A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g of ${\mathrm{CO}}_2$. The empirical formula of the hydrocarbon is

Mass of H in the obtained water is $\left(\frac{2M_{\mathrm{H}}}{M_{{\mathrm{H}}_2\mathrm{O}}}\right)m_{{\mathrm{H}}_2\mathrm{O}}=\frac{2}{18}\times 0.72\mathrm{g}=0.08\mathrm{g}$

Amount of H in the molecule is $n_{\mathrm{H}}=\frac{m_{\mathrm{H}}}{M_{\mathrm{H}}}=\frac{0.08\mathrm{g}}{1\mathrm{g}{\mathrm{mol}}^{-1}}=0.08\mathrm{mol}$

Mass of C in the obtained ${\mathrm{CO}}_2$ is $\left(\frac{M_{\mathrm{C}}}{M_{{\mathrm{CO}}_2}}\right)m_{{\mathrm{CO}}_2}=\frac{12}{44}\times 3.08\mathrm{g}=0.84\mathrm{g}$

Amount of C in the molecule is $n_{\mathrm{C}}=\frac{m_{\mathrm{C}}}{M_{\mathrm{C}}}=\frac{0.84\mathrm{g}}{12\mathrm{g}{\mathrm{mol}}^{-1}}=0.07\mathrm{mol}$

The ratio of C and H in the molecule is ${\mathrm{n}}_{\mathrm{C}}:{\mathrm{n}}_{\mathrm{H}}::0.07\mathrm{mol}:0.08\mathrm{mol}::0.07:0.08::7:8$ 

Hence, the empirical formula of the compound is ${\mathrm{C}}_7{\mathrm{H}}_8$.