OCR A Level Chemistry A: Atomic structure, compounds, formulae and equations, 2.1.1-2.1.2 (H432) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

The questions here practise sections 2.1.1 (Atomic structure and isotopes) and 2.1.2 (Compounds, formulae and equations) of Module 2 in the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. H432 is the full A level, and Module 2 is assessed in all three written components. Where a question needs relative atomic masses, they are given to you in the question. All data are invented for practice.

Read the 2.1.1 study guide, the 2.1.2 study guide and the revision notes first. See also the OCR chemistry hub, the printable checklist and the free diagnostics.

Questions

1. Give the number of protons, neutrons and electrons in each species.

(a) ²⁷Al³⁺ (atomic number 13) [3] (b) ⁷⁹Se²⁻ (atomic number 34) [3]

2. Carbon-13 and carbon-14 are isotopes of carbon (atomic number 6).

(a) State what is the same and what is different in the nuclei of these two isotopes. [2] (b) Explain why the two isotopes react in the same way chemically. [2]

3. Define each term.

(a) relative isotopic mass [2] (b) relative atomic mass [2]

4. A mass spectrum of an element X shows four peaks, from singly charged ions.

m/z	64	66	67	68
Relative abundance (%)	48.0	28.0	4.0	20.0

(a) State what the m/z value of each peak represents. [1] (b) Calculate the relative atomic mass of X. Give your answer to 3 significant figures. [3]

5. Element M has two isotopes, of relative isotopic mass 121 and 123. Its relative atomic mass is 121.9. Calculate the percentage abundance of each isotope. [3]

6. Ar values: P = 31.0, O = 16.0, Al = 27.0, S = 32.1.

(a) Calculate the relative molecular mass of P₄O₁₀. [2] (b) Calculate the relative formula mass of Al₂(SO₄)₃. [2] (c) State why the term “relative formula mass” is used for Al₂(SO₄)₃ and not “relative molecular mass”. [1]

7. (a) Predict the charge on the ion formed by each element: strontium, phosphorus and bromine. [3] (b) Write the formula of each compound.

• strontium phosphide [1]
• aluminium nitrate [1]
• ammonium carbonate [1]
• silver sulfate [1]

8. Give the formula, including the charge, of each ion: [4]

nitrate, hydroxide, zinc, ammonium

9. Balance each equation and add state symbols where they are missing.

(a) C₄H₁₀(g) + O₂(g) → CO₂(g) + H₂O(l) [2] (b) Mg(OH)₂(s) + HNO₃(aq) → Mg(NO₃)₂(aq) + H₂O(l) [2] (c) Al + HCl → AlCl₃ + H₂, where aluminium is a solid, the acid is aqueous, aluminium chloride is dissolved and hydrogen is a gas [2]

10. When aqueous lead(II) nitrate is added to aqueous potassium iodide, a yellow precipitate of lead(II) iodide forms. Potassium nitrate remains in solution. The lead(II) ion is Pb²⁺.

(a) Write the full balanced equation with state symbols. [2] (b) Write the ionic equation with state symbols. [2] (c) Identify the spectator ions. [1]

11. A Group 2 element M has atomic number 38. A sample has the following isotopes, from singly charged ions.

m/z	86	87	88
Relative abundance (%)	10.0	6.0	84.0

(a) Calculate the relative atomic mass of M to 3 significant figures. [3] (b) Deduce the number of protons, neutrons and electrons in an ion of the isotope of mass 88 in the compound M(NO₃)₂. [2] (c) Use your answer to (a) and the Ar values N = 14.0 and O = 16.0 to calculate the relative formula mass of M(NO₃)₂. [2] (d) Aqueous M(NO₃)₂ reacts with aqueous sodium sulfate to give a white precipitate of the sulfate of M. Write the ionic equation with state symbols, using M²⁺ for the metal ion. [2]

12. Early models of the atom pictured electrons embedded in a sphere of positive charge. In alpha-particle scattering experiments, most alpha particles passed straight through a thin metal foil, and a very small fraction were deflected through large angles.

