In Kjeldahl's method, of an organic compound is digested and the evolved ammonia is absorbed in of . — Practical Organic Chemistry and Purification Chemistry Question
Question
In Kjeldahl's method, $0.50\ \text{g}$ of an organic compound is digested and the evolved ammonia is absorbed in $50\ \text{mL}$ of $0.5\ \text{M}\ H_2SO_4$. The residual acid requires $60\ \text{mL}$ of $0.5\ \text{M}\ NaOH$ for complete neutralization. What is the percentage of nitrogen in the compound?
💡 Solution & Explanation
Milli-equivalents of initial acid = $50 \times 0.5 \times 2 = 50\ \text{mEq}$. Milli-equivalents of base used = $60 \times 0.5 \times 1 = 30\ \text{mEq}$. The acid consumed by ammonia = $50 - 30 = 20\ \text{mEq}$. Percentage of Nitrogen = $\frac{1.4 \times \text{mEq of acid used}}{W} = \frac{1.4 \times 20}{0.50} = 56\%$.