Mass loss from Group 2 nitrate decomposition identifies the metal via percentage calculation

Steps:

• Decomposition: 2M(NO3)2 → 2MO + 4NO2 + O2; mass loss per mole M(NO3)2 = 108 g (from 2×46 + 16).
• Molar mass of M(NO3)2 = atomic mass M + 124 g/mol.
• Observed % mass loss = (3.2 g / 5.00 g) × 100% = 64%.
• Predicted % = [108 / (M + 124)] × 100% = 64%; solve M + 124 = 108 / 0.64 ≈ 168.75, M ≈ 44.75 g/mol (closest to Ca at 40).

Why B is correct:

• For Ca (M=40), molar mass 164 g/mol; % loss = 108/164 ≈ 65.9%, matches observed 64% within experimental approximation.

Why the others are wrong:

• A. Magnesium (M=24) gives 108/148 ≈ 73% loss, exceeds 64%.
• C. Strontium (M=88) gives 108/212 ≈ 51% loss, below 64%.
• D. Barium (M=137) gives 108/261 ≈ 41% loss, far below 64%.

Final answer: B