Question

A certain hydrocarbon has the following percentage composition C= 83.27%, H= 16.27%. The vapour density of compound is 43. Find the molecular formula of compound.​

Answer 1

Answer: $C_{6$$H_{12}$

Explanation:

Atomic Ration of Carbon= 83.27/12 = 6.93

Atomic Ratio of Hydrogen = 16.27/1 = 16.27

Simple Ratio = 1:2

Emperical Formula = CH$_{2}$

Molecular Weight = 2*VD = 2*43 = 86

Emperical Formula Weight = 12+2 = 14

n = 86/14 = 6.14 (rounding off to 6)

Molecular Formula = 6(CH$_{2}$) = C$_{6}$H$_{12}$