Question

A gaseous compound is composed of $85.7\%$ by weight of $C$ and $14.3\%$ by weight of $H$. Its density is $2.28g/L$ at $300K$ and $1atm$ pressure. Calculate the molecular formula of the compound.

Answer 1

$d=2.28g/lit,T=300K,P=1atm$

$PM=dRT$

$M=\frac{2.28\times 0.0821\times 300}{1}=56.156g$

$n_C$ = $\frac{85.7}{12}=7.14$

$n_H$=$\frac{14.3}{1}=14.3$

$\therefore$ Empirical formula $=C{H}_2$

Empirical wt $=14$

Mol wt $=56$

$n=\frac{Mol.wt}{emp.wt}=\frac{56}{14}=4$

$\therefore$ Molecular formula $=(empiricalformula{)}_4=(C{H}_2{)}_4=C_4{H}_8$