These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. The elements E, Q, Lm, X, Dv and Rz in Questions 1, 2 and 4–7 are fictional, and their data are invented for practice. Question 3 uses real average atomic masses. In every question, all ions are singly charged (+1), so m/z equals the isotope mass in amu.
Question 1 (multiple choice · foundation)
The mass spectrum of a pure sample of element E shows three peaks. Their m/z values and relative heights are: m/z 28, height 100; m/z 29, height 5; m/z 30, height 3. Which statement is correct?
- (A) Element E has three isotopes, and the isotope of mass 28 is the most abundant.
- (B) The sample contains three different elements, with masses 28, 29 and 30.
- (C) The average atomic mass of E is 29, because that is the middle peak.
- (D) The isotope of mass 28 is the most abundant because it is the lightest.
Answer and explanation
Answer: (A). The sample is a single element, so each peak is one isotope. The tallest peak (height 100, at m/z 28) is the most abundant isotope.
- (B) The question says the sample is pure element E. Different peaks are isotopes of the same element, not different elements.
- (C) A simple middle value ignores abundance. The m/z 28 isotope is far more common, so the average is just above 28 (about 28.1), not 29.
- (D) The peak at 28 is tallest because more ions of that mass reached the detector. Being lightest does not make an isotope more common.
Question 2 (multiple choice · core)
Element Q has two isotopes. Its mass spectrum shows a peak at m/z 46.0 with height 100 and a peak at m/z 48.0 with height 25. What is the average atomic mass of Q?
- (A) 46.4 amu
- (B) 47.0 amu
- (C) 47.6 amu
- (D) 58.0 amu
Answer and explanation
Answer: (A). The heights add up to 125, so the fractions are 100 ÷ 125 = 0.80 and 25 ÷ 125 = 0.20. Average = (46.0 × 0.80) + (48.0 × 0.20) = 36.8 + 9.6 = 46.4 amu.
- (B) is the simple mean of 46.0 and 48.0. It ignores the abundances.
- (C) swaps the weights, using 0.20 for the larger peak and 0.80 for the smaller one.
- (D) treats the heights as percentages: (46.0 × 100 + 48.0 × 25) ÷ 100 = 58.0. This is heavier than both isotopes, which is impossible.
Question 3 (multiple choice · core)
The mass spectrum of a pure element shows exactly two peaks, at m/z 79 and m/z 81. The two peaks have almost the same height. Using the average atomic masses below, which element is it?
- (A) Selenium, 78.97 amu
- (B) Bromine, 79.90 amu
- (C) Krypton, 83.80 amu
- (D) Rubidium, 85.47 amu
Answer and explanation
Answer: (B). Two isotopes of almost equal abundance give an average close to the midpoint: (79 + 81) ÷ 2 = 80. Bromine’s average atomic mass, 79.90 amu, fits. It is just below 80, so the lighter isotope is very slightly more common.
- (A) 78.97 is at the lighter peak, not between the two. An average that close to 79 would need almost every atom to be the lighter isotope, but the two peaks are almost equally tall.
- (C) and (D) are above 81. An average can never be higher than the heaviest isotope present.
Question 4 (calculation · core)
A pure sample of the fictional element Lm gives this mass spectrum data:
| m/z | Abundance (%) |
|---|---|
| 63.0 | 12.0 |
| 64.0 | 55.0 |
| 66.0 | 33.0 |
Calculate the average atomic mass of Lm.
Worked solution
- Check the total: 12.0 + 55.0 + 33.0 = 100.0%, so these are already percentages.
- Multiply each mass by its fraction: 63.0 × 0.120 = 7.56; 64.0 × 0.550 = 35.20; 66.0 × 0.330 = 21.78.
- Add: 7.56 + 35.20 + 21.78 = 64.54 amu.
Suggested mark points (2): 1 for multiplying each mass by its fractional abundance; 1 for 64.54 amu.
Check: 64.54 is between 63.0 and 66.0 and nearest to 64.0, the most abundant isotope. Common error: the simple mean of the three masses gives 64.33 amu, which ignores abundance. Forgetting to divide the percentages by 100 gives 6454, which fails the “between the isotope masses” check.
Question 5 (calculation · core)
The fictional element X has two isotopes, with masses 120.0 amu and 123.0 amu. Its average atomic mass is 120.9 amu.
(a) Calculate the percentage abundance of each isotope. (b) The mass spectrum of X is drawn with the tallest peak at a height of 100. Calculate the height of the other peak.
Worked solution
(a) Let x be the fraction of the 123.0 isotope, so the fraction of the 120.0 isotope is 1 − x. 120.0(1 − x) + 123.0x = 120.9 120.0 + 3.0x = 120.9, so x = 0.30. Abundances: 70.0% of mass 120.0 and 30.0% of mass 123.0.
