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.

Answer 1

Element	Atomic mass	Percentage	Relative number of moles	Simplest  mole ratio	Whole number ratio
C	12	12.76	12.76/12 = 1.06	1.06/1.06 = 1	1
H	1	2.13	2.13/1 =2.13	2.13/1.06 = 2	2
Br	80	85.11	85.11/80 =1.06	1.06/1.06 = 1	1

So the Empirical formula of the compound will be CH₂Br.
Now the Empirical formula mass will be = Atomic mass of C + Atomic mass of H + Atomic mass of Br
= 12 + 1 x 2 + 80 = 94
Now as Molecular mass = 2 x vapour density
= 2 x 94 = 188.
So n = Molecular mass / empirical formula mass
= 188/ 94 = 2
Molecular formula of the compound is = n x empirical formula
= 2 x (CH₂Br) = C₂H₄Br₂