six carbon chain hence the name “nylon-6,6”
The commercial name of Nylon-6,6 is Nylon It is a a tough, elastic and durable plastic. It is used to make clothes, plastic ropes and carpets.
2.Formation of Terylene
Method 1: Terylene can be made from the condensation polymerization of ethan-1,2-diol with benzene-1,4-dicarboxylic acid.
Benzene-1,4-dicarboxylic acid a group of homologous series with a general formula R-COOH where R is a ring of six carbon atom called Benzene ring .The functional group is -COOH.
During the formation of Terylene:
(i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups.
O O
H- O – C – C6H5 – C – O – H + H –O – CH2 CH2 – O – H
(iii)from each functional group an element is removed so as to form a molecule of H2O and the two monomers join at the linkage .
O O
H- O – C – C6H5 – C – O – (CH2) 6 – N – H + H 2O
.
Polymer bond linkage of terylene
Method 2: Terylene can be made from the condensation polymerization of benzene-1,4-dioyl dichloride with ethan-1,2-diol.
Benzene-1,4-dioyl dichloride belong to a group of homologous series with a general formula R-OCl and thus -OCl as the functional group and R as a benzene ring.
The R-OCl is formed when the “OH” in R-OOH is replaced by Cl/chlorine/Halogen
During the formation of Terylene
(i)the two monomers are brought together by high pressure to reduce distance between them and realign themselves at the functional groups.
O O
Cl – C – C5H5 – C – Cl + H –O – CH2 CH2 – O – H
(iii)from each functional group an element is removed so as to form a molecule of HCl and the two monomers join at the linkage .
O O
Cl – C – C5H5 – C – O – CH2 CH2 – O – H + HCl
.
Polymer bond linkage of terylene
The commercial name of terylene is Polyester /polyster It is a a tough, elastic and durable plastic. It is used to make clothes, plastic ropes and sails and plastic model kits.
Practice questions Organic chemistry
- A student mixed equal volumes of Ethanol and butanoic acid. He added a few drops of concentrated Sulphuric (VI) acid and warmed the mixture
(i) Name and write the formula of the main products
Name………………………………….
Formula……………………………………..
(ii) Which homologous series does the product named in (i) above belong?
| HC = CH2 |
| O |
- The structure of the monomer phenyl ethene is given below:-
- a) Give the structure of the polymer formed when four of the monomers are added together
- b) Give the name of the polymer formed in (a) above
- Explain the environmental effects of burning plastics in air as a disposal method
- Write chemical equation to represent the effect of heat on ammonium carbonate
- Sodium octadecanoate has a chemical formula CH3(CH2)6 COO–Na+, which is used as soap.
Explain why a lot of soap is needed when washing with hard water
| O
CH3CH2CH C OH
NH2 |
- A natural polymer is made up of the monomer:
(a) Write the structural formula of the repeat unit of the polymer
(b) When 5.0 x 10-5 moles of the polymer were hydrolysed, 0.515g of the monomer
were obtained.
Determine the number of the monomer molecules in this polymer.
(C = 12; H = 1; N = 14; O =16)
- The formula below represents active ingredients of two cleansing agents A and B
Which one of the cleansing agents would be suitable to be used in water containing magnesium hydrogen carbonate? Explain
| H H H H
C C C C
O H O H n
|
- Study the polymer below and use it to answer the questions that follow:
(a) Give the name of the monomer and draw its structures
(b) Identify the type of polymerization that takes place
(c) State one advantage of synthetic polymers
- Ethanol and Pentane are miscible liquids. Explain how water can be used to separate a mixture
of ethanol and pentane
| GLUCOSE SOLUTION |
| CRUDE ETHANOL |
| 95% ETHANOL |
| ABSOLUTE ETHANOL |
| G |
| H |
(a) What is absolute ethanol?
(b) State two conditions required for process G to take place efficiently
- (a) (i) The table below shows the volume of oxygen obtained per unit time when hydrogen
peroxide was decomposed in the presence of manganese (IV) Oxide. Use it to answer
the questions that follow:-
| Time in seconds | Volume of Oxygen evolved (cm3) |
| 0
30 60 90 120 150 180 210 240 270 300 |
0
10 19 27 34 38 43 45 45 45 45 |
(i) Plot a graph of volume of oxygen gas against time
(ii) Determine the rate of reaction at time 156 seconds
(iii) From the graph, find the time taken for 18cm3 of oxygen to be produced
(iv) Write a chemical equation to show how hydrogen peroxide decomposes in the presence
of manganese (IV) Oxide
(b) The diagram below shows how a Le’clanche (Dry cell) appears:-
(i) What is the function of MnO2 in the cell above?
(ii) Write the equation of a reaction that occurs at the cathode
(iii) Calculate the mass of Zinc that is consumed when a current of 0.1amperes flows
through the above cell for 30minutes (1F =96500c Zn =65)
- (a) Give the IUPAC names of the following compounds:
(i) CH3COOCH2CH3 *
| CH2 = C – CHCH3
Br |
(ii)
(b) The structure below shows some reactions starting with ethanol. Study it and answer
| CH3COOH |
| CH3COONa |
| CH3CH2OH |
| CH2=CH2 |
| CH3CH3 |
| CH2 CH2 |
| n |
| S |
| P |
| CH4 |
| T |
| Na Metal |
| Compound U |
| Step II |
| Step I |
| Step III |
| CH3COOH
Drops of Conc. H2SO4 |
| Reagent R |
| NaOH(aq) |
| Heat |
| Excess Cl2 /U.V |
the questions that follow:
(i) Write the formula of the organic compounds P and S *
(ii) Name the type of reaction, the reagent(s) and condition for the reactions in the following steps :-
(I) Step I *
(II) Step II *
(III) Step III *
(iii) Name reagent R …………………………………………………………… *
(iv) Draw the structural formula of T and give its name *
(v) (I) Name compound U………………………………………………………..
(II) If the relative molecular mass of U is 42000, determine the value of n (C=12, H=1)
(c) State why C2H4 burns with a more smoky flame than C2H6 *
- a) State two factors that affect the properties of a polymer
- b) Name the compound with the formula below :
CH3CH2CH2ONa
| CH3CH2CH3 |
| P |
| Step
V |
| CH3CH = CH2 |
| Step
W |
| CH3CH2CH2OH |
| CH3CH2COOH |
| Step R |
| Gas F |
| CH3CH2COONa |
| NaOH
Step J |
| K2CO3 |
| Step X |
| NaOH
Heat |
| CH3CH2COOCH3
G |
| Product T
+ Na2CO3 |
| H H
C – C
CH3 H |
| n |
| K |
- c) Study the scheme below and use it to answer the questions that follow:-
- i) Name the following compounds:-
- Product T ………………………… II. K ………
- ii) State one common physical property of substance G
iii) State the type of reaction that occurred in step J
- iv) Give one use of substance K
- v) Write an equation for the combustion of compound P
- vi) Explain how compounds CH3CH2COOH and CH3CH2CH2OH can be distinguished chemically
vii) If a polymer K has relative molecular mass of 12,600, calculate the value of n (H=1 C =12)
- Study the scheme given below and answer the questions that follow:-
| H2 (g) Ni
High temp |
| Polymer Q |
| Polymerization |
| Compound P |
| CH3CH2CH3 |
| CH3CH2CH2ONa + H2
|
| Na(s) |
| Propan-l-ol |
| Step I |
| Propylethanoate |
| CH3CH2COOH
|
| Solution T + CO2 (g)
|
| Step III |
| Na2CO3(aq) |
| Conc. H2SO4 180oC |
| Step II |
(a) (i) Name compound P ……………………………………………………………………
(ii) Write an equation for the reaction between CH3CH2COOH and Na2CO3
(b) State one use of polymer Q
(c) Name one oxidising agent that can be used in step II …………………………………..
(d) A sample of polymer Q is found to have a molecular mass of 4200. Determine the number of
monomers in the polymer (H = 1, C = 12)
(e) Name the type of reaction in step I …………………………………………………………..
(f) State one industrial application of step III
(g)State how burning can be used to distinguish between propane and propyne. Explain your
answer
(h) 1000cm3 of ethene (C2H4) burnt in oxygen to produce Carbon (II) Oxide and water vapour.
Calculate the minimum volume of air needed for the complete combustion of ethene
(Air contains 20% by volume of oxygen)
- (a) Study the schematic diagram below and answer the questions that follow:-
| CH3CH2COOCH2CH2CH3 |
| CH3CHCH2 |
| CH3CH2CH2ONa + Gas P |
| CH3CH2CH2OH |
| X |
| V |
| HCl Step 5 |
| Step 1 R |
| Na
Step2 |
| H+ |
| Step 3 |
| Q + H2O |
| MnO–4
Step 4 |
| Ni H2 |
(i) Identify the following:
Substance Q ………………………………………………………………………………………………..
Substance R…………………………………………………………………………………………………
Gas P…………………………………………………………………………………………………………..
(ii) Name:
Step 1…………………………………………………………………………………….
Step 4…………………………………………………………………………………….
(iii) Draw the structural formula of the major product of step 5
(iv) State the condition and reagent in step 3
- Study the flow chart below and answer the questions that follow
| Ethyl Ethanoate |
| CH2CH2OH |
| L |
| Reagent Q
Step 3 |
| KMnO4/H+(aq) |
| M |
| KMnO4/H+ |
| CH2CH2 |
| J |
| K |
| CO2 (g) |
| STEP 2 |
| Ni/H2(g)
Step 4 |
| Reagent P |
(a) (i) Name the following organic compounds:
M……………………………………………………………..……..
L…………………………………………………………………..
(ii) Name the process in step:
Step 2 ………………………………………………………….….
Step 4 ………………………………………………………….…
(iii) Identify the reagent P and Q
(iv) Write an equation for the reaction between CH3CH2CH2OH and sodium
- a) Give the names of the following compounds:
- i) CH3CH2CH2CH2OH ……………………………………………………………………
- ii) CH3CH2COOH …………………………………………………………………
iii) CH3C – O- CH2CH3 ……………………………………………………………………
| Step I |
- Study the scheme given below and answer the questions that follow;
| Step V
Complete combustion |
| Products |
| CH ºCH |
| C2H5COONa |
| Step II |
| Step IV + Heat |
| CH2 = CH2 |
| C2H6 |
| Step III |
|
CH2 = CHCl |
| n |
- i) Name the reagents used in:
Step I: ………………………………………………………………………
Step II ……………………………………………………………………
Step III ………………………………………………………………………
- ii) Write an equation to show products formed for the complete combustion of CH = CH
iii) Explain one disadvantage of continued use of items made form the compound formed
in step III
- A hydrated salt has the following composition by mass. Iron 20.2 %, oxygen 23.0%,
sulphur 11.5%, water 45.3%
- i) Determine the formula of the hydrated salt (Fe=56, S=32, O=16, H=11)
- ii) 6.95g of the hydrated salt in c(i) above were dissolved in distilled water and the total
volume made to 250cm3 of solution. Calculate the concentration of the resulting salt solution
in moles per litre. (Given that the molecula mass of the salt is 278)
- Write an equation to show products formed for the complete combustion of CH = CH
iii) Explain one disadvantage of continued use of items made form the compound formed
in step III
- Give the IUPAC name for each of the following organic compounds;
- i) CH3 – CH – CH2 – CH3
OH
ii)CH3 – CH – CH2 – CH2 – CH3
C2H5
iii)CH3COOCH2CH2CH3
- The structure below represents a cleansing agent.
O
R – S – O–Na+
O
- a) State the type of cleansing agent represented above
- b) State one advantage and one disadvantage of using the above cleansing agent.
- The structure below shows part of polymer .Use it to answer the questions that follow.
CH3 CH3 CH3
ï ï ï
― CH – CH2 – CH- CH2 – CH – CH2 ―
- a) Derive the structure of the monomer
- b) Name the type of polymerization represented above
- The flow chart below represents a series of reactions starting with ethanoic acid:-
| Ethanol |
| B
process I |
| Ethanoic acid
|
| Na2CO3 |
| Salt A + CO2 + H2O |
(a) Identify substances A and B
(b) Name the process I
- a) Write an equation showing how ammonium nitrate may be prepared starting with
ammonia gas
(b) Calculate the maximum mass of ammonium nitrate that can be prepared using 5.3kg of
ammonia (H=1, N=14, O=16)
- (a) What is meant by the term, esterification?
(b) Draw the structural formulae of two compounds that may be reacted to form ethylpropanoate
- (a) Draw the structure of pentanoic acid
(b) Draw the structure and give the name of the organic compound formed when ethanol
reacts with pentanoic acid in presence of concentrated sulphuric acid
- The scheme below shows some reactions starting with ethanol. Study it and answer the questions
| Q |
| P |
| CH3COONa |
| Ethanol |
| CH3CH2ONa |
| C |
| H2SO4(l)
170oC step 1 |
| Cr2O7(aq) / H+(aq) |
| Na(s) |
| Step 2 |
| Step 4 |
| CH3CH2OH/H2SO4 |
| Step 3 |
| 2- |
that follow:-
(i) Name and draw the structure of substance Q
(ii) Give the names of the reactions that take place in steps 2 and 4
(iii) What reagent is necessary for reaction that takes place in step 3
- Substances A and B are represented by the formulae ROH and RCOOH respectively.
They belong to two different homologous series of organic compounds. If both A and B
react with potassium metal:
(a) Name the common product produced by both
(b) State the observation made when each of the samples A and B are reacted with sodium
hydrogen carbonate
(i) A
(ii) B
- Below are structures of particles. Use it to answer questions that follow. In each case only
electrons in the outermost energy level are shown
| W |
| U |
| V |
| 19P
20N |
| Z |
| Y |
key
P = Proton
N = Neutron
X = Electron
(a) Identify the particle which is an anion
- Plastics and rubber are extensively used to cover electrical wires.
(a) What term is used to describe plastic and rubbers used in this way?
(b) Explain why plastics and rubbers are used this way
- The scheme below represents the manufacture of a cleaning agent X
| R |
| Conc.
H2SO4 |
| R |
|
SO3H |
| Cleaning agent X |
(a) Draw the structure of X and state the type of cleaning agent to which X belong
(b) State one disadvantage of using X as a cleaning agent
- Y grams of a radioactive isotope take 120days to decay to 3.5grams. The half-life period
of the isotope is 20days
(a) Find the initial mass of the isotope
(b) Give one application of radioactivity in agriculture
| H H
-C – C –
CH3 H |
| n
n |
- The structure below represents a polymer. Study and answer the questions that follow:-
(i) Name the polymer above……………………………………………………………………….
(ii) Determine the value of n if giant molecule had relative molecular mass of 4956
- RCOO–Na+ and RCH2OSO3–Na+ are two types of cleansing agents;
- i) Name the class of cleansing agents to which each belongs
- ii) Which one of these agents in (i) above would be more suitable when washing with water
from the Indian ocean. Explain
iii) Both sulphur (IV) oxide and chlorine are used bleaching agents. Explain the difference
in their bleaching properties
| H H H H
C C C C
O H O H |
| n |
- The formula given below represents a portion of a polymer
(a) Give the name of the polymer
(b) Draw the structure of the monomer used to manufacture the polymer
|
UPGRADE
CHEMISTRY
FORM 3
The Chemistry of NITROGEN
Comprehensive tutorial notes
MUTHOMI S.G 0720096206
|
Contents
A NITROGEN …………………………………………………………………….3
Occurrence………………………………………………………..……..…..3
Isolation from air…………………………………………………………….3
Fractional distillation …………………………………………………….….4
Preparation of Nitrogen…………………………………………………..….6
Properties of Nitrogen…………………………………………………….…..6
B.OXIDES OFNITROGEN……………………………………………….……..7
Nitrogen(I)Oxide………………………..……………………………………7
Nitrogen(II) oxide……………………….…………………………………..9
Nitrogen(IV) oxide …………………………………………………………11
C.AMMONIA…..………………………….…………………………….……..13
Occurrence…………………………………………………………………15
Preparation………………………………………………………………….15
Properties……………………………………………………………………16
Haber process……………………………….………………………………12
- NITRIC(V) ACID………..………………………….………………………28
Preparation…………………………………….……………………………13
Properties…………………………………………………………………..33
Ostwalds process…………………………………………………………..12
E.NO3– and NO2– salts ……………………….……………………………….……27
- COMPREHENSIVE REVISION QUESTIONS…………………………..…32
A.NITROGEN
- a) Occurrence:
Nitrogen is found in the atmosphere occupying about 78% by volume of air.
Proteins, amino acids, polypeptides in living things contain nitrogen.
- b) Isolation of nitrogen from the air.
Nitrogen can be isolated from other gases present in air like oxygen, water (vapour), carbon (IV) oxide and noble gases as in the school laboratory as in the flow chart below:
Water is added slowly into an “empty flask” which forces the air out into another flask containing concentrated sulphuric (VI) acid. Concentrated sulphuric (VI) acid is hygroscopic. It therefore absorb/remove water present in the air sample.
More water forces the air into the flask containing either concentrated sodium hydroxide or potassium hydroxide solution. These alkalis react with carbon IV) oxide to form the carbonates and thus absorbs/remove carbon IV) oxide present in the air sample.
Chemical equation 2NaOH (aq) + CO2 (g) -> Na2CO3 (aq) + H2O(l)
Chemical equation 2KOH (aq) + CO2 (g) -> K2CO3 (aq) + H2O(l)
More water forces the air through a glass tube packed with copper turnings. Heated brown copper turnings react with oxygen to form black copper (II) oxide.
Chemical equation 2Cu (s) + O2 (g) -> CuO (s)
(brown) (black)
The remaining gas mixture is collected by upward delivery/downward displacement of water/over water. It contains about 99% nitrogen and 1% noble gases.
On a large scale for industrial purposes, nitrogen is got from fractional distillation of air.
c)Nitrogen from fractional distillation of air.
For commercial purposes nitrogen is got from the fractional of air.
Air is first passed through a dust precipitator/filter to remove dust particles.
The air is then bubbled through either concentrated sodium hydroxide or potassium hydroxide solution to remove/absorb Carbon(IV) oxide gas.
Chemical equation 2NaOH (aq) + CO2 (g) -> Na2CO3 (aq) + H2O(l)
Chemical equation 2KOH (aq) + CO2 (g) -> K2CO3 (aq) + H2O(l)
Air mixture is the cooled to -25oC.At this temperature, water (vapour ) liquidifies and then solidify to ice and thus removed.
The air is further cooled to -200oC during which it forms a blue liquid.
The liquid is then heated. Nitrogen with a boiling point of -196oC distils first then Argon at-186oC and then finally Oxygen at -183oC boils last.
- c) School laboratory preparation of Nitrogen.
The diagram below shows the set up of the school laboratory preparation of nitrogen gas.
d.Properties of Nitrogen gas(Questions)
1.Write the equation for the reaction for the school laboratory preparation of nitrogen gas.
Chemical equation NH4Cl (s) + NaNO2(s)->NaCl (g)+ NH4NO2 (s)
Chemical equation NH4NO2 (s) -> N2 (g) + H2O (l)
- State three physical properties of nitrogen gas.
– colourless, odourless, less dense than air ,neutral and slightly soluble in water
- State and explain the observation made when a burning magnesium ribbon is lowered in a gas jar containing nitrogen gas.
Observation; It continues burning with a blight blindening flame forming white ash.
ExplanationMagnesium burns to produce enough heat /energy to reacts with nitrogen to form white magnesium nitride.
Chemical equation3Mg (s) + N2 (g) -> Mg3N2 (s)
(white ash/solid)
- State two main uses of nitrogen gas
-manufacture of ammonia from Haber process
– As a refrigerant in storage of semen for Artificial insemination.
- OXIDES OF NITROGEN
Nitrogen forms three main oxides:
i)Nitrogen(I) oxide(N2O)
- ii) Nitrogen(II) oxide (NO)
iii) Nitrogen (IV) oxide( NO2)
- i) Nitrogen (I) oxide(N2O)
- a) Occurrence
Nitrogen (I) oxide does not occur naturally but prepared in a laboratory.
b)Preparation
The set up below shows the set up of apparatus that can be used to prepare Nitrogen (I) oxide in a school laboratory.
- c) Properties of nitrogen (I) oxide (Questions)
- Write the equation for the reaction for the school laboratory preparation of Nitrogen (I) oxide.
Chemical equation NH4NO2(s) -> H2O (l) + N2O (g)
2.a) State and explain three errors made in the above set up
–Oxygen is being generated instead of Nitrogen (I) oxide.
Ammonium Nitrate(V) should be used instead of potassium manganate(VI) and manganese(IV)oxide.
- b) State three physical properties of Nitrogen (I) oxide.
-slightly soluble in water.
-colourless
-odourless
-less dense than air
-slightly sweet smell
- State and explain the observation made when a burning magnesium ribbon is lowered in a gas jar containing Nitrogen (I) oxide.
Observation – Continues to burn with a bright flame
-White solid/residue is formed
Explanation-Magnesium burns in air to produce enough heat/energy split/break Nitrogen (I) oxide gas into free Nitrogen and oxygen then continues to burn in oxygen to form white solid/ash of Magnesium oxide.
Chemical equation
Mg(s) + N2O (g)-> MgO (s) + N2(g)
- State and explain the observation made when the following non metals are burnt then lowered in a gas jar containing Nitrogen (I) oxide.
- a) Carbon/charcoal
Observation – Continues to burn with an orange glow
-colorless gas is formed that forms white precipitate with lime water.
Explanation-Carbon/charcoal burns in air to produce enough heat/energy split/break Nitrogen (I) oxide gas into free Nitrogen and oxygen then continues to burn in oxygen to form carbon (IV) oxide gas. Carbon (IV) oxide gas reacts to form a white precipitate with lime water.
Chemical equation C(s) + 2N2O (g)-> CO2 (g) + 2N2(g)
- b) sulphur powder
Observation – Continues to burn with a blue flame
-colorless gas is formed that turn orange acidified potassium dichromate (VI) to green.
Explanation-Sulphur burns in air to produce enough heat/energy split/break Nitrogen (I) oxide gas into free Nitrogen and oxygen then continues to burn in oxygen to form sulphur (IV) oxide gas.Sulphur (IV) oxide gas turns orange acidified potassium dichromate (VI) to green.
Chemical equation S(s) + 2N2O (g)-> SO2 (g) + 2N2(g)
- State two uses of nitrogen (I) oxide
-As laughing gas because as anesthesia the patient regain consciousness laughing hysterically after surgery.
-improves engine efficiency.
- State three differences between nitrogen (I) oxide and oxygen
–Oxygen is odourless while nitrogen (I) oxide has faint sweet smell
-Both relight/rekindle a glowing wooden splint but Oxygen can relight a feeble glowing splint while nitrogen (I) oxide relights well lit splint.
-Both are slightly soluble in water but nitrogen (I) oxide is more soluble.
- ii) Nitrogen (II) oxide (NO)
- a) Occurrence
Nitrogen (II) oxide does not occur naturally but prepared in a laboratory.
b)Preparation
The set up below shows the set up of apparatus that can be used to prepare Nitrogen (II) oxide in a school laboratory.
- c) Properties of nitrogen (II) oxide (Questions)
- Write the equation for the reaction for the school laboratory preparation of Nitrogen (II) oxide.
Chemical equation 3Cu(s) + 8HNO3(aq) -> 4H2O (l)+2NO (g) +2Cu(NO3)2(aq)
Chemical equation 3Zn(s) + 8HNO3(aq) -> 4H2O (l)+2NO (g) +2Zn(NO3)2(aq)
Chemical equation 3Mg(s) + 8HNO3(aq) -> 4H2O (l)+2NO (g)+2Mg(NO3)2(aq)
- State three physical properties of Nitrogen (II) oxide.
-insoluble in water.
-colourless
-odourless
-denser dense than air
-has no effect on both blue and red litmus papers
- State and explain the observation made when a burning magnesium ribbon is lowered in a gas jar containing Nitrogen (II) oxide.
Observation – Continues to burn with a bright flame
-White solid/residue is formed
Explanation-Magnesium burns in air to produce enough heat/energy split/break Nitrogen (II) oxide gas into free Nitrogen and oxygen then continues to burn in oxygen to form white solid/ash of Magnesium oxide.
