How many moles of oxygen atoms are present in 25.0 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
C. 1.01 mol
Principle or equation
Moles of compound = mass / molar mass. Each formula unit of Mg(NO3)2 contains 6 oxygen atoms.
Why this answer is correct
Molar mass of Mg(NO3)2 = 24.3 + 2*(14.0 + 3*16.0) = 24.3 + 2*(62.0) = 24.3 + 124.0 = 148.3 g/mol. Moles of Mg(NO3)2 = 25.0 / 148.3 = 0.1686 mol. Moles of O = 0.1686 * 6 = 1.0116 ≈ 1.01 mol.
Example
For 10.0 g of CaCO3 (molar mass 100 g/mol), moles = 0.100 mol, and moles of O = 0.100 * 3 = 0.300 mol.
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