0.45g of an organic compound on combustion yielded 1.1g of ${\text{CO}}_{\text{2}}$ and 0.3g of water. Besides carbon and hydrogen, nitrogen is also present in the compound. The empirical formula of the compound would be: 

Weight of an organic compound =0.45 g

Weight of  $C{O}_2=1.1g$

Weight of $H_{2}O=0.3g$

Besides corbon & Hydrogen, $N_2$ is also present in the Compound.

The Empiral formula of the compound =?

$\because$Atomic weight of carbon $\left(M_C\right)=12g$

$\because$Molecular weight of  $Carbondioxide\left(M_{C{O}_2}\right)=44g$

$\because$Molecular weight of Hydrogen $\left(M_{H_2}\right)=2g$

$\because$Molecular weight of Hydrogen $\left(M_{H_{2}O}\right)=18g$

\Percent of $\text{C}=\left[\frac{M_{\text{C}}}{M_{{\text{CO}}_{\text{2}}}}\right]\text{(}{m}_{{\text{CO}}_{\text{2}}}\text{)}\left[\frac{\text{100}}{m\text{\hspace{0.33em}compound}}\right]$

$=\left[\frac{12}{44}\right]\left(1.1\right)\left[\frac{100}{0.45}\right]$

=66.67

\Percent of  $\text{H}=\left[\frac{2M_{\text{H}}}{M_{{\text{HO}}_{\text{2}}}}\right]\left(m_{{\text{H}}_{\text{2}}\text{O}}\right)\left[\frac{100}{m\text{\hspace{0.33em}compound}}\right]$

$=\left[\frac{2}{18}\right]\left(0.3\right)\left[\frac{100}{0.45}\right]$

$=7.4$

$\Rightarrow \text{C}:\text{H}:\text{N}=\frac{66.67}{12}:\frac{7.4}{1}:\frac{25.93}{14}$

$=5.56:7.4:1.85$

$=3:4:1$

\Empirical formula$={\text{C}}_{\text{3}}{\text{H}}_{\text{4}}\text{N}$