Question

A compound contains — Carbon 14.4%, hydrogen 1.2% and chlorine 84.5%. Determine the empirical formula of this compound. Work correct to 1 decimal place. The relative molecular mass of this compound is 168, so what is it’s molecular formula?
[C = 12; H = 1; Cl = 35.5]

August 5, 2024

Answer

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Carbon	14.4	12	14.4/12 = 1.2	1.2/1.2 = 1
Hydrogen	1.2	1	1.2/1 = 1.2	1.2/1.2 = 1
chlorine	84.5	35.5	84.5/35.5 = 2.38	2.38/1.2 = 2

Simplest ratio of whole numbers = C : H : Cl

= 1 : 1 : 2

Hence, empirical formula is CHCl₂

Empirical formula weight = 12 + 1 + 2(35.5)

= 84 g

Relative molecular mass = 168

n = $168\over84$ = 2

Molecular formula = [CHCl₂]₂ = C₂H₂Cl₄

∴ Molecular formula = C₂H₂Cl₄

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