In a Kjeldahl determination, $0.620 g$ of an organic compound was digested and the ammonia evolved was neutralised exactly by $27.5 mL$ of $0.100 M H X_{2} SO X_{4}$ . Calculate the percentage of nitrogen in the compound (two decimals; atomic mass of N = 14).

Numerical answer type, answer in %, tolerance 0.02.

Answer: 12.42 %

2 NH X_{3} + H X_{2} SO X_{4} (NH X_{4}) X_{2} SO X_{4}

n (NH X_{3}) = 2 \times 0.100 \times 0.0275 = 5.50 \times 1 0^{- 3}

mol. Mass of N

= 5.50 \times 1 0^{- 3} \times 14 = 0.0770

g. Percentage

= 0.0770/0.620 \times 100 = 12.42%