$10 \mathrm{mL}$ of gaseous hydrocarbon on combustion gives $40 \mathrm{mL}$ of ${\mathrm{CO}}_2\left(\mathrm{g}\right)$ and $50 \mathrm{mL}$ of water vapour. Total number of carbon and hydrogen atoms in the hydrocarbon is _______.

The combustion reaction can be represented as 

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

At STP, $1$ mole of any gas occupies $22.4 \mathrm{L}$

$10 \mathrm{mL}$ of hydrocarbon, $50 \mathrm{mL}$ of water and $40 \mathrm{mL}$ 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 ${\mathrm{C}}_4{\mathrm{H}}_{10}$ and the combustion reaction is 

${\mathrm{C}}_4{\mathrm{H}}_{10}+6.5{\mathrm{O}}_2\to 4{\mathrm{CO}}_2+5{\mathrm{H}}_2\mathrm{O}$

$\mathrm{x}+\mathrm{y}=4+10=14$

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