At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% 02 by volume for complete combustion. After combustion the gases occupy 330 mL. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is 

Volume of O, present in the given volume of air $=\left(\frac{20}{100}\right)\left(375\mathrm{mL}\right)=75\mathrm{mL}$

Volume of air without${\mathrm{O}}_2=375\mathrm{mL}-75\mathrm{mL}=300\mathrm{mL}$

The combustion reaction may be represented as 

$\begin{array}{r}\mathrm{C}\underset{15 \mathrm{mL}}{{ }_x{\mathrm{H}}_y } +\underset{75 \mathrm{mL}}{\left(x+\frac{y}{4}\right){\mathrm{O}}_2} \to x\underset{(330 - 300) \mathrm{mL}}{{\mathrm{CO}}_2+\frac{y}{2}{\mathrm{H}}_2\mathrm{O}}\end{array}$

                                              30mL

Obviously, the reaction is 

${\mathrm{C}}_x{\mathrm{H}}_y+5{\mathrm{O}}_2\to 2{\mathrm{CO}}_2+\frac{y}{2}{\mathrm{H}}_2\mathrm{O}$ 

Thus  $x=2$ and $y/4=5-x=5-2=3\text{ i.e. }y=12$

The hydrocarbon is ${\mathrm{C}}_2{\mathrm{H}}_{12 }$, which theoretically, is not possible.