Chemical equation 2Mg(s) + 2NO (g)-> 2MgO (s) + N2(g)
- State and explain the observation made when the following non metals are burnt then lowered in a gas jar containing Nitrogen (II) oxide.
- a) Carbon/charcoal
Observation – Continues to burn with an orange glow
-colorless gas is formed that forms white precipitate with lime water.
Explanation-Carbon/charcoal burns in air to produce enough heat/energy split/break Nitrogen (II) oxide gas into free Nitrogen and oxygen then continues to burn in oxygen to form carbon (IV) oxide gas.Carbon (IV) oxide gas reacts to form a white precipitate with lime water.
Chemical equation C(s) + 2NO (g)-> CO2 (g) + N2(g)
- b) sulphur powder
Observation – Continues to burn with a blue flame
-colorless gas is formed that turn orange acidified potassium dichromate (VI) to green.
Explanation-Sulphur burns in air to produce enough heat/energy split/break Nitrogen (II) oxide gas into free Nitrogen and oxygen then continues to burn in oxygen to form sulphur (IV) oxide gas.Sulphur (IV) oxide gas turns orange acidified potassium dichromate (VI) to green.
Chemical equation S(s) + N2O (g)-> SO2 (g) + N2(g)
- c) Phosphorus
Observation – Continues to produce dense white fumes
Explanation-Phosphorus burns in air to produce enough heat/energy split/break Nitrogen (II) oxide gas into free Nitrogen and oxygen then continues to burn in oxygen to form dense white fumes of phosphorus (V) oxide gas.
Chemical equation 4P(s) + 10NO (g)-> 2P2O5(g) + 5N2(g)
- State one use of nitrogen (II) oxide
As an intermediate gas in the Ostwalds process for manufacture of nitric(V) gas.
- State and explain the observation made when nitrogen (II) oxide is exposed to the atmosphere.
Observation–brown fumes produced/evolved that turn blue litmus paper red.
Explanation– Nitrogen (II) oxide gas on exposure to air is quickly oxidized by the air/ oxygen to brown nitrogen (IV) oxide gas. Nitrogen (IV) oxide gas is an acidic gas.
Chemical equation 2NO (g)+ O2(g)-> 2NO2 (g)
(colorless) (brown)
- ii) Nitrogen (IV) oxide (NO2)
- a) Occurrence
Nitrogen (IV) oxide occurs -naturally from active volcanic areas.
-formed from incomplete combustion of the internal combustion engine of motor vehicle exhaust fumes.
-from lightening
b)Preparation
The set up below shows the set up of apparatus that can be used to prepare Nitrogen (IV) oxide in a school laboratory.
- c) Properties of nitrogen (IV)oxide (Questions)
- Write the equation for the reaction for the school laboratory preparation of Nitrogen (II) oxide.
Chemical equation Cu(s) + 4HNO3(aq) -> 2H2O (l)+2NO 2(g) +Cu(NO3)2(aq)
Chemical equation Zn(s) + 4HNO3(aq) -> 2H2O (l)+2NO 2(g) +Zn(NO3)2(aq)
Chemical equation Fe(s) + 4HNO3(aq) -> 2H2O (l)+2NO 2(g) +Fe(NO3)2(aq)
- State three physical properties of Nitrogen (IV) oxide.
-soluble/dissolves in water.
-brown in colour
-has pungent irritating poisonous odour/smell
-denser dense than air
-turns blue litmus papers to red
- State and explain the observation made when Nitrogen (IV) oxidegas is bubbled in water.
Observation–The gas dissolves and thus brown colour of the gas fades
-A colourless solution is formed
-solution formed turns blue litmus papers to red
-solution formed has no effect on red
Explanation-Magnesium burns in air to produce enough heat/energy split/break Nitrogen (IV) oxide gas dissolves then react with water to form an acidic mixture of nitric(V) acid andnitric(III) acid.
Chemical equationH2O (l) + 2NO 2(g)->HNO3(aq) + HNO2(aq)
(nitric(V) acid) (nitric(III) acid)
- State and explain the observation made when a test tube containing Nitrogen (IV) oxide is cooled then heated gently then strongly.
Observation on cooling
-Brown colour fades
-Yellow liquid formed
Observation on gentle heating
–Brown colour reappears
–Yellow liquid formed changes to brown fumes/gas
Observation on gentle heating
–Brown colour fades
–brown fumes/gas changes to a colourless gas
Explanation-Brown nitrogen (IV) oxide gas easily liquefies to yellow dinitrogen tetraoxide liquid.When the yellow dinitrogen tetraoxide liquid is gently heated it changes back to the brown nitrogen (IV) oxidegas.When the brown nitrogen (IV) oxide gas is strongly heated it decomposes to colourless mixture of Nitrogen (II) oxide gas and Oxygen.
Chemical equation O2(s) + 2NO (g) ===== 2NO2 (g) ===== N2O4(l)
(colourless gases) (brown gas) (yellow liquid)
- State and explain the observation made when a burning magnesium ribbon is lowered in a gas jar containing Nitrogen (IV) oxide.
Observation – Continues to burn with a bright flame
-White solid/residue is formed
-Brown fumes/colour fades
Explanation-Magnesium burns in air to produce enough heat/energy split/break brown Nitrogen (IV) oxide gas into free colourless Nitrogen and oxygen then continues to burn in oxygen to form white solid/ash of Magnesium oxide.
Chemical equation 4Mg(s) + 2NO 2(g)-> 4MgO (s) + N2(g)
- State and explain the observation made when the following non metals are burnt then lowered in a gas jar containing Nitrogen (IV) oxide.
- a) Carbon/charcoal
Observation – Continues to burn with an orange glow
-Brown fumes/colour fades
-colorless gas is formed that forms white precipitate with lime water.
Explanation-Carbon/charcoal burns in air to produce enough heat/energy split/break brown Nitrogen (IV) oxide gas into free colourless Nitrogen and oxygen then continues to burn in oxygen to form carbon (IV) oxide gas.Carbon (IV) oxide gas reacts to form a white precipitate with lime water.
Chemical equation2C(s) + 2NO 2(g)-> 2CO2 (g) + N2(g)
- b) sulphur powder
Observation – Continues to burn with a blue flame
-Brown fumes/colour fades
-colorless gas is formed that turn orange acidified potassium dichromate (VI) to green.
Explanation-Sulphur burns in air to produce enough heat/energy split/break brown Nitrogen (IV) oxide gas into free colourless Nitrogen and oxygen then continues to burn in oxygen to form sulphur (IV) oxide gas.Sulphur (IV) oxide gas turns orange acidified potassium dichromate (VI) to green.
Chemical equation2S(s) + 2NO2 (g)-> 2SO2 (g) + N2(g)
- c) Phosphorus
Observation– Continues to produce dense white fumes
-Brown fumes/colour fades
Explanation-Phosphorus burns in air to produce enough heat/energy split/break brown Nitrogen (IV) oxide gas into free colourless Nitrogen and oxygen then continues to burn in oxygen to form dense white fumes of phosphorus (V) oxide gas.
Chemical equation 8P(s) + 10NO2 (g)-> 4P2O5(g) + 5N2(g)
- State two uses of nitrogen (IV) oxide
-In theOstwald process for industrial manufacture of nitric (V) gas.
-In the manufacture of T.N.T explosives
- State and explain the observation made when nitrogen (II) oxide is exposed to the atmosphere.
Observation–brown fumes produced/evolved that turn blue litmus paper red.
Explanation– Nitrogen (II) oxide gas on exposure to air is quickly oxidized by the air/ oxygen to brown nitrogen (IV) oxide gas. Nitrogen (IV) oxide gas is an acidic gas.
Chemical equation 2NO (g) + O2(g) -> 2NO2 (g)
(colourless) (brown)
- AMMONIA (NH3)
Ammonia is a compound of nitrogen and hydrogen only. It is therefore a hydride of nitrogen.
- a) Occurrence
Ammonia gas occurs -naturally from urine of mammals and excretion of birds
-formed in the kidney of human beings
b)Preparation
The set up below shows the set up of apparatus that can be used to prepare dry Ammonia gas in a school laboratory.
Set up method 1
- c) P
- Write the equation for the reaction taking place in:
- Method 1
Chemical equation
Ca (OH)2(s) + NH4 Cl(s)->CaCl2 (aq) + H2O(l) + 2NH3(g)
b)Method 2
Chemical equation
NaOH (aq) + NH4 Cl(aq) -> NaCl (aq) + H2O(l) + NH3(g)
- State three physical properties of ammonia.
-has a pungent choking smell of urine
-Colourless
-Less dense than air hence collected by upward delivery
-Turns blue litmus paper blue thus is the only naturally occurring basic gas (at this level)
- Calcium oxide is used as the drying agent. Explain why calcium chloride and concentrated sulphuric(VI) acid cannot be used to dry the gas.
-Calcium chloride reacts with ammonia forming the complex compound CaCl2.8H2O.
Chemical equation CaCl2 (s) + 8NH3(g) -> CaCl2 .8NH3(g)
-Concentrated sulphuric(VI) acid reacts with ammonia forming ammonium sulphate(VI) salt compound
Chemical equation 2NH3(g) +H2SO4(l) ->(NH4)2SO4(aq)
4.Describe the test for the presence of ammonia gas.
Using litmus paper:
Dip moist/damp/wet blue and red litmus papers in a gas jar containing a gas suspected to be ammonia.The blue litmus paper remain blue and the red litmus paper turns blue.Ammonia is the only basic gas.(At this level)
Using hydrogen chloride gas
Dip a glass rod in concentrated hydrochloric acid. Bring the glass rod near the mouth of a gas jar suspected to be ammonia. White fumes (of ammonium chloride)are produced/evolved.
- Describe the fountain experiment to show the solubility of ammonia.
Ammonia is very soluble in water.
When a drop of water is introduced into flask containing ammonia, it dissolves all the ammonia in the flask. If water is subsequently allowed into the flask through a small inlet, atmospheric pressure forces it very fast to occupy the vacuum forming a fountain. If the water contains three/few drops of litmus solution,the litmus solution turns blue because ammonia is an alkaline/basic gas. If the water contains three/few drops of phenolphthalein indicator,the indicator turns pink because ammonia is an alkaline/basic gas. Sulphur(IV)oxide and hydrogen chloride gas are also capable of the fountain experiment . If the water contains three/few drops of phenolphthalein indicator, the indicator turns colourless because both Sulphur(IV) oxide and hydrogen chloride gas are acidic gases.
6.State and explain the observation made when hot platinum /nichrome wire is placed over concentrated ammonia solution with Oxygen gas bubbled into the mixture.
Observations
Hot platinum /nichrome wire continues to glow red hot.
Brown fumes of a gas are produced.
Explanation
Ammonia reacts with Oxygen on the surface of the wire .This reaction is exothermic producing a lot of heat/energy that enables platinum wire to glow red hot. Ammonia is oxidized to Nitrogen(II)oxide gas and water. Hot platinum /nichrome wire acts as catalyst to speed up the reaction. Nitrogen(II)oxide gas is further oxidized to brown Nitrogen(IV)oxide gas on exposure to air.
Chemical equation
(i)4NH3(g) + 5O2(g) -Pt-> 4NO(g) + 6H2O(l)
(ii)2NO(g) + O2(g) -> 2NO2(g)
- Ammonia gas was ignited in air enriched with Oxygen gas. State and explain the observations made
Observations
– Ammonia gas burns with a green flame
-Colourless gas produced
Explanation
Ammonia gas burns with a green flame in air enriched with Oxygen to from Nitrogen gas and water.
Chemical equation
2NH3(g) + O2(g) -> N2(g) + 3H2O(l)
- Dry ammonia was passed through heated copper(II)Oxide as in the set up below.
(a)State the observations made in tube K
-Colour changes from black to brown
-Colourless liquid droplet form on the cooler parts of tube K
(b)(i)Identify liquid L.
-Water/ H2O(l)
(ii)Explain a chemical and physical test that can be used to identify liquid L.
Chemical test
(i) Add three/few drops of liquid L into anhydrous copper(II)sulphate(VI).
Colour changes from white to blue.
Explanation-Water changes white anhydrous copper(II)sulphate(VI) to blue hydrated copper(II)sulphate(VI)
(ii) Add three/few drops of liquid L into anhydrous cobalt(II)Chloride.
Colour changes from blue to pink.
Explanation-Water changes blue anhydrous cobalt(II)Chloride to pink hydrated cobalt(II)Chloride.
Physical test
(i)Heat the liquid. It boils at 100oC at sea level (1atmosphere pressure/760mmHg pressure, 101300Pa,101300Nm-2).
(ii)Cool the liquid. It freezes at 0.0oC .
(iii)Determine the density. It is 1.0gcm-3
(c)Write the equation for the reaction that take place.
2NH3(g) + 3CuO(s) -> N2(g) + 3H2O(l) + 3Cu(s)
(black) (brown)
2NH3(g) + 3PbO(s) -> N2(g) + 3H2O(l) + 3Pb(s)
(brown when hot) (grey)
8.(a)What is aqueous ammonia
Aqueous ammonia is formed when ammonia gas is dissolved in water.
NH3(g) + (aq) -> NH3(aq)
A little NH3(aq) reacts with ammonia water to form ammonia solution(NH4OH)
NH3 (aq) + H2O(l) OH–(aq) + NH4+(aq)
This makes a solution of aqueous ammonia is a weak base /alkali unlike other two alkalis.
9.Using dot and cross to represent outer electrons show the bonding in:
(a) NH3
(b) NH4+
(c)NH4Cl
10.Name four uses of ammonia
(i)In the manufacture of nitrogenous fertilizers.
(ii) In the manufacture of nitric(V)acid from Ostwalds process.
(iii)As a refrigerant in ships and warehouses.
(iv)In softening hard water.
(v)In the solvay process for the manufacture of sodium carbonate.
(vi)In the removal of grease and stains.
11.(a)Calculate the percentage of Nitrogen in the following fertilizers:
(i) (NH4)2SO4
Molar mass of (NH4)2SO4 = 132g
Mass of N in (NH4)2SO4= 28g
% of N => 28 x 100 = 21.2121%
132
(ii) (NH4)3PO4
Molar mass of (NH4)3PO4 = 149g
Mass of N in (NH4)3PO4= 42g
% of N => 42 x 100 = 28.1879%
149
(b)State two advantages of fertilizer a (i) over a (ii) above.
(i)Has higher % of Nitrogen
(ii)Has phosphorus which is necessary for plant growth.
(c) Calculate the mass of Nitrogen in a 50kg bag of:
(i) (NH4)2SO4
% of N in (NH4)2SO4 = 21.2121%
Mass of N in 50 kg (NH4)2SO4= 21.2121 x 50 = 10.6 kg
100
(ii) NH4NO3
Molar mass of NH4NO3 = 80g
Mass of N in (NH4)3PO4= 28g
% of N => 28 x 100 = 35%
80
% of N in NH4NO3 = 35%
Mass of N in 50 kg (NH4)2SO4= 35 x 50 = 17.5 kg
100
NH4NO3 therefore has a higher mass of Nitrogen than (NH4)2SO4
d).Manufacture of Ammonia /Haber process
Most of the Ammonia produced for industrial purposes uses the Haber process developed by the German Scientist Fitz Haber.
(i)Raw materials
The raw materials include:
(i)Nitrogen from Fractional distillation of air from the atmosphere.
(ii)Hydrogen from:
- Water gas-passing steam through heated charcoal
C(s) + H2O(l) -> CO(g) + H2 (g)
II .Passing natural gas /methane through steam.
CH4(g)+ H2O(l) -> CO(g) + 3H2 (g)
(ii)Chemical process
Hydrogen and Nitrogen are passed through a purifier to remove unwanted gases like Carbon(IV)oxide,Oxygen,sulphur(IV)oxide, dust, smoke which would poison the catalyst.
Hydrogen and Nitrogen are then mixed in the ratio of 3:1 respectively. The mixture is compressed to 200-250atmoshere pressure to liquidify. The liquid mixture is then heated to 400- 450oC.The hot compressed gases are then passed over finely divided Iron catalyst promoted/impregnated with Al2O3 /K2O .Promoters increase the efficiency of the catalyst.
Optimum conditions in Haber processs
Chemical equation
N2 (g) + 3H2 (g) ===Fe/Pt=== 2NH3 (g) ΔH = -92kJ
Equilibrium/Reaction rate considerations
(i)Removing ammonia gas once formed shift the equilibrium forward to the right to replace the ammonia. More/higher yield of ammonia is attained.
(ii)Increase in pressure shift the equilibrium forward to the right where there is less volume/molecules . More/higher yield of ammonia is attained. Very high pressures raises the cost of production because they are expensive to produce and maintain. An optimum pressure of about 200atmospheres is normally used.
(iii)Increase in temperature shift the equilibrium backward to the left because the reaction is exothermic(ΔH = -92kJ) . Ammonia formed decomposes back to Nitrogen and Hydrogen to remove excess heat therefore a less yield of ammonia is attained. Very low temperature decrease the collision frequency of Nitrogen and Hydrogen and thus the rate of reaction too slow and uneconomical.
An optimum temperature of about 450oC is normally used.
(iv)Iron and platinum can be used as catalyst. Platinum is a better catalyst but more expensive and easily poisoned by impurities than Iron. Iron is promoted /impregnated with AluminiumOxide(Al2O3) to increase its surface area/area of contact with reactants and thus efficiency.The catalyst does not increase the yield of ammonia but it speed up its rate of formation.
- e) Nitric(V)acid (HNO3)
a)Introduction.
Nitric(V)acid is one of the mineral acids .There are three mineral acids; Nitric(V)acid, sulphuric(VI)acid and hydrochloric acid.Mineral acids do not occur naturally but are prepared in a school laboratory and manufactured at industrial level.
- b) School laboratory preparation
Nitric(V)acid is prepared in a school laboratory from the reaction of Concentrated sulphuric(VI)acid and potassium nitrate(V) below.
(c)Properties of Concentrated Nitric (V)acid(Questions)
1.Write an equation for the school laboratory preparation of nitric(V)acid.
KNO3(s) + H2SO4(l) -> KHSO4(s) + HNO3(l)
2.Sodium nitrate(V)can also be used to prepare nitric(V)acid. State two reasons why potassium nitrate(V) is preferred over Sodium nitrate(V).
(i) Potassium nitrate(V) is more volatile than sodium nitrate(V) and therefore readily displaced from the less volatile concentrated sulphuric(VI)acid
(ii) Sodium nitrate(V) is hygroscopic and thus absorb water . Concentrated sulphuric(VI)acid dissolves in water. The dissolution is a highly exothermic process.
- An all glass apparatus /retort is used during the preparation of nitric(V) acid. Explain.
Hot concentrated nitric(V) acid vapour is highly corrosive and attacks rubber cork apparatus if used.
- Concentrated nitric(V) acid is colourless . Explain why the prepared sample in the school laboratory appears yellow.
Hot concentrated nitric(V) acid decomposes to brown nitrogen(IV)oxide and Oxygen gases.
4HNO3(l/g) -> 4NO2(g) + H2O (l) +O2(g)
Once formed the brown nitrogen(IV)oxide dissolves in the acid forming a yellow solution .
- State and explain the observation made when concentrated nitric (V) acid is heated.
Observation
Brown fumes are produced.
Colourless gas that relights/rekindles glowing splint
Explanation
Hot concentrated nitric(V) acid decomposes to water, brown nitrogen(IV)oxide and Oxygen gases. Oxygen gas is not visible in the brown fumes of nitrogen (IV) oxide.
4HNO3(g) -> 4NO2(g) + H2O (l) +O2(g)
- Explain the observations made when:
(a) About 2cm3 of Iron(II)sulphate(VI) solution is added about 5 drops of concentrated nitric(V) acid and the mixture then heated/warmed in a test tube.
Observation
(i)Colour changes from green to brown.
(ii)brown fumes /gas produced on the upper parts of the test tube.
Explanation
Concentrated nitric(V) acid is a powerful/strong oxidizing agent. It oxidizes green Fe2+ ions in FeSO4 to brown/yellow Fe3+ .The acid is reduced to colourless Nitrogen(II)oxide.
Chemical equation:
6FeSO4(aq) + 3H2SO4 (aq) + 2HNO3(aq) -> 3Fe2(SO4) 3 (aq)+ 4H2O + 2NO(g)
Colourless Nitrogen(II)oxide is rapidly further oxidized to brown Nitrogen(IV)oxide by atmospheric oxygen.
Chemical equation:
2NO(g) + O(g) -> 2NO2 (g)
(colourless) (brown)
(b) A spatula full of sulphur powder in a claen dry beaker was added to 10cm3 concentrated nitric (V) acid and then heated gently/warmed.
Observation
(i)Yellow colour of sulphur fades .
(ii)brown fumes /gas produced.
Explanation
Concentrated nitric(V) acid is a powerful/strong oxidizing agent. It oxidizes yellow sulphur to colourless concentrated sulphuric(VI)acid. The acid is reduced to brown Nitrogen(IV)oxide gas.
Chemical equation:
S(s) + 6HNO3(l) -> 4NO2(g) + H2O (l) +H2SO4(l)
(c) A few/about 1.0g pieces of copper turnings/Zinc granules/ Magnesium ribbon are added 10cm3 of concentrated nitric(V) acid in a beaker.
Observation
(i) brown fumes /gas produced.
(ii) blue solution formed with copper turnings
(iii) colourless solution formed with Zinc granules/Magnesium ribbon
Explanation
Concentrated nitric (V) acid is a powerful/strong oxidizing agent. It oxidizes metals to their metal nitrate (VI) salts. The acid is reduced to brown Nitrogen (IV) oxide gas.
Chemical equation:
Cu(s) + 4HNO3(l) -> 2NO2(g) + H2O (l) + Cu(NO3) 2 (aq)
Zn(s) + 4HNO3(l) -> 2NO2(g) + H2O (l) + Zn(NO3) 2 (aq)
Mg(s) + 4HNO3(l) -> 2NO2(g) + H2O (l) + Mg(NO3) 2 (aq)
Pb(s) + 4HNO3(l) -> 2NO2(g) + H2O (l) + Pb(NO3) 2 (aq)
Ag(s) + 2HNO3(l) -> NO2(g) + H2O (l) + AgNO3 (aq)
(d)Properties of Dilute Nitric (V)acid(Questions)
(i)What is dilute nitric(v)acid
When concentrated nitric(v)acid is added to over half portion of water ,it is relatively said to be dilute. A dilute solution is one which has more solvent/water than solute/acid. The number of moles of the acid are present in a large amount/volume of the solvent.This makes the molarity /number of moles present in one cubic decimeter of the solution to be low e.g. 0.02M.
If more water is added to the acid until the acid is too dilute to be diluted further then an infinite dilute solution if formed.
(ii))1cm length of polished Magnesium ribbon was put is a test tube containing 0.2M dilute nitric(v)acid. State and explain the observation made.
Observation
-Effervescence/bubbling/fizzing
-Colourless gas produced that extinguish burning splint with an explosion/pop sound
-Colourless solution formed
-Magnesium ribbon dissolves/decrease in size
Explanation
Dilute dilute nitric(v)acid reacts with Magnesium to form hydrogen gas.
Mg(s) + 2HNO3(aq) -> H2 (g) + Mg(NO3) 2 (aq)
With other reactive heavy metals, the hydrogen gas produced is rapidly oxidized to water.
Chemical equation 3Pb(s) + 8HNO3(aq) -> 4H2O (l)+2NO (g) +2Pb(NO3)2(aq)
Chemical equation 3Zn(s) + 8HNO3(aq) -> 4H2O (l)+2NO (g) +2Zn(NO3)2(aq)
Chemical equation 3Fe(s) + 8HNO3(aq) -> 4H2O (l)+2NO (g) +2Fe(NO3)2(aq)
Hydrogen gas therefore is usually not prepared in a school laboratory using dilute nitric (v)acid.
(iii)A half spatula full of sodium hydrogen carbonate and Copper(II) carbonate were separately to separate test tubes containing 10cm3 of 0.2M dilute nitric (V) acid.
Observation
-Effervescence/bubbling/fizzing
-Colourless gas produced that forms a white precipitate with lime water.
-Colourless solution formed with sodium hydrogen carbonate.
