Some Basic Concepts of Chemistry · Question 34

Quantitative analysis of an organic compound (X) shows following % composition. C : 14.5\% Cl : 64.46\% H : 1.8\% Empirical formula mass of the compound (X) is 10-1 (Given molar mass in g mol-1 of C : 12, H : 1, O : 16, Cl : 35.5)

Answer

1655

Solution

% of O (by difference) = 100 - (14.5 + 64.46 + 1.8) = 19.24\%

Relative moles:

C: 14.512 = 1.208; Cl: 64.4635.5 = 1.816; H: 1.81 = 1.8; O: 19.2416 = 1.2025

Dividing by the smallest (1.2025):

C : Cl : H : O = 1 : 1.5 : 1.5 : 1 = 2 : 3 : 3 : 2

Empirical formula = C2H3Cl3O2

Empirical formula mass = 2(12) + 3(1) + 3(35.5) + 2(16)

= 24 + 3 + 106.5 + 32 = 165.5 = 1655 10-1

Answer: 1655