3A+B
C
(6 marks)
mc = 30.1 % x 100 g = 30.1 g
mH = 3.16 % x 100 g = 3.16 g
nCl = 66.7 % x 100 g = 66.7 g
nc = 30.1 g /12.01 g/mol = 2.51 mol C
nH = 3.16/1.01 g/mol = 3.13 mol H
nCl = 66.7/35.45 g/mol = 1.88
Mole Ratio
2.51/1.88 = 1.34 x3
3.13/1.88 = 1.66 x3
1.88/1.88 = 1
x3
Whole Ratio
= 4
= 5
= 3
The empirical formula is C4H5Cl3. If the molecular formula is twice the empirical then the molecular formula must be C8H10Cl6.
(2 marks)
MM C8H10Cl6 = 318.9 g/mol
nC8H10Cl6 = 38.0 g / 318.9 g/mol = 0.119 mol C8H10Cl6 (2 marks)
% yield = (actual yield/theoretical yield) x 100%
theoretical yield = actual yield / (% yield / 100%) = 0.119 mol / (68% / 100%) = 0.175 mol (2 marks)
OR
% yield = (actual yield/theoretical yield) x 100%
theoretical yield = actual yield / (% yield / 100%) 38.0 g / (68% / 100%) = 55.9 g
n C8H10Cl6 = 55.9 g / 318.9 g/mol = 0.175 mol
nA = 3 mol A/ 2 mol C x 0.175 mol C = 0.263 mol A (2 marks)
MMA = 70.91 g/mol
mB = n x MM = 0.263 mol x 70.91 g/mol = 18.6 g A (1 mark)
BONUS: Since we know that compound B doesn’t contain any chlorine, compound A must, by inspection we can see that the
molar
mass for A corresponds to Cl2. With this value we can see that our chemical equation would be:
3 Cl2 + B -> C8H10Cl6 In order to balance this equation B must be C8H10 (2 marks)
Benefits
Drawbacks