Question

Calculate the percentage composition of Nitrogen in Urea. Urea formula is${\mathrm{NH}}_2{\mathrm{CONH}}_2$. (Atomic mass of $\mathrm{N}=14,\mathrm{H}=1,\mathrm{C}=12,\mathrm{O}=16$)

Answer 1

Step 1: Given data:-

The molecular formula of Urea= ${\mathrm{NH}}_2{\mathrm{CONH}}_2$

Atomic mass of $\mathrm{N}=14,\mathrm{H}=1,\mathrm{C}=12,\mathrm{O}=16$

Step 2: Calculation of molecular mass for urea:-

$$\begin{gathered} \mathrm{Molecular}\mathrm{mass}\mathrm{of}\mathrm{Urea}=\left(2\times \mathrm{atomic}\mathrm{mass}\mathrm{of}\mathrm{N}\right)+\left(\mathrm{atomic}\mathrm{mass}\mathrm{of}\mathrm{O}+4\right)\times \left(\mathrm{atomic}\mathrm{mass}\mathrm{of}\mathrm{H}\right)+\left(\mathrm{atomic}\mathrm{mass}\mathrm{of}\mathrm{C}\right) \\ =\left(2\times 14\right)+\left(16+4\right)\times \left(1+12\right) \\ \mathrm{Molecular}\mathrm{mass}\mathrm{of}\mathrm{Urea}=60\mathrm{gm} \end{gathered}$$

Step 3: Calculation of % composition of N:-

$$\begin{gathered} \%\mathrm{composition}\mathrm{of}\mathrm{N}=\frac{\mathrm{mass}\mathrm{of}\mathrm{nitogen}\mathrm{in}\mathrm{urea}}{\mathrm{mass}\mathrm{of}\mathrm{urea}}\times 100 \\ =\frac{28}{60}\times 100 \\ \%\mathrm{composition}\mathrm{of}\mathrm{N}=46.6\% \end{gathered}$$

Therefore, the percentage composition of Nitrogen in Urea is calculated to be $46.6\%$.