A gaseous hydrocarbon gives upon combustion,0.72 g of water and 3.08 g of CO₂ The empirical formula of the hydrocarbon is  

18 g H₂O contains 2 g H. $\therefore 0.72{\mathrm{g} \mathrm{H}}_2\mathrm{O}$ contains 0.08 g H
44 g CO₂ contains 12 g C. $\therefore 3.08\mathrm{g} {\mathrm{CO}}_2$ contains 0.84 g C.

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

$\therefore$ Empirical formula $={\mathrm{C}}_7{\mathrm{H}}_8$