A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine.

A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulae?

Answer»

Element	% of element	Atomic mass	Relative no. of moles	Simplest molar ratio	Simplest whole number molar ratio
C	24.27	12	\(\frac{24.27}{12}\)= 0.02	\(\frac{2.02}{2.02}\) = 1	1
H	4.07	1	\(\frac{4.07}{1}\) = 4.07	\(\frac{4.07}{2.02}\) = 2	2
Cl	71.65	35.5	\(\frac{71.65}{35.5}\) = 2.02	\(\frac{2.02}{2.02}\) = 1	1

\(\therefore\) Empirical formula of compound = CH₂Cl

n = \(\frac{Molecular\,mass}{Empirical\,formula\,mass}\)

Molecular mass = 98.96 g

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

= 49.5

n = \(\frac{98.96}{49.5}\) = 2

\(\therefore\) Molecular formula = n × Empirical formula

= 2 x CH₂Cl

= C₂H₄Cl₂