Question

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

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

Answer 1

Answer: (C)

$C_7{H}_8$

$18$ g of $H_{2}O$ contains $2$ g of $H$.

So, $0.72$ g of water contains $0.72\times \frac{2}{18}$ $=$ $0.08g$ of $H$.

$44$ g of $C{O}_2$ contains $12$ g of $C$.

So, $3.08$ g of $C{O}_2$ contains $3.08\times \frac{12}{44}$ $=$ $0.84g$ of $C$.

Ratio of atoms of carbon and hydrogen can be found now.

$\therefore \frac{C}{H}=\frac{\frac{0.84}{12}}{\frac{0.08}{1}}$ $=\frac{0.07}{0.08}$ $=\frac{7}{8}$.

Hence, the empirical formula is $C_7{H}_8$.