Exactly of an organic compound containing nitrogen was digested according to Kjeldahl's method. The — Qualitative and Quantitative Analysis Chemistry Question
Question
Exactly $29.5 \text{ mg}$ of an organic compound containing nitrogen was digested according to Kjeldahl's method. The evolved ammonia was passed into $20 \text{ mL}$ of $0.1 \text{ M}$ $HCl$ solution. The unreacted excess acid required exactly $15 \text{ mL}$ of $0.1 \text{ M}$ $NaOH$ solution for complete neutralization. What is the precise percentage of nitrogen in the compound?
💡 Solution & Explanation
Milli-moles of $HCl$ initially taken = $20 \text{ mL} \times 0.1 \text{ M} = 2.0 \text{ mmol}$. Milli-moles of $NaOH$ used for back titration = $15 \text{ mL} \times 0.1 \text{ M} = 1.5 \text{ mmol}$. Therefore, milli-moles of $HCl$ neutralized by $NH_3 = 2.0 - 1.5 = 0.5 \text{ mmol}$. Since 1 mole of $NH_3$ neutralizes 1 mole of $HCl$, moles of $NH_3$ evolved = $0.5 \text{ mmol} = 0.5 \times 10^{-3} \text{ mol}$. Weight of Nitrogen = $0.5 \times 10^{-3} \times 14 = 7 \times 10^{-3} \text{ g} = 7 \text{ mg}$. Percentage of $N = (\frac{7 \text{ mg}}{29.5 \text{ mg}}) \times 100 \approx 23.7\%$.