Mass, moles & particles

12 practice questions

Choose the entity and formula multiplier

Learning goalTranslate a formula into the exact atoms, ions or electrons requested.

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Before using a formula, decide what the question wants you to count. A molecule, an atom and an ion are not interchangeable labels.

Use molecules for substances made of separate molecules, such as water. For an ionic compound such as NaCl, use formula units rather than molecules.

Then look inside the formula. One H₂O molecule contains two H atoms and one O atom. One formula unit of an ionic compound may contain more than one of a particular ion. Apply that multiplier only when the question asks for those component particles.

Formula	One-entity statement
CO₂	One molecule contains one C atom and two O atoms.
Al₂(SO₄)₃	One formula unit represents two Al³⁺ and three SO₄²⁻ ions, five ions in total.
Na⁺	With atomic number 11, a Na⁺ ion has 11 −1 =10 electrons.

Amount of requested particles = amount of substance × number per entity

Worked example

Sulfate ions from a formula unit

Find the amount in moles of sulfate ions represented by 0.120 mol Al₂(SO₄)₃.

Worked answer

There are three sulfate ions per formula unit.

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

0.360 mol sulfate ions.

EXAM TRAPFor total ions, add the cation and anion subscripts. For Al₂(SO₄)₃ use 2 +3 =5, not 2 ×3. A sulfate ion remains one ion.

🧪 TEST IT

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

Exam-style question 5

(a) Name the entity represented by each of Fe, O₂ and NaCl. (3 marks)

(b) Calculate the amount in moles of oxygen atoms represented by 0.120 mol O₂. (1 mark)

Total: 6 marks

Show answer & marking points

(a) Fe: atoms [1]; O₂: molecules [1]; NaCl: formula units. [1]

(b) Moles of oxygen atoms = moles of O₂ × oxygen atoms per molecule
n(O atoms) = 2 × 0.120 = 0.240 mol. [1]

(c) Moles of chloride ions = moles of MgCl₂ × chloride ions per formula unit
n(Cl⁻) = 2 × 0.0800 = 0.160 mol [1]; Total moles of ions = moles of MgCl₂ × total ions per formula unit
n(all ions) = (1 + 2) × 0.0800 = 0.240 mol. [1]

✓ DONE WHENTranslate a formula into the exact atoms, ions or electrons requested.

Mass and amount

Learning goalUse n = m/M in both directions with mass in grams and molar mass in g mol⁻¹.

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Molar mass tells you the mass of one mole. It is the bridge between the mass you can weigh and the amount in moles.

Find M by adding the Aᵣ contributions in the formula. Use g mol⁻¹ for molar mass.
For grams to moles, divide by M: n = m/M.
For moles to grams, multiply by M: m = nM.

Check the units before substituting. When M is in g mol⁻¹, mass must be in grams.

A quick check is to compare with one mole. If M = 100.1 g mol⁻¹, then 0.350 mol must have a mass below 100.1 g.

n = m/M; m = nM; M = m/n

M molar mass of the specified substance, in g mol⁻¹; m mass of the specified sample or substance, in g; n amount of the specified substance or particles, in mol.

Worked example

Mass to amount

Find the amount in moles of H₂SO₄ in 9.81 g. M(H₂SO₄) =98.1 g mol⁻¹.

Worked answer

n = m/M
n =9.81/98.1

0.100 mol H₂SO₄.

Worked example

Amount to mass

Find the mass of 0.125 mol Na₂CO₃. M =106 g mol⁻¹.

Worked answer

m = n × M
m =0.125 ×106 =13.25 g

13.3 g Na₂CO₃ to three significant figures.

EXAM TRAPM must describe the entire chemical formula, not one element inside it. A mass about one-tenth of the molar mass corresponds to about 0.1 mol.

WHEN DENSITY IS GIVEN

For a pure liquid, mass = density × volume. Match the volume unit to the density unit: a density in g cm⁻³ multiplied by a volume in cm³ gives mass in g. Then use n = m/M.

