An organic compound contains C = 48% and H = 8%. Further 0.48 gm of the compound was

 kjeldalised and the liberated ammonia required 19.2 m $\text{mL\hspace{0.17em}}\frac{\text{N}}{\text{2}}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$. The empirical formula of

the compound is  

$\text{\%\hspace{0.17em}N\hspace{0.17em}=\hspace{0.17em}}\frac{{\text{1.4\hspace{0.17em}×\hspace{0.17em}no\hspace{0.17em}of\hspace{0.17em}meq\hspace{0.17em}of\hspace{0.17em}H}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{\hspace{0.17em}used\hspace{0.17em}to\hspace{0.17em}neutralise\hspace{0.17em}NH}}_{\text{3}}}{\text{Mass\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}compound}}$

$\text{\%\hspace{0.17em}N=\hspace{0.17em}}\frac{\text{1.4\hspace{0.17em}×\hspace{0.17em}9.6}}{\text{0.48}}\text{\hspace{0.17em}×\hspace{0.17em}=\hspace{0.17em}28\%}$

$\begin{array}{cccc}\text{Element}& \text{\%\hspace{0.17em}of\hspace{0.17em}element}& \begin{array}{l}\text{Relative\hspace{0.17em}number}\\ \text{of\hspace{0.17em}atoms}\end{array}& \begin{array}{l}\text{Simplest}\\ \text{ratio}\end{array}\\ \text{C}& \text{48}& \frac{\text{48}}{\text{12}}\text{\hspace{0.17em}=\hspace{0.17em}4}& \text{4}\\ \text{H}& \text{8}& \frac{\text{8}}{\text{1}}\text{\hspace{0.17em}=\hspace{0.17em}8}& \text{8}\\ \text{N}& \text{28}& \frac{\text{28}}{\text{14}}\text{\hspace{0.17em}=\hspace{0.17em}2}& \text{2}\\ \text{O}& \text{16}& \frac{\text{16}}{\text{16}}\text{\hspace{0.17em}=\hspace{0.17em}1}& \text{1}\end{array}$

Emperical formula is ${\text{C}}_{\text{4}}{\text{H}}_{\text{8}}{\text{N}}_{\text{2}}\text{O}$