Mass–mole–particle calculations

8 practice questions

Convert between mass and moles

Learning goalUse n = m/Mᵣ and m = nMᵣ with the correct units.

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Use the relative molecular or formula mass, M_r, in mass-to-moles calculations. M_r has no unit. Its numerical value is the same as the mass of one mole in grams: M_r(H₂SO₄) = 98.1 means one mole of H₂SO₄ has a mass of 98.1 g.

Add the Aᵣ contributions in the formula to find M_r.
With mass in grams, divide by M_r to find amount in moles: n = m/M_r.
To find mass in grams, multiply the amount in moles by M_r: m = nM_r.

These are numerical calculation rules using mass in g and amount in moles. For atoms, use Aᵣ instead. Molar mass is the mass per mole and has unit g mol⁻¹; M_r itself remains unitless.

Compare with one mole to check your answer. If M_r(CaCO₃) = 100.1, then 0.350 mol must have a mass below 100.1 g.

n = m/M_r; m = nM_r; M_r = m/n

M_r relative molecular or formula mass; no unit; m mass of the specified sample or substance, in g; n amount of the specified substance or particles, in mol.

Worked example

Mass to moles

Calculate the number of moles in 9.81 g H₂SO₄; M_r = 98.1.

Worked answer

n = m/Mᵣ
n = 9.81 / 98.1 = 0.100 mol

0.100 mol H₂SO₄.

Worked example

Moles to mass

Calculate the mass of 0.350 mol CaCO₃; M_r = 100.1.

Worked answer

m = n × Mᵣ
m = 0.350 × 100.1 = 35.035 g

35.0 g CaCO₃ to three significant figures.

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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 6

Calculate the number of moles in 11.7 g NaCl. Mᵣ(NaCl) = 58.5. (2 marks)
Calculate the mass of 0.0750 mol CO₂. Mᵣ(CO₂) = 44.0. (2 marks)
A molecular substance has mass 7.30 g and contains 0.125 mol. Calculate its relative molecular mass. (2 marks)

Total: 6 marks

Show answer & marking points

(a) n = m/Mᵣ
n = 11.7/58.5 [1] = 0.200 mol. [1]

(b) m = n × Mᵣ
m = 0.0750 × 44.0 [1] = 3.30 g. [1]

✓ DONE WHENUse n = m/Mᵣ and m = nMᵣ with the correct units.

Convert between moles and particles

Learning goalMultiply or divide by the Avogadro constant in the correct direction.

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The Avogadro constant, L, connects amount in moles with particle number. Use L = 6.022 × 10²³ mol⁻¹ unless a question supplies another value.

Moles to particles: multiply by L, using N = nL.
Particles to moles: divide by L, using n = N/L.

N is the number of particles and has no unit. The amount of substance, n, is measured in mol. Label the answer with the entity, such as molecules or chloride ions. L has unit mol⁻¹ because it gives the number of particles per mole.

If the target is an atom or ion within a formula, include its formula multiplier as well.

N = nL; n = N/L

L Avogadro constant, 6.022 × 10²³ mol⁻¹; N number of the specified particles; a count with no unit; n amount of the specified substance or particles, in mol.

Worked example

Count ammonia molecules

How many molecules are in 0.0750 mol NH₃?

Use L = 6.022 × 10²³ mol⁻¹.

Worked answer

N = nL
N = 0.0750 × 6.022 × 10²³ = 4.5165 × 10²²

4.52 × 10²² NH₃ molecules.

Worked example

Count an amount of neon

Find the amount in moles represented by 1.204 × 10²⁴ Ne atoms.

Use L = 6.022 × 10²³ mol⁻¹.

Worked answer

n = (1.204 × 10²⁴) / (6.022 × 10²³)

2.000 mol Ne atoms to four significant figures.

EXAM TRAPUse brackets around the full denominator when entering a power of ten. A small fraction of a mole still contains an enormous number of particles.