M molar mass of the specified substance, in g mol⁻¹; m mass of the specified sample or substance, in g; n amount of the specified substance or particles, in mol.

🧪 TEST IT

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

Exam-style question 6

(a) Calculate the amount in moles of HCl in 7.30 g. M(HCl) = 36.5 g mol⁻¹. (2 marks)

(b) Calculate the mass of 0.0750 mol Na₂CO₃. M = 106 g mol⁻¹. (2 marks)

(c) A pure liquid has density 0.800 g cm⁻³ and molar mass 60.0 g mol⁻¹. Calculate the number of moles in 15.0 cm³ of liquid. (2 marks)

Total: 6 marks

Show answer & marking points

(a) n = m/M
n = 7.30/36.5 [1] = 0.200 mol. [1]

(b) m = n × M
m = 0.0750 × 106 [1] = 7.95 g. [1]

✓ DONE WHENUse n = m/M in both directions with mass in grams and molar mass in g mol⁻¹.

Amount and particle number

Learning goalConvert between amount and a named particle count using the Avogadro constant.

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The Avogadro constant tells you how many specified entities are in one mole. Use it to move between moles and particle number.

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

N is a count, so it has no unit. Label the answer with the entity, such as molecules or chloride ions. Use the supplied value of N_A; these questions normally use 6.02 × 10²³ mol⁻¹.

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

N = nN_A; n = N/N_A

N number of the specified particles; a count with no unit; N_A Avogadro constant, 6.02 × 10²³ mol⁻¹; n amount of the specified substance or particles, in mol.

Worked example

Moles to formula units

How many MgCl₂ formula units are represented by 0.0450 mol?

Use N_A = 6.02 × 10²³ mol⁻¹.

Worked answer

N = nN_A
N =0.0450 ×6.02 ×10²³ =2.709 ×10²²

2.71 ×10²² MgCl₂ formula units.

Worked example

Atoms to moles

Find the amount in moles represented by 1.204 ×10²³ Ne atoms.

Use N_A = 6.02 × 10²³ mol⁻¹.

Worked answer

n =(1.204 ×10²³)/(6.02 ×10²³)

0.200 mol Ne.

EXAM TRAPN is a count with no unit, but still needs the label atoms, molecules, formula units, ions or electrons.

🧪 TEST IT

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

Quick check 7

Use N_A = 6.02 × 10²³ mol⁻¹ where needed.

(a) Calculate the number of MgCl₂ formula units in 0.0350 mol MgCl₂. [2 marks]

(b) Calculate the amount in moles of Ne represented by 9.03 × 10²² Ne atoms. [2 marks]

Total: 4 marks

Show answer & marking points

(a) N = nN_A = 0.0350 × 6.02 × 10²³ [1] = 2.11 × 10²² MgCl₂ formula units [1].

(b) n(Ne) = N/N_A = (9.03 × 10²²)/(6.02 × 10²³) [1] = 0.150 mol Ne [1].

✓ DONE WHENConvert between amount and a named particle count using the Avogadro constant.

Count atoms, ions and electrons

Learning goalApply formula multipliers once and keep polyatomic ions together.

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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 = nN_A 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 N_A.

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

Worked example

Oxygen atoms in molecules

Find the number of O atoms in 0.250 mol CO₂.

Use N_A = 6.02 × 10²³ mol⁻¹.

Worked answer

Moles of oxygen atoms = moles of CO₂ × oxygen atoms per molecule
n(O atoms) =0.250 ×2 =0.500 mol

N(O atoms) = nN_A
N(O atoms) =0.500 ×6.02 ×10²³

3.01 ×10²³ O atoms.

Worked example

Total ions

Find the total number of ions represented by 0.0300 mol Al₂(SO₄)₃.

Use N_A = 6.02 × 10²³ mol⁻¹.

Worked answer

Each formula unit represents 2 +3 =5 ions.

