Chemistry • Year 11 • Module 1 • Lesson 14
Isotopes and Relative Atomic Mass
Build HSC Band 5–6 extended-response technique by evaluating mass spectrum data, designing an inquiry, and applying the weighted-average model to unfamiliar element contexts.
1. Multi-step calculation — lead mass spectrum (Band 4–5)
8 marks
Data. Lead (Pb) has four naturally occurring isotopes. A mass spectrometer analysis gives the following data:
| Isotope | Mass number | Percentage abundance (%) | Fractional abundance | Product (mass × fraction) |
|---|---|---|---|---|
| ²⁰⁴Pb | 204 | 1.4 | ||
| ²⁰⁶Pb | 206 | 24.1 | ||
| ²⁰⁷Pb | 207 | 22.1 | ||
| ²⁰⁸Pb | 208 | 52.4 | ||
| Check | — | Sum = _____ | Sum = _____ | Ar = _____ |
Data based on IUPAC atomic weights (2021). Percentages are approximate.
Q1a. Complete the “Fractional abundance” and “Product” columns in the table. Calculate Ar(Pb) to 4 significant figures. Verify that percentages and fractional abundances both sum correctly before finalising your answer. 4 marks
Q1b. Explain why your calculated Ar(Pb) is not a whole number and explain what it would mean for the abundances if Ar(Pb) were exactly equal to 207 (the value listed in some older periodic tables). 2 marks
Q1c. A geologist analysing old rock samples finds that the ratio of ²⁰⁶Pb to ²⁰⁴Pb in a sample is significantly higher than in fresh lead ore. Suggest a reason for this finding. (Hint: consider radioactive decay of uranium isotopes.) 2 marks
2. Experimental design — measuring the isotopic composition of boron (Band 5–6)
7 marks Band 5–6
Research question. A chemist suspects that a sample of boron extracted from the Atacama Desert (Chile) may have a slightly different isotopic composition from standard boron. Standard boron has Ar = 10.81, arising from approximately 19.9% ¹⁰B and 80.1% ¹¹B. The chemist has access to a mass spectrometer and standard laboratory equipment.
Your task: Design an investigation to determine whether the Atacama boron sample has the same or different isotopic composition as standard boron. You must address variables, procedure, use of the mass spectrometer, and how you would interpret the results.
Q2. Design the investigation and present it in the format below.
- State a testable hypothesis including the independent and dependent variables.
- Identify the independent variable, dependent variable, and at least two controlled variables.
- Describe the procedure in at least four numbered steps, including how you will use the mass spectrometer to obtain quantitative data and how you will calculate Ar from the spectrum.
- Explain what result would support your hypothesis and what result would falsify it.
- State two limitations of your design and suggest one improvement to increase reliability.
3. Data + scenario: the meaning of Ar (Band 5–6)
7 marks Band 5–6
Scenario. A Year 11 student reads on the internet that “chlorine has an atomic mass of 35.5, which means some chlorine atoms have a mass of 35.5.” Another student in the class says: “Ar is just an average — it doesn’t mean any real atom has that mass. The actual atoms are either 35 or 37.” A third student adds: “Since chlorine is mostly Cl-35, maybe we should just round the periodic table value to 35 and ignore Cl-37.”
Data for reference. ³⁵Cl: 75.77%; ³⁷Cl: 24.23%.
Q3. Critically evaluate each of the three students’ statements. In your response you must:
- Identify whether Student 1’s claim is correct or incorrect, and explain what Ar actually represents using the weighted-average formula.
- Assess Student 2’s statement and explain why a single chlorine atom cannot have a mass of 35.5.
- Evaluate Student 3’s suggestion: calculate what error (in Ar and in grams per mole) would result from ignoring ³⁷Cl and rounding to 35. Discuss when this approximation would and would not be acceptable.
- State what additional information would allow a student to predict whether Ar is closer to the lower or higher isotope mass, without doing the full calculation.