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Quick check 7

Use L = 6.022 × 10²³ mol⁻¹. Calculate the amount in moles represented by 9.03 × 10²² atoms. (2 marks)

Total: 2 marks

Show answer & marking points

n = N/L = (9.03 × 10²²)/(6.022 × 10²³) [1] = 0.150 mol of atoms. [1]

✓ DONE WHENMultiply or divide by the Avogadro constant in the correct direction.

Count atoms, ions and electrons

Learning goalApply formula subscripts once and use atomic number and charge to count electrons.

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The amount of a compound is not automatically the amount of every particle inside it. Use the formula to count how many of the requested particles occur in one molecule or formula unit.

Find the amount in moles of the stated substance.
Multiply by the number of target atoms or ions in each formula.
Use N = nL if the question asks for the actual particle number.

For electron counts, use atomic number and ionic charge first. Al³⁺ has 13 - 3 = 10 electrons; Cl⁻ has 17 + 1 = 18. Multiply the amount of those ions by their electron count, then by L.

Write the entity beside intermediate amounts. This makes it easier to spot an incorrect multiplier.

One formula unit of Al₂(SO₄)₃ represents	Count
Al³⁺ ions	2
SO₄²⁻ ions	3
All ions	5
O atoms	12
All atoms	2 + 3 + 12 = 17

Worked example

Ions from aluminium sulfate

For 0.120 mol Al₂(SO₄)₃, find the amounts in moles of each ion and the total number of ions.

Use L = 6.022 × 10²³ mol⁻¹.

Worked answer

Moles of aluminium ions = moles of Al₂(SO₄)₃ × aluminium ions per formula unit
n(Al³⁺) = 2 × 0.120 = 0.240 mol

Moles of sulfate ions = moles of Al₂(SO₄)₃ × sulfate ions per formula unit
n(SO₄²⁻) = 3 × 0.120 = 0.360 mol

N(total ions) = nL
N(total ions) = (0.240 + 0.360) × 6.022 × 10²³

3.61 × 10²³ ions in total.

EXAM TRAPApply a subscript multiplier once. If you already have moles of oxygen atoms, do not multiply by the oxygen subscript again.

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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 8

Use L = 6.022 × 10²³ mol⁻¹.

Calculate the amounts in moles of Al³⁺ and SO₄²⁻ ions represented by 0.0800 mol Al₂(SO₄)₃. (2 marks)
Calculate the total number of ions represented by this sample. (2 marks)
Sulfur has atomic number 16. Calculate the number of electrons in 0.0150 mol S²⁻ ions. (3 marks)

Total: 7 marks

Show answer & marking points

(a) Moles of aluminium ions = moles of Al₂(SO₄)₃ × aluminium ions per formula unit
n(Al³⁺) = 2 × 0.0800 = 0.160 mol [1]; Moles of sulfate ions = moles of Al₂(SO₄)₃ × sulfate ions per formula unit
n(SO₄²⁻) = 3 × 0.0800 = 0.240 mol. [1]

(b) Total moles of ions = moles of Al₂(SO₄)₃ × total ions per formula unit
n(all ions) = (2 + 3) × 0.0800 = 0.400 mol [1]; N = nL
N = 0.400 × 6.022 × 10²³ = 2.41 × 10²³ ions. [1]

(c) Each S²⁻ has 16 + 2 = 18 electrons. [1] n(electrons) = electrons per ion × n(ions)
n(electrons) = 18 × 0.0150 = 0.270 mol. [1] N = nL
N = 0.270 × 6.022 × 10²³ = 1.63 × 10²³ electrons. [1]

✓ DONE WHENApply formula subscripts once and use atomic number and charge to count electrons.

Use the mass–mole–particle route

Learning goalConnect a measured mass to a specified particle count.

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For a mass-to-particles question, make two clear moves: mass to moles, then moles to particles.

Divide mass in grams by M_r to find moles of the substance. For atoms, use Aᵣ.
If needed, use the formula to find moles of the particular atom or ion being counted.
Multiply that mole amount by L.

