The question was popular among candidates and the performance was good.
In part(a)(i), candidates gave two reasons why soda lime was used instead of caustic soda in
the preparation of methane thus:
- soda lime does not attack glass apparatus unlike caustic soda
- soda lime is not deliquescent unlike caustic soda
In (a)(ii), candidates listed two physical properties of methane from among the following:
- it is a gas at room temperature;
- colourless gas;
- odourless gas;
- slightly soluble in water;
- less dense than air.
In (a)(iii), candidates determined correctly both empirical and molecule formulae of the
hydrocarbon thus:
|
hydrogen |
Carbon |
% |
17.24 |
82.76 |
|
17.24
1.00 |
82.76
12 |
|
|
|
|
17.24 |
6.90 |
|
17.24
6.90 |
6.90
6.90 |
|
2.5
5 |
1 |
|
|
2 |
|
|
|
1. Empirical formula C2H5
II.
Molecular formula (C2 H5)n = V.D x 2
→(12 x 2 + 1 x 5)n = 58
29n = 58
n = 2
→(C2 H5)2 = C4 H10
In part (b), candidates stated what was meant by isomerism, drew the structures of the two
isomers of the compound C2H60, gave the name of the two isomers in (i) - (v) respectively.
However, some of them lost marks because of spelling errors, omission of bonds etc.
The expected responses from candidates were as follows:
(i) The existence of two or more compounds with the same molecular formula but
different molecular structures/different arrangements of atoms.
(ii) H H
I I
H- C- C- O -H
I I
H H
H H
I I
H- C- O- C -H
I I
H H
[All bonds must be shown however, accept -OH was accepted]
(iii) Ethanol and methoxymethane respectively.
(iv) The two isomers belong to different homologous series/they have different functional
groups
In part (c), most candidates could not give three deductions made from qualitative and
quantitative analysis of a given organic compound hence lost marks. The required answers
from candidates were:
- functional groups
- number of atoms of different elements
- types of elements
- types of bonds
- percentage composition of the elements in compound
- spatial arrangements of atoms in a molecule.
In parts( d), only very few candidates could correctly give one chemical test to distinguish
between propene and propane. The expected answer from candidates was as written below:
- Pass each of the gases into (acidified) KMnO4J bromine water/bromine in CCl4/ bromine,
propene decolourizes bromine water/KMnO4/bromine in CCl4/bromine whereas propane
does not.
