Question

While estimating the nitrogen present in an organic compound by Kjeldahl’s method, the ammonia evolved from 0.25 g of the compound neutralized 2.5 mL of 2 M H₂SO₄. The percentage of nitrogen present in organic compound is ________

Answer 1

Given : Mass of compound = \(0.25\,g\)

Volume of \(H_2SO_4 = 2.5\,mL = 2.5 \times 10^{-3}\,L\)

Molarity of \(H_2SO_4 = 2\,M\)

Moles of \(H_2SO_4\)

\(\begin{array}{l} = 2 \times 2.5 \times 10^{-3} = 5 \times 10^{-3} \end{array}\)

Reaction :

\(\begin{array}{l} H_2SO_4 + 2NH_3 \rightarrow (NH_4)_2SO_4 \end{array}\)

\(\begin{array}{l} 1\;mol\;H_2SO_4 \equiv 2\;mol\;NH_3 \end{array}\)

Moles of \(NH_3\)

\(\begin{array}{l} = 2 \times 5 \times 10^{-3} = 10 \times 10^{-3} = 0.01 \end{array}\)

Moles of nitrogen = moles of \(NH_3 = 0.01\)

Mass of nitrogen

\(\begin{array}{l} = 0.01 \times 14 = 0.14\,g \end{array}\)

Percentage of nitrogen

\(\begin{array}{l} = \frac{0.14}{0.25} \times 100 = 56\% \end{array}\)

Hence, percentage of nitrogen = 56%