Question

An organic compound on analysis gave the following % composition C- 57.6%, H-3.6% and the rest is oxygen. The vapour density of compound was 83. Find empirical formula and molecular formula of the compound.

Answer 1

Element % amount Relative number of atoms Ratio of number of atoms Simplest ratio

C 57.8 (57.8/12) = 4.82 (4.82/2.41) = 2 2x2 = 4

H 3.6 (3.6/1) = 3.6 (3.6/2.41) = 1.5 1.5x2 = 3

O (100-57.8-3.6) = 38.6 (38.6/16) = 2.41 (2.41/2.41) = 1 1x2=2

So, we can deduce the empirical formula of the organic compound = C H O .

The empirical Formula mass = 4x12 + 3x1 + 2x16 = 83 u.

The molecular mass of compound can be calculated as:

Molecular mass = 2 x Vapour Density = 2 x 83 = 166 u.

n = Molecular mass / Empirical Formula Mass = 166/83 = 2.

Molecular Formula of the compound = n(Empirical Formula) = 2(C H O ) = C H O .