Total moles of ions = moles of Al₂(SO₄)₃ × total ions per formula unit
n(ions) =0.0300 ×5 =0.150 mol

N(ions) = nN_A
N(ions) =0.150 ×6.02 ×10²³

9.03 ×10²² ions.

EXAM TRAPSO₄²⁻ contains five atoms but is one polyatomic ion. Do not split it when counting ions.

🧪 TEST IT

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

Exam-style question 8

Use N_A = 6.02 × 10²³ mol⁻¹.

A sample contains 0.240 mol ethanol, C₂H₅OH.

(a) Calculate the number of ethanol molecules. (2 marks)

(b) Calculate the number of oxygen atoms. (1 mark)

Total: 5 marks

Show answer & marking points

(a) N = nN_A
N = 0.240 × 6.02 × 10²³ [1] = 1.44 × 10²³ molecules. [1]

(b) One O atom per molecule, so 1.44 × 10²³ oxygen atoms. [1]

(c) Each molecule contains 2 + 6 + 1 = 9 atoms. [1] Total number of atoms = number of C₂H₅OH molecules × total atoms per molecule
N(all atoms) = 9 × 1.4448 × 10²³ = 1.30 × 10²⁴ atoms. [1]

Exam-style question 9

Use N_A = 6.02 × 10²³ mol⁻¹.

A sample contains 0.0200 mol Al₂(SO₄)₃.

(a) Calculate the number of sulfate ions. (2 marks)

(b) Calculate the total number of ions. (2 marks)

Total: 5 marks

Show answer & marking points

(a) Moles of sulfate ions = moles of Al₂(SO₄)₃ × sulfate ions per formula unit
n(SO₄²⁻) = 3 × 0.0200 = 0.0600 mol [1]; N = 3.61 × 10²² ions. [1]

(b) Total moles of ions = moles of Al₂(SO₄)₃ × total ions per formula unit
n(all ions) = (2 + 3) × 0.0200 = 0.100 mol [1]; N = 6.02 × 10²² ions. [1]

Exam-style question 10

Use N_A = 6.02 × 10²³ mol⁻¹.

A sample contains 0.0150 mol Al³⁺ ions. Aluminium has atomic number 13.

(a) Determine the number of electrons in one Al³⁺ ion. (1 mark)

(b) Calculate the amount in moles of electrons represented by the sample. (1 mark)

Total: 4 marks

Show answer & marking points

(a) 13 − 3 = 10 electrons. [1]

(b) n(electrons) = electrons per ion × n(ions)
n(electrons) = 10 × 0.0150 = 0.150 mol. [1]

✓ DONE WHENApply formula multipliers once and keep polyatomic ions together.

The mass–mole–particle route

Learning goalConnect measured mass to the exact particle requested.

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For a mass-to-particles question, make two clear moves: mass to moles, then moles to particles. The mole stage keeps the units and the target entity visible.

Divide the mass by the molar mass to find moles of the substance.
If needed, use the formula to find moles of the particular atom or ion being counted.
Multiply the appropriate mole amount by N_A.

The combined expression N = (m/M)N_A is valid when N counts the same basic entity described by M. If you need a component particle, remember its formula multiplier. Writing both stages can make that easier to see.

Mass → divide by M → amount → formula multiplier → multiply by N_A → particle count

Worked example

Mass to chloride ions

Find the number of chloride ions represented by 5.55 g CaCl₂. M(CaCl₂) =111 g mol⁻¹.

Use N_A = 6.02 × 10²³ mol⁻¹.

Worked answer

n(CaCl₂) = m/M
n(CaCl₂) =5.55/111 =0.0500 mol

Moles of chloride ions = moles of CaCl₂ × chloride ions per formula unit
n(Cl⁻) =0.0500 ×2 =0.100 mol

N(Cl⁻) = nN_A
N(Cl⁻) =0.100 ×6.02 ×10²³

6.02 ×10²² chloride ions.

