10mL of gaseous hydrocarbon on combustion gives 40mL of CO _ 2 (g)...

The combustion reaction can be represented as 

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

At STP, $1$ mole of any gas occupies $22 . 4L$

$10mL$ of hydrocarbon, $50mL$ of water and $40mL$ of carbon dioxide corresponds to $0 . 45, 2 . 23$,  and $1 . 79$moles respectively.

Hence, the combustion of $1$ mole of hydrocarbon will produce $5$ moles of water and $4$ moles of carbon dioxide.

Thus, the molecular formula contains $4$ carbon atoms and $10$ hydrogen atoms.

The hydrocarbon is ${C}_{4}{H}_{10}$ and the combustion reaction is 

${C}_{4}{H}_{10} + 6 . 5{O}_{2} \to 4{CO}_{2} + 5{H}_{2}O$

$x + y = 4 + 10 = 14$

Hence the Total number of carbon and hydrogen atoms in the hydrocarbon is $14$.