Chemistry • Year 11 • Module 1 • Lesson 14

Isotopes and Relative Atomic Mass

Apply your understanding of isotopic calculations, mass spectrum interpretation, and the meaning of relative atomic mass to real data and chemical contexts.

Apply · Data & Reasoning

1. Interpret a mass spectrum — neon

The mass spectrum below shows the three naturally occurring isotopes of neon (Ne). Use it to answer the questions. 8 marks

1.1 State the number of naturally occurring isotopes of neon shown by the mass spectrum and write the nuclide symbol for each. 2 marks

1.2 Using the data from the mass spectrum, show that the fractional abundances sum to approximately 1.000. 1 mark

1.3 Calculate the relative atomic mass of neon. Show full working. 3 marks

1.4 Predict, without calculating, whether A_r(Ne) will be closer to 20, 21, or 22. Justify your prediction by referring to the mass spectrum. 2 marks

Stuck? Revisit Card 2 (Calculating Relative Atomic Mass) and Worked Example 1 in the lesson. Convert percentages to fractions first.

2. Interpret isotope data — silicon

Silicon (Si) has three naturally occurring stable isotopes. The table below shows data from a mass spectrometer. 9 marks

Isotope	Nuclide notation	Protons	Neutrons	Percentage abundance (%)
Silicon-28	²⁸Si	14	14	92.23
Silicon-29	²⁹Si	14	15	4.67
Silicon-30	³⁰Si	14	16	3.10

2.1 Complete the “Fractional abundance” column in the table. Verify that the fractional abundances sum to 1.000. 2 marks

2.2 Calculate A_r(Si). Show full working including the formula. 3 marks

2.3 Explain why all three silicon isotopes have identical chemical properties, even though they have different masses. 2 marks

2.4 A student suggests that the relative atomic mass of silicon should be approximately 28, not 28.09. Evaluate this claim. 2 marks

Stuck? Revisit Card 2, the formula box (A_r = Σ(isotopic mass × fractional abundance)), and the Common Mistakes section in the lesson.

3. Compare isotopes across five features

The table below compares ¹²C and ¹⁴C, two isotopes of carbon. Complete every blank cell. 10 marks (1 per cell)

Feature	Carbon-12 (¹²C)	Carbon-14 (¹⁴C)
Protons (Z)
Neutrons	6
Mass number (A)		14
Nuclear stability	Stable
Natural abundance	98.89%
Chemical properties (same or different?)
One physical property that differs
Application / use	Reference standard: ¹²C = exactly 12 u

Stuck? Revisit Card 1 (isotope table), the Key Definitions panel, and the Think First answer about carbon dating.

4. Predict and justify — copper isotopes

Copper (Cu) has two stable isotopes: ⁶³Cu and ⁶⁵Cu. A student is told that the relative atomic mass of copper is 63.55. 5 marks

4.1 Before calculating, predict which isotope (⁶³Cu or ⁶⁵Cu) is more abundant in nature. Justify your prediction by referring to the A_r value. 2 marks

4.2 Using algebra, calculate the percentage abundance of each isotope. Let x = fractional abundance of ⁶³Cu. Show all steps. 3 marks

Stuck? Revisit Worked Example 2 in the lesson: set up the equation A_r = (63 × x) + (65 × (1−x)) = 63.55, then solve for x. Verify your answer by substituting back in.

Answers — Do not peek before attempting

Q1.1 — Isotopes of neon

There are three naturally occurring isotopes: ²⁰Ne (90.48%), ²¹Ne (0.27%), and ²²Ne (9.25%). The three peaks on the spectrum at m/z = 20, 21, and 22 represent these three isotopes. Award 1 mark for the correct count and 1 mark for all three nuclide symbols written correctly.

Q1.2 — Sum of fractional abundances

Convert each: ²⁰Ne = 90.48 ÷ 100 = 0.9048; ²¹Ne = 0.27 ÷ 100 = 0.0027; ²²Ne = 9.25 ÷ 100 = 0.0925. Sum = 0.9048 + 0.0027 + 0.0925 = 1.0000 ✓. Award 1 mark.

