How many moles of oxygen atoms are present in 25.0 g of copper(II) nitrate, Cu(NO3)2? (Molar masses: Cu = 63.5 g/mol, N = 14.0 g/mol, O = 16.0 g/mol)
Show the answer and explanation
Correct answer
B. 0.533 mol
Principle or equation
Moles of compound = mass / molar mass. Multiply by the number of oxygen atoms per formula unit.
Why this answer is correct
Molar mass of Cu(NO3)2 = 63.5 + 2*(14.0 + 3*16.0) = 63.5 + 2*(62.0) = 187.5 g/mol. Moles of Cu(NO3)2 = 25.0 / 187.5 = 0.1333 mol. Each formula unit has 6 O atoms, so moles of O = 0.1333 * 6 = 0.800 mol.
Example
For 0.100 mol of Ca(NO3)2, moles of O = 0.100 * 6 = 0.600 mol.
This published item includes a stored explanation and passed the platform’s publication workflow. It is independent preparation material, not a claim of an official or recalled examination question.