When a hydrocarbon A undergoes combustion in the presence of air,...

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

Number of equivalents of ${O}_{2}$ = Number of equivalents of ${H}_{2}O$

Number of equivalents of ${H}_{2}O=\dfrac{y}{2} = 3 ⇒ y = 6$

Number of equivalents of ${O}_{2}=x + \dfrac{y}{4} = 9 . 5$

$⇒ x + \dfrac{6}{4} = 9 . 5$

$⇒ x + 1 . 5 = 9 . 5$

$⇒ x = 9 . 5 - 1 . 5 = 8$

$⇒ x = 8, y = 6$.

So, the formula is ${C}_{8}{H}_{6}$.