Q1.3 — A_r(Ne) calculation

A_r(Ne) = (20 × 0.9048) + (21 × 0.0027) + (22 × 0.0925) [1 mark — correct formula with fractional abundances]
= 18.096 + 0.0567 + 2.035 [1 mark — correct intermediate products]
= 20.19 (to 4 s.f.) [1 mark]. Acceptable range: 20.18–20.19; matches IUPAC value of 20.18.

Q1.4 — Predict A_r without calculating

A_r will be closest to 20 [1 mark]. Justification: ²⁰Ne is by far the most abundant isotope (~90.5% of all neon atoms). The weighted average is strongly pulled toward the value of the most abundant isotope [1 mark]. The small contributions from ²¹Ne and ²²Ne raise the average only slightly above 20.

Q2.1 — Fractional abundances (Si)

²⁸Si = 92.23 ÷ 100 = 0.9223; ²⁹Si = 4.67 ÷ 100 = 0.0467; ³⁰Si = 3.10 ÷ 100 = 0.0310. Sum = 0.9223 + 0.0467 + 0.0310 = 1.0000 ✓. Award 1 mark for all three fractions correct; 1 mark for confirming the sum = 1.000.

Q2.2 — A_r(Si) calculation

A_r(Si) = (28 × 0.9223) + (29 × 0.0467) + (30 × 0.0310) [1 mark]
= 25.824 + 1.354 + 0.930 [1 mark]
= 28.11 [1 mark] (IUPAC = 28.09; small rounding differences acceptable).

Q2.3 — Same chemical properties

All three silicon isotopes have 14 protons, giving them 14 electrons in identical electron configurations (neutral atoms). Chemical properties — bonding, reactivity, valence — are determined solely by electron configuration. Because the electron arrangements are the same, the isotopes react identically [1 mark]. The different number of neutrons does not affect electron configuration [1 mark].

Q2.4 — Evaluate the student’s claim

The student’s claim is approximately correct but not exact [1 mark]. ²⁸Si is the dominant isotope at 92.23%, so the weighted average is strongly pulled toward 28. However, the small contributions from ²⁹Si (4.67%) and ³⁰Si (3.10%) increase the average slightly above 28, yielding A_r ≈ 28.09. The relative atomic mass is the weighted average and is not equal to the most abundant isotope’s mass number unless all other isotopes have zero abundance [1 mark].

Q3 — Compare ¹²C and ¹⁴C

Protons (Z): Both 6. Neutrons: ¹²C = 6 (given); ¹⁴C = 8. Mass number: ¹²C = 12; ¹⁴C = 14 (given). Nuclear stability: ¹²C = Stable (given); ¹⁴C = Radioactive (t½ = 5730 yr). Natural abundance: ¹²C = 98.89% (given); ¹⁴C = trace (cosmogenic; artificially produced). Chemical properties: Identical (same 6 electrons, same configuration). Physical property that differs: Mass / density (¹⁴C is heavier by 2 neutrons; reactions involving bond breaking / forming are slightly slower for ¹⁴C due to kinetic isotope effect). Application (¹⁴C): Radiocarbon (carbon) dating — the known decay rate of ¹⁴C is used to date organic materials up to ~50,000 years old.

Q4.1 — Predict more abundant isotope

⁶³Cu is more abundant [1 mark]. A_r(Cu) = 63.55 is much closer to 63 than to 65, indicating the weighted average is pulled strongly toward the lighter isotope. If both were equally abundant, A_r would be exactly 64 [1 mark].

Q4.2 — Abundance calculation

Let x = fractional abundance of ⁶³Cu; then (1 − x) = fractional abundance of ⁶⁵Cu [1 mark].
A_r = (63 × x) + (65 × (1 − x)) = 63.55
63x + 65 − 65x = 63.55
−2x = −1.45
x = 0.725 [1 mark]
Therefore: ⁶³Cu = 72.5%; ⁶⁵Cu = 27.5% [1 mark].
Verify: (63 × 0.725) + (65 × 0.275) = 45.675 + 17.875 = 63.55 ✓