– Blue solution formed with Copper(II) carbonate.
Explanation
Dilute dilute nitric (v)acid reacts with Carbonates and hydrogen carbonates to form Carbon(IV)oxide, water and nitrate(V)salt
CuCO3 (s) + 2HNO3(aq) -> H2O (l) + Cu(NO3) 2 (aq) + CO2 (g)
ZnCO3 (s) + 2HNO3(aq) -> H2O (l) + Zn(NO3) 2 (aq) + CO2 (g)
CaCO3 (s) + 2HNO3(aq) -> H2O (l) + Ca(NO3) 2 (aq) + CO2 (g)
PbCO3 (s) + 2HNO3(aq) -> H2O (l) + Pb(NO3) 2 (aq) + CO2 (g)
FeCO3 (s) + 2HNO3(aq) -> H2O (l) + Fe(NO3) 2 (aq) + CO2 (g)
NaHCO3 (s) + HNO3(aq) -> H2O (l) + NaNO3 (aq) + CO2 (g)
KHCO3 (s) + HNO3(aq) -> H2O (l) + KNO3 (aq) + CO2 (g)
NH4HCO3 (aq) + HNO3(aq) -> H2O (l) + NH4NO3 (aq) + CO2 (g)
Ca(HCO3) 2 (aq) + 2HNO3(aq) -> 2H2O (l) + Ca(NO3) 2 (aq) + 2CO2 (g)
Mg(HCO3) 2 (aq) + 2HNO3(aq) -> 2H2O (l) + Mg(NO3) 2 (aq) + 2CO2 (g)
(iii) 25.0cm3 of 0.1M Nitric(V) acid was titrated with excess 0.2M sodium hydroxide solution using phenolphthalein indicator.
- State the colour change at the end point
Colourless
- What was the pH of the solution at the end point. Explain.
pH 1/2/3
A little of the acid when added to the base changes the colour of the indicator to show the end point. The end point therefore is acidic with low pH of Nitric(V) acid. Nitric(V) acid is a strong acid with pH 1/2/3.
III. Calculate the number of moles of acid used.
Number of moles = molarity x volume => 0.1 x 25 = 2.5 x 10-3moles
1000 1000
- Calculate the volume of sodium hydroxide used
Volume of sodium hydroxide in cm3
= 1000 x Number of moles => 1000x 2.5 x 10-3 = 12.5cm3
Molarity 0.2
(e)Industrial large scale manufacture of Nitric (V)acid
(i)Raw materials
- Air/Oxygen
Oxygen is got from fractional distillation of air
Ammonia from Haber process.
- 2. Chemical processes
Air from the atmosphere is passes through electrostatic precipitators/filters to remove unwanted gases like Nitrogen,Carbon(IV)oxide,dust,smoke which may poison the catalyst.The ammonia -air mixture is compressed to 9 atmospheres to reduce the distance between reacting gases.
The mixture is passed through the heat exchangers where a temperature of 850oC-900oC is maintained.
The first reaction take place in the catalytic chamber where Ammonia reacts with the air to form Nitrogen(II)Oxide and water.
Optimum condition in Ostwalds process
Chemical equation
4NH3(g) + 5O2(g) === Pt/Rh === 4NO(g) + 6H2O(g) ΔH = –950kJ
The reaction is reversible and exist in dynamic equilibrium where the products reform back the reactants. The following factors are used to increase the yield/amount of Notrogen(II)oxide:
(i)Removing Nitrogen(II)oxide gas once formed shift the equilibrium forward to the right to replace the Nitrogen(II) oxide.
More/higher yield of Nitrogen(II) oxide is attained as reactants try to return the equilibrium balance.
(ii)Increase in pressure shift the equilibrium backward to the left where there is less volume/molecules .
Less/lower yield of Nitrogen(II)oxide is attained.
Very low pressures increases the distance between reacting NH3 and O2 molecules.
An optimum pressure of about 9 atmospheres is normally used.
Cooling the mixture condenses the water vapour to liquid water
(iii)Increase in temperature shift the equilibrium backward to the left because the reaction is exothermic(ΔH = -950kJ).
Nitrogen(II)oxide and water vapour formed decomposes back to Ammonia and Oxygen to remove excess heat therefore a less yield of Nitrogen(II)oxide is attained.
Very low temperature decrease the collision frequency of Ammonia and Oxygen and thus the rate of reaction too slow and uneconomical.
An optimum temperature of about 900oC is normally used.
(iv)Platinum can be used as catalyst.
Platinum is very expensive. It is:
-promoted with Rhodium to increase the surface area/area of contact.
-added/coated on the surface of asbestos to form platinized –asbestos to reduce the amount/quantity used.
The catalyst does not increase the yield of Nitrogen (II)Oxide but it speed up its rate of formation.
Nitrogen(II)oxide formed is passed through an oxidation reaction chamber where more air oxidizes the Nitrogen(II)Oxide to Nitrogen(IV)Oxide gas.
Chemical equation
2NO(g) + O2(g) -> 2NO2(g)
Nitrogen(IV)Oxide gas is passed up to meet a downward flow of water in the absorption chamber. The gas react with water to form a mixture of Nitric(V) and Nitric(III)acids
Chemical equation.
2NO2(g) + H2O (l) -> HNO2(aq) + HNO3(aq)
Excess air is bubbled through the mixture to oxidize Nitric(III)/ HNO2(aq) to Nitric(V)/HNO3(aq)
Chemical equation.
O2(g) + 2HNO2(aq) -> 2HNO3(aq)
Overall chemical equation in the absorption chamber.
O2(g) + 4NO2(g) + 2H2O (l) -> 4HNO3(aq)
The acid is 65% concentrated.It is made 100% concentrated by either:
(i)fractional distillation or
(ii)added to concentrated sulphuric(VI) acid to remove the 35% of water.
A factory uses 63.0 kg of 68% pure nitric(V)acid per day to produce an ammonium fertilizer for an agricultural county. If the density of the acid is 1.42 gcm-3 , calculate :
(i)the concentration of the acid used in moles per litre.
Molar mass HNO3 = 63
Method 1
Moles of HNO3 in 1cm3 = Mass in 1cm3 1.42 => 1.42 = 0.0225 moles
Molar mass HNO3 63
Molarity = Moles x 1000=>0.0225 moles x10000 = 22.5molesdm-3/M
1 cm3
100% = 22.5molesdm-3/M
68% = 68 x 22.5 = 15.3M/ molesdm-3
100
Method 2
Moles of HNO3 in 1000cm3 = Mass in 1000cm3 =>1.42 x1000
Molar mass HNO3 63
=22.5397 molesdm-3/M
100% = 22.5397 molesdm-3/M
68% = 68 x 22.5397 = 15.327 molesdm-3
100
(ii)the volume of ammonia gas at r.t.p used. (H=1.0,N=14.0,O=16.0,one mole of gas = 24 dm-3 at r.t.p)
Chemical equation
HNO3 (aq) + NH3 (g) -> NH4NO3(aq)
Mole ratio HNO3 (aq) : NH3 (g) = 1 : 1
1 mole HNO3 (aq) -> 24dm3 NH3 (g)
15.327 mole HNO3 (aq) ->15.327 mole x 24 dm3 = 367.848dm3
1dm3
(iii)the number of crops which can be applied the fertilizer if each crop require 4.0g.
HNO3 (aq) + NH3 (g) -> NH4NO3(aq)
Molar mass NH4NO3 =80 g
Mole ratio HNO3 : NH4NO3 = 1 : 1
Mass of HNO3 in 63.0 kg = 68% x 63 =42.84kg
1 mole HNO3 (aq)=63g -> 80g NH4NO3
(42.84×1000)g HNO3 (aq) -> (42.84×1000)g x 80
63
= 54400g
Mass of fertilizer = 54400g = 13600 crops
Mass per crop 4.0
- NITRATE(V) NO3– and NITRATE(III) NO2– Salts
Nitrate(V) /NO3– and Nitrate(III) /NO2– are salts derived from Nitric(V)/HNO3 and Nitric(III)/HNO2 acids. Both HNO3 and HNO2 are monobasic acids with only one ionizable hydrogen in a molecule.
Only KNO2 ,NaNO2 and NH4NO2 exist. All metallic nitrate(V)salts exist.
All Nitrate(V) /NO3– and Nitrate(III) /NO2– are soluble/dissolve in water.
(a)Effect of heat on Nitrate(V) /NO3– and Nitrate(III) /NO2– salts(Test for presence of Nitrate(V) /NO3– ions in solid state)
- All Nitrate(III) /NO2– salts are not affected by gentle or strong heating except ammonium nitrate(III) NH4NO2 .
Ammonium nitrate(III) NH4NO2 is a colourless solid that decompose to form Nitrogen gas and water.
Chemical equation
NH4NO2 (s) -> H2O(l) + N2(g)
This reaction is used to prepare small amounts of Nitrogen in a school laboratory.
- All Nitrate(V) /NO3– salts decompose on strong heating:
Experiment
Put ½ spatula full of sodium nitrate(V) into a a test tube. Place moist blue/red litmus papers on the mouth of the test tube. Heat strongly when test tube is slanted.
Test the gases produced using glowing splint
Caution (i) Wear safety gas mask and hand gloves
(ii)Lead(II)nitrate(V)decomposes to Lead(II)oxide that react and fuses with the test tube permanently.
Repeat with potassium nitrate(V), copper(II) nitrate(V), Lead(II)nitrate(V), silver nitrate(V), Zinc nitrate(V), Magnesium nitrate(V) and Ammonium nitrate(V).
Observations
Cracking sound
Brown fumes/gas produced except in potassium nitrate(V) and Sodium nitrate(V)
Glowing splint relights/rekindles but feebly in Ammonium nitrate(V).
Black solid residue with copper(II) nitrate(V)
White residue/solid with sodium nitrate(V), potassium nitrate(V),silver nitrate(V), Magnesium nitrate(V)
Yellow residue/solid when hot but white on cooling with Zinc nitrate(V)
Brown residue/solid when hot but yellow on cooling with Lead(II)nitrate(V)
Explanation
- Potassium nitrate(V) and Sodium nitrate(V) decomposes on strong heating to form potassium nitrate(III) and Sodium nitrate(III) producing Oxygen gas. Oxygen gas relights/rekindles a glowing splint.
Chemical equation.
2KNO3(s) -> 2KNO2(s) + O2 (g)
2NaNO3(s) -> 2NaNO2(s) + O2 (g)
2.Heavy metal nitrate(V)salts decomposes to form the oxide, brown nitrogen (IV) oxide and Oxygen gas.
Copper(II)oxide is black.Zinc oxide is yellow when hot and white when cool/cold. Lead(II)oxide is yellow when cold/cool and brown when hot/heated.
Hydrated copper(II)nitrate is blue. On heating it melts and dissolves in its water of crystallization to form a green solution. When all the water of crystallization has evaporated,the nitrate(V)salt decomposes to black Copper(II)oxide and a mixture of brown nitrogen(IV)oxide gas and colourless Oxygen gas.
Chemical equation
2Cu(NO3)2(s) -> 2CuO (s) + 4NO2(g) + O2(s)
2Ca(NO3)2(s) -> 2CaO (s) + 4NO2(g) + O2(s)
2Zn(NO3)2(s) -> 2ZnO (s) + 4NO2(g) + O2(s)
2Mg(NO3)2(s) -> 2MgO (s) + 4NO2(g) + O2(s)
2Pb(NO3)2(s) -> 2PbO (s) + 4NO2(g) + O2(s)
2Fe(NO3)2(s) -> 2FeO (s) + 4NO2(g) + O2(s)
Silver nitrate(V)and Mercury(II)nitrate decomposes to the corresponding metal and a mixture of brown nitrogen(IV)oxide gas and colourless Oxygen gas.
Chemical equation
2AgNO3 (s) -> 2Ag (s) + 2NO2(g) + O2(s)
Hg(NO3)2(s) -> Hg(l) + 2NO2(g) + O2(s)
The production/evolution of brown fumes of Nitrogen(IV)oxide gas on heating a salt is a confirmatory test for presence of NO3– ions of heavy metals
(b)Brown ring test (Test for presence of Nitrate(V) /NO3– ions in aqueous/ solution state)
Experiment
Place 5cm3 of Potassium nitrate(V)solution onto a clean test tube. Add 8 drops of freshly prepared Iron(II)sulphate(VI)solution. Swirl/ shake.
Using a test tube holder to firmly slant and hold the test tube, carefully add 5cm3 of Concentrated sulphuric (VI) acid down along the side of test tube.Do not shake the test tube contents.
Caution: Concentrated sulphuric (VI) acid is highly corrosive.
Observation.
Both Potassium nitrate(V)and freshly prepared Iron(II)sulphate (VI)do not form layers
On adding Concentrated Sulphuric(VI)acid,two layers are formed.
A brown ring is formed between the layers.
Explanation
All nitrate(V)salts are soluble. They form a miscible mixture when added freshly prepared Iron(II)sulphate(VI)solution. Concentrated sulphuric(VI)acid is denser than the miscible mixture thus settle at the bottom.
At the junction of the layers, the acid reacts with nitrate(V)salts to form Nitric(V)acid/HNO3. Nitric(V)acid/HNO3 is reduced to Nitrogen (II)oxide by the Iron(II)sulphate(VI) salt to form the complex compound Nitroso-iron(II)sulphate(VI)/FeSO4.NO . Nitroso-iron(II)sulphate(VI) is brown in colour.It forms a thin layer at the junction between concentrated sulphuric (VI)acid and the miscible mixture of freshly prepared Iron(II) sulphate(VI) and the nitrate(V)salts as a brown ring.
Chemical equation
FeSO4(aq) + NO(g) -> FeSO4.NO(aq)
(Nitroso-iron(II)sulphate(VI)complex)
The brown ring disappear if shaken because concentrated sulphuric (VI)acid mixes with the aqueous solution generating a lot of heat which decomposes Nitroso-iron(II)sulphate(VI)/FeSO4.NO to iron(II)sulphate(VI) and Nitrogen(II)oxide.
Chemical equation
FeSO4.NO(aq) ->FeSO4(aq) + NO(g) ->
Iron(II)sulphate(VI) solution is easily/readily oxidized to iron(III)sulphate(VI) on exposure to air/oxygen. The brown ring test thus require freshly prepared Iron(II) sulphate(VI) solution
(c)Devardas alloy test (Test for presence of Nitrate(V) /NO3– ions in aqueous/ solution state)
Experiment
Place 5cm3 of Potassium nitrate(V)solution onto a clean test tube. Add 5 drops of sodium hydroxide solution. Swirl/ shake. Add a piece of aluminium foil to the mixture.Heat.Test any gases produced using both blue and red litmus papers.
Observation. Inference
Effervescence/bubbles/fizzing
colourless gas that has a pungent smell of urine NO3–
Blue limus paper remain blue
Red litmus paper turn red.
Explanation
The Devardas alloy test for NO3– ions in solution was developed by the Italian scientist Artulo Devarda(1859-1944)
When a NO3–salt is added sodium hydroxide and aluminium foil, effervescence of ammonia gas is a confirmatory test for NO3– ions.
- ENVIRONMENTAL EFFECTS OF NITROGENOUS COMPOUNDS
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UPGRADE CHEMISTRY
FORM 3
Chemistry of SULPHUR
Comprehensive tutorial notes
MUTHOMI S.G 0720096206
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A.SULPHUR (S)
Sulphur is an element in Group VI Group 16)of the Periodic table . It has atomic number 16 and electronic configuration 16 and valency 2 /divalent and thus forms the ion S2-
- Occurrence.
Sulphur mainly occurs :
(i) as free element in Texas and Louisiana in USA and Sicily in Italy.
(ii)Hydrogen sulphide gas in active volcanic areas e.g. Olkaria near Naivasha in Kenya
(iii)as copper pyrites(CuFeS2) ,Galena (PbS,Zinc blende(ZnS))and iron pyrites(FeS2) in other parts of the world.
- Extraction of Sulphur from Fraschs process
Suphur occurs about 200 metres underground. The soil structure in these areas is usually weak and can easily cave in.
Digging of tunnels is thus discouraged in trying to extract the mineral.
Sulphur is extracted by drilling three concentric /round pipes of diameter of ratios 2:8: 18 centimeters.
Superheated water at 170oC and 10atmosphere pressure is forced through the outermost pipe.
The high pressures ensure the water remains as liquid at high temperatures instead of vapour of vapour /gas.
The superheated water melts the sulphur because the melting point of sulphur is lower at about at about 115oC.
A compressed air at 15 atmospheres is forced /pumped through the innermost pipe.
The hot air forces the molten sulphur up the middle pipe where it is collected and solidifies in a large tank.
It is about 99% pure.
Diagram showing extraction of Sulphur from Fraschs Process
- Allotropes of Sulphur.
- Sulphur exist as two crystalline allotropic forms:
(i)Rhombic sulphur
(ii)Monoclinic sulphur
| Rhombic sulphur | Monoclinic sulphur |
| Bright yellow crystalline solid
Has a melting point of 113oC Has a density of 2.06gcm-3 Stable below 96oC Has octahedral structure |
Pale yellow crystalline solid
Has a melting point of 119oC Has a density of 1.96gcm-3 Stable above 96oC Has a needle-like structure |
Rhombic sulphur and Monoclinic sulphur have a transition temperature of 96oC.This is the temperature at which one allotrope changes to the other.
- Sulphur exists in non-crystalline forms as:
(i)Plastic sulphur-
Plastic sulphur is prepared from heating powdered sulphur to boil then pouring a thin continuous stream in a beaker with cold water. A long thin elastic yellow thread of plastic sulphur is formed .If left for long it turn to bright yellow crystalline rhombic sulphur.
(ii)Colloidal sulphur-
Colloidal sulphur is formed when sodium thiosulphate (Na2S2O3) is added hydrochloric acid to form a yellow precipitate.
- Heating Sulphur.
A molecule of sulphur exists as puckered ring of eight atoms joined by covalent bonds as S8.
On heating the yellow sulphur powder melts at 113oC to clear amber liquid with low viscosity and thus flows easily.
On further heating to 160oC the molten liquid darkens to a brown very viscous liquid that does not flow easily.
This is because the S8 rings break into S8 chain that join together to form very long chains made of over 100000 atoms of Sulphur.
The long chains entangle each other reducing their mobility /flow and hence increases their viscosity.
On continued further heating to above 160oC, the viscous liquid darkens but becomes more mobile/flows easily and thus less viscous.
This is because the long chains break to smaller/shorter chains.
At 444oC, the liquid boils and forms brown vapour of a mixture of S8 ,S6 ,S2 molecules that solidifies to S8 ring of “flowers of sulphur” on the cooler parts.
Summary of changes on heating sulphur
| Observation on heating | Explanation/structure of Sulphur |
| Solid sulphur
Heat to 113oC Amber yellow liquid
Heat to 160oC Liquid darkens
Heat to 444oC Liquid boils to brown vapour
Cool to room temperature Yellow sublimate (Flowers of Sulphur) |
Puckered S8 ring
Puckered S8 ring in liquid form (low viscosity/flow easily)
Puckered S8 ring break/opens then join to form long chains that entangle (very high viscosity/very low rate of flow)
Mixture of S8 ,S6 ,S2 vapour
Puckered S8 ring |
- Physical and Chemical properties of Sulphur.(Questions)
- State three physical properties unique to Sulphur
Sulphur is a yellow solid, insoluble in water, soluble in carbon disulphide/tetrachloromethane/benzene, poor conductor of heat and electricity. It has a melting point of 115oC and a boiling point of 444oC.
- Moist/damp/wet blue and red litmus papers were put in a gas jar containing air/oxygen. Burning sulphur was then lowered into the gas jar. State and explain the observation made.
Observations
-Sulphur melts then burns with a blue flame
Colourless gas produced that has a pungent smell
Red litmus paper remains red. Blue litmus paper turns red.
Explanation
Sulphur burns in air and faster in Oxygen to form Sulphur(IV)Oxide gas and traces/small amount of Sulphur(VI)Oxide gas. Both oxides react with water to form the corresponding acidic solution i.e
(i) Sulphur(IV)Oxide gas reacts with water to form sulphuric(IV)acid
(ii) Sulphur(VI)Oxide gas reacts with water to form sulphuric(VI)acid
Chemical equation
S(s) + O2(g) -> SO2(g) (Sulphur(IV)Oxide gas)
2S(s) + 3O2(g) -> 2SO3(g) (Sulphur(VI)Oxide gas traces)
SO2(g) + H2O(l) -> H2 SO3 (aq) ( sulphuric(IV)acid) SO3(g) + H2O(l) -> H2 SO4 (aq) ( sulphuric(VI)acid).
- Iron filings were put in a test tube containing powdered sulphur then heated on a Bunsen flame. Stop heating when reaction starts. State and explain the observations made. Test the effects of a magnet on the mixture before and after heating. Explain.
Observations
Before heating, the magnet attracts iron filings leaving sulphur
After heating, the magnet does not attract the mixture.
After heating, a red glow is observed that continues even when heating is stopped..
Black solid is formed.
Explanation Iron is attracted to a magnet because it is ferromagnetic.
When a mixture of iron and sulphur is heated, the reaction is exothermic giving out heat energy that makes the mixture to continue glowing even after stopping heating.
Black Iron(II)sulphide is formed which is a compound and thus not ferromagnetic.
Chemical equation Fe(s) + S(s) -> FeS(s) (Exothermic reaction/ –∆H)
Heated powdered heavy metals combine with sulphur to form black sulphides.
Cu(s) + S(s) -> CuS(s)
Zn(s) + S(s) -> ZnS(s)
Pb(s) + S(s) -> PbS(s)
4.The set up below show the reaction of sulphur on heated concentrated sulphuric(VI)acid.
(i)State and explain the observation made.
Observation
Yellow colour of sulphur fades
Orange colour of potassium dichromate(VI)paper turns to green.
Explanation
Hot concentrated sulphuric(VI)acid oxidizes sulphur to sulphur (IV)oxide gas. The oxide is also reduced to water. Traces of sulphur (VI)oxide is formed.
Chemical equation
S(s) + 3H2 SO4 (l) -> 3SO2(g) + 3H2O(l) +SO3(g)
Sulphur (IV)oxide gas turns Orange potassium dichromate(VI)paper to green.
(ii)State and explain the observation made if concentrated sulphuric (VI) acid is replaced with concentrated Nitric (V) acid in the above set up.
Observation
Yellow colour of sulphur fades
Colurless solution formed
Brown fumes/gas produced.
Explanation
Hot concentrated Nitric(V)acid oxidizes sulphur to sulphuric (VI)acid. The Nitric (V) acid is reduced to brown nitrogen(IV)oxide gas.
Chemical equation
S(s) + 6HNO3 (l) -> 6NO2(g) + 2H2O(l) +H2SO4 (l)
NB:
Hydrochloric acid is a weaker oxidizing agent and thus cannot oxidize sulphur like the other mineral acids.
5.State three main uses of sulphur .
Sulphur is mainly used in:
(i)Contact process for the manufacture/industrial/large scale production of concentrated sulphuric(VI)acid.
(ii)Vulcanization of rubber to make it harder, tougher, stronger, and more durable.
(iii)Making gun powder and match stick heads
(iv) As ointments to treat fungal infections
- Revision Practice
The diagram below represents the extraction of sulphur by Fraschs process. Use it to answer the questions that follow.
| N |
| M |
| L |
(a)Name the substances that passes through:
M Superheated water at 170oC and 10 atmosphere pressure
L Hot compressed air
N Molten sulphur
(b)What is the purpose of the substance that passes through L and M?
M- Superheated water at 170oC and 10 atmosphere pressure is used to melt the sulphur
L- Hot compressed air is used to force up the molten sulphur.
(c) The properties of the two main allotropes of sulphur represented by letters A and B are given in the table below. Use it to answer the questions that follow.
| A | B | |
| Appearance | Bright yellow | Pale yellow |
| Density(gcm-3) | 1.93 | 2.08 |
| Melting point(oC) | 119 | 113 |
| Stability | Above 96oC | Below 96oC |
I.What are allotropes?
Different forms of the same element existing at the same temperature and pressure without change of state.
