A compound formed during the incomplete combustion of plastics shows the following percentage composition: 44.76 % C
1.25 % H
44.05 % CI 994 % O
The molar mass is known to be 321.97 g/mol. Determine the empirical formula and the molecular formula for this compound

The empirical formula : C₆H₂Cl₂O

The molecular formula : C₁₂H₄Cl₄O₂

Given

The percentage composition

Required

The empirical formula and the molecular formula

Solution

The mol ratio of the components :

C : H : Cl : O

=44.76/12 : 1.25/1 : 44.05/35.5 : 9.94/16

=3.73 : 1.25 : 1.241 : 0.621 divide by 1.241

= 3 : 1 : : 1 : 0.5 x 2

= 6 : 2 : 2 : 1

The empirical formula : C₆H₂Cl₂O

(Empirical formula)n=molecular formula

(C₆H₂Cl₂O)n=321.97

(160.986)n=321.97

n=2

(C₆H₂Cl₂O)₂=C₁₂H₄Cl₄O₂

The molecular formula : C₁₂H₄Cl₄O₂