Q1a — Lead Ar calculation (4 marks)
Fractional abundances: ²⁰⁴Pb = 1.4 ÷ 100 = 0.014; ²⁰⁶Pb = 0.241; ²⁰⁷Pb = 0.221; ²⁰⁸Pb = 0.524. Sum of % = 100.0 ✓; sum of fractions = 0.014 + 0.241 + 0.221 + 0.524 = 1.000 ✓ [1 mark]. Products: (204 × 0.014) = 2.856; (206 × 0.241) = 49.646; (207 × 0.221) = 45.747; (208 × 0.524) = 108.992 [1 mark]. Ar(Pb) = 2.856 + 49.646 + 45.747 + 108.992 = 207.241 ≈ 207.2 [1 mark]. Award the 4th mark for correctly verified sums and no arithmetic errors throughout.
Marking criteria. 1 = all four fractional abundances correct and sum verified; 1 = all four products calculated correctly; 1 = correct Ar (207.2 ± 0.1); 1 = clear working laid out in logical order (formula stated, fractions, products, sum).
Q1b — Interpretation of Ar (2 marks)
Ar(Pb) is not a whole number because it is the weighted average of four isotopes with fractional abundances that do not produce an exactly integer result [1 mark]. If Ar(Pb) were exactly 207, this would mean the specific combination of the four isotope masses and their fractional abundances averages to precisely 207. This could in theory happen if, for example, the four abundances were exactly chosen so that the weighted sum = 207. In practice, natural isotope abundances at the 0.01% level mean the average rarely hits a whole number. It would not mean “all lead atoms have mass 207” — individual atoms still have masses 204, 206, 207, or 208 [1 mark].
Q1c — Elevated ²⁰⁶Pb/²⁰⁴Pb ratio (2 marks)
The uranium isotope ²³⁸U undergoes radioactive decay and ultimately produces ²⁰⁶Pb as its stable end product [1 mark]. In old rocks, ²³⁸U present since the rock formed has been gradually decaying over billions of years, continuously adding ²⁰⁶Pb to the system. ²⁰⁴Pb is non-radiogenic (not produced by radioactive decay) and its abundance stays roughly constant. Therefore, the ²⁰⁶Pb/²⁰⁴Pb ratio increases over geological time. An elevated ratio in old rocks compared to fresh ore indicates the sample has accumulated radiogenic ²⁰⁶Pb — this is the principle behind uranium–lead radiometric dating [1 mark].
Q2 — Sample Band 6 experimental design (7 marks), annotated
Hypothesis (IV + DV): If the Atacama boron sample has a different isotopic composition from standard boron, its mass spectrum will show peaks at m/z = 10 and m/z = 11 with different relative heights (and thus different percentage abundances) compared to standard boron (¹⁰B: 19.9%, ¹¹B: 80.1%), resulting in a different calculated Ar. Independent variable: the source of the boron sample (Atacama vs standard reference). Dependent variable: the percentage abundances of ¹⁰B and ¹¹B (peak heights at m/z = 10 and 11) and the calculated Ar. Controlled variables: (1) sample mass used (e.g. 1.0 mg of boron per run); (2) mass spectrometer operating conditions (ionisation energy, accelerating voltage, detector sensitivity settings). [1 mark]
Procedure: (1) Prepare two separate boron samples: 1.0 mg from the Atacama source and 1.0 mg from a certified NIST standard boron reference material. (2) Introduce the standard boron sample into the mass spectrometer first; record the mass spectrum (peak positions at m/z = 10 and 11 and their relative intensities). (3) Repeat with the Atacama boron sample under identical instrument settings. (4) Read the percentage abundance of ¹⁰B and ¹¹B from each spectrum (peak heights proportional to abundance). Calculate Ar for each sample using Ar = (10 × f¹⁰B) + (11 × f¹¹B), where f = fractional abundance. Compare Ar(Atacama) to Ar(standard) and to the IUPAC value of 10.81. [1 mark]
Support / falsification: The hypothesis is supported if Ar(Atacama) differs significantly from 10.81 (i.e. ¹⁰B abundance significantly different from 19.9%). It is falsified if Ar(Atacama) = 10.81 ± instrument precision, indicating the same isotopic composition as standard boron. [1 mark]
Limitations: (1) A single run of the mass spectrometer may contain instrument noise, causing slightly imprecise peak height measurements [1 mark]. (2) The boron sample may contain contaminants (e.g. other elements that produce peaks near m/z = 10 or 11), interfering with the measurement [1 mark].
