0.2 g of an organic compound was subjected to estimation of nitrogen by Dumas method in which volume — Practical Organic Chemistry and Purification Chemistry Question
Question
0.2 g of an organic compound was subjected to estimation of nitrogen by Dumas method in which volume of N evolved (at STP) was found to be 22.400 mL. The percentage of nitrogen in the compound is _______ [nearest integer] (Given: Molar mass of N is 28 g mol , Molar volume of N at STP: 22.4L) 2 2 –1 2
💡 Solution & Explanation
**Step 1: Convert volume of N₂ to moles** Volume of N₂ evolved = 22.400 mL = 0.0224 L Using molar volume at STP (22.4 L/mol): Moles of N₂ = 0.0224 L ÷ 22.4 L/mol = 0.001 mol **Step 2: Calculate mass of nitrogen** Each N₂ molecule contains 2 nitrogen atoms. Moles of N = 2 × 0.001 = 0.002 mol Mass of N = 0.002 mol × 28 g/mol = 0.056 g **Step 3: Calculate percentage of nitrogen** Percentage of N = (Mass of N / Mass of compound) × 100 Percentage of N = (0.056 g / 0.2 g) × 100 Percentage of N = 0.28 × 100 = 28% **Step 4: Check calculation** 0.056/0.2 = 0.28 = 28% Wait - the correct answer is 14%. Let me recalculate: if molar mass of N₂ is 28 g/mol (as stated), then mass of N₂ = 0.001 mol × 28 = 0.028 g Mass of N atoms in N₂ = 0.028 g (since N₂ mass = mass of N) Percentage = (0.028/0.2) × 100 = 14% Therefore, the answer is 14.