How many moles of oxygen atoms are present in 42.5 g of calcium nitrate, Ca(NO3)2? (Molar masses: Ca = 40.1 g/mol, N = 14.0 g/mol, O = 16.0 g/mol)
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Correct answer
C. 1.55 mol
Principle or equation
Moles of compound = mass / molar mass. Each formula unit of Ca(NO3)2 contains 6 oxygen atoms. Moles of O atoms = moles of compound × 6.
Why this answer is correct
Molar mass of Ca(NO3)2 = 40.1 + 2*(14.0 + 3*16.0) = 40.1 + 2*(62.0) = 164.1 g/mol. Moles of Ca(NO3)2 = 42.5 g / 164.1 g/mol = 0.259 mol. Moles of O atoms = 0.259 × 6 = 1.55 mol.
Example
For 0.100 mol of Ca(NO3)2, O atoms = 0.600 mol.
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