- Identify allotrope:
- Monoclinic sulphur
B . Rhombic sulphur
III. State two main uses of sulphur.
-Manufacture of sulphuric(VI)acid
-as fungicide
-in vulcanization of rubber to make it harder/tougher/ stronger
-manufacture of dyes /fibres
(d)Calculate the volume of sulphur (IV)oxide produced when 0.4 g of sulphur is completely burnt in excess air.(S = 32.0 ,I mole of a gas occupies 24 dm3 at room temperature)
Chemical equation
S(s) + O2(g) -> SO2(g)
Mole ratio S: SO2 = 1:1
Method 1
32.0 g of sulphur -> 24 dm3 of SO2(g)
0.4 g of sulphur -> 0.4 g x 24 dm3 = 0.3 dm3
32.0 g
Method 2
Moles of sulphur used = Mass of sulphur => 0.4 = 0.0125 moles
Molar mass of sulphur 32
Moles of sulphur used = Moles of sulphur(IV)oxide used=>0.0125 moles
Volume of sulphur(IV)oxide used = Moles of sulphur(IV)oxide x volume of one mole of gas =>0.0125 moles x 24 dm3 = 0.3 dm3
B.COMPOUNDS OF SULPHUR
The following are the main compounds of sulphur:
(i) Sulphur(IV)oxide
(ii) Sulphur(VI)oxide .
(iii) Sulphuric(VI)acid
(iv) Hydrogen Sulphide
(v) Sulphate(IV)/SO32- and Sulphate(VI)/ SO42- salts
(i) Sulphur(IV)oxide(SO2)
(a) Occurrence
Sulphur (IV)oxide is found in volcanic areas as a gas or dissolved in water from geysersand hot springs in active volcanic areas of the world e.g. Olkaria and Hells gate near Naivasha in Kenya.
(b) School laboratory preparation
In a Chemistry school laboratory Sulphur (IV)oxide is prepared from the reaction of
Method 1:Using Copper and Sulphuric(VI)acid.
Method 2:Using Sodium Sulphate(IV) and hydrochloric acid.
(c)Properties of Sulphur(IV)oxide(Questions)
- Write the equations for the reaction for the formation of sulphur (IV)oxide using:
(i)Method 1
Cu(s) + 2H2SO4(l) -> CuSO4(aq) + SO2(g) + 2H2O(l)
Zn(s) + 2H2SO4(l) -> ZnSO4(aq) + SO2(g) + 2H2O(l)
Mg(s) + 2H2SO4(l) -> MgSO4(aq) + SO2(g) + 2H2O(l)
Fe(s) + 2H2SO4(l) -> FeSO4(aq) + SO2(g) + 2H2O(l)
Calcium ,Lead and Barium will form insoluble sulphate(VI)salts that will cover unreacted metals stopping further reaction thus producing very small amount/quantity of sulphur (IV)oxide gas.
(ii)Method 2
Na2SO3(aq) + HCl(aq) -> NaCl(aq ) + SO2(g) + 2H2O(l)
K2SO3(aq) + HCl(aq) -> KCl(aq ) + SO2(g) + 2H2O(l)
BaSO3(s) + 2HCl(aq) -> BaCl2(aq ) + SO2(g) + H2O(l)
CaSO3(s) + 2HCl(aq) -> CaCl2(aq ) + SO2(g) + H2O(l)
PbSO3(s) + 2HCl(aq) -> PbCl2(s ) + SO2(g) + H2O(l)
Lead(II)chloride is soluble on heating thus reactants should be heated to prevent it coating/covering unreacted PbSO3(s)
2.State the physical properties unique to sulphur (IV)oxide gas.
Sulphur (IV)oxide gas is a colourless gas with a pungent irritating and choking smell which liquidifies easily. It is about two times denser than air.
- The diagram below show the solubility of sulphur (IV)oxide gas. Explain.
Sulphur(IV) oxide is very soluble in water.
One drop of water dissolves all the Sulphur (IV) oxide in the flask leaving a vacuum.
If the clip is removed, atmospheric pressure forces the water up through the narrow tube to form a fountain to occupy the vacuum.
An acidic solution of sulphuric (IV)acid is formed which turns litmus solution red.
Chemical equation
SO2(g) + H2O(l) -> H2 SO3 (aq) ( sulphuric(IV)acid turn litmus red)
4.Dry litmus papers and wet/damp/moist litmus papers were put in a gas jar containing sulphur(IV) oxide gas. State and explain the observations made.
Observations
(i)Dry Blue litmus paper remains blue.
Dry red litmus paper remains red.
(ii) Wet/damp/moist blue litmus paper turns red.
Moist/damp/wet red litmus paper remains red.
Both litmus papers are then bleached /decolorized.
Explanation
Dry sulphur(IV) oxide gas is a molecular compound that does not dissociate/ionize to release H+(aq)ions and thus has no effect on dry blue/red litmus papers.
Wet/damp/moist litmus papers contain water that dissolves /react with dry sulphur(IV) oxide gas to form a solution of weak sulphuric(IV)acid (H2 SO3 (aq)).
Weak sulphuric(IV)acid(H2 SO3 (aq)) dissociates /ionizes into free H+(aq)ions:
H2 SO3 (aq) -> 2H+(aq) + SO32- (aq)
The free H+(aq)ions are responsible for turning blue litmus paper turns red showing the gas is acidic.
The SO32- (aq) ions in wet/damp/moist sulphur(IV) oxide gas is responsible for many reactions of the gas.
It is easily/readily oxidized to sulphate(VI) SO42- (aq) ions making sulphur(IV) oxide gas act as a reducing agent as in the following examples:
(a)Bleaching agent
Wet/damp/moist coloured flowers/litmus papers are bleached/decolorized when put in sulphur(IV) oxide gas.
This is because sulphur(IV) oxide removes atomic oxygen from the coloured dye/ material to form sulphuric(VI)acid.
Chemical equations
(i)Formation of sulphuric(IV)acid
SO2(g) + H2O(l) -> H2 SO3 (aq)
(ii)Decolorization/bleaching of the dye/removal of atomic oxygen.
Method I. H2 SO3 (aq) + (dye + O) -> H2 SO4 (aq) + dye
(coloured) (colourless)
Method II. H2 SO3 (aq) + (dye) -> H2 SO4 (aq) + (dye – O)
(coloured) (colourless)
Sulphur(IV) oxide gas therefore bleaches by reduction /removing oxygen from a dye unlike chlorine that bleaches by oxidation /adding oxygen.
The bleaching by removing oxygen from Sulphur(IV) oxide gas is temporary.
This is because the bleached dye regains the atomic oxygen from the atmosphere/air in presence of sunlight as catalyst thus regaining/restoring its original colour. e.g.
Old newspapers turn brown on exposure to air on regaining the atomic oxygen.
The bleaching through adding oxygen by chlorine gas is permanent.
(b)Turns Orange acidified potassium dichromate(VI) to green
Experiment:
(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing acidified potassium dichromate(VI) solution. or;
(ii)Dip a filter paper soaked in acidified potassium dichromate(VI) into a gas jar containing Sulphur(IV) oxide gas.
Observation:
Orange acidified potassium dichromate(VI) turns to green.
Explanation:
Sulphur(IV) oxide gas reduces acidified potassium dichromate(VI) from orange Cr2O72- ions to green Cr3+ ions without leaving a residue itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid.
Chemical/ionic equation:
(i)Reaction of Sulphur(IV) oxide gas with water
SO2(g) + H2O(l) -> H2 SO3 (aq)
(ii)Dissociation /ionization of Sulphuric(IV)acid.
H2 SO3 (aq) -> 2H+(aq) + SO32- (aq)
(iii)Oxidation of SO32- (aq)and reduction of Cr2O72-(aq)
3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l)
This is a confirmatory test for the presence of Sulphur(IV) oxide gas.
Hydrogen sulphide also reduces acidified potassium dichromate(VI) from orange Cr2O72- ions to green Cr3+ ions leaving a yellow residue.
(c)Decolorizes acidified potassium manganate(VII)
Experiment:
(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing acidified potassium manganate(VII) solution. or;
(ii)Dip a filter paper soaked in acidified potassium manganate(VII) into a gas jar containing Sulphur(IV) oxide gas.
Observation:
Purple acidified potassium manganate(VII) turns to colourless/ acidified potassium manganate(VII) is decolorized.
Explanation:
Sulphur(IV) oxide gas reduces acidified potassium manganate(VII) from purple MnO4– ions to green Mn2+ ions without leaving a residue itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid.
Chemical/ionic equation:
(i)Reaction of Sulphur(IV) oxide gas with water
SO2(g) + H2O(l) -> H2 SO3 (aq)
(ii)Dissociation /ionization of Sulphuric(IV)acid.
H2 SO3 (aq) -> 2H+(aq) + SO32- (aq)
(iii)Oxidation of SO32- (aq)and reduction of MnO4– (aq)
5SO32-(aq) + 2MnO4– (aq) +6H+(aq) -> 5SO42-(aq) + 2Mn2+(aq) + 3H2O(l)
(purple) (colourless)
This is another test for the presence of Sulphur(IV) oxide gas.
Hydrogen sulphide also decolorizes acidified potassium manganate(VII) from purple MnO4– ions to colourless Mn2+ ions leaving a yellow residue.
(d)Decolorizes bromine water
Experiment:
(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing bromine water . or;
(ii)Put three drops of bromine water into a gas jar containing Sulphur(IV) oxide gas. Swirl.
Observation:
Yellow bromine water turns to colourless/ bromine water is decolorized.
Explanation:
Sulphur(IV) oxide gas reduces yellow bromine water to colourless hydrobromic acid (HBr) without leaving a residue itself oxidized from SO32- ions in sulphuric (IV) acid to SO42- ions in sulphuric(VI) acid.
Chemical/ionic equation:
(i)Reaction of Sulphur(IV) oxide gas with water
SO2(g) + H2O(l) -> H2 SO3 (aq)
(ii)Dissociation /ionization of Sulphuric(IV)acid.
H2 SO3 (aq) -> 2H+(aq) + SO32- (aq)
(iii)Oxidation of SO32- (aq)and reduction of MnO4– (aq)
SO32-(aq) + Br2 (aq) + H2O(l) -> SO42-(aq) + 2HBr(aq)
(yellow) (colourless)
This can also be used as another test for the presence of Sulphur(IV) oxide gas.
Hydrogen sulphide also decolorizes yellow bromine water to colourless leaving a yellow residue.
(e)Reduces Iron(III) Fe3+ salts to Iron(II) salts Fe2+
Experiment:
(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing about 3 cm3 of Iron (III)chloride solution. or;
(ii)Place about 3cm3 of Iron (III)chloride solution into a gas jar containing Sulphur(IV) oxide gas.Swirl.
Observation:
Yellow/brown Iron (III)chloride solution turns to green
Explanation:
Sulphur(IV) oxide gas reduces Iron (III)chloride solution from yellow/brown Fe3+ ions to green Fe2+ ions without leaving a residue itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid.
Chemical/ionic equation:
(i)Reaction of Sulphur(IV) oxide gas with water
SO2(g) + H2O(l) -> H2 SO3 (aq)
(ii)Dissociation /ionization of Sulphuric(IV)acid.
H2 SO3 (aq) -> 2H+(aq) + SO32- (aq)
(iii)Oxidation of SO32- (aq)and reduction of Fe3+ (aq)
SO32-(aq) + 2Fe3+ (aq) +3H2O(l) -> SO42-(aq) + 2Fe2+(aq) + 2H+(aq)
(yellow) (green)
(f)Reduces Nitric(V)acid to Nitrogen(IV)oxide gas
Experiment:
(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing about 3 cm3 of concentrated nitric(V)acid. or;
(ii)Place about 3cm3 of concentrated nitric(V)acid into a gas jar containing Sulphur(IV) oxide gas. Swirl.
Observation:
Brown fumes of a gas evolved/produced.
Explanation:
Sulphur(IV) oxide gas reduces concentrated nitric(V)acid to brown nitrogen(IV)oxide gas itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid.
Chemical/ionic equation:
SO2(g) + 2HNO3 (l) -> H2 SO4 (l) + NO2 (g)
(brown fumes/gas)
(g)Reduces Hydrogen peroxide to water
Experiment:
(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing about 3 cm3 of 20 volume hydrogen peroxide. Add four drops of Barium nitrate(V)or Barium chloride followed by five drops of 2M hydrochloric acid/ 2M nitric(V) acid.
Observation:
A white precipitate is formed that persist /remains on adding 2M hydrochloric acid/ 2M nitric(V) acid.
Explanation:
Sulphur(IV) oxide gas reduces 20 volume hydrogen peroxide and itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in sulphuric(VI) acid.
When Ba2+ ions in Barium Nitrate(V) or Barium chloride solution is added, a white precipitate of insoluble Barium salts is formed showing the presence of of either SO32- ,SO42- ,CO32- ions. i.e.
Chemical/ionic equation:
SO32-(aq) + Ba2+ (aq) -> BaSO3(s)
white precipitate
SO42-(aq) + Ba2+ (aq) -> BaSO4(s)
white precipitate
CO32-(aq) + Ba2+ (aq) -> BaCO3(s)
white precipitate
If nitric(V)/hydrochloric acid is added to the three suspected insoluble white precipitates above, the white precipitate:
(i) persist/remains if SO42-(aq)ions (BaSO4(s)) is present.
(ii)dissolves if SO32-(aq)ions (BaSO3(s)) and CO32-(aq)ions (BaCO3(s))is present. This is because:
- BaSO3(s) reacts with Nitric(V)/hydrochloric acid to produce acidic SO2 gas that turns Orange moist filter paper dipped in acidified Potassium dichromate to green.
Chemical equation
BaSO3(s) +2H+(aq) -> Ba2+ (aq) + SO2(g) + H2O(l)
- BaCO3(s) reacts with Nitric(V)/hydrochloric acid to produce acidic CO2 gas that forms a white precipitate when bubbled in lime water.
Chemical equation
BaCO3(s) +2H+(aq) -> Ba2+ (aq) + CO2(g) + H2O(l)
5.Sulphur(IV)oxide also act as an oxidizing agent as in the following examples.
(a)Reduction by burning Magnesium
Experiment
Lower a burning Magnesium ribbon into agas jar containing Sulphur(IV)oxide gas
Observation
Magnesium ribbon continues to burn with difficulty.
White ash and yellow powder/speck
Explanation
Sulphur(IV)oxide does not support burning/combustion. Magnesium burns to produce enough heat energy to decompose Sulphur(IV)oxide to sulphur and oxygen.
The metal continues to burn on Oxygen forming white Magnesium oxide solid/ash.
Yellow specks of sulphur residue form on the sides of reaction flask/gas jar.
During the reaction, Sulphur(IV)oxide is reduced(oxidizing agent)while the metal is oxidized (reducing agent)
Chemical equation
SO2(g) + 2Mg(s) -> 2MgO(s) + S(s)
(white ash/solid) (yellow speck/powder)
(b)Reduction by Hydrogen sulphide gas
Experiment
Put two drops of water into a gas jar containing dry Sulphur(IV)oxide gas
Bubble hydrogen sulphide gas into the gas jar containing Sulphur(IV)oxide gas.
Or
Put two drops of water into a gas jar containing dry Sulphur(IV)oxide gas
Invert a gas jar full of hydrogen sulphide gas over the gas jar containing Sulphur(IV)oxide gas. Swirl
Observation
Yellow powder/speck
Explanation
Sulphur(IV)oxide oxidizes hydrogen sulphide to yellow specks of sulphur residue and itself reduced to also sulphur that form on the sides of reaction flask/gas jar.
A little moisture/water act as catalyst /speeds up the reaction.
Chemical equation
SO2(g) + 2H2S(g) -> 2H2O(l) + 3S(s)
(yellow speck/powder)
6.Sulphur(IV)oxide has many industrial uses. State three.
(i)In the contact process for the manufacture of Sulphuric(VI)acid
(ii)As a bleaching agent of pulp and paper.
(iii)As a fungicide to kill microbes’
(iv)As a preservative of jam, juices to prevent fermentation
(ii) Sulphur(VI)oxide(SO3)
(a) Occurrence
Sulphur (VI)oxide is does not occur free in nature/atmosphere
(b) Preparation
In a Chemistry school laboratory Sulphur (VI)oxide may prepared from:
Method 1;Catalytic oxidation of sulphur(IV)oxide gas.
Sulphur(IV)oxide gas and oxygen mixture are first dried by being passed through Concentrated Sulphuric(VI)acid .
The dry mixture is then passed through platinised asbestos to catalyse/speed up the combination to form Sulphur (VI)oxide gas.
Sulphur (VI)oxide gas readily solidify as silky white needles if passed through a freezing mixture /ice cold water.
The solid fumes out on heating to a highly acidic poisonous gas.
Chemical equation
2SO2(g) + O2(g) –platinised asbestos–> 2SO3 (g)
Method 2; Heating Iron(II)sulphate(VI) heptahydrate
When green hydrated Iron(II)sulphate(VI) heptahydrate crystals are heated in a boiling tube ,it loses the water of crystallization and colour changes from green to white.
Chemical equation
FeSO4.7H2O(s) -> FeSO4(s) + 7H2O(l)
(green solid) (white solid)
On further heating ,the white anhydrous Iron(II)sulphate(VI) solid decomposes to a mixture of Sulphur (VI)oxide and Sulphur (IV)oxide gas.
Sulphur (VI) oxide readily / easily solidify as white silky needles when the mixture is passed through a freezing mixture/ice cold water.
Iron(III)oxide is left as a brown residue/solid.
Chemical equation
2FeSO4 (s) -> Fe2O3(s) + SO2 (g) + SO3(g)
(green solid) (brown solid)
Caution
On exposure to air Sulphur (VI)oxide gas produces highly corrosive poisonous fumes of concentrated sulphuric(VI)acid and thus its preparation in a school laboratory is very risky.
(c) Uses of sulphur(VI)oxide
One of the main uses of sulphur(VI)oxide gas is as an intermediate product in the contact process for industrial/manufacture/large scale/production of sulphuric(VI)acid.
(iii) Sulphuric(VI)acid(H2SO4)
(a) Occurrence
Sulphuric (VI)acid(H2SO4) is one of the three mineral acids. There are three mineral acids;
Nitric(V)acid
Sulphuric(VI)acid
Hydrochloric acid.
Mineral acids do not occur naturally but are prepared in a school laboratory and manufactured at industrial level.
(b)The Contact process for industrial manufacture of H2SO4 .
- Raw materials
The main raw materials for industrial preparation of Sulphuric(VI)acid include:
(i)Sulphur from Fraschs process or from heating metal sulphide ore like Galena(PbS),Zinc blende(ZnS)
(ii)Oxygen from fractional distillation of air
(iii)Water from rivers/lakes
- Chemical processes
The contact process involves four main chemical processes:
(i)Production of Sulphur (IV)oxide
As one of the raw materials, Sulphur (IV)oxide gas is got from the following sources;
- Burning/roasting sulphur in air.
Sulphur from Fraschs process is roasted/burnt in air to form Sulphur (IV)oxide gas in the burners
Chemical equation
S(s) + O2(g) –> SO2 (g)
- Burning/roasting sulphide ores in air.
Sulphur (IV)oxide gas is produced as a by product in extraction of some metals like:
– Lead from Lead(II)sulphide/Galena,(PbS)
– Zinc from zinc(II)sulphide/Zinc blende, (ZnS)
– Copper from Copper iron sulphide/Copper pyrites, (CuFeS2)
On roasting/burning, large amount /quantity of sulphur(IV)oxide is generated/produced.
Chemical equation
(i)2PbS (s) + 3O2 (g) -> 2PbO(s) + 2SO2 (g)
(ii)2ZnS (s) + 3O2 (g) -> 2ZnO(s) + 2SO2 (g)
(ii)2CuFeS2 (s) + 4O2 (g) -> 2FeO(s) + 3SO2 (g) + Cu2O(s)
Sulphur(IV)oxide easily/readily liquefies and thus can be transported to a far distance safely.
(ii)Purification of Sulphur(IV)oxide
Sulphur(IV)oxide gas contain dust particles and Arsenic(IV)oxide as impurities. These impurities “poison”/impair the catalyst by adhering on/covering its surface.
The impurities are removed by electrostatic precipitation method .
In the contact process Platinum or Vanadium(V)oxide may be used. Vanadium(V)oxide is preferred because it is :
(i) cheaper/less expensive
(ii) less easily poisoned by impurities
(iii)Catalytic conversion of Sulphur(IV)oxide to Sulphur(VI)oxide
Pure and dry mixture of Sulphur (IV)oxide gas and Oxygen is heated to 450oC in a heat exchanger.
The heated mixture is passed through long pipes coated with pellets of Vanadium (V)oxide catalyst.
The close “contact” between the reacting gases and catalyst give the process its name.
Vanadium (V)oxide catalyse the conversion/oxidation of Sulphur(IV)oxide to Sulphur(VI)oxide gas.
Chemical equation
2SO2 (g) + O2(g) — V2O5 –> 2SO2 (g)
This reaction is exothermic (-∆H) and the temperatures need to be maintained at around 450oC to ensure that:
(i)reaction rate/time taken for the formation of Sulphur(VI)oxide is not too slow/long at lower temperatures below 450oC
(ii) Sulphur(VI)oxide gas does not decompose back to Sulphur(IV)oxide gas and Oxygen gas at higher temperatures than 450oC.
(iv)Conversion of Sulphur(VI)oxide of Sulphuric(VI)acid
Sulphur(VI)oxide is the acid anhydride of concentrated Sulphuric(VI)acid. Sulphur(VI)oxide reacts with water to form thick mist of fine droplets of very/highly corrosive concentrated Sulphuric(VI)acid because the reaction is highly exothermic.
To prevent this, Sulphur (VI)oxide is a passed up to meet downward flow of 98% Sulphuric(VI)acid in the absorption chamber/tower.
The reaction forms a very viscous oily liquid called Oleum/fuming Sulphuric (VI) acid/ pyrosulphuric (VI) acid.
Chemical equation
H2SO4 (aq) + SO3 (g) -> H2S2O7 (l)
Oleum/fuming Sulphuric (VI) acid/ pyrosulphuric (VI) acid is diluted carefully with distilled water to give concentrated sulphuric (VI) acid .
Chemical equation
H2S2O7 (l) + H2O (l) -> 2H2SO4 (l)
The acid is stored ready for market/sale.
III. Environmental effects of contact process
Sulphur(VI)oxide and Sulphur(IV)oxide gases are atmospheric pollutants that form acid rain if they escape to the atmosphere.
In the Contact process, about 2% of these gases do not form sulphuric (VI) acid.
The following precautions prevent/minimize pollution from Contact process:
(i)recycling back any unreacted Sulphur(IV)oxide gas back to the heat exchangers.
(ii)dissolving Sulphur(VI)oxide gas in concentrated sulphuric (VI) acid instead of water.
This prevents the formation of fine droplets of the corrosive/ toxic/poisonous fumes of concentrated sulphuric (VI) acid.
(iii)scrubbing-This involves passing the exhaust gases through very tall chimneys lined with quicklime/calcium hydroxide solid.
This reacts with Sulphur (VI)oxide gas forming harmless calcium(II)sulphate (IV) /CaSO3
Chemical equation
Ca(OH)2 (aq) + SO2(g) –> CaSO3 (aq) + H2O (g)
- Uses of Sulphuric(VI)acid
Sulphuric (VI) acid is used:
(i) in making dyes and paint
(ii)as acid in Lead-acid accumulator/battery
(iii) for making soapless detergents
(iv) for making sulphate agricultural fertilizers
- Sketch chart diagram showing the Contact process
(c) Properties of Concentrated sulphuric(VI)acid (i)Concentrated sulphuric(VI)acid is a colourless oily liquid with a density of 1.84gcm-3.It has a boiling point of 338oC.
(ii) Concentrated sulphuric(VI)acid is very soluble in water.
The solubility /dissolution of the acid very highly exothermic.
The concentrated acid should thus be diluted slowly in excess water.
Water should never be added to the acid because the hot acid scatters highly corrosive fumes out of the container.
(iii) Concentrated sulphuric (VI)acid is a covalent compound. It has no free H+ ions.