(b) The 120.0 peak is tallest (70.0%), so it is drawn at 100. The other peak: (30.0 ÷ 70.0) × 100 = 42.9.
Suggested mark points (3): 1 for a correct equation using x and 1 − x; 1 for both percentages; 1 for a relative height of 42.9.
Check: 120.0 × 0.700 + 123.0 × 0.300 = 120.9 amu. The average is closer to 120.0, so the 120.0 isotope must be more common, as found.
Question 6 (constructed response · core)
A mass spectrum of the fictional element Dv shows three peaks. The tallest peak is scaled to 100.
| m/z | Relative height |
|---|---|
| 84.0 | 8 |
| 86.0 | 100 |
| 87.0 | 42 |
(a) Calculate the percentage abundance of each isotope. (b) Calculate the average atomic mass of Dv. (c) A student writes: “Most atoms of Dv have a mass of 86.17 amu, because that is the average atomic mass.” Evaluate this claim, using the spectrum.
Worked solution and suggested Marlbridge rubric
(a) Total height = 8 + 100 + 42 = 150. Mass 84.0: 8 ÷ 150 = 5.33%. Mass 86.0: 100 ÷ 150 = 66.7%. Mass 87.0: 42 ÷ 150 = 28.0%.
(b) Average = (84.0 × 0.05333) + (86.0 × 0.6667) + (87.0 × 0.2800) = 86.17 amu.
(c) The claim is incorrect. The spectrum has peaks only at m/z 84.0, 86.0 and 87.0, so every Dv atom has one of these three masses. No peak appears at 86.17, so no atom has that mass. The value 86.17 is a weighted average of the three isotope masses. The most common mass is 86.0 amu (about two-thirds of the atoms). The average is slightly above 86.0 because the 87.0 isotope (28.0%) outweighs the 84.0 isotope (5.33%).
| Point | What earns it |
|---|---|
| 1 | Divides each height by the total (150) to get all three percentages |
| 1 | Average atomic mass of 86.17 amu, with working (accept answers carried forward from (a)) |
| 1 | States the claim is wrong and uses the spectrum as evidence: atoms have only the masses at the peaks (84.0, 86.0 or 87.0) |
| 1 | Explains that 86.17 is a weighted average, and identifies 86.0 amu as the most common mass |
Do not award the point in (c) for “the claim is wrong” without evidence from the spectrum. Common error in (b): dividing by 100 instead of 150 gives 129.26 amu, which is heavier than every isotope.
Question 7 (explanation · stretch)
The fictional element Rz has two isotopes, of mass 50.0 amu and 52.0 amu. In samples from Earth, the abundances are 80.0% and 20.0%, giving an average atomic mass of 50.40 amu. A sample of Rz from a meteorite gives a mass spectrum with peaks at the same two m/z values, but the abundances are 60.0% and 40.0%.
(a) Explain why the peaks appear at m/z 50.0 and 52.0 in both spectra. (b) Without a full calculation, predict whether the meteorite sample has a higher or lower average atomic mass than 50.40 amu. Justify your prediction, then confirm it with a calculation. (c) A chemist uses the periodic-table value, 50.40 g mol⁻¹, to find the number of moles in a mass of the meteorite sample. State the assumption that makes this incorrect.
Model answer and suggested Marlbridge rubric
(a) Each peak comes from ions of one isotope with a charge of +1, so m/z equals the isotope’s mass. The isotopes of Rz have the same masses wherever the sample comes from, so the peak positions do not change. Only the peak heights (abundances) differ.
(b) Higher. The meteorite sample has a larger share of the heavier isotope (40.0% instead of 20.0%), so the weighted average moves towards 52.0. Calculation: (50.0 × 0.600) + (52.0 × 0.400) = 50.80 amu, which is higher than 50.40 amu.
(c) The periodic-table value assumes the natural (Earth) mix of isotopes. The meteorite sample has a different mix, so its average mass per atom is 50.80 amu, not 50.40 amu, and using 50.40 g mol⁻¹ would give a slightly wrong number of moles.
| Point | What earns it |
|---|---|
| 1 | Peaks come from singly charged ions, so m/z = isotope mass, and isotope masses do not depend on the source |
| 1 | Predicts “higher” with the reason (more of the heavier isotope) and confirms 50.80 amu |
| 1 | Identifies the natural-abundance assumption behind the periodic-table value |
How did you do?
- Q1 or Q3 wrong: re-read “Reading a mass spectrum” and the two quick checks in the study guide.
- Q2, Q4 or Q6(a)–(b) wrong: practise converting heights to fractions, as in Worked examples 1 and 2 of the study guide.
- Q5 wrong: follow the x and 1 − x method in Worked example 3.
- Q6(c) or Q7 incomplete: your reasoning needs evidence from the spectrum (peak positions and heights), not only the numbers.
Then tick off the topic checklist.