How many moles of oxygen atoms are present in 49.5 g of magnesium nitrate, Mg(NO3)2? (Molar masses: Mg = 24.3 g/mol, N = 14.0 g/mol, O = 16.0 g/mol)
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Correct answer
B. 2.00 mol
Principle or equation
The molar mass of Mg(NO3)2 is calculated by summing atomic masses. The number of moles of the compound is mass divided by molar mass. Each formula unit contains 6 oxygen atoms, so moles of O = 6 × moles of compound.
Why this answer is correct
Molar mass of Mg(NO3)2 = 24.3 + 2*(14.0 + 3*16.0) = 24.3 + 2*(62.0) = 148.3 g/mol. Moles of compound = 49.5 / 148.3 ≈ 0.334 mol. Moles of O = 6 × 0.334 ≈ 2.00 mol.
Example
For 85.0 g of NaNO3 (molar mass 85.0 g/mol), moles = 1.00, and O atoms = 3.00 mol.
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