Question

What is the percentage of Nitrogen in urea $\left({\mathrm{NH}}_2{\mathrm{CONH}}_2\right)$?

Answer 1

Step 1: Calculation of molecular mass of urea

$$\begin{gathered} \mathrm{Molecular}\mathrm{mass}\mathrm{of}{\mathrm{NH}}_2{\mathrm{CONH}}_2=\left(\mathrm{Molar}\mathrm{mass}\mathrm{of}\mathrm{N}\times 2\right)+\left(\mathrm{Molar}\mathrm{mass}\mathrm{of}\mathrm{H}\times 4\right)+\left(\mathrm{Molar}\mathrm{mass}\mathrm{of}\mathrm{O}\times 1\right)+\left(\mathrm{Molar}\mathrm{mass}\mathrm{of}\mathrm{C}\times 1\right) \\ =\left(14\mathrm{g}{\mathrm{mol}}^{-1}\times 2\right)+\left(1\mathrm{g}{\mathrm{mol}}^{-1}\times 4\right)+\left(16\mathrm{g}{\mathrm{mol}}^{-1}\times 1\right)+\left(12\mathrm{g}{\mathrm{mol}}^{-1}\times 1\right) \\ =60\mathrm{g}{\mathrm{mol}}^{-1} \end{gathered}$$

Step 2: Calculation of percentage of nitrogen in urea

$$\begin{gathered} \mathrm{Percentage}\mathrm{of}\mathrm{Nitrogen}\mathrm{in}{\mathrm{NH}}_2{\mathrm{CONH}}_2=\frac{\mathrm{Mass}\mathrm{of}\mathrm{N}\mathrm{in}{\mathrm{NH}}_2{\mathrm{CONH}}_2}{\mathrm{Molecular}\mathrm{mass}\mathrm{of}{\mathrm{NH}}_2{\mathrm{CONH}}_2}\times 100 \\ =\frac{28\mathrm{g}{\mathrm{mol}}^{-1}}{60\mathrm{g}{\mathrm{mol}}^{-1}}\times 100 \\ =46.66\% \end{gathered}$$

Therefore, the required percentage is $46.66\%$.