A sample of 0.125 g of an organic compound when analysed by Duma’s method yields 22.78 mL of nitroge — Practical Organic Chemistry and Purification Chemistry Question
Question
A sample of 0.125 g of an organic compound when analysed by Duma’s method yields 22.78 mL of nitrogen gas collected over KOH solution at 280K and 759 mm Hg. The percentage of nitrogen in the given organic compound is ____. (Nearest integer). (a) The vapour pressure of water at 280 K is 14.2 mm Hg (b) R = 0.082 L atm K mol –1 –1
💡 Solution & Explanation
**Step 1: Correct the gas volume for water vapor pressure** Nitrogen is collected over KOH solution, so we must subtract water vapor pressure from total pressure. P(N₂) = 759 - 14.2 = 744.8 mm Hg = 0.9800 atm **Step 2: Convert volume and temperature to SI units** V(N₂) = 22.78 mL = 0.02278 L T = 280 K **Step 3: Apply ideal gas law to find moles of nitrogen** Using PV = nRT: n(N₂) = PV/RT = (0.9800 × 0.02278)/(0.082 × 280) n(N₂) = 0.02232/22.96 = 0.000972 mol **Step 4: Convert moles of N₂ to mass of nitrogen** Mass of N = 0.000972 mol × 28 g/mol = 0.0272 g **Step 5: Calculate percentage of nitrogen** % N = (mass of N / mass of compound) × 100 % N = (0.0272/0.125) × 100 = 21.76% **Step 6: Round to nearest integer** % N ≈ 22% Therefore, the answer is 22.00.