Answer
Molecular formula: $C_{4}H_{8}$
Work Step by Step
$\omega(C) = 85.63 \%$
$\omega(H) = 14.37 \%$
$\rho = 2.5 \frac{g}{L}$
$T=273.15K$
$p=1atm=101.3kPa$
---Molecular formula?
$pV=nRT \implies pV=\frac{m}{M}RT \implies M=\frac{m}{V}\frac{RT}{p}=\frac{\rho RT}{p}=\frac{2.5\frac{g}{dm^{3}}\times 8.314\frac{J}{K\times mol} \times 273.15K}{101.3kPa}=56\frac{g}{mol}$
Let's suppose that the molecular formula of this gas is $C_{x}H_{y}$, Since $A(C)=12\frac{g}{mol}$ and $A(H)=1\frac{g}{mol}$, we have $12\frac{g}{mol}\times x=0.8563\times 56\frac{g}{mol}=48\frac{g}{mol}$. Hence, $x=4$. The molecular weight of this gas is equal to $(12x+y) \frac{g}{mol}$, so we can obtain the following equation: $12x+y=56$. Using $x=4$, we can conclude that $y=8$, so the molecular formula of this gas is $C_{4}H_{8}$.
