A compound on analysis was found to contain 53.33%C, 15.5%H, and the rest nitrogen. The formula of the compound is

Total percent composition of the molecule is 100% or the total mass of the molecule is 100 g. Mass of C in the molecule is 53.33 g (53.33%), mass of hydrogen is 15.5 g ( or 15.5%).

Mass of nitrogen is calculated as:

$\begin{array}{ccc}\mathrm{Mass} \mathrm{of} \mathrm{nitrogen}& =& 100 \mathrm{g}-(15.5 \mathrm{g}+53.33 \mathrm{g})\\ & =& 31.17 \mathrm{g}\end{array}$

Number of moles of C is calculated as:

$\begin{array}{ccc}\mathrm{Moles} \mathrm{of} \mathrm{C}& =& \frac{\mathrm{Weight}}{\mathrm{Molar} \mathrm{mass}}\\ & =& \frac{53.33 \mathrm{g}}{12 \mathrm{g}/\mathrm{mol}}\\ & =& 4.44 \mathrm{mol}\end{array}$

Mole of H is calculated as:

$\begin{array}{ccc}\mathrm{Mole} \mathrm{of} \mathrm{H}& =& \frac{15.5 \mathrm{g}}{1 \mathrm{g}/\mathrm{mol}}\\ & =& 15.5 \mathrm{mol}\end{array}$

Mole of N is calculated as:

$\begin{array}{ccc}\mathrm{Mole} \mathrm{of} \mathrm{N}& =& \frac{31.17 \mathrm{g}}{14 \mathrm{g}/\mathrm{mol}}\\ & =& 2.22 \mathrm{mol}\end{array}$

Formula of the compound determine by diving the moles of C, N, and H with 2.22.

$\begin{array}{ccc}\mathrm{Ratio} \mathrm{of} \mathrm{C}:\mathrm{H}:\mathrm{N}& =& {\mathrm{C}}_{\frac{4.44}{2.22}}{\mathrm{H}}_{\frac{15.5}{2.22}}{\mathrm{N}}_{\frac{2.22}{2.22}}\\ & =& {\mathrm{C}}_2{\mathrm{H}}_7\mathrm{N}\end{array}$