Improvement: Run each sample at least three times and use the average of the peak heights. This reduces random measurement error and improves reliability of the calculated Ar [1 mark].
Additional mark: Uses precise terminology throughout (isotopic abundance, m/z, fractional abundance, Ar formula, mass spectrometer) [1 mark].
Marking criteria (7 marks): 1 = testable hypothesis with IV and DV; 1 = four numbered procedure steps including use of mass spectrometer and Ar calculation; 1 = correct support and falsification conditions; 1 = first valid limitation; 1 = second valid limitation; 1 = specific improvement to increase reliability; 1 = precise chemical terminology throughout.
Q3 — Sample Band 6 evaluation of student statements (7 marks), annotated
Student 1 (incorrect) [2 marks]: Student 1’s claim is incorrect. Ar = 35.5 does not mean any individual chlorine atom has a mass of 35.5. No chlorine atom can have a mass of 35.5 because the nucleus contains only whole numbers of protons and neutrons; each real chlorine atom is either ³⁵Cl (A = 35) or ³⁷Cl (A = 37) [1 mark]. Ar is the weighted average of all naturally occurring isotopes: Ar(Cl) = (35 × 0.7577) + (37 × 0.2423) = 26.52 + 8.97 = 35.45 ≈ 35.5. This is a statistical population average representing what an “average chlorine atom” would weigh if you sampled millions of atoms; it does not correspond to any physical atom [1 mark].
Student 2 (correct) [1 mark]: Student 2 is correct. A single chlorine atom cannot have a mass of 35.5; it is either ³⁵Cl or ³⁷Cl. Ar is a mathematical construct — it describes the composition of a macroscopic sample of chlorine atoms, not any individual atom [1 mark].
Student 3 (error quantification + discussion) [3 marks]: If we rounded Ar to 35 and ignored ³⁷Cl, the error would be: 35.45 − 35 = 0.45 u per atom, or 0.45 g mol−1 in molar mass calculations [1 mark]. This error compounds in stoichiometric calculations: for a compound like HCl, the molar mass error is 0.45 g mol−1 out of ~36.5 g mol−1 (~1.2% error). For large-scale industrial chemistry (e.g. calculating chlorine requirements in water treatment), a 1.2% error could represent meaningful mass discrepancies [1 mark]. The approximation might be acceptable for rough order-of-magnitude estimates but not for precise analytical work or HSC examination questions where marks depend on the correct Ar [1 mark].
Additional direction criterion [1 mark]: To predict whether Ar is closer to the lower or higher isotope mass without doing a full calculation, compare the abundances: whichever isotope has the higher percentage abundance will pull the weighted average closer to its mass. For chlorine, ³⁵Cl (~76%) is more abundant, so Ar will be much closer to 35 than to 37 [1 mark].
Marking criteria (7 marks): 1 = correctly identifies Student 1 as wrong, with reason (Ar is a weighted average, no atom has mass 35.5); 1 = uses the formula correctly to show Ar = 35.45; 1 = correctly assesses Student 2 as right with explanation (atoms are discrete integer masses); 1 = quantifies the error from Student 3’s rounding (0.45 u / 0.45 g mol−1); 1 = discusses impact in a chemistry context (stoichiometry / molar mass); 1 = reaches a balanced judgement on when rounding is acceptable; 1 = states the rule for predicting Ar direction (highest abundance isotope pulls Ar toward its mass).