Question

An organic compound has the following percentage composition: C = 12.76%, H = 2.13%, Br = 85.11%. The vapour density of the compound is 94. Find out its molecular formula [C = 12, H = 1, Br = 80].

Answer 1

Element	Atomic Mass	Percentage	Relative Number of Moles	Simplest Mole Ratio
C	12	12.76	$\frac{12.76}{12}=1.06$	$\frac{1.06}{1.06}=1$
H	1	2.13	$\frac{2.13}{1}=2.13$	$\frac{2.13}{1.06}=2$
Br	80	85.11	$\frac{85.11}{80}=1.06$	$\frac{1.06}{1.06}=1$

So, the empirical formula of the compound is CH₂Br.
Empirical formula mass = (12 + 2 + 80) g = 94 g
We know that the molecular formula of a compound can be determined with the help of the following relation:
Molecular formula = (Empirical formula)_n
Molecular mass is calculated as:
Vapour density $\times$ 2 = Molecular mass
Vapour density $\times$ 2 = Empirical formula mass $\times$ n
94 $\times$ 2 = 94 $\times$ n
n = 2
Hence, the molecular formula of the compound is C₂H₄Br₂.