EXAM TRAPThe first particle count from mass/M ×N_A counts basic entities. Apply a further multiplier only when the question asks for component atoms, ions or electrons.

🧪 TEST IT

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

Exam-style question 11

Use N_A = 6.02 × 10²³ mol⁻¹.

Calculate the number of sulfate ions represented by 3.42 g Al₂(SO₄)₃. M = 342 g mol⁻¹. (3 marks)

Total: 3 marks

Show answer & marking points

n(salt) = m/M
n(salt) = 3.42/342 = 0.0100 mol. [1] Moles of sulfate ions = moles of Al₂(SO₄)₃ × sulfate ions per formula unit
n(SO₄²⁻) = 3 × 0.0100 = 0.0300 mol. [1] N = nN_A
N = 0.0300 × 6.02 × 10²³ = 1.81 × 10²² sulfate ions, to 3 significant figures. [1]

Exam-style question 12

Use N_A = 6.02 × 10²³ mol⁻¹.

Separate 5.00 g samples of NaCl and MgCl₂ are compared. M(NaCl) = 58.5 g mol⁻¹; M(MgCl₂) = 95.3 g mol⁻¹. Determine which sample represents more chloride ions. Support your conclusion with calculations. (5 marks)

Total: 5 marks

Show answer & marking points

NaCl: n = m/M
n = 5.00/58.5 = 0.0854701… mol, so n(Cl⁻) = 0.0854701… mol. [1] N(Cl⁻) = 5.15 × 10²² ions. [1]

MgCl₂: n = m/M
n = 5.00/95.3 = 0.0524659… mol, so Moles of chloride ions = moles of MgCl₂ × chloride ions per formula unit
n(Cl⁻) = 2 × 0.0524659… = 0.104932… mol. [1] N(Cl⁻) = 6.32 × 10²² ions. [1] The MgCl₂ sample represents more chloride ions. [1]

✓ DONE WHENConnect measured mass to the exact particle requested.

Reverse and unfamiliar calculations

Learning goalWork from a particle count to mass or an unknown molar mass.

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A reverse calculation uses the same relationships in the opposite direction. Start by writing down the target and the information you have.

For a particle count to moles, divide by N_A.
Undo any formula multiplier if you need moles of the whole compound rather than its component particles.
For moles to mass, use m = nM.
If molar mass is the unknown, rearrange to M = m/n.

Label the substance and entity at every step. The worked examples show how the route changes when the question starts with a particle count or asks for an unfamiliar molar mass.

Worked example

Particles to mass

Find the mass of 1.204 ×10²³ CO₂ molecules. M =44.0 g mol⁻¹.

Use N_A = 6.02 × 10²³ mol⁻¹.

Worked answer

n =(1.204 ×10²³)/(6.02 ×10²³) =0.200 mol

m = n × M
m =0.200 ×44.0

8.80 g CO₂.

Worked example

Identify an element from chloride-ion count

A 9.53 g sample of XCl₂ represents 1.204 ×10²³ chloride ions. Use A_r(Cl) =35.5.

Calculate the relative atomic mass of X and identify the element. Use N_A = 6.02 × 10²³ mol⁻¹.

Worked answer

n(Cl⁻) =0.200 mol; Moles of XCl₂ = moles of chloride ions ÷ chloride ions per formula unit
n(XCl₂) =0.200/2 =0.100 mol

M(XCl₂) = m/n
M(XCl₂) =9.53/0.100 =95.3 g mol⁻¹

A_r(X) =95.3 −2(35.5) =24.3

X is magnesium, Mg (A_r =24.3).

EXAM TRAPReverse the formula multiplier before finding the mass of the whole compound. Keep unrounded intermediate amounts.

🧪 TEST IT

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

Exam-style question 13

Use N_A = 6.02 × 10²³ mol⁻¹ where needed.

A sample contains 3.01 × 10²² carbon dioxide molecules. M(CO₂) = 44.0 g mol⁻¹.

