$25\mathrm{g}$ of an unknown hydrocarbon upon burning produces $88g$ of ${\mathrm{CO}}_2$ and $9\mathrm{g}$ of ${\mathrm{H}}_2\mathrm{O}$. This unknown hydrocarbon contains

$$\begin{gathered} {\mathrm{C}}_{\mathrm{X}}{\mathrm{H}}_{\mathrm{Y}} + \left(\mathrm{x}+\frac{\mathrm{y}}{4}\right){\mathrm{O}}_2⟶{\mathrm{xCO}}_2+ \frac{\mathrm{Y}}{2} {\mathrm{H}}_2\mathrm{O} \\ \left(\frac{25}{\mathrm{M}}\right) \mathrm{xx}\frac{25}{\mathrm{M}} \frac{\mathrm{y}}{2}\mathrm{x}\frac{25}{\mathrm{M}} \\ =2 =0.5 \\ \mathrm{C} \mathrm{xx}\frac{25}{\mathrm{M}}=2 \\ \mathrm{H} \mathrm{yx}\frac{25}{\mathrm{M}}=1 \end{gathered}$$

${\mathrm{C}}_{2\mathrm{y}}{\mathrm{H}}_{\mathrm{y}}\equiv 24\mathrm{y} \mathrm{gm} \mathrm{C}+\mathrm{y} \mathrm{gm} \mathrm{H}$ or $24 : 1$ ratio by mass.