Free H+ ions are responsible for turning the blue litmus paper red. Concentrated sulphuric (VI) acid thus do not change the blue litmus paper red.
(iv) Concentrated sulphuric (VI)acid is hygroscopic. It absorbs water from the atmosphere and do not form a solution.
This makes concentrated sulphuric (VI) acid very suitable as drying agent during preparation of gases.
(v)The following are some chemical properties of concentrated sulphuric (VI) acid:
- As a dehydrating agent
Experiment I;
Put about four spatula end fulls of brown sugar and glucose in separate 10cm3 beaker.
Carefully add about 10cm3 of concentrated sulphuric (VI) acid .Allow to stand for about 10 minutes.
Observation;
Colour( in brown sugar )change from brown to black.
Colour (in glucose) change from white to black.
10cm3 beaker becomes very hot.
Explanation
Concentrated sulphuric (VI) acid is strong dehydrating agent.
It removes chemically and physically combined elements of water(Hydrogen and Oxygen in ratio 2:1)from compounds.
When added to sugar /glucose a vigorous reaction that is highly exothermic take place.
The sugar/glucose is charred to black mass of carbon because the acid dehydrates the sugar/glucose leaving carbon.
Caution
This reaction is highly exothermic that start slowly but produce fine particles of carbon that if inhaled cause quick suffocation by blocking the lung villi.
Chemical equation
Glucose: C6H12O6(s) –conc.H2SO4–> 6C (s) + 6H2O(l)
(white) (black)
Sugar: C12H22O11(s) –conc.H2SO4–> 12C (s) +11H2O(l)
(brown) (black)
Experiment II;
Put about two spatula end full of hydrated copper(II)sulphate(VI)crystals in a boiling tube .Carefully add about 10cm3 of concentrated sulphuric (VI) acid .Warm .
Observation;
Colour change from blue to white.
Explanation
Concentrated sulphuric (VI) acid is strong dehydrating agent.It removes physically combined elements of water(Hydrogen and Oxygen in ratio 2:1)from hydrated compounds.
The acid dehydrates blue copper(II)sulphate to white anhydrous copper(II)sulphate .
Chemical equation
CuSO4.5H2O(s) –conc.H2SO4–> CuSO4 (s) + 5H2O(l)
(blue) (white)
Experiment III;
Put about 4cm3 of absolute ethanol in a boiling tube .Carefully add about 10cm3 of concentrated sulphuric (VI) acid.
Place moist/damp/wet filter paper dipped in acidified potassium dichromate(VI)solution on the mouth of the boiling tube. Heat strongly.
Caution:
Absolute ethanol is highly flammable.
Observation;
Colourless gas produced.
Orange acidified potassium dichromate (VI) paper turns to green.
Explanation
Concentrated sulphuric (VI) acid is strong dehydrating agent.
It removes chemically combined elements of water(Hydrogen and Oxygen in ratio 2:1)from compounds.
The acid dehydrates ethanol to ethene gas at about 170oC.
Ethene with =C=C= double bond turns orange acidified potassium dichromate (VI) paper turns to green.
Chemical equation
C2H5OH(l) –conc.H2SO4/170oC –> C2H4 (g) + H2O(l)
NB: This reaction is used for the school laboratory preparation of ethene gas
Experiment IV;
Put about 4cm3 of methanoic acid in a boiling tube .Carefully add about 6 cm3 of concentrated sulphuric (VI) acid. Heat gently
Caution:
This should be done in a fume chamber/open
Observation;
Colourless gas produced.
Explanation
Concentrated sulphuric (VI) acid is strong dehydrating agent. It removes chemically combined elements of water (Hydrogen and Oxygen in ratio 2:1)from compounds.
The acid dehydrates methanoic acid to poisonous/toxic carbon(II)oxide gas.
Chemical equation
HCOOH(l) –conc.H2SO4 –> CO(g) + H2O(l)
NB: This reaction is used for the school laboratory preparation of small amount carbon (II)oxide gas
Experiment V;
Put about 4cm3 of ethan-1,2-dioic/oxalic acid in a boiling tube .Carefully add about 6 cm3 of concentrated sulphuric (VI) acid. Pass any gaseous product through lime water.Heat gently
Caution:
This should be done in a fume chamber/open
Observation;
Colourless gas produced.
Gas produced forms a white precipitate with lime water.
Explanation
Concentrated sulphuric (VI) acid is strong dehydrating agent.
It removes chemically combined elements of water (Hydrogen and Oxygen in ratio 2:1)from compounds.
The acid dehydrates ethan-1,2-dioic/oxalic acid to a mixture of poisonous/toxic carbon(II)oxide and carbon(IV)oxide gases.
Chemical equation
HOOCCOOH(l) –conc.H2SO4 –> CO(g) + CO2(g) + H2O(l)
NB: This reaction is also used for the school laboratory preparation of small amount carbon (II) oxide gas.
Carbon (IV) oxide gas is removed by passing the mixture through concentrated sodium/potassium hydroxide solution.
- As an Oxidizing agent
Experiment I
Put about 2cm3 of Concentrated sulphuric (VI) acid into three separate boiling tubes. Place a thin moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution on the mouth of the boiling tube. Put about 0.5g of Copper turnings, Zinc granule and Iron filings to each boiling tube separately.
Observation;
Effervescence/fizzing/bubbles
Blue solution formed with copper,
Green solution formed with Iron
Colourless solution formed with Zinc
Colourless gas produced that has a pungent irritating choking smell.
Gas produced turn orange moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution to green.
Explanation
Concentrated sulphuric (VI) acid is strong oxidizing agent.
It oxidizes metals to metallic sulphate(VI) salts and itself reduced to sulphur(IV)oxide gas.
Sulphur (IV) oxide gas turn orange moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution to green.
CuSO4(aq) is a blue solution. ZnSO4(aq) is a colourless solution. FeSO4(aq) is a green solution.
Chemical equation
Cu(s) + 2H2SO4(aq) –> CuSO4(aq) + SO2(g) + 2H2O(l)
Zn(s) + 2H2SO4(aq) –> ZnSO4(aq) + SO2(g) + 2H2O(l)
Fe(s) + 2H2SO4(aq) –> FeSO4(aq) + SO2(g) + 2H2O(l)
Experiment II
Put about 2cm3 of Concentrated sulphuric (VI) acid into two separate boiling tubes. Place a thin moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution on the mouth of the boiling tube.
Put about 0.5g of powdered charcoal and sulphur powder to each boiling tube separately.
Warm.
Observation;
Black solid charcoal dissolves/decrease
Yellow solid sulphur dissolves/decrease
Colourless gas produced that has a pungent irritating choking smell.
Gas produced turn orange moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution to green.
Explanation
Concentrated sulphuric (VI) acid is strong oxidizing agent. It oxidizes non-metals to non metallic oxides and itself reduced to sulphur(IV)oxide gas. Sulphur (IV) oxide gas turn orange moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution to green.
Charcoal is oxidized to carbon(IV)oxide. Sulphur is oxidized to Sulphur(IV)oxide .
Chemical equation
C(s) + 2H2SO4(aq) –> CO2(aq) + 2SO2(g) + 2H2O(l)
S(s) + 2H2SO4(aq) –> 3SO2(g) + 2H2O(l)
III. As the least volatile acid
Study the table below showing a comparison in boiling points of the three mineral acids
| Mineral acid | Relative molecula mass | Boiling point(oC) |
| Hydrochloric acid(HCl) | 36.5 | 35.0 |
| Nitric(V)acid(HNO3) | 63.0 | 83.0 |
| Sulphuric(VI)acid(H2SO4) | 98.0 | 333 |
1.Which is the least volatile acid? Explain
Sulphuric(VI)acid(H2SO4) because it has the largest molecule and joined by Hydrogen bonds making it to have the highest boiling point/least volatile.
- Using chemical equations, explain how sulphuric(VI)acid displaces the less volatile mineral acids.
(i)Chemical equation
KNO3(s) + H2SO4(aq) –> KHSO4(l) + HNO3(g)
NaNO3(s) + H2SO4(aq) –> NaHSO4(l) + HNO3(g)
This reaction is used in the school laboratory preparation of Nitric(V) acid (HNO3).
(ii)Chemical equation
KCl(s) + H2SO4(aq) –> KHSO4(s) + HCl(g)
NaCl(s) + H2SO4(aq) –> NaHSO4(s) + HCl(g)
This reaction is used in the school laboratory preparation of Hydrochloric acid (HCl).
(d) Properties of dilute sulphuric(VI)acid.
Dilute sulphuric(VI)acid is made when about 10cm3 of concentrated sulphuric
(VI) acid is carefully added to about 90cm3 of distilled water.
Diluting concentrated sulphuric (VI) acid should be done carefully because the reaction is highly exothermic.
Diluting concentrated sulphuric (VI) acid decreases the number of moles present in a given volume of solution which makes the acid less corrosive.
On diluting concentrated sulphuric(VI) acid, water ionizes /dissociates the acid fully/wholly into two(dibasic)free H+(aq) and SO42-(aq)ions:
H2SO4 (aq) -> 2H+(aq) + SO42-(aq)
The presence of free H+(aq)ions is responsible for ;
(i)turn litmus red because of the presence of free H+(aq)ions
(ii)have pH 1/2/3 because of the presence of many free H+(aq)ions hence a strongly acidic solution.
(iii)Reaction with metals
Experiment:
Place 5cm3 of 0.2M dilute sulphuric(VI)acid into four separate clean test tubes. Add about 0.1g of Magnesium ribbon to one test tube. Cover the mixture with a finger as stopper. Introduce a burning splint on top of the finger and release the finger “stopper”. Repeat by adding Zinc, Copper and Iron instead of the Magnesium ribbon.
Observation:
No effervescence/ bubbles/ fizzing with copper
Effervescence/ bubbles/ fizzing with Iron ,Zinc and Magnesium
Colourless gas produced that extinguishes burning splint with a “pop” sound.
Colourless solution formed with Zinc and Magnesium.
Green solution formed with Iron
Explanation:
When a metal higher than hydrogen in the reactivity/electrochemical series is put in a test tube containing dilute sulphuric(VI)acid, effervescence/ bubbling/ fizzing takes place with evolution of Hydrogen gas.
Impure hydrogen gas extinguishes burning splint with a “pop” sound.
A sulphate (VI) salts is formed. Iron, Zinc and Magnesium are higher than hydrogen in the reactivity/electrochemical series.
They form Iron (II)sulphate(VI), Magnesium sulphate(VI) and Zinc sulphate(VI).
. When a metal lower than hydrogen in the reactivity/electrochemical series is put in a test tube containing dilute sulphuric(VI)acid, there is no effervescence/ bubbling/ fizzing that take place.
Copper thus do not react with dilute sulphuric(VI)acid.
Chemical/ionic equation
Mg(s) + H2SO4(aq) –> MgSO4(aq) + H2(g)
Mg(s) + 2H+(aq) –> Mg2+ (aq) + H2(g)
Zn(s) + H2SO4(aq) –> ZnSO4(aq) + H2(g)
Zn(s) + 2H+(aq) –> Zn2+ (aq) + H2(g)
Fe(s) + H2SO4(aq) –> FeSO4(aq) + H2(g)
Fe(s) + H+(aq) –> Fe2+ (aq) + H2(g)
NB:(i) Calcium,Lead and Barium forms insoluble sulphate(VI)salts that cover/coat the unreacted metals.
(ii)Sodium and Potassium react explosively with dilute sulphuric(VI)acid
(iv)Reaction with metal carbonates and hydrogen carbonates
Experiment:
Place 5cm3 of 0.2M dilute sulphuric(VI)acid into four separate clean boiling tubes. Add about 0.1g of sodium carbonate to one boiling tube. Introduce a burning splint on top of the boiling tube. Repeat by adding Zinc carbonate, Copper (II)carbonate and Iron(II)Carbonate in place of the sodium hydrogen carbonate.
Observation:
Effervescence/ bubbles/ fizzing.
Colourless gas produced that extinguishes burning splint.
Colourless solution formed with Zinc carbonate, sodium hydrogen carbonate and sodium carbonate.
Green solution formed with Iron(II)Carbonate
Blue solution formed with Copper(II)Carbonate
Explanation:
When a metal carbonate or a hydrogen carbonates is put in a test tube containing dilute sulphuric(VI)acid, effervescence/ bubbling/ fizzing takes place with evolution of carbon(IV)oxide gas. carbon(IV)oxide gas extinguishes a burning splint and forms a white precipitate when bubbled in lime water.
A sulphate (VI) salts is formed.
Chemical/ionic equation
ZnCO3(s) + H2SO4(aq) –> ZnSO4(aq) + H2O(l) + CO2(g)
ZnCO3(s) + 2H+(aq) –> Zn2+ (aq) + H2O(l) + CO2(g)
CuCO3(s) + H2SO4(aq) –> CuSO4(aq) + H2O(l) + CO2(g)
CuCO3(s) + 2H+(aq) –> Cu2+ (aq) + H2O(l) + CO2(g)
FeCO3(s) + H2SO4(aq) –> FeSO4(aq) + H2O(l) + CO2(g)
FeCO3(s) + 2H+(aq) –> Fe2+ (aq) + H2O(l) + CO2(g)
2NaHCO3(s) + H2SO4(aq) –> Na2SO4(aq) + 2H2O(l) + 2CO2(g)
NaHCO3(s) + H+(aq) –> Na+ (aq) + H2O(l) + CO2(g)
Na2CO3(s) + H2SO4(aq) –> Na2SO4(aq) + H2O(l) + CO2(g)
NaHCO3(s) + H+(aq) –> Na+ (aq) + H2O(l) + CO2(g)
(NH4)2CO3(s) + H2SO4(aq) –> (NH4)2SO4 (aq) + H2O(l) + CO2(g)
(NH4)2CO3 (s) + H+(aq) –> NH4+ (aq) + H2O(l) + CO2(g)
2NH4HCO3(aq) + H2SO4(aq) –> (NH4)2SO4 (aq) + H2O(l) + CO2(g)
NH4HCO3(aq) + H+(aq) –> NH4+ (aq) + H2O(l) + CO2(g)
NB:
Calcium, Lead and Barium carbonates forms insoluble sulphate(VI)salts that cover/coat the unreacted metals.
(v)Neutralization-reaction of metal oxides and alkalis/bases
Experiment I:
Place 5cm3 of 0.2M dilute sulphuric(VI)acid into four separate clean boiling tubes. Add about 0.1g of copper(II)oxide to one boiling tube. Stir.
Repeat by adding Zinc oxide, calcium carbonate and Sodium (II)Oxide in place of the Copper(II)Oxide.
Observation:
Blue solution formed with Copper(II)Oxide
Colourless solution formed with other oxides
Explanation:
When a metal oxide is put in a test tube containing dilute sulphuric(VI)acid, the oxide dissolves forming a sulphate (VI) salt.
Chemical/ionic equation
ZnO(s) + H2SO4(aq) –> ZnSO4(aq) + H2O(l)
ZnO(s) + 2H+(aq) –> Zn2+ (aq) + H2O(l)
CuO(s) + H2SO4(aq) –> CuSO4(aq) + H2O(l)
CuO(s) + 2H+(aq) –> Cu2+ (aq) + H2O(l)
MgO(s) + H2SO4(aq) –> MgSO4(aq) + H2O(l)
MgO(s) + 2H+(aq) –> Mg2+ (aq) + H2O(l)
Na2O(s) + H2SO4(aq) –> Na2SO4(aq) + H2O(l)
Na2O(s) + 2H+(aq) –> 2Na+ (aq) + H2O(l)
K2CO3(s) + H2SO4(aq) –> K2SO4(aq) + H2O(l)
K2O(s) + H+(aq) –> 2K+ (aq) + H2O(l)
NB:
Calcium, Lead and Barium oxides forms insoluble sulphate(VI)salts that cover/coat the unreacted metals oxides.
Experiment II:
Fill a burette wuth 0.1M dilute sulphuric(VI)acid. Pipette 20.0cm3 of 0.1Msodium hydroxide solution into a 250cm3 conical flask. Add three drops of phenolphthalein indicator.Titrate the acid to get a permanent colour change.Repeat with0.1M potassium hydroxide solution inplace of 0.1Msodium hydroxide solution
Observation:
Colour of phenolphthalein changes from pink to colourless at the end point.
Explanation
Like other (mineral) acids dilute sulphuric(VI)acid neutralizes bases/alkalis to a sulphate salt and water only.
Colour of the indicator used changes when a slight excess of acid is added to the base at the end point
Chemical equation:
2NaOH(aq) + H2SO4(aq) –> Na2SO4(aq) + H2O(l)
OH–(s) + H+(aq) –> H2O(l)
2KOH(aq) + H2SO4(aq) –> K2SO4(aq) + H2O(l)
OH–(s) + H+(aq) –> H2O(l)
2NH4OH(aq) + H2SO4(aq) –> (NH4)2SO4(aq) + H2O(l)
OH–(s) + H+(aq) –> H2O(l)
(iv) Hydrogen sulphide(H2S)
(a) Occurrence
Hydrogen sulphide is found in volcanic areas as a gas or dissolved in water from geysers and hot springs in active volcanic areas of the world e.g. Olkaria and Hells gate near Naivasha in Kenya.
It is present in rotten eggs and human excreta.
(b) Preparation
Hydrogen sulphide is prepared in a school laboratory by heating Iron (II) sulphide with dilute hydrochloric acid.
(c) Properties of Hydrogen sulphide(Questions)
- Write the equation for the reaction for the school laboratory preparation of Hydrogen sulphide.
Chemical equation: FeS (s) + 2HCl (aq) -> H2S (g) FeCl2 (aq)
- State three physical properties unique to Hydrogen sulphide.
Hydrogen sulphide is a colourless gas with characteristic pungent poisonous smell of rotten eggs. It is soluble in cold water but insoluble in warm water. It is denser than water and turns blue litmus paper red.
- Hydrogen sulphide exist as a dibasic acid when dissolved in water. Using a chemical equation show how it ionizes in aqueous state.
H2S(aq) -> H+(aq) + HS–(aq)
H2S(aq) -> 2H+(aq) + S2- (aq)
Hydrogen sulphide therefore can form both normal and acid salts e.g
Sodium hydrogen sulphide and sodium sulphide both exist
- State and explain one gaseous impurity likely to be present in the gas jar containing hydrogen sulphide above.
Hydrogen/ H2
Iron(II)sulphide contains Iron as impurity .The iron will react with dilute hydrochloric acid to form iron(II)chloride and produce hydrogen gas that mixes with hydrogen sulphide gas.
- State and explain the observations made when a filter paper dipped in Lead(II) ethanoate /Lead (II) nitrate(V) solution is put in a gas jar containing hydrogen sulphide gas.
Observations
Moist Lead(II) ethanoate /Lead (II) nitrate(V) paper turns black.
Explanation
When hydrogen sulphide is bubbled in a metallic salt solution, a metallic sulphide is formed.
All sulphides are insoluble black salts except sodium sulphide, potassium sulphide and ammonium sulphides.
Hydrogen sulphide gas blackens moist Lead (II) ethanoate /Lead (II) nitrate(V) paper .
The gas reacts with Pb2+ in the paper to form black Lead(II)sulphide.
This is the chemical test for the presence of H2S other than the physical smell of rotten eggs.
Chemical equations
Pb2+(aq) + H2S -> PbS + 2H+(aq)
(black)
Fe2+(aq) + H2S -> FeS + 2H+(aq)
(black)
Zn2+(aq) + H2S -> ZnS + 2H+(aq)
(black)
Cu2+(aq) + H2S -> CuS + 2H+(aq)
(black)
2Cu+(aq) + H2S -> Cu2S + 2H+(aq)
(black)
- Dry hydrogen sulphide was ignited as below.
| Dry Hydrogen sulphide gas |
| Flame A |
(i) State the observations made in flame A
Hydrogen sulphide burns in excess air with a blue flame to form sulphur(IV)oxide gas and water.
Chemical equation: 2H2S(g) + 3O2(g) -> 2H2O(l) + 2SO2(g)
Hydrogen sulphide burns in limited air with a blue flame to form sulphur solid and water.
Chemical equation: 2H2S(g) + O2(g) -> 2H2O(l) + 2S(s)
- Hydrogen sulphide is a strong reducing agent that is oxidized to yellow solid sulphur as precipitate. The following experiments illustrate the reducing properties of Hydrogen sulphide.
(a)Turns Orange acidified potassium dichromate(VI) to green
Experiment:
(i)Pass a stream of Hydrogen sulphide gas in a test tube containing acidified potassium dichromate (VI) solution. or;
(ii)Dip a filter paper soaked in acidified potassium dichromate (VI) into a gas jar containing Hydrogen sulphide gas.
Observation:
Orange acidified potassium dichromate (VI) turns to green.
Yellow solid residue.
Explanation:
Hydrogen sulphide gas reduces acidified potassium dichromate(VI) from orange Cr2O72- ions to green Cr3+ ions leaving a yellow solid residue as itself is oxidized to sulphur.
Chemical/ionic equation:
4H2S(aq) + Cr2O72-(aq) +6H+(aq) -> 4S(aq) + 2Cr3+(aq) + 7H2O(l)
This test is used for differentiating Hydrogen sulphide and sulphur (IV)oxide gas.
Sulphur(IV)oxide also reduces acidified potassium dichromate(VI) from orange Cr2O72- ions to green Cr3+ ions without leaving a yellow residue.
(b)Decolorizes acidified potassium manganate(VII)
Experiment:
(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing acidified potassium manganate(VII) solution. or;
(ii)Dip a filter paper soaked in acidified potassium manganate(VII) into a gas jar containing Hydrogen Sulphide gas.
Observation:
Purple acidified potassium manganate(VII) turns to colourless/ acidified potassium manganate(VII) is decolorized.
Yellow solid residue.
Explanation:
Hydrogen sulphide gas reduces acidified potassium manganate(VII) from purple MnO4– ions to green Mn2+ ions leaving a residue as the gas itself is oxidized to sulphur.
Chemical/ionic equation:
5H2S(g) + 2MnO4– (aq) +6H+(aq) -> 5S (s) + 2Mn2+(aq) + 8H2O(l)
(purple) (colourless)
This is another test for differentiating Hydrogen sulphide and Sulphur(IV) oxide gas.
Sulphur(IV) oxide also decolorizes acidified potassium manganate(VII) from purple MnO4– ions to colourless Mn2+ ions leaving no yellow residue.
(c)Decolorizes bromine water
Experiment:
(i)Pass a stream of Hydrogen sulphide gas in a test tube containing bromine water . or;
(ii)Put three drops of bromine water into a gas jar containing Hydrogen sulphide gas. Swirl.
Observation:
Yellow bromine water turns to colourless/ bromine water is decolorized.
Yellow solid residue
Explanation:
Hydrogen sulphide gas reduces yellow bromine water to colourless hydrobromic acid (HBr) leaving a yellow residue as the gas itself is oxidized to sulphur.
Chemical/ionic equation:
H2 S(g) + Br2 (aq) -> S (s) + 2HBr(aq)
(yellow solution) (yellow solid) (colourless)
This is another test for differentiating Hydrogen sulphide and Sulphur(IV) oxide gas.
Sulphur(IV) oxide also decolorizes acidified potassium manganate(VII) from purple MnO4– ions to colourless Mn2+ ions leaving no yellow residue.
(d)Reduces Iron(III) Fe3+ salts to Iron(II) salts Fe2+
Experiment:
(i)Pass a stream of Hydrogen sulphide gas in a test tube containing about 3 cm3 of Iron (III)chloride solution. or;
(ii)Place about 3cm3 of Iron (III)chloride solution into a gas jar containing Hydrogen sulphide gas. Swirl.
Observation:
Yellow/brown Iron (III)chloride solution turns to green.
Yellow solid solid
Explanation:
Hydrogen sulphide gas reduces Iron (III)chloride solution from yellow/brown Fe3+ ions to green Fe2+ ions leaving a yellow residue.The gas is itself oxidized to sulphur.
