Question

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

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

Answer 1

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

18 g of $H_{2}O$ has 2g of Hydrogen

0.72g of $H_{2}O$ has = $\frac{0.72\times 2}{18}=0.08g$ of hydrogen

44g of $C{O}_2$ has 12 g of carbon.

3.08g of $C{O}_2$ has = $\frac{3.08\times 12}{44}=0.84g$ of carbon

$C:H=\frac{0.84}{12}:\frac{0.08}{1}=0.07:0.08\Rightarrow 7:8$

$\therefore$ Empirical formula = $C_7{H}_8$