When a hydrocarbon A undergoes combustion in the presence of air, it requires 9 . 5 equivalents of oxygen…

${\mathrm{C}}_{\mathrm{x}}{\mathrm{H}}_{\mathrm{y}}+\left(\mathrm{x}+\frac{\mathrm{y}}{4}\right){\mathrm{O}}_2\to {\mathrm{xCO}}_2+\frac{\mathrm{y}}{2}{\mathrm{H}}_2\mathrm{O}$

Number of equivalents of ${\mathrm{O}}_2$ = Number of equivalents of ${\mathrm{H}}_2\mathrm{O}$

Number of equivalents of ${\mathrm{H}}_2\mathrm{O}$ = $\frac{\mathrm{y}}{2}=3 \Rightarrow \mathrm{y} = 6$

Number of equivalents of ${\mathrm{O}}_2$ = $\mathrm{x}+\frac{\mathrm{y}}{4}= 9.5$

$\Rightarrow \mathrm{x} + \frac{6}{4} = 9.5$

$\Rightarrow \mathrm{x} + 1.5 = 9.5$

$\Rightarrow \mathrm{x} = 9.5 - 1.5 = 8$

$\Rightarrow \mathrm{x}=8, \mathrm{y}=6$.

So, the formula is ${\mathrm{C}}_8{\mathrm{H}}_6$.