Chemical/ionic equation:
H2S(aq) + 2Fe3+ (aq) -> S (s) + Fe2+(aq) + 2H+(aq)
(yellow solution) (yellow residue) (green)
(e)Reduces Nitric(V)acid to Nitrogen(IV)oxide gas
Experiment:
(i)Pass a stream of Hydrogen sulphide gas in a test tube containing about 3 cm3 of concentrated nitric(V)acid. or;
(ii)Place about 3cm3 of concentrated nitric(V)acid into a gas jar containing Hydrogen sulphide gas. Swirl.
Observation:
Brown fumes of a gas evolved/produced.
Yellow solid residue
Explanation:
Hydrogen sulphide gas reduces concentrated nitric(V)acid to brown nitrogen(IV)oxide gas itself oxidized to yellow sulphur.
Chemical/ionic equation:
H2S(g) + 2HNO3 (l) -> 2H2O(l) + S (s) + 2NO2 (g)
(yellow residue) (brown fumes)
(f)Reduces sulphuric(VI)acid to Sulphur
Experiment:
(i)Pass a stream of Hydrogen sulphide gas in a test tube containing about 3 cm3 of concentrated sulphuric(VI)acid. or;
(ii)Place about 3cm3 of concentrated sulphuric (VI) acid into a gas jar containing Hydrogen sulphide gas. Swirl.
Observation:
Yellow solid residue
Explanation:
Hydrogen sulphide gas reduces concentrated sulphuric(VI)acid to yellow sulphur.
Chemical/ionic equation:
3H2S(g) + H2SO4 (l) -> 4H2O(l) + 4S (s)
(yellow residue)
(g)Reduces Hydrogen peroxide to water
Experiment:
(i)Pass a stream of Hydrogen sulphide gas in a test tube containing about 3 cm3 of 20 volume hydrogen peroxide.
Observation:
Yellow solid residue
Explanation:
Hydrogen sulphide gas reduces 20 volume hydrogen peroxide to water and itself oxidized to yellow sulphur
Chemical/ionic equation:
H2S(g) + H2O2 (l) -> 2H2O(l) + S (s)
(yellow residue)
8.Name the salt formed when:
(i)equal volumes of equimolar hydrogen sulphide neutralizes sodium hydroxide solution:
Sodium hydrogen sulphide
Chemical/ionic equation:
H2S(g) + NaOH (l) -> H2O(l) + NaHS (aq)
(ii) hydrogen sulphide neutralizes excess concentrated sodium hydroxide solution:
Sodium sulphide
Chemical/ionic equation:
H2S(g) + 2NaOH (l) -> 2H2O(l) + Na2S (aq)
Practice
Hydrogen sulphide gas was bubbled into a solution of metallic nitrate(V)salts as in the flow chart below
| Hydrogen sulphide |
| Blue solution |
| Brown solution |
| Black solid |
| Green solution |
(a)Name the black solid Copper(II)sulphide
(b)Identify the cation responsible for the formation of:
- Blue solution Cu2+(aq)
- Green solution Fe2+(aq)
III. Brown solution Fe3+(aq)
(c)Using acidified potassium dichromate(VI) describe how you would differentiate between sulphur(IV)Oxide and hydrogen sulphide
-Bubble the gases in separate test tubes containing acidified Potassium dichromate(VI) solution.
-Both changes the Orange colour of acidified Potassium dichromate(VI) solution to green.
-Yellow solid residue/deposit is formed with Hydrogen sulphide
Chemical/ionic equation:
4H2S(aq) + Cr2O72-(aq) +6H+(aq) -> 4S(aq) + 2Cr3+(aq) + 7H2O(l)
3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l)
(d)State and explain the observations made if a burning splint is introduced at the mouth of a hydrogen sulphide generator.
ObservationGas continues burning with a blue flame
Explanation: Hydrogen sulphide burns in excess air with a blue flame to form sulphur(IV)oxide gas and water.
Chemical equation: 2H2S(g)+ 3O2(g) -> 2H2O(l) + 2SO2 (g)
(v)Sulphate (VI) (SO42-)and Sulphate(IV) (SO32-) salts
- Sulphate (VI) (SO42-) salts are normal and acid salts derived from Sulphuric (VI)acid H2SO4.
- Sulphate(IV) (SO32-) salts are normal and acid salts derived from Sulphuric (IV)acid H2SO3.
- Sulphuric (VI)acid H2SO4 is formed when sulphur(VI)oxide gas is bubbled in water.
The acid exist as a dibasic acid with two ionisable hydrogen. It forms therefore the Sulphate (VI) (SO42-) and hydrogen sulphate (VI) (HSO4–) salts.
i.e.
H2SO4 (aq) -> 2H+(aq) + SO42-(aq)
H2SO4 (aq) -> H+(aq) + HSO4 –(aq)
All Sulphate (VI) (SO42-) salts dissolve in water/are soluble except Calcium (II) sulphate (VI) (CaSO4), Barium (II) sulphate (VI) (BaSO4) and Lead (II) sulphate (VI) (PbSO4)
All Hydrogen sulphate (VI) (HSO3–) salts exist in solution/dissolved in water. Sodium (I) hydrogen sulphate (VI) (NaHSO4), Potassium (I) hydrogen sulphate (VI) (KHSO4) and Ammonium hydrogen sulphate (VI) (NH4HSO4) exist also as solids.
Other Hydrogen sulphate (VI) (HSO4–) salts do not exist except those of Calcium (II) hydrogen sulphate (VI) (Ca (HSO4)2) and Magnesium (II) hydrogen sulphate (VI) (Mg (HSO4)2).
- Sulphuric (IV)acid H2SO3 is formed when sulphur(IV)oxide gas is bubbled in water.
The acid exist as a dibasic acid with two ionisable hydrogen. It forms therefore the Sulphate (IV) (SO32-) and hydrogen sulphate (VI) (HSO4–) salts.
i.e.
H2SO3 (aq) -> 2H+(aq) + SO32-(aq)
H2SO3 (aq) -> H+(aq) + HSO3 –(aq)
All Sulphate (IV) (SO32-) salts dissolve in water/are soluble except Calcium (II) sulphate (IV) (CaSO3), Barium (II) sulphate (IV) (BaSO3) and Lead (II) sulphate (IV) (PbSO3)
All Hydrogen sulphate (IV) (HSO3–) salts exist in solution/dissolved in water. Sodium (I) hydrogen sulphate (IV) (NaHSO3), Potassium (I) hydrogen sulphate (IV) (KHSO3) and Ammonium hydrogen sulphate (IV) (NH4HSO3) exist also as solids.
Other Hydrogen sulphate (IV) (HSO3–) salts do not exist except those of Calcium (II) hydrogen sulphate (IV) (Ca (HSO3)2) and Magnesium (II) hydrogen sulphate (IV) (Mg (HSO3)2).
5.The following experiments show the effect of heat on sulphate(VI) (SO42-)and sulphate(IV) (SO32-) salts:
Experiment:
In a clean dry test tube place separately about 1.0g of :
Zinc(II)sulphate (VI), Iron(II)sulphate(VI), Copper(II)sulphate(VI),Sodium (I) sulphate (VI), Sodium (I) sulphate (IV).Heat gently then strongly. Test any gases produced using litmus papers.
Observations:
-Colourless droplets of liquid forms on the cooler parts of the test tube in all cases.
-White solid residue is left in case of Zinc (II)sulphate(VI),Sodium (I) sulphate (VI) and Sodium (I) sulphate (IV).
-Colour changes from green to brown /yellow in case of Iron (II)sulphate(VI)
-Colour changes from blue to white then black in case of Copper (II) sulphate (VI)
-Blue litmus paper remain and blue and red litmus paper remain red in case of Zinc(II)sulphate(VI), Sodium (I) sulphate (VI) and Sodium (I) sulphate (IV)
-Blue litmus paper turns red and red litmus paper remain red in case of Iron (II)sulphate(VI) and Copper (II) sulphate (VI).
Explanation
(i)All Sulphate (VI) (SO42-) salts exist as hydrated salts with water of crystallization that condenses and collects on cooler parts of test tube as a colourless liquid on gentle heating. e.g.
K2SO4.10H2O(s) -> K2SO4(s) + 10H2O(l)
Na2SO4.10H2O(s) -> Na2SO4(s) + 10H2O(l)
MgSO4.7H2O(s) -> MgSO4(s) + 7H2O(l)
CaSO4.7H2O(s) -> CaSO4(s) + 7H2O(l)
ZnSO4.7H2O(s) -> ZnSO4(s) + 7H2O(l)
FeSO4.7H2O(s) -> FeSO4(s) + 7H2O(l)
Al2(SO4)3.6H2O(s) -> Al2(SO4)3 (s) + 6H2O(l)
CuSO4.5H2O(s) -> CuSO4(s) + 5H2O(l)
All Sulphate (VI) (SO42-) salts do not decompose on heating except Iron (II) sulphate (VI) and Copper (II) sulphate (VI).
(i)Iron (II) sulphate (VI) decomposes on strong heating to produce acidic sulphur (IV)oxide and sulphur(VI)oxide gases. Iron(III)oxide is formed as a brown /yellow residue.
Chemical equation
2FeSO4 (s) -> Fe2O3(s) + SO2(g) + SO3(g)
This reaction is used for the school laboratory preparation of small amount of sulphur(VI)oxide gas.
Sulphur (VI) oxide readily /easily solidifies as white silky needles when the mixture is passed through freezing mixture/ice cold water.
Sulphur (IV) oxide does not.
(ii) Copper(II)sulphate(VI) decomposes on strong heating to black copper (II) oxide and Sulphur (VI) oxide gas.
Chemical equation
2CuSO4 (s) -> CuO(s) + SO3(g)
This reaction is used for the school laboratory preparation of small amount of sulphur(VI)oxide gas.
- The following experiments show the test for the presence of sulphate (VI) (SO42-)and sulphate(IV) (SO32-) ions in a sample of a salt/compound:
Experiments/Observations:
(a)Using Lead(II)nitrate(V)
- To about 5cm3 of a salt solution in a test tube add four drops of Lead(II)nitrate(V)solution. Preserve.
| Observation | Inference |
| White precipitate/ppt | SO42- , SO32- , CO32- , Cl– ions |
- To the preserved sample in (I) above, add six drops of 2M nitric(V) acid . Preserve.
Observation 1
| Observation | Inference |
| White precipitate/ppt persists | SO42- , Cl– ions |
Observation 2
| Observation | Inference |
| White precipitate/ppt dissolves | SO32- , CO32- , ions |
III.(a)To the preserved sample observation 1 in (II) above, Heat to boil.
Observation 1
| Observation | Inference |
| White precipitate/ppt persists on boiling | SO42- ions |
Observation 2
| Observation | Inference |
| White precipitate/ppt dissolves on boiling | Cl – ions |
.(b)To the preserved sample observation 2 in (II) above, add 4 drops of acidified potassium manganate(VII) /dichromate(VI).
Observation 1
| Observation | Inference |
| (i)acidified potassium manganate(VII)decolorized
(ii)Orange colour of acidified potassium dichromate(VI) turns to green |
SO32- ions
|
Observation 2
| Observation | Inference |
| (i)acidified potassium manganate(VII) not decolorized
(ii)Orange colour of acidified potassium dichromate(VI) does not turns to green
|
CO32- ions
|
Experiments/Observations:
(b)Using Barium(II)nitrate(V)/ Barium(II)chloride
- To about 5cm3 of a salt solution in a test tube add four drops of Barium(II) nitrate (V) / Barium(II)chloride. Preserve.
| Observation | Inference |
| White precipitate/ppt | SO42- , SO32- , CO32- ions |
- To the preserved sample in (I) above, add six drops of 2M nitric(V) acid . Preserve.
Observation 1
| Observation | Inference |
| White precipitate/ppt persists | SO42- , ions |
Observation 2
| Observation | Inference |
| White precipitate/ppt dissolves | SO32- , CO32- , ions |
III.To the preserved sample observation 2 in (II) above, add 4 drops of acidified potassium manganate(VII) /dichromate(VI).
Observation 1
| Observation | Inference |
| (i)acidified potassium manganate(VII)decolorized
(ii)Orange colour of acidified potassium dichromate(VI) turns to green |
SO32- ions
|
Observation 2
| Observation | Inference |
| (i)acidified potassium manganate(VII) not decolorized
(ii)Orange colour of acidified potassium dichromate(VI) does not turns to green
|
CO32- ions
|
Explanations
Using Lead(II)nitrate(V)
(i)Lead(II)nitrate(V) solution reacts with chlorides(Cl–), Sulphate (VI) salts (SO42- ), Sulphate (IV)salts (SO32-) and carbonates(CO32-) to form the insoluble white precipitate of Lead(II)chloride, Lead(II)sulphate(VI), Lead(II) sulphate (IV) and Lead(II)carbonate(IV).
Chemical/ionic equation:
Pb2+(aq) + Cl– (aq) -> PbCl2(s)
Pb2+(aq) + SO42+ (aq) -> PbSO4 (s)
Pb2+(aq) + SO32+ (aq) -> PbSO3 (s)
Pb2+(aq) + CO32+ (aq) -> PbCO3 (s)
(ii)When the insoluble precipitates are acidified with nitric(V) acid,
– Lead(II)chloride and Lead(II)sulphate(VI) do not react with the acid and thus their white precipitates remain/ persists.
– Lead(II) sulphate (IV) and Lead(II)carbonate(IV) reacts with the acid to form soluble Lead(II) nitrate (V) and produce/effervesces/fizzes/bubbles out sulphur(IV)oxide and carbon(IV)oxide gases respectively.
. Chemical/ionic equation:
PbSO3 (s) + 2H+(aq) -> H2 O (l) + Pb2+(aq) + SO2 (g)
PbCO3 (s) + 2H+(aq) -> H2 O (l) + Pb2+(aq) + CO2 (g)
(iii)When Lead(II)chloride and Lead(II)sulphate(VI) are heated/warmed;
– Lead(II)chloride dissolves in hot water/on boiling(recrystallizes on cooling)
– Lead(II)sulphate(VI) do not dissolve in hot water thus its white precipitate persists/remains on heating/boiling.
(iv)When sulphur(IV)oxide and carbon(IV)oxide gases are produced;
– sulphur(IV)oxide will decolorize acidified potassium manganate(VII) and / or Orange colour of acidified potassium dichromate(VI) will turns to green. Carbon(IV)oxide will not.
Chemical equation:
5SO32-(aq) + 2MnO4– (aq) +6H+(aq) -> 5SO42-(aq) + 2Mn2+(aq) + 3H2O(l)
(purple) (colourless)
3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l)
(Orange) (green)
– Carbon(IV)oxide forms an insoluble white precipitate of calcium carbonate if three drops of lime water are added into the reaction test tube when effervescence is taking place. Sulphur(IV)oxide will not.
Chemical equation:
Ca(OH)2(aq) + CO2 (g) -> CaCO3(s) + H2O(l)
These tests should be done immediately after acidifying to ensure the gases produced react with the oxidizing agents/lime water.
Using Barium(II)nitrate(V)/ Barium(II)Chloride
(i)Barium(II)nitrate(V) and/ or Barium(II)chloride solution reacts with Sulphate (VI) salts (SO42- ), Sulphate (IV)salts (SO32-) and carbonates(CO32-) to form the insoluble white precipitate of Barium(II)sulphate(VI), Barium(II) sulphate (IV) and Barium(II)carbonate(IV).
Chemical/ionic equation:
Ba2+(aq) + SO42+ (aq) -> BaSO4 (s)
Ba2+(aq) + SO32+ (aq) -> BaSO3 (s)
Ba2+(aq) + CO32+ (aq) -> BaCO3 (s)
(ii)When the insoluble precipitates are acidified with nitric(V) acid,
– Barium (II)sulphate(VI) do not react with the acid and thus its white precipitates remain/ persists.
– Barium(II) sulphate (IV) and Barium(II)carbonate(IV) reacts with the acid to form soluble Barium(II) nitrate (V) and produce /effervesces /fizzes/ bubbles out sulphur(IV)oxide and carbon(IV)oxide gases respectively.
. Chemical/ionic equation:
BaSO3 (s) + 2H+(aq) -> H2 O (l) + Ba2+(aq) + SO2 (g)
BaCO3 (s) + 2H+(aq) -> H2 O (l) + Ba2+(aq) + CO2 (g)
(iii) When sulphur(IV)oxide and carbon(IV)oxide gases are produced;
– sulphur(IV)oxide will decolorize acidified potassium manganate(VII) and / or Orange colour of acidified potassium dichromate(VI) will turns to green. Carbon(IV)oxide will not.
Chemical equation:
5SO32-(aq) + 2MnO4– (aq) +6H+(aq) -> 5SO42-(aq) + 2Mn2+(aq) + 3H2O(l)
(purple) (colourless)
3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l)
(Orange) (green)
– Carbon(IV)oxide forms an insoluble white precipitate of calcium carbonate if three drops of lime water are added into the reaction test tube when effervescence is taking place. Sulphur(IV)oxide will not.
Chemical equation:
Ca(OH)2(aq) + CO2 (g) -> CaCO3(s) + H2O(l)
These tests should be done immediately after acidifying to ensure the gases produced react with the oxidizing agents/lime water.
Summary test for Sulphate (VI) (SO42-)and Sulphate(IV) (SO32-) salts
| Heat to boil |
| White ppt persist
on heating in SO42- in CO32-
|
| White ppt dissolves on heating in Cl–
|
| Acidified KMnO4 decolorized in SO32- |
| White ppt with lime water in CO32- |
| Acidified KMnO4
K2Cr2O7 / Lime water |
| white ppt persist /remains in SO32- and CO32-
|
| white ppt dissolves in SO32- and CO32- |
| Dilute nitric(V) acid |
| Lead(II)nitrate(V) |
| White precipitates of Cl–, SO42- , SO32- and CO32- |
| Unknown salt |
Practice revision question
- Study the flow chart below and use it to answer the questions that follow
| Sodium salt solution |
| Acidified K2Cr2O7 |
| Gas G and colour of solution
changes orange to green |
| Colourless solution B |
| Barium nitrate(VI) (VI)(aq) |
| White precipitate |
| Dilute HCl |
(a)Identify the:
I: Sodium salt solution
Sodium sulphate(IV)/Na2SO3
II: White precipitate
Barium sulphate(IV)/BaSO3
III: Gas G
Sulphur (IV)Oxide /SO2
IV: Colourless solution H
Barium chloride /BaCl2
(b)Write an ionic equation for the formation of: I.White precipitate
Ionic equation Ba2+(aq) + SO32-(aq) -> BaSO3(s)
II.Gas G
Ionic equation BaSO3(s)+ 2H+(aq) -> SO2 (g) + H2O (l) + Ba2+(aq)
III. Green solution from the orange solution
3SO32-(aq) + Cr2O72-(aq) +8H+(aq) -> 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l)
(Orange) (green)
- Study the flow chart below and answer the questions that follow.
(i)Write equation for the reaction taking place at:
I.The roasting furnace (1mk)
2FeS2 (s) + 5O2 (g) -> 2FeO(s) + 4SO2 (g)
II.The absorption tower (1mk)
H2SO4 (l) + SO3 (g) -> H2S2O7(l)
III.The diluter (1mk)
H2S2O7(l) + H2 O(l) -> 2H2SO4 (l)
(ii)The reaction taking place in chamber K is
SO2 (g) + 1/2O2 (g) SO3 (g)
- Explain why it is necessary to use excess air in chamber K
To ensure all the SO2 reacts
II.Name another substance used in chamber K
Vanadium(V)oxide
3.(a)Describe a chemical test that can be used to differentiate between sodium sulphate (IV) and sodium sulphate (VI).
Add acidified Barium nitrate(V)/chloride.
White precipitate formed with sodium sulphate (VI)
No white precipitate formed with sodium sulphate (IV)
(b)Calculate the volume of sulphur (IV) oxide formed when 120 kg of copper is reacted with excess concentrated sulphuric(VI)acid.(Cu = 63.5 ,1 mole of a gas at s.t.p =22.4dm3)
Chemical equation
Cu(s) + 2H2SO4(l) -> CuSO4(aq)+ H2O(l) + SO2 (g)
Mole ratio Cu(s: SO2 (g) = 1:1
Method 1
1 Mole Cu =63.5 g -> 1 mole SO2 = 22.4dm3
(120 x 1000) g -> (120 x 1000) g x 22.4.dm3)
63.5 g
= 42330.7087
Method 2
Moles of Cu = ( 120 x 1000 ) g =1889.7639 moles
63.5
Moles SO2 = Moles of Cu = 1889.7639 moles
Volume of SO2 = Mole x molar gas volume = (1889.7639 moles x 22.4)
= 42330.7114
(a)Identify the:
(i)cation responsible for the green solution T
Cr3+
(ii)possible anions present in white precipitate R
CO32-, SO32-, SO42-
(b)Name gas V
Sulphur (IV)oxide
(c)Write a possible ionic equation for the formation of white precipitate R.
Ba2+ (aq) + CO32- (aq) -> BaCO3(s)
Ba2+ (aq) + SO32- (aq) -> BaSO3(s)
Ba2+ (aq) + SO42- (aq) -> BaSO4(s)
|
UPGRADE CHEMISTRY
FORM 3
Chemistry of CHLORINE
Comprehensive tutorial notes
MUTHOMI S.G 0720096206
|
CONTENTS
- Chlorine.
- The Halogens.
- Hydrogen chloride.
- Chloride (Cl–) salts.
- Comprehensive revision questions.
A.CHLORINE
Chlorine is a non-metallic element in group VII (Group 17) of the periodic table. It has electronic configuration 2:8:7. It gains one valence election to form stable Cl–ion, it belongs to the chemical family of halogens.
Occurrence
-As Brine-concentration sodium chloride solution dissolved in salty seas water, oceans and lakes e.g. Lake Magadi in Kenya is very salty.
-As rock-salt solid sodium chloride crystals in the earths crust all over the world.
- B) Preparation
Chlorine gas may be prepared in the school laboratory from the following:
a)Heating solid Manganese (iv) Oxide and Concentrated Hydrochloric acid.
- b) Heating Lead (IV) Oxide and concentrated hydrochloric acid.
c)Reacting Potassium Manganate (VII) with concentrated Hydrochloric acid
d)Reacting Potassium /sodium Dichromate (VI) Acid with Concentrated Hydrochloric acid.
Set up of school laboratory preparation of chlorine.
- c) Properties of chlorine. (Questions)
- What is the colour of chlorine?
Pale green.
- Describe the smell of chlorine.
Pungent irritating smell.
- What method is used in collection of chlorine gas explain.
-Downward delivery.
-Chlorine is 11/2 denser than air.
4.(i) What is the purpose of concentrated sulphuric (VI) acid.
-To dry the gas.
(ii) Name two other substances that can be used in place of concentrated sulphuric (VI) acid.
-Calcium chloride
-Silica gel
(iii) Name a substance that cannot be used in place of concentrated sulphuric (VI) acid explain.
-Calcium oxide reacts with chlorine.
5.(a)Write three possible reactions between concentrated hydrochloric acid and the oxidizing agents.
- 2KMnO4(s) +16HCl(aq) → 2KCl(aq)+2MnCl2(aq) + 8H2O(l) + 5Cl2(g)
2.K2Cr2O7(s) +14HCl(aq) → 2KCl(aq) + 2CrCl3(aq) + 7H2O(l) + 3Cl2(g)
3.Na2Cr2O7(s) + 14HCl(aq) → 2NaCl(aq) + CrCl3(aq) + 7H2O(l) + 3Cl2(g)
4.PbO2(s) + 4HCl(aq) → PbCl2(aq) + Cl2(g) + 2H2O(l)
5.MnO2(s)+ 4HCl(aq) → MnCl2(aq) + Cl2(g) + 2H2O(l)
(b) Why is Hydrochloric acid used in all the above cases?
Oxidizing agents KMnO4/PbO2/MnO2/K2Cr2O/Na2Cr2O7 readily oxidize hydrochloric acid to chlorine themselves reduced to their chlorides.
Generally:
2HCl (aq) + [O] → Cl2 (g) + H2O (l)
(From oxidizing agent)
- State and explain the observation made when chlorine is bubbled in water.
Observation
-Pale yellow colour of chlorine fades.
-yellow solution formed.
Explanation
Chlorine dissolves then reacts with water to form yellow chlorine water. Chlorine water is chemically a mixture of hydrochloric acid and chloric(I)acid (hypochlorous acid).
