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Chemical Calculations

The mole, the Avogadro constant and molar mass: moles and percentage composition by mass

  • 3 things to know
  • AS Unit 1

Overview

Why a mole of any element holds the same number of atoms, calculating amounts in moles and masses from them, and working out the percentage by mass of an element in a compound and the mass extractable from an ore.

What you need to know

Answer each one before you reveal it.

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Numbers

What mass of iron contains the same number of atoms as 12.0 g of carbon?

Numbers

Calculate the amount, in moles, of sodium ions in 5.85 g of sodium chloride.

Key term

What percentage by mass of iron(III) oxide is iron?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Explain why one mole of magnesium has a greater mass than one mole of carbon even though both contain the same number of atoms.

    Model answer

    Both contain 6.02 × 10²³ atoms, but a magnesium atom has a greater relative atomic mass than a carbon atom, 24.3 against 12.0. The mole fixes the number of particles, not the mass, so the heavier atoms give the greater molar mass.

  2. 2

    Calculate the amount, in moles, of sodium ions in 5.85 g of sodium chloride.

    Model answer

    M(NaCl) = 23.0 + 35.5 = 58.5 g mol⁻¹, so n(NaCl) = 5.85 ÷ 58.5 = 0.100 mol. Each formula unit provides one sodium ion, so there is 0.100 mol of Na⁺.

  3. 3

    What mass of iron contains the same number of atoms as 12.0 g of carbon?

    Model answer

    12.0 g of carbon is 12.0 ÷ 12.0 = 1.00 mol, so 1.00 mol of iron is needed. With Aᵣ(Fe) = 55.8 the mass required is 1.00 mol × 55.8 g mol⁻¹ = 55.8 g.

  4. 4

    Why does the value of the Avogadro constant have the unit mol⁻¹ rather than no unit at all?

    Model answer

    It is a number of particles per mole, so it converts an amount in moles into a count of particles. Writing it as 6.02 × 10²³ mol⁻¹ makes the units cancel correctly when it multiplies an amount in mol.

  5. 5

    Calculate the amount in moles of calcium carbonate in 100 g, and comment on the answer.

    Model answer

    M(CaCO₃) = 100.1 g mol⁻¹, so n = 100 ÷ 100.1 = 0.999 mol. It is very slightly less than one mole, because the molar mass is a little above 100 g mol⁻¹, and rounding it to 100 would hide that.

  6. 6

    Determine the percentage by mass of nitrogen in ammonium nitrate.

    Model answer

    M(NH₄NO₃) = (2 × 14.0) + (4 × 1.0) + (3 × 16.0) = 28.0 + 4.0 + 48.0 = 80.0 g mol⁻¹. The nitrogen contributes 28.0 of this, so the percentage is (28.0 ÷ 80.0) × 100 = 35.0%.

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