(a) Explain how this evidence led to the model being rejected and replaced. [3] (b) Name one phenomenon that a model with electrons in shells can explain. [1]

Answers

1. (a) Protons = atomic number = 13. [1] Neutrons = 27 − 13 = 14. [1] Electrons = 13 − 3 = 10. [1] 13 p, 14 n, 10 e [3] (b) Protons = 34. [1] Neutrons = 79 − 34 = 45. [1] Electrons = 34 + 2 = 36. [1] 34 p, 45 n, 36 e [3] Examiner insight: Answers tend to gain credit for showing the subtraction, and lose it by treating the charge the wrong way round: cations have lost electrons, so they hold fewer electrons than protons, while anions hold extra.

2. (a) The number of protons is the same (6). [1] The number of neutrons differs: 7 in carbon-13 and 8 in carbon-14. [1] [2] (b) Chemical reactions depend on the electrons, particularly the outer ones. [1] Equal proton numbers mean the two isotopes have the same electron arrangement. [1] [2] Examiner insight: A reason that mentions electron arrangement or number of electrons tends to be needed. Saying only that they are the same element gives little to build on.

3. (a) The mass of an atom of an isotope [1] compared with 1/12th the mass of an atom of carbon-12. [1] [2] (b) The mean mass of an atom of the element, weighted by the abundance of each isotope, [1] compared with 1/12th the mass of an atom of carbon-12. [1] [2] Examiner insight: The two definitions share the same standard, so answers are distinguished by “isotope” in the first and “weighted mean” in the second.

4. (a) The relative isotopic mass of that isotope, since the ions carry a single charge. [1] [1] (b) Sum of mass × abundance = 64 × 48.0 + 66 × 28.0 + 67 × 4.0 + 68 × 20.0 = 3072 + 1848 + 268 + 1360 = 6548. [1] Divide by the total abundance, 100. [1] 6548 / 100 = 65.48, which is 65.5 to 3 s.f. [1] [3] Examiner insight: Show each product. A value outside 64 to 68 signals an error.

5. Let the abundance of the 123 isotope be x%. Then the abundance of the 121 isotope is (100 − x)%. [1] (121 × (100 − x) + 123 × x) / 100 = 121.9, so 12100 + 2x = 12190. [1] x = 45, so 45% of the 123 isotope and 55% of the 121 isotope. [1] [3] Examiner insight: Define the unknown before solving. A quick check helps: 121.9 is nearer 121 than 123, so the lighter isotope should be the more abundant.

6. (a) 4 × 31.0 + 10 × 16.0 = 124.0 + 160.0. [1] 284.0 [1] [2] (b) 2 × 27.0 + 3 × (32.1 + 4 × 16.0) = 54.0 + 3 × 96.1. [1] 54.0 + 288.3 = 342.3 [1] [2] (c) Al₂(SO₄)₃ is an ionic compound with a giant structure, not made of discrete molecules. [1] [1] Examiner insight: Multiplying the contents of a bracket by the subscript outside it is the point that most often goes wrong. Linking the term to “giant structure” gains credit that “it is ionic” alone may not.

7. (a) Strontium is in Group 2: Sr²⁺. [1] Phosphorus is in Group 15: P³⁻. [1] Bromine is in Group 17: Br⁻. [1] [3] (b) Sr²⁺ and P³⁻: three Sr²⁺ (+6) balance two P³⁻ (−6): Sr₃P₂. [1] Al³⁺ and NO₃⁻: Al(NO₃)₃. [1] NH₄⁺ and CO₃²⁻: (NH₄)₂CO₃. [1] Ag⁺ and SO₄²⁻: Ag₂SO₄. [1] [4] Examiner insight: Brackets are part of the formula. “AlNO33” does not show three nitrate ions, so a formula needs to be unambiguous.

8. NO₃⁻ [1], OH⁻ [1], Zn²⁺ [1], NH₄⁺ [1] [4] Examiner insight: Both formula and charge are asked for, so an uncharged formula such as NO₃ leaves out part of the answer.

