Question

An organic compound with vapour density = 94 contains C = 12.67%, H = 2.13%, and Br = 85.11%. Find the molecular formula.
[Atomic mass: C = 12, H = 1, Br = 80]

Answer 1

Element	Atomic mass	Percentage	Relative number of moles	Simplest mole ratio	Whole number ratio
C	12	12.67	$\frac{12.67}{12}=1.05$	$\frac{1.05}{1.05}=1$	1
H	1	2.13	$\frac{2.13}{1}=2.13$	$\frac{2.13}{1.05}=2.03$	2
Br	80	85.11	$\frac{85.11}{80}=1.06$	$\frac{1.06}{1.05}=1.01$	1

Empirical formula of the compound is CH₂Br.
Empirical formula mass = (12 + 1 + 1 + 80) = 94
Molecular mass of the gas = 2 $\times$ Vapour density of the gas
Molecular mass of the gas = (2 $\times$ 94) = 188
Molecular formula of the compound = (Empirical formula of the compound)_n
Molecular mass of the compound = n $\times$ Empirical formula mass of the compound
$n=\frac{\mathrm{Molecular}\mathrm{mass}}{\mathrm{Empirical}\mathrm{formula}\mathrm{mass}}=\frac{188}{94}=2$
Molecular formula of the compound = (CH₂Br)₂ = C₂H₄Br₂