In an estimation of bromine by Carius method, 1.6 g of an organic compound gave 1.88 g of AgBr. The — Practical Organic Chemistry and Purification Chemistry Question

💡 Solution & Explanation

**Step 1: Find moles of AgBr produced** Molar mass of AgBr = 108 + 80 = 188 g/mol Moles of AgBr = 1.88 g ÷ 188 g/mol = 0.01 mol **Step 2: Determine moles of Br in the compound** From the formula AgBr, the mole ratio is: - 1 mol AgBr contains 1 mol Br Therefore, moles of Br = 0.01 mol **Step 3: Calculate mass of Br** Mass of Br = 0.01 mol × 80 g/mol = 0.8 g **Step 4: Apply the mass percentage formula** Mass percentage of Br = (mass of Br / mass of compound) × 100 Mass percentage of Br = (0.8 g / 1.6 g) × 100 **Step 5: Calculate final answer** Mass percentage of Br = 0.5 × 100 = 50.00% Therefore, the answer is 50.00.