9. (a) 2C₄H₁₀(g) + 13O₂(g) → 8CO₂(g) + 10H₂O(l) [1] correct carbon and hydrogen balance, [1] oxygen balanced (26 on each side) [2] (b) Mg(OH)₂(s) + 2HNO₃(aq) → Mg(NO₃)₂(aq) + 2H₂O(l) [1] for 2HNO₃, [1] for 2H₂O [2] (c) 2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g) [1] balance, [1] state symbols [2] Examiner insight: Balance by adding coefficients only. Check each element on both sides at the end, particularly oxygen in (a) and hydrogen in (c), and make sure every species carries a state symbol.

10. (a) Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq) [1] correct formulae, [1] balance and state symbols [2] (b) Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s) [1] correct species and states, [1] balance of atoms and charge [2] (c) K⁺ and NO₃⁻ [1] [1] Examiner insight: The ionic equation must balance in charge as well as atoms: here 2+ and 2− on the left total zero, matching the uncharged solid. Spectator ions are those that appear unchanged on both sides.

11. (a) 86 × 10.0 + 87 × 6.0 + 88 × 84.0 = 860 + 522 + 7392 = 8774. [1] 8774 / 100 = 87.74. [1] 87.7 to 3 s.f. [1] [3] (b) Protons = 38 and neutrons = 88 − 38 = 50. [1] The ion is M²⁺, so electrons = 38 − 2 = 36. [1] 38 p, 50 n, 36 e [2] (c) M(NO₃)₂: 87.7 + 2 × (14.0 + 3 × 16.0) = 87.7 + 124.0. [1] 211.7 [1] [2] (d) M²⁺(aq) + SO₄²⁻(aq) → MSO₄(s) [1] correct species, [1] balance and state symbols [2] Examiner insight: Part (c) uses your part (a) value, so write it clearly. In (b), state the 2+ charge, because the compound contains the ion and not the neutral atom.

12. (a) Most alpha particles passed straight through, so the atom must be mostly empty space. [1] A few were deflected through large angles, so there must be a small region of concentrated positive charge. [1] An evenly spread positive sphere would not deflect them like this, so the model was rejected and replaced by one with a small, dense, positive nucleus. [1] [3] (b) Any one of: periodic properties, such as the repeating pattern in ionisation energies or in reactivity across a period or down a group. [1] [1] Examiner insight: Link each observation to a conclusion about the structure. Listing what was seen without saying what it shows tends to gain less credit.

Where marks are usually lost

• Adding the charge to the electron count of a cation, or subtracting it for an anion.
• Giving the mass number as the number of neutrons.
• Writing a definition of relative atomic mass without “weighted mean” or without carbon-12.
• Dividing a weighted sum by the number of isotopes instead of the total abundance.
• Not multiplying the contents of a bracket by the subscript when finding relative formula mass.
• Using “relative molecular mass” for an ionic compound.
• Forgetting brackets in formulae such as Al(NO₃)₃ and (NH₄)₂CO₃.
• Writing an ionic equation that balances in atoms but not in charge.
• Omitting state symbols when the question asks for them.

Next steps

• Review the revision notes for anything you missed.
• Reread the study guides for 2.1.1 and 2.1.2.
• Go to the OCR A Level chemistry hub and tick off the printable checklist.
• Try the free 10-minute diagnostics to find your next gap.
• Book a free trial class.

Frequently asked questions

Are these OCR A Level Chemistry 2.1.1 and 2.1.2 questions from past papers? No. They are original practice questions with worked answers on atomic structure, isotopes, compounds, formulae and equations in OCR A Level Chemistry A (H432).

What must a definition of relative atomic mass include? It must say “weighted mean” and refer to carbon-12. Writing “average mass” or leaving out carbon-12 loses marks.

How do you calculate relative atomic mass from mass spectrum data? Multiply each relative isotopic mass by its abundance, add the products, and divide by the total abundance. Dividing by the number of isotopes is a common error.

What is a common mistake when writing ionic equations? Writing an equation that balances in atoms but not in charge. Check that the total charge is the same on both sides, and include state symbols when they are asked for.

Official syllabus

OCR, OCR Level 3 Advanced GCE in Chemistry A (H432) specification, Version 3.1, May 2026, first assessment 2017. Sections 2.1.1 “Atomic structure and isotopes” and 2.1.2 “Compounds, formulae and equations”.