A mixture of hydrochloric acid and chloric(I)acid (hypochlorous acid) is commonly called Chlorine water
Chemical equation:
Cl2(g) + H2O(l) → HCl(aq) + HClO(aq)
- Chlorine water in a boiling tube inverted into a trough was exposed to sunlight for two hours. Using a well labeled diagram show and explain the observations made.
Chlorine (I) acid is an unstable compound.
After two hours the chloric (I) acid in chlorine water decomposes to hydrochloric acid and releases oxygen gas. This reaction takes place in sunlight.
Chemical equation
2HOCl(aq) → 2HCl(aq) + O2 (g)
- State and explain the observation made when chlorine gas is bubbled in gas jar containing damp/wet/moist litmus papers.
Observation
The blue litmus turns red then both the red/blue litmus papers are bleached/decolourized.
Explanation
Chlorine reacts with water in the litmus papers to form acidic hydrochloric acid and chloric (l) acid that turns blue litmus papers red.
Chemical Equation
Cl2(g) + H2O(l) → HCl(aq) + HClO(aq)
Explanation
Unstable chloric (I) acid oxidizes the dye/colured litmus paper to colourless material
Chemical Equation
HClO(aq) + dye → HCl(aq) + (dye + O)
(coloured) (colourless)
Or:
HClO(aq) + dye-O → HCl(aq) + dye
(coloured) (colourless)
NB Chlorine does not therefore bleach/decolourize dry litmus paper/dye because chloric(I) acid cannot be formed in absence of water.
- Blue litmus papers were put in a flask containing cold dilute sodium hydroxide. Chlorine gas was bubbled into the solution. State and explain the observations made.
Observation
blue litmus papers were bleached /decolorized.
Pale green colour of chlorine fades.
Explanation
-Sodium hydroxide reacts with chlorine to form sodium chloride and sodium hypochlorite. Sodium hypochlorite bleaches dyes by oxidation.
Chemical Equation
Cl2 + 2NaOH(aq) → NaCl(aq) + NaClO(aq) + H2O
NaClO(aq) + dye → NaCl(aq) + (dye + O)
(coloured) (Colourless)
NaClO(aq) + (dye-O) → NaCl(aq) + dye
(Coloured) (Colourless)
10.Blue litmus papers were put in flask containing hot concentrated sodium hydroxide. Chlorine gas was bubbled into the solution. State and explain the observations made.
Observation.
blue litmus papers were bleached.
Pale green colour of chlorine fades.
Explanation
Hot concentrated sodium hydroxide reacts with chlorine to form sodium chloride and sodium chloride (V).Sodium chlorate (V) bleaches by oxidation.
Chemical equation
2Cl2(g) + 4NaOH(aq) → 3NaCl(aq) + NaClO3(aq) + H2O(l)
NaClO3(aq) + 3(dyes) → NaCl(aq) + 3(dye + O)
NaClO3(aq) + 3(dyes-O) → NaCl(aq) + 3 dyes
NaClO3 is also a weed killer
- State three main use of chlorine gas.
-Manufacture of polyvinyl chloride (P.V.C) // polychloroethene pipes.
-Manufacture of hydrochloric acid used in “Pickling” of metals.
-Manufacture of bleaching agents
-Chlorination of water to kill germs.
- The diagram below shows the effect of chlorine on heated iron wool.
Method I
Method II
| Suction pump |
| HEAT |
| Chlorine gas |
| Concentrated sodium/potassium hydroxide. |
| Iron/Aluminium |
| Aluminium(III)oxide/iron(III)oxide |
- a) Identify a suitable drying agent to dry chlorine gas.
-Conc. H2SO4 / Concentrated sulphuric (VI) acid.
-Anhydrous Calcium (II) Chloride.
-Silica gel
- b) State and explain the observations made in combustion tube in method I and II
Observation
Iron glows red hot
Brown crystals are formed
Explanation
Iron reacts with chlorine to form dark brown crystals of iron (III) Chloride.
This reaction is exothermic and requires no farther heating once started.
Iron (III) Chloride sublimes away ensuring the unreacted Iron completely reacts with chlorine gas.
Chemical equation
2Fe(s) + 3Cl2 (g) → 2FeCl3(g)
- c) (i) Why is the brown solid collected at the point as shown in method I and II.
-Heated iron (III) Chloride crystals sublime to gas and solidify on the cooler parts.
(ii) Name another metal that can be used in place of iron to react with chlorine and collected at similar point on heating explain.
Metal Aluminum
Explanation
Aluminum reacts with chlorine to form a while sublimate of aluminum (III) chloride at the cooler parts
Chemical equation
2Al(s) + 3Cl2(g) → 2AlCl3(s/g)
- d) What is the purpose of suction pump?
To pull the gaseous products into the set up.
- e) What is the function of:
(i) Sodium hydroxide in method II. Explain.
To absorb poisonous/toxic excess unreacted chlorine gas.
Sodium hydroxide reacts with chlorine to form sodium chloride, Sodium hypochlorite and water.
Chemical equation:
2NaOH(aq) + Cl2(g) → NaCl(aq) + NaClO(aq) + H2O(l)
2KOH(aq) + Cl2(g) → KCl(aq) + KClO(aq) + H2O(l)
(ii) Anhydrous calcium chloride/calcium oxide in method I. Explain.
To absorb moisture/water in the set up to prevent it from hydrolyzing iron (III) chloride/aluminium oxide.
Explanation
Iron (III) chloride and Aluminium chloride fumes and reacts with small traces of water to form a solution of iron (III) hydroxide/aluminium hydroxide and hydrogen chloride gas.
Chemical equation
FeCl3(s) + 3HCl(aq) → Fe(OH)3(aq) + 3HCl(g)
AlCl3(s) + 3HCl(aq) → Al(OH)3(aq) + 3HCl(g)
- f) Based on e (i) and (ii) above what precaution should be made in:
(i) method II to ensure correct results.
-Tube B should be completely dry to prevent hydrolysis of iron (III) Chloride to iron (III) hydroxide.
(ii) Carrying out method I
-Should be done in a fume chamber or in the open because chlorine gas is poisonous/toxic.
(g) Name another substance that can be used place of Sodium hydroxide in method I
Potassium hydroxide
(h) Calcium oxide cannot be used in place of calcium chloride during preparation of chlorine. Explain.
Calcium oxide is a base. It reacts /absorbs water to form calcium hydroxide solution.
Calcium hydroxide reacts with chlorine to form a mixture of calcium chloride and calcium hypochlorite.
Chemical equation
2Ca (OH)2(aq) + 2Cl2(g) → CaCl2(aq) + CaOCl2(aq) + H2O(l)
- (a)State and explain the observation made when a piece of burning magnesium ribbon is lowered in a gas jar containing chlorine gas.
-Magnesium ribbon continues burning with a bright flame.
-White solid formed.
-Pale yellow colour of chlorine fades
Explanation:
Magnesium reacts with chlorine forming a white solid of magnesium chloride.
Chemical equation
Mg(s) + Cl2(g) → MgCl2(s)
(b) Write the equation for the reaction that takes place if zinc is used.
Zn(s) + Cl2(g) → ZnCl2(s)
- Burning phosphorus was lowered in a gas jar containing chlorine gas.
- a) State the observations made.
-Phosphorus continues to burn.
-Dense white fumes formed.
-Pale green colour of chlorine fades.
- b) Write two possible equations that take place.
P4(s) + 6Cl2(g) → 4 PCl3(s)
P4(s) + 10Cl2(g) → 4 PCl3(s)
(c) State two reasons why the deflagrating spoon with rid/cover should be used.
-Chlorine in the gas jar is poisonous/toxic.
-Burning phosphorus produces poisonous/toxic phosphorus (III) chloride // phosphorus (V) chloride.
-Ensure the reaction is not affected by air/oxygen from the atmosphere.
(d) After the reaction is complete, 2cm3 of distilled water were added. The solution formed was tested with both blue and red litmus papers.
(i) State the observations made.
-Blue litmus paper turns red
-Red litmus paper remain red
(ii) Explain the observation made in d(i) above
-Phosphoric (V) Chloride hydrolyze in water to phosphoric (V) acid and produce hydrogen chloride gas. Both hydrogen chloride and phosphoric (V) acid are acidic.
Chemical equation
PCl5 (l) + 4H2O(l) → H3PO4 (aq) + 5HCl(g)
- State and explain the observations made when gas jar containing chlorine is inverted over another containing hydrogen sulphide gas.
Observation
Yellow solid formed.
Pale colour of chlorine fades
Explanation
Chlorine oxidizes hydrogen sulphide to sulphur itself reduced to hydrogen chloride gas. A little water catalyzes the reaction.
Chemical equation
H2S(g) + Cl2(g) → S(s) + HCl(g)
(yellow solid) (White Fume)
- Chlorine was bubbled in aqueous ammonia solution in a beaker state and explain the observation made.
Observation:
White fumes evolved.
Pale green colour of chlorine fades.
Explanation
Chlorine reacts with ammonia gas to form a dense white fume of ammonia chloride and Nitrogen gas is produced.
Chemical equation
8NH3(g) + 3Cl2(g) → 6NH4Cl(s) + N2(g)
- (a) Dry gas was bubbled in cold dilute sodium hydroxide solution. Explain the observations made:
Observation
Pale green colour of chlorine fades.
Pale yellow solution is formed.
Explanation
Chlorine reacts withhot concentrated sodiumsodium hydroxide / Potassium hydroxide solution to form pale yellow solution of metal chlorate (V) and chlorides of the metal
Chemical equation
Cl2(g) + 2NaOH → NaClO(aq) + NaCl(aq) + H2O(l)
(sodium hydroxide) (Sodium Chlorate (I))
Cl2(g) + 2KOH → KClO(aq) + NaCl(aq) + H2O(l)
(Potassium hydroxide) (Potassium Chlorate (I))
(b)The experiment in 17(a) was repeated with hot concentrated sodium hydroxide solution. Explain the observation made.
Observation
Pale green colour of chlorine fades.
Pale yellow solution is formed.
Explanation
Chlorine reacts with hot concentrated Sodium hydroxide/Potassium hydroxide solution to form pale yellow solution of metal chlorate (v) and chlorides of metals.
Chemical equation
3Cl2(g) + 6NaOH(aq) → NaClO3 (aq) + 5NaCl(aq) + 3H2O(l)
(Sodium hydroxide) (Sodium Chlorate (V))
3Cl2(g) + 6KOH(aq) → KClO3 (aq) + 5KCl(aq) + 3H2O(l)
(Potassium hydroxide) (Potassium Chlorate (V))
The products formed when chlorine reacts with alkalis depend thus on temperature and the concentration of alkalis.
(c) (i) Write the equation for the formation of calcium chlorite (I) and calcium chlorate (V).
2Ca (OH)2(aq) + 2Cl2(g) → CaCl2(aq) + CaOCl2(aq) + H2O(l)
(Calcium hydroxide) (Calcium Chlorate(I))
(Cold/dilute)
Ca (OH)2(aq) + Cl2(g) → CaCl2(aq) + Ca(ClO3)2(aq) + H2O(l)
(Calcium Chlorate(V))
B: THE HALOGENS
- a) What are halogens?
These are elements in group VII of the periodic table. They include:
| Element | Symbol | Atomic number | Electric configuration | Charge of ion | Valency | State at Room Temperature |
| Fluorine
Chlorine
Bromine
Iodine
Astatine |
F
Cl
Br
I
At |
9
17
35
53
85 |
2:7
2:8:7
2:8:18:7
2:8:18:18:7
2:8:18:32:18:7 |
F–
Cl–
Br–
I–
At– |
1
1
1
1
1 |
Pale yellow gas
Pale green gas Red liquid
Grey Solid
Radioactive |
- b) Compare the atomic radius and ionic radius of chloride ion and chlorine. Explain.
The radius of chlorine is smaller than the ionic radius o the chloride ion.
Effective nucleus attraction on outer energy level in chloride ion is less than chlorine atom because of extra gained electron gained electron that repelled thus causes the outer energy level to expand/increase.
- c) Compare the atomic radius of chlorine and fluorine Explain.
Atomic radius of Fluorine is smaller than that of chlorine.
Chlorine has more energy levels than fluorine occupied by more electrons.
- d) Chlorine is a gas, Bromine is a liquid, Iodine is a solid. Explain the above observations.
–Bromine, Chlorine and iodine exists as diatomic molecules bonded by strong covalent bond. Each molecule is joined to the other by weak intermolecular forces/ Van-der-waals forces.
-The strength of intermolecular/Van-der-waals forces of attraction increase with increase in molecular size/atomic radius.Iodine has therefore the largest atomic radius and thus strongest intermolecular forces to make it a solid.
- e) (i) What is electronegativity?
Electronegativity is the tendency/ease of acquiring /gaining electrons by an element during chemical reaction.
It is measured using Pauling’s scale.
Fluorine with Pauling scale 4.0 is the most electronegative element in the periodic table and thus the highest tendency to acquire/gain extra electron.
(ii) The table below shows the electronegativity of the halogens.
| Halogen | F | Cl | Br | I | At |
| Electronegativity (Pauling’s scale) | 4.0 | 3.0 | 2.8 | 2.5 | 2.2 |
Explain the trend in electronegativity of the halogens.
Decrease down the group from fluorine to Astatine
Atomic radius increase down the group decreasing electron – attracting power down the group from fluorine to astatine.
(f) (i)What is electron affinity
Electron affinity is the energy required to gain an electron in an atom of an element in its gaseous state.
(ii) Study the table below showing the election affinity of halogens for the process x + e → x–
| Halogen | F | Cl | Br | I |
| Electron affinity kJmole-1 | -333 | -364 | -342 | -295 |
(iii) Explain the trend in electron affinity of the halogens.
-Decrease down the group
-Atomic radius of halogens increase down the group thus incoming/gained electron is attracted less strongly by the progressively larger atoms with a decreasing effective nuclear charge on outer energy level
(iv) Which is a move stable ion Cl– or Br – explain?
-Cl– ion.
-Has a more negative/exothermic electron affinity than Br–
(v) Differentiate between electron affinity and:
- Ionization energy.
Ionization energy is the energy required to lose /donate an electron in an atom of an element in its gaseous state while electron affinity is the energy required to gain/acquire extra electron by an atom of an element in its gaseous state.
Both are measured in kilojoules per mole.
- Electronegativity.
-Electron affinity is the energy required to gain an electron in an atom of an element in gaseous state. It involves the process:
X(g) + e → X–(g)
Electronegativity is the ease/tendency of gaining/ acquiring electrons by an element during chemical reactions.
It does not involve use of energy but theoretical arbitrary Pauling’s scale of measurements.
(g) (i) 5cm3 of sodium chloride, Sodium bromide and Sodium iodide solutions were put separately in test tubes. 5 drops of chlorine water was added to each test tube: state and explain the observation made.
Observation
Yellow colour of chlorine water fades in all test tubes expect with sodium chloride.
-Coloured Solution formed.
Explanation
Chlorine is more electronegative than bromine and iodine. On adding chlorine water, bromine and Iodine are displaced from their solutions by chlorine.
(ii) The experiment in g (i) was repeated with 5 drops of bromine water instead of chlorine water .explain the observation made.
Observation
Yellow colour of bromine water fades in test tube containing sodium iodide.
- Brown solution formed in test tube containing sodium iodide
Explanation
Bromine is more electronegative than iodide but less 6than chlorine.
On adding Bromine water, iodide displaced from its solution but not chlorine.
(iii) Using the knowledge in g(i) and (ii) above,
- Complete the table below using (X) to show no reaction and (√) to show a reaction.
| Halide ion Halogen
ion in solution Halogen |
F– | Cl– | Br– | I– |
| F2 | X | √ | √ | √
|
| Cl2 | X | X | √ | √
|
| Br2 | X | X | X | √
|
| I2 | X | X | X | √ |
Write an ionic equation for the reaction where there is (V)
F2 (g)+ 2Cl– (aq) -> 2F–(aq) + Cl2(g)
F2 (g)+ 2Br– (aq) -> 2F–(aq) + Br2(aq)
F2 (g)+ 2I– (aq) -> 2F–(aq) + I2(aq)
Cl2 (g) + 2Br– (aq) -> 2Cl–(aq) + Br2(aq)
Cl2 (g)+ 2I– (aq) -> 2Cl–(aq) + I2(aq)
Br2 (aq)+ 2I– (aq)- -> 2Br–(aq) + I2(aq)
(h) State one uses of:
- Fluorine
Manufacture of P.T.F.E (Poly tetra fluoroethene) synthetic fiber.
Reduce tooth decay when added in small amounts/equations in tooth paste.
Note: large small quantities of fluorine /fluoride ions in water cause browning of teeth/flourosis.
Hydrogen fluoride is used to engrave word pictures in glass.
- Bromine
Silver bromide is used to make light sensitive photographic paper/films.
- Iodide
Iodine dissolved in alcohol is used as medicine to kill bacteria in skin cuts. It is called tincture of iodine.
- The table below to show some compounds of halogens.
| Element
Halogen |
H | Na | Mg | Al | Si | C | P |
| F | HF | NaF | MgH2 | AlF3 | SiF4 | CF4 | PF3
|
| Cl | HCl | NaCl | MgCl | AlCl3 | SiCl3 | CCl4 | PCl3
|
| Br | HBr | NaBr | MgBr2 | AlBo3 | SiBr4 | CBr4 | PBr3
|
| I | Hl | Nal | Mgl2 | All3 | SiL4 | Cl2 | Pb3
|
(j) (i) Using dot (.) and Cross (x) to represent electrons, show the bonding in chlorine molecule.
- Name the type of bond formed.
Covalent.
- Below is the table showing the bond energy of four halogens.
Bond Bond energy k J mole-1
Cl-Cl 242
Br-Br 193
I-I 151
- What do you understand by the term “bond energy”
Bond energy is the energy required to break/ form one mole of chemical bond
- Explain the trend in bond Energy of the halogens above:
–Decrease down the group from chlorine to Iodine
-Atomic radius increase down the group decreasing the energy required to break the covalent bonds between the larger atom with reduced effective nuclear charge an outer energy level that take part in bonding.
(k) Some compounds of chlorine are in the table below the oxidation state of chlorine in each compound.
Compound Oxidation state Name of compound
NaClO3 +5 Sodium chlorate (V)
ClO2 +4 Chloric (IV) oxide
KClO2 +3 Potassium chlorate (III)
NaClO +1 Sodium Chlorite (I)
Cl2 0 Chlorine Molecule
NaCl -1 Sodium Chloride (I)
MgCl2 -1 Magnesium Chloride (I)
- HYDROGEN CHLORIDE
- Occurrence
Hydrogen Chloride does not occur free in the atmosphere or in nature
- Preparation
Hydrogen chloride may be prepared in the school laboratory by reacting solid sodium/potassium chloride crystals with concentrated sulphuric (Vi) acid as in the set up below.
- Properties of hydrogen chloride gas(questions)
- What precautions should be taken when handling concentrated sulphuric acid? Explain.
-Wear protective clothing/gloves to avoid accidental contact with skin.
-Concentrated sulphuric (VI) acid is highly corrosive-it causes painful wounds when in contact with skin.
- What method of gas collection is used? Explain.
-Downward delivery// upward displacement of water
-Hydrogen chloride is denser than air.
- a) Write the equation for the reaction that takes place.
NaCl(s) + H2SO4(l) -> NaHSO4(aq) + HCl(g)
KCl(s) + H2SO4(l) -> KHSO4(aq) + HCl(g)
NaCl is commonly used because it is cheaper than KCl
- b) What property of concentrated sulphuric (VI) acid is used during the above reaction
-is the least volatile mineral acid, thus displace the more volatile hydrogen chloride from its salt (KCl/NaCl)
- i)What is the purpose of concentrated sulphuric (VI) acid.
-Drying agent / to dry the gas.
- ii) What property of concentrated sulphuric (VI) acid is used during the above use.
-Is hygroscopic – absorbs water but do not form solution.
iii) Name another substance which can be used to dry chlorine gas.
-anhydrous Calcium chloride
– silica gel
iv)Using a chemical equation, explain why anhydrous calcium oxide cannot be used in flask B
-Calcium oxide reacts with water /moisture to form calcium hydroxide. The calcium hydroxide formed reacts with chlorine to form calcium hypochlorite.
Chemical equations:
CaO(s) + 2H2O(l) -> Ca(OH)2(aq) + H2O(l)
Ca(OH)2(aq) + Cl2 (g) -> CaOCl2(aq) + H2O(l)
This reduces the amount of Chlorine produced.
d)Blue and red litmus papers were dipped in the hydrogen chloride prepared above. The Procedure was repeated with damp/wet/moist litmus papers. Explain the differences in observations made.
–Dry blue litmus papers remain blue
-Dry red litmus papers remain red
-Damp/moist/wet blue litmus papers turn red
-Damp/moist/wet red litmus paper turns red.
-Dry hydrogen chloride is a molecular compound that is joined by covalent bonds between the atoms. The gas is polar thus dissolves in water and ionize completely to free H+ that are responsible to turning blue litmus paper red.
- Dry hydrogen chloride gas was bubbled in two separately beakers containing water and in methylbenzene.
- Classify the two solvents as either “polar” or “non-polar”
Water – polar
Methylbenzene – non-polar
(ii) State and explain the observations made in the beaker containing:
(i)Methylbenzene
Colour of litmus solution remain.
Hydrogen chloride is a molecular substance. When dissolved in non-polar solvent, it does not dissociate / ionize to release H+ ions that changes the colour of litmus solution.
(ii)Water
Colour of litmus solution change to red.
Hydrogen chloride is a molecular substance. When dissolved in polar solvent like water, it dissociate/ionize to release H+ ions that changes litmus solution to red.
(iii)Why should an inverted filter funnel be used to dissolve hydrogen chloride.
– The filter funnel is dipped just below the water surface to increase the surface area of dissolving the gas and prevent suck back.
(iv)Name the solution formed when hydrogen chloride dissolves in water.
Hydrochloric acid
(f) Describe the test for presence of hydrogen chloride gas.
-Dip a glass rod in ammonia. Bring it to the mouth of a gas jar containing a gas suspected to be hydrogen chloride
-White fumes of ammonia chloride are formed.
(g) Place 5cm3 of dilute hydrochloric acid into a four separate test tubes. To separate test tube add zinc, magnesium iron and copper metals. State and explain the observations made.
Observation
– Effervescence/bubbles/fizzing in all cases except copper
- Colourless solution formed with zinc and magnesium.
- Green solution formed with ion.
- Gas produced that extinguishes splint with explosion.
Explanation
Metals above hydrogen in reactivity series react with hydrochloric and liberating hydrogen gas.
Chemical Equation:
Concentrated hydrochloric acid is a weak oxidizing agent than other concentrated acids i.eSulphuric (VI) acid and nitric (V) acid that react with all metals even those lower in the reactivity series.
(h) Place 5cm3 of dilute hydrochloric acid into five separate test tubes. To separate test tubes, add calcium carbonate, silver carbonate, copper carbonate, iron (II) carbonate and Sodium hydrogen carbonate. Explain the observations made.
Observation
Effervescence/bubbles/fizzing vigorously except in silver carbonate and lead (II) carbonate that stop later.
- Colourless solution formed except with iron (II) carbonate and copper (II) carbonate
- Green solution formed with iron (II) carbonate
- Blue solution formed with copper (II) carbonate
Explanation.
Carbonates and hydrogen carbonate react with dilute hydrochloric acid to produce carbon (IV) oxide, water and form chlorides.
All chlorides formed are soluble Except Lead (II) Chloride (soluble on heating/warming) and silver chloride.
Chemical equation:
CaCO3 (s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
(Colourless solution)
Chemical equation:
Ag2CO3 (s) + 2HCl(aq) → AgCl(s) + H2O(l) + CO2(g)
(Coats/Cover Ag2CO3)
Chemical equation:
CuCO3 (s) + 2HCl(aq) → CuCl2(aq) + H2O(l) + CO2(g)
(Blue Solution)
Chemical equation:
FeCO3 (s) + 2HCl(aq) → FeCl2(aq) + H2O(l) + CO2(g)
Chemical equation:
NaHCO3 (s) + HCl(aq) → NaCl(aq) + H2O(l) + CO2(g)
- Place 5cm3 of dilute sodium hydroxide, Potassium hydroxide and aqueous ammonia solution into three separate test tubes. Add one drop of phenolphthalein indicator drop wise, add dilute hydrochloric acid. Explain the observations made.