(a) Calculate the amount in moles of CO₂. [2 marks]

(b) Calculate the mass of the sample. [2 marks]

Total: 4 marks

Show answer & marking points

(a) n(CO₂) = N/N_A = (3.01 × 10²²)/(6.02 × 10²³) = 0.0500 mol. [2]

(b) m = nM = 0.0500 × 44.0 = 2.20 g CO₂. [2]

Exam-style question 14

Use N_A = 6.02 × 10²³ mol⁻¹ where needed.

A 4.40 g sample of molecular substance X contains 6.02 × 10²² molecules.

(a) Calculate the amount in moles of X. [2 marks]

(b) Calculate the molar mass of X, including the unit. [2 marks]

Total: 4 marks

Show answer & marking points

(a) n(X) = (6.02 × 10²²)/(6.02 × 10²³) = 0.100 mol. [2]

(b) M(X) = m/n = 4.40/0.100 = 44.0 g mol⁻¹. [2]

Exam-style question 15

Use N_A = 6.02 × 10²³ mol⁻¹.

A 15.96 g sample of X₂O₃ represents 1.806 × 10²³ oxygen atoms. Use A_r(O) = 16.0.

(a) Calculate the amount in moles of oxygen atoms. (2 marks)

(b) Calculate the amount in moles of X₂O₃. (1 mark)

(d) Calculate A_r(X) and identify X. Use A_r: Mg 24.3, Al 27.0, Ca 40.1, Fe 55.8, Zn 65.4. (3 marks)

Total: 8 marks

Show answer & marking points

(a) n(O atoms) = (1.806 × 10²³)/(6.02 × 10²³) = 0.300 mol. [2]

(b) Moles of X₂O₃ = moles of oxygen atoms ÷ oxygen atoms per formula unit
n(X₂O₃) = 0.300/3 = 0.100 mol. [1]

(c) M = m/n
M = 15.96/0.100 = 159.6 g mol⁻¹. [2] Retain the calculator value for identification. A final value of 160 g mol⁻¹ to three significant figures is also appropriate.

(d) 2A_r(X) + 3(16.0) = 159.6 [1]; A_r(X) = 55.8 [1]; X is Fe. [1]

✓ DONE WHENWork from a particle count to mass or an unknown molar mass.

Precision and final checks

Learning goalFinish with correct units, standard form, significant figures and a named entity.

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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.

Check	What good working shows
Units	m in g; n in mol; M in g mol⁻¹; N has no unit.
Standard form	One non-zero digit before the decimal point.
Significant figures	Round the final result using the data precision or the instruction.
Entity	Name atoms, molecules, formula units, ions or electrons.
Sense check	Less than 1 mol of specified entities contains fewer of those entities than 1 mol.²³ basic entities.

EXAM TRAPDo not round an intermediate amount such as 0.04996 mol to 0.050 mol and then reuse it. Store the full value.

🧪 TEST IT

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

Exam-style question 16

Use N_A = 6.02 × 10²³ mol⁻¹.

A student writes the following calculation for the number of molecules in 2.50 g CO₂: N = (2.50/44.0)/(6.02 × 10²³). M(CO₂) = 44.0 g mol⁻¹.

(a) Identify the error and calculate the correct number of molecules. (3 marks)

(b) Explain why intermediate values should not be rounded, and give an appropriate final precision here. (2 marks)

Total: 6 marks

Show answer & marking points

(a) Multiply by N_A when converting moles to particle number. [1] n = m/M
n = 2.50/44.0 = 0.05681818… mol [1]; N = nN_A
N = 0.05681818… × 6.02 × 10²³ = 3.42 × 10²² molecules. [1]

(b) Keeping the calculator value avoids an extra rounding error in subsequent operations. [1] Give the final answer to 3 significant figures, matching the supplied data. [1]

✓ DONE WHENFinish with correct units, standard form, significant figures and a named entity.

Mass, moles & particles: OCR A A-level Chemistry