Edexcel A-Level Chemistry: Atomic Structure and Mass Spectrometry — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Atomic Structure and Mass Spectrometry study guide | Atomic Structure and Mass Spectrometry revision notes

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Section A

1. State the relative mass and relative charge of a proton, a neutron and an electron. [3]

2. Define the term “isotope”. [2]

Section B

3. An ion has atomic number 16 and mass number 32, and carries a charge of 2−. State the number of protons, neutrons and electrons it contains. [3]

4. Bromine has two naturally occurring isotopes: ⁷⁹Br (51% abundance) and ⁸¹Br (49% abundance). Calculate the relative atomic mass of bromine, showing your working. [3]

5. An element has isotopes of mass 24 (79% abundance), 25 (10% abundance) and 26 (11% abundance). Calculate its relative atomic mass to one decimal place. [4]

6. Bromine exists as a diatomic molecule, Br₂. State the number of distinct molecular ion peaks its mass spectrum would show, and explain why. [3]

7. Sketch, in words, the relative heights of the three molecular ion peaks you would expect for Cl₂’s mass spectrum (isotopes ³⁵Cl 75%, ³⁷Cl 25%), explaining the reasoning rather than simply stating equal heights. [5]

8. A mass spectrum shows a molecular ion peak at m/z = 58 for an unknown compound believed to contain only carbon, hydrogen and oxygen. Suggest two possible molecular formulae consistent with this mass, and explain why the molecular ion peak alone cannot distinguish between them. [5]

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Answers

1. Proton: relative mass 1, relative charge +1 [1]. Neutron: relative mass 1, relative charge 0 [1]. Electron: relative mass negligible, relative charge −1 [1].

2. Isotopes are atoms of the same element (identical proton/atomic number) with different numbers of neutrons, and therefore different mass numbers [1] [1].

3. Protons = 16 (equal to the atomic number) [1]. Neutrons = mass number − atomic number = 32 − 16 = 16 [1]. Electrons: a neutral atom would have 16, but a 2− charge means 2 extra electrons, giving 18 electrons [1].

4. (79 × 51 + 81 × 49) ÷ 100 [1] = (4029 + 3969) ÷ 100 [1] = 7998 ÷ 100 = 79.98 [1].

5. (24 × 79) + (25 × 10) + (26 × 11) [1] = 1896 + 250 + 286 = 2432 [1]. 2432 ÷ 100 = 24.32 [1], rounded to 24.3 [1].

6. Three distinct molecular ion peaks [1], because a diatomic molecule formed from an element with two isotopes has three possible isotope combinations — both atoms the lighter isotope, both the heavier isotope, or one of each — not simply one peak per isotope [1] [1].

7. The three peaks correspond to ³⁵Cl–³⁵Cl (mass 70), ³⁵Cl–³⁷Cl (mass 72) and ³⁷Cl–³⁷Cl (mass 74) [1]. Their relative heights follow the probability of each combination occurring, calculated from the isotopic abundances, not equal heights [1]. The homonuclear combinations have probabilities 0.75 × 0.75 = 0.5625 (³⁵Cl–³⁵Cl) and 0.25 × 0.25 = 0.0625 (³⁷Cl–³⁷Cl) [1], while the mixed combination has probability 2 × 0.75 × 0.25 = 0.375 (the factor of 2 because either atom could be the ³⁵Cl one) [1]. So the peak at mass 70 is tallest, mass 72 is the next tallest, and mass 74 is the smallest — a ratio of roughly 9:6:1 [1].

8. Two candidates of relative molecular mass 58: C₃H₆O (e.g. propanone/acetone: (12 × 3) + (1 × 6) + 16 = 58) and C₄H₁₀ (butane: (12 × 4) + (1 × 10) = 58) [1] [1]. The molecular ion peak only gives the relative molecular mass, not the arrangement of atoms within the molecule [1], so any formula (or combination of elements) that sums to the same relative molecular mass is equally consistent with a single m/z value [1]; distinguishing between them requires further evidence, such as a fragmentation pattern or other spectroscopic data [1].

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Where marks are usually lost

• Calculating relative atomic mass as an unweighted average instead of weighting each isotope mass by its percentage abundance.
• Forgetting to adjust the electron count for a charged ion.
• Assuming a diatomic element’s mass spectrum shows only as many molecular ion peaks as it has isotopes.
• Assuming the three peaks of a two-isotope diatomic spectrum are equal in height rather than following combination probability.
• Claiming a single molecular ion mass uniquely identifies a compound.

Approaching atomic structure and mass spectrometry questions

Always set out the relative-atomic-mass calculation as an explicit weighted sum divided by 100, rather than jumping straight to a final number — examiners award method marks for showing (mass × abundance) terms even if an arithmetic slip changes the final answer. For diatomic mass spectrum questions, work out the full set of isotope combinations first (there are always three for a two-isotope diatomic element), then use combination probability, not intuition, to rank the peak heights. When asked to identify a compound from a single m/z value, resist the urge to state one formula with confidence — the specification explicitly tests whether you recognise that relative molecular mass alone is insufficient without further evidence.

Before starting any subatomic-particle question, write out the atomic number and mass number as two separate labelled values first, since most errors come from mixing the two up under time pressure rather than from a genuine misunderstanding of what each represents. For ion questions specifically, decide the sign of the charge before adjusting the electron count — a common slip is adding electrons for a positive ion or removing them for a negative one, which is the opposite of the correct rule. Treat every isotopic-abundance calculation as a two-step process worth checking independently: first confirm the abundances sum to 100%, then confirm the final relative atomic mass falls between the lowest and highest isotope masses given, since a result outside that range always signals an arithmetic error.