Question

A compound on analysis gave the following percentage composition by mass : 

H = 9.09; O = 36.36; C = 54.55. Mol mass of compound is 88. Find its molecular formula.

Answer 1

Given : 

Percentage of H, O, C = 9.09%, 36.36%, 54.55% respectively. 

Molar mass of the compound = 88 g mol 

To find : 

The molecular formula of the compound

Calculation :

Hence,

The ratio of number of moles of C:H:O is

Hence,

Empirical formula is C₂H₄O. 

Empirical formula mass = 24 + 4 + 16 

= 44 g mol^-1

Hence,

r = \(\frac{Molar\,mass}{Empirical\,formula\,mass}\)

= \(\frac{88\,g\,mol^{-1}}{44\,g\,mol^{-1}}\) 

= 2

Molecular formula = r × empirical formula 

Molecular formula = 2 × C₂H₂O = C₄H₈O₂

∴ Molecular formula of the compound = C₄H₈O₂