A 0.166 g sample of an organic compound was digested with conc. H SO and then distilled with NaOH. T — Practical Organic Chemistry and Purification Chemistry Question
Question
A 0.166 g sample of an organic compound was digested with conc. H SO and then distilled with NaOH. The ammonia gas evolved was passed through 50.0 mL of 0.5 N H SO . The used acid required 30.0 mL of 0.25 N NaOH for complete neutralization. The mass percentage of nitrogen in the organic compound is_____. 2 4 2 4
💡 Solution & Explanation
**Step 1: Calculate moles of H₂SO₄ neutralized by NaOH** Moles of NaOH used = 0.25 N × 0.030 L = 0.0075 mol Since H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O Moles of H₂SO₄ neutralized = 0.0075 ÷ 2 = 0.00375 mol **Step 2: Calculate moles of H₂SO₄ consumed by NH₃** Initial moles of H₂SO₄ = 0.5 N × 0.050 L = 0.025 mol Moles of H₂SO₄ used by NH₃ = 0.025 - 0.00375 = 0.02125 mol **Step 3: Calculate moles of NH₃ produced** Reaction: 2NH₃ + H₂SO₄ → (NH₄)₂SO₄ Moles of NH₃ = 2 × 0.02125 = 0.0425 mol **Step 4: Calculate mass of nitrogen** Molar mass of N = 14 g/mol Mass of N = 0.0425 mol × 14 g/mol = 0.595 g **Step 5: Calculate mass percentage of nitrogen** Mass percentage = (0.595 g ÷ 0.166 g) × 100% = 63.00% Therefore, the answer is 63.00.