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A compound has the following percentage composition by mass, carbon 14.4%. hydrogen 1.2% and chlorine 84.5%. Determine the empirical formula of this compound. (H=1, C=12 and Cl=35.5)

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Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Carbon	14.4	12	$\dfrac{14.4}{12}$ = 1.2	$\dfrac{1.2}{1.2}$ = 1
Hydrogen	1.2	1	$\dfrac{1.2}{1}$ = 1.2	$\dfrac{1.2}{1.2}$ = 1
chlorine	84.5	35.5	$\dfrac{84.5}{35.5}$ = 2.38	$\dfrac{2.38}{1.2}$ = 1.98 = 2

Simplest ratio of whole numbers = C : H : Cl = 1 : 1 : 2

Hence, empirical formula is CHCl₂

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