Question

An organic compound contains 4.09% hydrogen and 71.19% of chlorine and the rest carbon. If its molecular mass is 148.5 g, find its empirical formula and chemical formula.

Answer 1

1. Find the Percentage of Carbon:

% of Carbon (C) = 100 − (4.09% + 71.19%)

= 100 − 75.28

= 24.72%

2. Create the Atomic Ratio Table:

Element	% by Mass	Atomic Mass	Atomic Ratio (Moles)	Simplest Molar Ratio
Carbon (C)	24.72%	12	`24.72/12` = 2.06	`2.06/2.01` ≈ 2
Hydrogen (H)	4.09%	1	`4.09/1` = 4.09	`4.09/2.01` ≈ 2
Chlorine (Cl)	71.19%	35.5	`71.19/35.5` = 2.01	`2.01/2.01` = 1

Empirical Formula = CH₂Cl

3. Determine the Chemical (Molecular) Formula:

Empirical Formula Mass = 12 + 2(1) + 35.5 = 49.5 g

Value of n = `"Molecular mass"/"Empirical mass"`

= `148.5/49.5`

= 3

Chemical Formula = 3 × (CH₂Cl) = C₆H₆Cl₃