question_answer 3)

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

Answer:

Element % Atomic mass Relative no. of atoms Simplest ratio Whole no. ratio Carbon 24.27 12 \[\frac{24.27}{12}=2.022\] \[\frac{2.022}{2.018}\approx 1\] 1 Hydrogen 4.07 1 \[\frac{4.07}{1}=4.07\] \[\frac{4.07}{2.018}\approx 2\] 2 Chlorine 71.65 35.5 \[\frac{71.65}{355}=2018\] \[\frac{2.018}{2.018}=1\] 1   Empirical formula = \[C{{H}_{2}}Cl\] Molecular formula = \[{{\left( C{{H}_{2}}Cl \right)}_{n}}\] \[n=\frac{Molecular\text{ }mass}{Empirical\text{ }formula\text{ }mass~}=\frac{98.96}{49.5}\approx 2\] \[\therefore \]Molecular formula = \[{{(C{{H}_{2}}Cl)}_{2}}={{C}_{2}}{{H}_{4}}C{{l}_{2}}\]