Question

A compound contains $4.07$% hydrogen, $24.27$% carbon and $71.65$% chlorine. It molar mass is $98g$. What are its empirical and molecular formulas? (Atomic mass of C=12,H=1,Cl=35)

Answer 1

Let us assume the mass to be $100g$

$\therefore$ No. of moles of $H=\frac{4.07}{1}=4.07\approx 4$

No of moles of $Cl=\frac{71.65}{35.5}=2.01\approx 2$

No. of moles of $C=\frac{24.27}{12}=2.02\approx 2$

$\therefore n_c:n_H:n_{Cl}=2:4:2=1:2:1=C:H:Cl$

$therefore$ Empirical formula$={CH}_{2}Cl$

But Molecular formula $=n\times$ Empirical formula

$n=\frac{Molarmass}{EmpiricalFormulamass}$

$=\frac{98}{12+2+35}$

$=2.1$

$n\approx 2$

$\therefore$ Molecular formula$=2\times \left({CH}_{2}Cl\right)$

$=C_2{H}_4{Cl}_2$