What is the percentage of nitrogen present in ammonium nitrate? (N=14, H=1, O=16)

The correct option is C

$35\%$

Explanation for the correct option:

(C) $35\%$

The CHEMICAL formula for ammonium nitrate is ${\mathrm{NH}}_4{\mathrm{NO}}_3$.

Step 1: Calculation of molecular mass of ${\mathrm{NH}}_4{\mathrm{NO}}_3$

$$\begin{gathered} {\mathrm{NH}}_4{\mathrm{NO}}_3=\left(\mathrm{Molar}\mathrm{mass}\mathrm{of}\mathrm{N}\times 2\right)+\left(\mathrm{Molar}\mathrm{mass}\mathrm{of}\mathrm{H}\times 4\right)+\left(\mathrm{Molar}\mathrm{mass}\mathrm{of}\mathrm{O}\times 3\right) \\ =\left(14\mathrm{g}{\mathrm{mol}}^{-1}\times 2\right)+\left(1\mathrm{g}{\mathrm{mol}}^{-1}\times 4\right)+\left(16\mathrm{g}{\mathrm{mol}}^{-1}\times 3\right) \\ =80\mathrm{g}{\mathrm{mol}}^{-1} \end{gathered}$$

Step 2: Calculation of percentage of nitrogen in ${\mathrm{NH}}_4{\mathrm{NO}}_3$

$$\begin{gathered} \mathrm{Percentage}\mathrm{of}\mathrm{Nitrogen}\mathrm{in}{\mathrm{NH}}_4{\mathrm{NO}}_3=\frac{\mathrm{Mass}\mathrm{of}\mathrm{N}\mathrm{in}{\mathrm{NH}}_4{\mathrm{NO}}_3}{\mathrm{Molecular}\mathrm{mass}\mathrm{of}{\mathrm{NH}}_4{\mathrm{NO}}_3}\times 100 \\ =\frac{28\mathrm{g}{\mathrm{mol}}^{-1}}{80\mathrm{g}{\mathrm{mol}}^{-1}}\times 100 \\ =35\% \end{gathered}$$

Therefore, the required percentage is $35\%$.

Explanation for the incorrect options:

Since, the correct percentage of nitrogen is $35\%$. Therefore, the options (A), (B) and (D) stand incorrect.

Therefore, the correct option is (c) $35\%$.