For molecules or formula units, N = (m/M_r)L when m is in grams. If you need a component particle, include its formula multiplier. Writing both stages makes the target easier to track.

Mass → divide by M_r → moles → multiply by L → particles

Worked example

Chloride ions in sodium chloride

How many chloride ions are represented by 5.85 g NaCl? M_r(NaCl) = 58.5; L = 6.022 × 10²³ mol⁻¹.

Worked answer

n(NaCl) = m/Mᵣ
n(NaCl) = 5.85 / 58.5 = 0.100 mol.
Each formula unit represents one Cl⁻ ion, so n(Cl⁻) = 0.100 mol.
N(Cl⁻) = nL
N(Cl⁻) = 0.100 × 6.022 × 10²³.

6.02 × 10²² chloride ions.

EXAM TRAPA mass comparison alone does not tell you which sample contains more molecules, atoms or ions. First account for molar mass, then the target-entity multiplier.

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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 9

Use L = 6.022 × 10²³ mol⁻¹.

Calculate the number of H₂O molecules in 9.00 g water. Mᵣ(H₂O) = 18.0. (2 marks)
Calculate the number of oxygen atoms in 8.80 g CO₂. Mᵣ(CO₂) = 44.0. (3 marks)

Total: 5 marks

Show answer & marking points

(a) n(H₂O) = m/Mᵣ
n(H₂O) = 9.00/18.0 = 0.500 mol [1]; N = nL
N = 0.500 × 6.022 × 10²³ = 3.01 × 10²³ molecules. [1]

(b) n(CO₂) = m/Mᵣ
n(CO₂) = 8.80/44.0 = 0.200 mol [1]; Moles of oxygen atoms = moles of CO₂ × oxygen atoms per molecule
n(O atoms) = 2 × 0.200 = 0.400 mol [1]; N = nL
N = 0.400 × 6.022 × 10²³ = 2.41 × 10²³ oxygen atoms. [1]

Exam-style question 10

Sample A contains 3.20 g O₂, Mᵣ = 32.0. Sample B contains 4.40 g CO₂, Mᵣ = 44.0. Determine which contains more oxygen atoms. Support your answer with calculations. (4 marks)

Total: 4 marks

Show answer & marking points

A: n(O₂) = m/Mᵣ
n(O₂) = 3.20/32.0 = 0.100 mol. [1] B: n(CO₂) = m/Mᵣ
n(CO₂) = 4.40/44.0 = 0.100 mol. [1] Each molecule contains two O atoms, so each sample contains 0.200 mol O atoms. [1] The numbers of oxygen atoms are equal. [1] Multiplication by L is valid but unnecessary for the comparison.

✓ DONE WHENConnect a measured mass to a specified particle count.

Reverse the route and find an unknown relative molecular mass

Learning goalMove from particles to moles before finding mass or relative molecular mass.

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A reverse calculation follows the same relationships in the opposite direction. Start with the target and the information you have.

Divide a particle count by L to find moles.
Undo any formula multiplier if you need moles of the whole compound.
To find mass in grams, use m = nM_r.
To find an unknown M_r, divide mass in grams by amount in moles: M_r = m/n. Give no unit for M_r.

Label the substance and entity at each step. Keep the full calculator value of the amount until the final answer.

Worked example

Particles to mass

Find the mass of 3.01 × 10²² CO₂ molecules. M_r(CO₂) = 44.0.

Use L = 6.022 × 10²³ mol⁻¹.

Worked answer

n = (3.01 × 10²²) / (6.022 × 10²³) = 0.0499834… mol

m = n × Mᵣ
m = 0.0499834… × 44.0

2.20 g CO₂.

Worked example

Find an unknown relative molecular mass

A 1.83 g molecular sample contains 2.50 × 10²² molecules.

Calculate the relative molecular mass, Mᵣ, of the substance. Use L = 6.022 × 10²³ mol⁻¹.

