Question

A gaseous hydrocarbon upon combustion gives $0.72$ g of $H_{2}O$ and $3.08$ g of $C{O}_2$. The empirical formula of the hydrocarbon is :

• A. $C_6{H}_5$
• B. $C_7{H}_8$
• C. $C_2{H}_4$
• D. $C_3{H}_4$

Answer 1

The correct option is B $C_7{H}_8$

Let the formula of hydrocarbon be $C_a{H}_b$.
$C_a{H}_b+O_2=aC{O}_2+\frac{b}{2}{H}_{2}O$
Number of moles of $C{O}_2\left(a\right)$ formed =$\frac{3.08}{44}=0.07$
Number of moles of $H_{2}O$ formed $(\frac{b}{2})=\frac{0.72}{18}=0.04$
$\frac{a}{b/2}=\frac{0.07}{0.04}$
$\frac{a}{b}=\frac{0.07}{0.08}=\frac{7}{8}$
Therefore, mole ratio of $C$ and $H$ is $7:8$.
Hence, the empirical formula is $C_7{H}_8$.