A sample of ammonium phosphate, (NH4)3PO4, has a mass of 74.5 g. How many moles of nitrogen atoms are present in this sample? (Molar masses: N = 14.0 g/mol, H = 1.0 g/mol, P = 31.0 g/mol, O = 16.0 g/mol)
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Correct answer
C. 1.50 mol
Principle or equation
The molar mass of a compound is the sum of the atomic masses of its constituent atoms. The number of moles of a specific element in a sample is the moles of the compound multiplied by the number of atoms of that element per formula unit.
Why this answer is correct
Molar mass of (NH4)3PO4 = 3*(14.0 + 4*1.0) + 31.0 + 4*16.0 = 3*18.0 + 31.0 + 64.0 = 54.0 + 95.0 = 149.0 g/mol. Moles of compound = 74.5 g / 149.0 g/mol = 0.500 mol. Each formula unit contains 3 nitrogen atoms, so moles of N = 0.500 mol * 3 = 1.50 mol.
Example
For 0.500 mol of (NH4)3PO4, there are 0.500 * 3 = 1.50 mol of N atoms.
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