Observation
Colour of Phenophthalein indicator change from pink to colourless.
Explanation
Hydrochloric acid neutralizes alkalis to salt and water
When all the alkali has reacted with the acid, An extra slight excess acid turns the indicator used to colourless.
Chemical equation:
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
Chemical equation:
KOH(aq) + HCl(aq) → KCl(aq) + H2O(l)
Chemical equation:
NH4OH(aq) + HCl(aq) → NHaCl(aq) + H2O(l)
(j) Place 5cm3 of hydrochloric acid into four separate test tube tubes Separately add about 1g of each of copper (II) Oxide, Zinc (II) Oxide, Lead (II) Oxide< Calcium (II) Oxide. What happens to each test tube? Explain.
Observation:
All Solid dissolves except Lead (II) Oxide
Colourless solution formed with zinc Oxide and calcium (II) Oxide blue solution formed with copper (II) Oxide.
Explanation:
Metal oxides dissolves in dilute hydrochloric acid to form water and chloride salt Insoluble Lead (II) chloride and silver chloride once formed cover/coat unreacted oxides stopping further reaction.
Chemical equation: CuO(s) + HCl (aq) → CaCl2(aq) + H2O(l)
Chemical equation: CaO(s) + HCl (aq) → CaCl2 (aq) + H2O (l)
Chemical equation: PbO(s) + 2HCl (aq) → PbCl2 (aq) + H2O (l)
Chemical equation: ZnO(s) + HCl (aq) → ZCl2 (aq) + H2O (l)
(k) Manufacture of Hydrochloric acid.
(i) Raw Materials
- Hydrogen
(i) During electrolysis of Brine from the flowing mercury-cathode cell during the manufacture of sodium hydroxide solution.
(ii)From water gas by passing steam in heated charcoal.
C(s) + H2O → CO(g) + H2(g)
(iii)From partial oxidation of natural gas/methane
CH4(g) + O2(g) → CO(g) + 3H2(g)
2.Chlorine
(i)From electrolysis of fused/solid sodium chloride in the downs process during extraction of sodium
(ii)From electrolysis of brine/concentrated sodium chloride solution in the flowing mercury-cathode during the manufacture of sodium hydroxide solution.
(ii)Chemical processes.
- Hydrogen and chlorine gases are separately passed through concentrated sulphuric(VI) acid to act as a drying
- Small amount of pure hydrogen is continuously ignited in a chamber with continous supply of pure dry chlorine.
Large amount of hydrogen explodes.
- Hydrogen burns in chlorine to form hydrogen chloride gas.
Chemical Equation
H2(g) + Cl(g) → 2HCl(g)
- The hydrogen chloride produced is then passed up to meet a downward flow of water in the absorbtion Hydrogen chloride is very soluble in water and dissolves to form 35% concentrated hydrochloric acid.
Chemical Equation
HCl(g) + (aq) → HCl(aq)
The absorption chamber is shelved and packed with broken glass beads to
(i)Slow down the downward flow of water.
(ii)Increase surface area over which the water dissolves
The hydrochloric acid is then transported in steel tanks lined with rubber for market
(iii)Uses of Hydrochloric Acid
- To standardize the pH of (alcohol and wines)
- Regenerating ion-exchange resin during removal of hardness of water.
- Pickling of metals to remove oside layers on their surfaces.
- In the manufacture of dyes and drugs.
- Making zinc chloride for making dry cells.
| Cold water |
(iv)Diagram Showing Industrial manufacture.
| Dry Hydrogen gas |
| Dry chlorine gas |
| Burning Hydrogen gas |
| Absorbtion
chamber |
| 35% Conc. HCl |
(ii)Environmental effects of manufacturing HCl.
- Hydrochloric acid is acidic. Any leakage from a manufacturing plant to nearby rivers/lake causes exess acidity that lowers pH of water killing marine life.
- Hydrogen chloride leakage into atmosphere dissolves to form “acidic rain” that accelerate corrosion in buildings, Breathing problems to human beings and kill fauna and flora around the paint.
- Chlorine leakage causes breathing and sight problems to human being. It accelerates bleaching of dyed metals.
- Hydrogen leakage can cause an explosion because impure hydrogen explodes on ignition.
- Factors considered in setting hydrochloric acid manufacturing plant.
- Nearness to the manufacturing of sodium hydroxide because the by products of electrolysis of brine are the raw materials for hydrochloric acid plant.
- Availability of natural gas for extraction of hydrogen.
- Nearness/Availability of water to dissolve the hydrogen chloride gas.
- Availability of labour, market, capital and good means of transport.
D: CHLORIDE (Cl–) SALTS
- Chlorides are salts derived from hydrochloric acid. Hydrochloric acid is a monobasic (HX) salt with only one ionazable/replaceable “H” in its molecule. All chlorides are therefore normal salts.
- All metals exist as chloride salt except platinum and gold as below
| Metal | K | Na | Li | Mg | Ca | Al | Zn | Fe | Pb | H. | Cu | Ag | Hg |
| Formula of chloride | KCl | NaCl | LiCl | MgCl2 | CaCl2 | AlCl3 | ZnCl2 | FeCl2
FeCl3 |
PbCl
PbCl4 |
HCl | CuCl
CuCl2 |
AgCl | Hg2Cl2
HgCl2 |
(i)Both FeCl2 and FeCl3 exists but FeCl2 is readily oxidized to FeCl3 because it is more stable.
(ii)PbCl2 and PbCl4 exist but PbCl4 is only oxidized to form PbCl2 by using excess chlorine. It is less stable.
(iii)CuCl and CuCl2 exists but CuCl2 is (thermodynamically) more stable than CuCl. CuCl disproportionate to Cu and CuCl2..
(iv)HgCl and HgCl2exists as molecular compounds.
- All chlorides are soluble/dissolves in water except silver chloride(AgCl), Copper (I) chloride CuCl, Mercury (I) Chloride Hg2Cl2 and Lead (II) Chloride PbCl2 that dissolves in warm
- Most chlorides are very stable compounds. They do not decompose on gentle or strong bunsen burner heating in a school laboratory except Ammonium Chloride.
- Heating ammonium chloride
Place about 2g of solid ammonium chloride crystals in a clean dry boiling tube.Heat gently then strongly.
Observation
–red litmus paper turn blue
-blue litmus paper remains blue
Then later:
-both blue litmus papers turn red
Explanation:
Ammonium chloride on heating decomposes through chemical sublimation to ammonia and hydrogen chloride gas. Ammonia gas is less dense than hydrogen chloride. It is a basic gas and diffuses out faster to turn red litmus paper to blue. Hydrogen chloride is an acidic gas .It is denser than ammonia gas and thus diffuses slower than ammonia gas to turn the already both blue litmus paper to red.
Chemical equation
NH4Cl(s) -> HCl(g) + NH3 (g)
(acidic gas) (basic/alkaline gas)
- Test for Cl– ions
- The following experiment shows the test for the presence of Cl– ions in solids chloride salts.
- Procedure:
Place about 1g of sodium chloride, Zinc chloride and copper (II) chloride in separate boiling tubes. Place moist blue and red litmus papers on the mouth of the test tube. Carefully, add three drops of concentrated sulphuric (VI) acid.
Dip a glass rod in aqueous ammonia solution then bring it to the mouth of the boiling tube.
| observation | inference |
| -red litmus paper remain red
-blue litmus paper turn red
-vigorous effervescence/fizzing /bubbling
-white fumes produced on |
H+ ions
Cl– ions
HCl gas suspected |
(b)Explanation:
Concentrated sulphuric (VI) acid is the less volatile mineral acid.
It vigorously displaces chlorine in metal chlorides to evolve acidic hydrogen chloride gas fumes.
Chemical equation
NaCl(s) + H2SO4(l) -> NaHSO4(aq) + HCl(g)
KCl(s) + H2SO4(l) -> KHSO4(aq) + HCl(g)
CuCl2(s) + H2SO4(l) -> CuSO4(aq) + 2HCl(g)
ZnCl2(s) + H2SO4(l) -> ZnSO4(aq) + 2HCl(g)
Hydrogen chloride and ammonia gases react and form white fumes of ammonium chloride that confirms presence of Cl– ions in the solid substance.
Chemical equation
NH3(g) + HCl(g) -> NH4Cl(s)
- The following experiment shows the test for the presence of Cl– ions in solution /aqueous chloride salts.
(i)Using aqueous Lead (II) nitrate(V)
(a)Procedure:
I.Place about 5cm3 of sodium chloride, Iron (III) chloride and copper (II) chloride in separate boiling tubes. Add four drops of Lead (II) nitrate(V) solution to each. Preserve.
| Observation | Inference |
| White precipitate/ppt | SO42-, SO32-, Cl–,CO32-
|
II.To the preserved sample, add six drops of nitric (V) acid. Preserve.
| Observation | Inference |
| White precipitate/ppt persist | SO42-, Cl–
|
III. To the preserved sample, heat the mixture to boil
| Observation | Inference |
| White precipitate/ppt dissolves on boiling/warming | Cl–
|
Explanation:
I.When Lead(II) nitrate(V) solution is added to an unknown salt , a white precipitate/ppt of Lead(II) sulphate(VI) Lead(II) carbonate(IV) Lead(II) sulphate(IV) Lead(II) chloride(I) are formed.
Ionic equation:
Pb2+ (aq) + SO42-(aq) -> PbSO4(s)
Pb2+ (aq) + SO32-(aq) -> PbSO3(s)
Pb2+ (aq) + CO32-(aq) -> PbCO3(s)
Pb2+ (aq) + Cl–(aq) -> PbCl2(s)
II.When the white precipitate/ppt formed is acidified with dilute nitric(V) acid, the white precipitate of Lead(II) sulphate(VI) and Lead(II) chloride(I) persist/remain while that of Lead(II) carbonate(IV) and Lead(II) sulphate(IV) dissolves.
III.On heating /warming Lead (II) chloride (I) dissolves but on cooling it recrystallizes.This shows the presence of Cl– ions in aqueous solutions
.
(ii)Using aqueous silver (I) nitrate(V)
Procedure
- Place about 5cm3 of sodium chloride, Iron (III) chloride and copper (II) chloride in separate boiling tubes. Add four drops of silver(I) nitrate(V) solution to each. Preserve.
| Observation | Inference |
| White precipitate/ppt | Cl–, CO32-
|
- To the preserved sample, add six drops of nitric (V) acid. Preserve.
| Observation | Inference |
| White precipitate/ppt persist | Cl–
|
Explanation:
I.When silver(I) nitrate(V) solution is added to an unknown salt , a white precipitate /ppt of silver(I) carbonate(IV) and silver(I) chloride(I) are formed.
Ionic equation:
2Ag+ (aq) + CO32-(aq) -> Ag2CO3(s)
Ag+ (aq) + Cl–(aq) -> AgCl(s)
- When the white precipitate/ppt formed is acidified with dilute nitric (V) acid, the white precipitate of silver (I) chloride (I) persist/remain. This shows the presence of Cl–ions in aqueous solutions.
Silver (I) carbonate (IV) dissolves when reacted with nitric (V) acid.
COMPREHENSIVE REVISION QUESTIONS
- In an experiment ,dry hydrogen chloride gas was passed through heated zinc turnings as in the set up below.The gas produced was the passed through copper(II) oxide
- Write the equation for the reaction :
(i)For the school laboratory preparation of hydrogen chloride gas.
NaCl(s) + H2SO4(l) -> NaHSO4(aq) + HCl(g)
(ii)in tube S
Zn(s) + 2HCl(aq) -> ZnCl2(aq) + H2(g)
b)State and explain the observation made in tube V.
Observations-colour of solid changes from black to brown
-colourless liquid forms on the cooler parts of tube V
Explanation-Hydrogen produced in tube S reduces black copper(II) oxide to brown copper metal and the gas oxidized to water vapour that condense on cooler parts..
Chemical equation.
CuO(s) +H2(g) ->Cu(s) + H2O(l)
(c)How would the total mass of tube S and tube V and their contents compare before and after the experiment.
Tube S- Mass increase/rise because Zinc combine with chlorine to form heavier Zinc Chloride.
Tube V- Mass decrease/falls/lowers because copper (II) oxide is reduced to lighter copper and oxygen combine with hydrogen to form water vapour that escape.
- Chlorine is prepared by using solid sodium chloride, concentrated sulphuric(VI) acid and potassium manganate(VII)
a)What is the role of the following in the reaction;
(i) concentrated sulphuric(VI)
To produce hydrogen chloride gas by reacting with the solid sodium chloride.
(ii)potassium manganate(VII)
To oxidize hydrogen chloride gas to chlorine
3.Use the flow chart below to answer the questions that follow.
a)(i) Name:
gas X Hydrogen chloride
solution W hydrochloric acid
gas Q chlorine
bleaching agent Z sodium chlorate(V)
b)Write the chemical equation for the formation of :
(i) gas X
NaCl(s) + H2SO4(l) -> NaHSO4(aq) + HCl(g)
(ii)solution W
HCl(g) + (aq) -> HCl(aq)
(iii)gas Q
2KMnO4 + 16HCl(aq) -> 2KCl(aq) + 2MnCl2(aq) + 8H2O(l) + 5Cl2(g)
(iv)bleaching agent Z
6NaOH(aq) + 3Cl2(g) ->NaCl(aq) + NaClO3(aq) + 3H2O(l)
c)State and explain the following observations;
(i) a glass rod dipped in aqueous ammonia is brought near gas X
Observation: Dense white fumes
Explanation:Ammonia gas reacts with hydrogen chloride gas to form dense white fumes of ammonium chloride.
Chemical equation: NH3(g) +HCl(g) -> NH4Cl(s)
(ii)Wet blue and red litmus papers were dipped into gas Q
Observations: Blue litmus paper turned red the both are bleached
/decolorized.
Explanations: chlorine reacts with water to form both acidic hydrochloric and chloric (I) acids that turn blue litmus paper red. Unstable chloric (I) acid oxidizes the dye in the papers to colourless.
Chemical equations
Cl2(g) + HCl(aq) ->HCl(aq) + HClO(aq)
Coloured dye +HClO(aq) ->HCl(aq) + (Colourless dye +O)//
(Coloured dye-O) + HClO(aq) ->HCl(aq) + Colourless dye
4.Use the flow chart below to answer the questions that follow
- Name
Liquid A Concentrated sulphuric(VI) acid
Process Z Neutralization
White solid X Ammonium chloride
b)Write the equation for the formation of:
(i) Hydrogen chloride
NaCl(s) + H2SO4(l) -> NaHSO4(aq) + HCl(g)
(ii) B
HCl(g) + (aq) -> HCl(aq)
(iii)process Z (using ionic equation)
H+ (aq) + OH–(aq) -> H2O(l)
(iv)C (using ionic equation)
Ag+ (aq) + Cl–(aq) -> AgCl(s)
c)Describe how solution B is obtained.
Bubbling hydrogen chloride gas through inverted funnel into distilled water until no more dissolve.
5 The results obtained when halogens are bubbled into test tubes containing solutions of halide A,B and C is as in the table below. Tick(v) means a reaction took place.Cross(x) means no reaction took place.
|
Halogens |
Halide ions in solution | ||
| A | B | C | |
| I2 | x | – | x |
| Br2 | x | v | – |
| Cl2 | – | v | v |
a)Identify the halide ions represented by letter
A Cl–
B I–
C Br–
b)Write the ionic equation for the reaction that take place with halide:
(i) C
Cl2(g) + 2Br–(aq) -> 2Cl–(aq) + Br2(aq)
(ii) B
Cl2(g) + 2Br–(aq) -> 2Cl–(aq) + Br2(aq)
Cl2(g) + 2I–(aq) -> 2Cl–(aq) + I2(aq)
6.The diagram below shows a set up of apparatus for the school laboratory collection of dry chlorine gas.
a)Name:
(i) substance Q
Concentrated hydrochloric acid
(ii)suitable drying agent L
-Concentrated sulphuric(VI) acid
-anhydrous calcium chloride
-silica gel
- b) State a missing condition for the reaction to take place faster.
-Heat/Heating
c)Red and blue litmus papers were dipped into the chlorine gas from the above set up .State and explain the observations made.
Observation: Blue litmus paper remain blue. Red litmus paper remain red.
Explanation: Dry chlorine has no effect on dry litmus papers.
d)Write the equation for the reaction taking place in the conical flask
MnO4 (s) + 4HCl(aq) -> MnCl2(aq) + 2H2O(l) + Cl2(g)
e)Name two other substances that can be used in place of MnO2
Lead(IV) oxide (PbO2)
Potassium manganate(VI)(KMnO4)
Potassium dichromate(K2Cr2O4)
Bleaching powder(CaOCl2)
7.The set up below shows the apparatus used to prepare and collect anhydrous iron(III) chloride.
a)Name salt K
Iron(III)cchloride
- Write the equation for the reaction for the formation of salt K
2Fe(s) + 3Cl2 (g) -> 2FeCl3 (s/g)
- State and explain the following
(i)Small amount of water is added to iron (II) chloride in a test tube then shaken
Solid dissolves to form a green solution. Iron(II) chloride is soluble in water
(ii)I.Three drops of aqueous sodium hydroxide is added to aqueous iron(II) chloride and then added excess of the alkali.
Observation:
Green precipitate is formed that persist/remain /insoluble in excess akali.
Explanation:
Iron(II) chloride reacts with aqueous sodium hydroxide to form a green precipitate of iron(II) hydroxide.
Ionic equation:
Fe2+(aq) + OH–(aq) -> Fe(OH)2(s)
II.Six drops of hydrogen peroxide is added to the mixture in d(ii) above.
Observation:
Effervescence/bubbling/fizzing take place and the green precipitate dissolve to form a yellow/brown solution.
Explanation:
hydrogen peroxide oxidizes green Fe2+to yellow/ brown Fe3+solution.
9.Use the flow chart below to answer the questions that follow.
a)Write the chemical equation for the formation of gas A
NaCl(s) + H2SO4(l) -> NaHSO4(aq) + HCl(g)
b)Identify:
(i) four possible ions that can produce white precipitate B
SO42-,SO32-, CO32-, Cl–
(ii)two possible ions that can produce;
I.White precipitate C
SO42-,Cl–
II.colourless solution D
SO32-, CO32-
(iii)possible ions present in
I.White precipitate E
SO42-
II.colourless solution F
Cl–
- Below is a set up in the preparation of a particular salt. Study it and answer the questions that follow.
State the observation made when aluminium wool is heated.
Glows red hot.
b)(i) Identify salt A
aluminium(III) chloride// AlCl3
(ii)Write the equation for the formation of salt A
2Al(s) + 3Cl2(g) -> 2AlCl3(s/g)
(iii)What property of salt A is exhibited as shown in the experiment.
It sublimes//sublimation.
(iv)Calculate the minimum volume of chlorine required to form 700kg of iron(III) chloride at room temperature.(Fe= 56.0, Cl=35.5, 1 mole of a gas =24000cm3, 1000g = 1kg)
Mole ratio Fe : Cl2 = 2: 3 molar mass FeCl3 = 162.5g
Method 1
2 x 162.5 g FeCl3 -> 3x 22400 cm3 Cl2
700 x1000 gFeCl3 -> (700 x1000 x3 x22400)/(2 x 162.5)
=1.4474 x 10-8 cm3
Method 2
Moles of FeCl3= mass/ molar mass
=> (700 x 1000) / 162.5 = 4307.6923 moles
Moles of Cl2= 3/2 moles of FeCl3
=>3/2 x 4307.6923 = 6461.5385 moles
Volume of chlorine= moles x molar gas volume
=>6461.5385 x 24000 = 1.5508 x 10-8 cm3
- c) Name another metal that can produce similar results as salt K.
Iron
d)(i) What is the purpose of anhydrous calcium chloride.
-ensure the apparatus are water free.
-prevent water from the atmosphere from entering and altering//hydrolysing salt A
(ii) Write the equation for the reaction that take place if anhydrous calcium chloride is not used in the above set up.
AlCl3(s) + 3H2O(l) -> Al(OH)3(aq) + 3HCl(g)
(iii) Write the equation for the reaction that take place when Iron metal is reacted with dry hydrogen chloride gas.
Fe(s) + 2HCl(g) -> FeCl2(s) + H2(g)
(iv)Calculate the mass of Iron(II)chloride formed when 60cm3 of hydrogen chloride at r.t.p is completely reacted. (1 mole of a gas =24dm3 at r.t.p, Fe = 56.O, Cl= 35.5)
Chemical equation Fe(s) + 2HCl(g) -> FeCl2(s) + Cl2(g)
Mole ratio HCl: FeCl2 = 1:1
Molar mass FeCl2 = 127g
Moles of HCl used = 60cm3 /24000cm3 = 2.5 x 10 -3 moles
Moles of FeCl2 = Moles of HCl => 2.5 x 10 -3 moles
Mass of FeCl2 = moles x molar mass => 2.5 x 10 -3 x 127 =0.3175g
12.Study the flow chart below and use it to answer the questions that follow
a)Identify substance:
P Iron(II) chloride//FeCl2
Q Chlorine // Cl2
R Iron(III) chloride//FeCl3
b)Write the equation for the reaction for the formation of:
(i) gas Q
2KMnO4 (s) + 16HCl(aq) -> 2KCl(aq) + 2MnCl2(aq) + 8H2O(l) + 5Cl2(g)
(ii) the green precipitate (using ionic equation)
Ionic equation:
Fe2+(aq) + 2OH–(aq) -> Fe(OH)2(s)
(ii) the brown precipitate (using ionic equation)
Ionic equation:
Fe3+(aq) + 3OH–(aq) -> Fe(OH)3(s)
c)A glass rod was dipped in aqueous ammonia. The rod was then brought near hydrogen chloride. State and explain the observation made.
Observation:
White fumes
Explanation:
Ammonia gas reacts with hydrogen chloride gas to form white fumes of ammonium chloride.
Chemical equation:
NH3(g) + HCl(g) -> NH4Cl(s)
- Using dot(.)and cross(x)to represent electrons,show the bonding in aluminium chloride in vapour phase.
(b)How many electrons in :
(i)aluminium atoms are used in bonding.
Six electrons(three valence electrons in each aluminium atom)
(ii)chlorine atoms atoms are used in dativebonding.
four electrons(two lone pairs of valence electrons in two chlorine atoms)
(iii)the molecule are used in bonding.
Sixteen electrons
-six valence electrons from aluminium atom through covalent bond
-six valence electrons from chlorine atoms through covalent bond.
– four valence electrons from chlorine atoms through dative bond
(c)How many lone pair of electrons do not take part in bonding within the molecule.
Sixteen(16) lone pairs from six chlorine atoms(32 electrons)
(d)Aluminium chloride does not conduct electricity in molten state but Magnesium chloride conduct.
Aluminium chloride is a molecular compound that has no free mobile Al3+ and Cl– ions which are responsible for conducting electricity. Magnesium chloride has free mobile Mg2+ and Cl– ions because it is an ionic compound.
- Use the flow chart below to answer the questions that follow:
a)Write an equation for the school laboratory formation of hydrogen chloride gas
NaCl(s) + H2SO4(l) -> NaHSO4(aq) + HCl(g)
KCl(s) + H2SO4(l) -> KHSO4(aq) + HCl(g)
b)Name:
- solid Y Iron (II) chloride (FeCl2)
II green precipitateIron (II) hydroxide (Fe (OH)2
III Gas Y Chlorine (Cl2)
- Bleaching agent A Sodium hypochlorite (NaOCl)
c)Blue and red litmus papers were dipped into bleaching agent A. Write the equation for the reaction that takes place.
Coloured dye +NaOCl(aq) ->NaCl(aq) + (Colourless dye + O)//
(Coloured dye-O) + NaOCl(aq) ->NaCl(aq) + Colourless dye
d)State four uses of gas Z
- Bleaching agent
- Manufacture of hydrochloric acid
- Chlorination of water to kill germs
- Manufacture of PVC pipes
