Chemistry - Nomenclature Question with Solution | TestHub

${\mathrm{C}}_{\mathrm{x}}{\mathrm{H}}_{\mathrm{y}}+\left(\mathrm{x}+\frac{\mathrm{y}}{4}\right){\mathrm{O}}_2\to \mathrm{x}\mathrm{C}{\mathrm{O}}_2+\frac{\mathrm{y}}{2}{\mathrm{H}}_2\mathrm{O}$
Number of moles of$\mathrm{C}{\mathrm{O}}_2=\frac{3.08}{44}=0.07$
So$\mathrm{x}=0.07$
Number of moles of${\mathrm{H}}_2\mathrm{O}=\frac{0.72}{18}=0.04$
So$\mathrm{y}=0.08$
$\mathrm{x}:\mathrm{y}=7:8$
Therefore, empirical formula of hydrocarbon is${\mathrm{C}}_7{\mathrm{H}}_8$.