Worked answer

n = (2.50 × 10²²) / (6.022 × 10²³) = 0.0415144… mol

M_r = m/n = 1.83 / 0.0415144…

M_r = 44.1 with no unit.

EXAM TRAPRetain the full calculator value of n. Molar mass is mass per mole, not the mass of the sample.

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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 11

A sample contains 7.525 × 10²² O₂ molecules. Mᵣ(O₂) = 32.0; L = 6.022 × 10²³ mol⁻¹.

Calculate the mass of the sample. (3 marks)
State the appropriate number of significant figures for the final answer and justify it. (1 mark)

Total: 4 marks

Show answer & marking points

(a) n = (7.525 × 10²²)/(6.022 × 10²³) = 0.124958… mol. [1] m = n × 32.0 [1] = 3.99867… g, giving 4.00 g. [1]

(b) Three significant figures, limited by the supplied Mᵣ = 32.0. [1]

Exam-style question 12

A 2.30 g molecular sample contains 3.01 × 10²² molecules. Use L = 6.022 × 10²³ mol⁻¹.

Calculate its relative molecular mass. (3 marks)
State the mass of one mole of the substance. (1 mark)
Use an approximate calculation to check that your relative molecular mass is reasonable. (1 mark)

Total: 5 marks

Show answer & marking points

(a) n = (3.01 × 10²²)/(6.022 × 10²³) = 0.0499834… mol. [1] Mᵣ = 2.30/n [1] = 46.0, with no unit. [1]

(b) 46.0 g. [1]

(c) The sample is approximately 0.0500 mol, one-twentieth of a mole, so one mole should have mass about 20 × 2.30 = 46 g. [1]

✓ DONE WHENMove from particles to moles before finding mass or relative molecular mass.

Standard form, units and final checks

Learning goalFinish calculations with a correct unit, particle label and appropriate precision.

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A good method can still lose marks through units, powers of ten or early rounding. Make a final check part of every calculation.

Keep the calculator's full value during intermediate steps. Round only the final answer.
Use sensible significant figures from the measured or supplied data. Exact subscripts and equation coefficients do not limit them.
Check the unit and the entity: mol, g, atoms and molecules describe different answers.
Check the direction. Dividing by a large molar mass should give fewer moles for a fixed mass.

In standard form, one non-zero digit comes before the decimal point. For example, 0.003045 mol = 3.045 × 10⁻³ mol. At three significant figures, 7.849 × 10²² becomes 7.85 × 10²².

EXAM TRAPIf dividing by L produces a tiny particle count from an ordinary laboratory sample, recheck the direction. Changing the calculation to multiplication is necessary when moving from moles to particles.

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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.

Exam-style question 13

A student calculates the number of CO₂ molecules in 2.50 g CO₂, Mᵣ = 44.0, using L = 6.022 × 10²³ mol⁻¹.

Student working: n = 2.50/44.0 = 0.0568 mol; N = 0.0568/(6.022 × 10²³) = 9.44 × 10⁻²⁶.

Identify the error in converting amount to particle number. (1 mark)
Calculate the correct number of CO₂ molecules. (3 marks)
Explain why the unrounded amount should be retained and state the appropriate final precision. (2 marks)
Write 0.004062 mol in standard form. (1 mark)

Total: 7 marks

Show answer & marking points

(a) Moving from moles to particle number requires multiplication by L. [1]

(b) n = m/Mᵣ
n = 2.50/44.0 = 0.056818… mol [1]; N = nL [1] = 3.42 × 10²² CO₂ molecules. [1]

(c) Retaining the calculator value avoids an extra rounding error. [1] The final answer is to three significant figures, matching the mass and supplied Mᵣ. [1]

(d) 4.062 × 10⁻³ mol. [1]

✓ DONE WHENFinish calculations with a correct unit, particle label and appropriate precision.

Mass–mole–particle calculations: AQA A-level Chemistry