ChemistryYear 11 · Module 1 · IQ2⏱ ~25 min

Isotopes and Relative Atomic Mass

Carbon dating tells archaeologists whether a bone is 500 years old or 50,000 years old. It works because carbon has isotopes — atoms of carbon with different numbers of neutrons. ¹⁴C is slightly unstable and decays at a known rate; ¹²C is stable. The ratio of ¹⁴C to ¹²C in organic material decreases predictably over time. Without isotopes, there would be no carbon dating, no nuclear medicine, no PET scans. Understanding isotopes is also the key to understanding why chlorine's atomic mass is 35.5 — not a whole number.

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Think First

Chlorine has two naturally occurring isotopes: chlorine-35 and chlorine-37. Both are chlorine atoms, but one is slightly heavier than the other. If you look up chlorine on the periodic table, its relative atomic mass is listed as 35.45 — not a whole number. Why isn't it simply 35 or 37?

Before reading on, write your best answer. What would need to be true for the relative atomic mass to be exactly 36 (halfway between 35 and 37)?

📚 Know

  • Definition of isotope and how isotopes differ
  • How to write nuclide notation for isotopes
  • Definition and meaning of relative atomic mass (Ar)

🔗 Understand

  • Why isotopes of the same element have identical chemical properties
  • How mass spectrometry data gives isotopic abundance
  • How to calculate Ar from isotopic masses and percentage abundances

✅ Can Do

  • Write nuclide notation and identify differences between isotopes
  • Calculate relative atomic mass from isotopic abundance data
  • Interpret mass spectrometry data for an element
📋

Key Definitions

isotopesAtoms of the same element (same Z) that have different numbers of neutrons (different A). Same element, different mass number. Example: ¹²C and ¹⁴C are both carbon (Z=6) but have 6 and 8 neutrons respectively.
relative atomic mass (Ar)The weighted average mass of all naturally occurring isotopes of an element, relative to 1/12 the mass of a ¹²C atom. Dimensionless. Accounts for the abundance of each isotope in nature.
mass spectrometryAn analytical technique that separates ions by their mass-to-charge ratio (m/z). For elements, it gives the mass and relative abundance of each isotope, allowing Ar to be calculated.
isotopic abundanceThe percentage (or fractional) occurrence of each isotope in a naturally occurring sample of an element. Abundances of all isotopes of an element sum to 100%.

Misconceptions to Fix

Wrong: Isotopes have different chemical properties because they have different masses.

Right: Isotopes have identical chemical properties because chemistry is determined by electron configuration, which is the same for all isotopes of an element. Isotopes differ in physical properties (density, boiling point) and nuclear stability due to different neutron numbers.

01

Isotopes: Same Element, Different Mass

Isotopes of the same element share the same number of protons (Z) but differ in neutron number. Since chemical behaviour is determined by the number and arrangement of electrons (which equals protons in neutral atoms), isotopes have identical chemical properties. They do differ in:

  • Physical properties: slightly different mass → slightly different density, melting/boiling points, reaction rates (kinetic isotope effect)
  • Nuclear stability: some isotopes are radioactive (unstable nucleus) while others are stable
IsotopeProtons (Z)Neutrons (A−Z)Mass number (A)StabilityNatural abundance
¹H (protium)101Stable99.985%
²H (deuterium)112Stable0.015%
³H (tritium)123RadioactiveTrace (artificial)
¹²C6612Stable98.89%
¹³C6713Stable1.11%
¹⁴C6814Radioactive (t½ = 5730 yr)Trace
³⁵Cl171835Stable75.77%
³⁷Cl172037Stable24.23%
Why Cl has Ar = 35.5: Chlorine is ~75.77% ³⁵Cl and ~24.23% ³⁷Cl. The weighted average: (35 × 0.7577) + (37 × 0.2423) = 26.52 + 8.97 = 35.49 ≈ 35.5. It's not a whole number because it reflects the weighted average of two isotopes, not a single mass.
⚖️

Calculating Relative Atomic Mass

The relative atomic mass (Ar) is the weighted average of all isotopic masses, weighted by their fractional abundance. The standard reference is ¹²C = exactly 12.

Formula:
Ar = Σ (isotopic mass × fractional abundance)
= (m₁ × f₁) + (m₂ × f₂) + (m₃ × f₃) + ...
where fractional abundance = percentage ÷ 100

Mass Spectrometry and Isotope Data

A mass spectrometer ionises atoms and accelerates them through a magnetic field. Heavier ions deflect less, lighter ions more — they separate by mass. A detector measures the relative number of ions at each mass value, producing a mass spectrum showing peaks at each isotope's mass, with peak height proportional to abundance.

Mass Spectrum of Chlorine (atomic Cl isotope peaks shown) 100 75 50 25 0 34 35 36 37 38 75.77% 24.23% ³⁵Cl 75.77% ³⁷Cl 24.23% Mass-to-charge ratio (m/z) Relative Abundance (%) Ar(Cl) = (35 × 75.77 + 37 × 24.23) / 100 = 35.45 Weighted average → not a whole number (reflects isotope mixture)
🔄

Mass Spectrum of Chlorine

Mass Spectrum of Chlorine Mass / charge ratio (m/z) Relative abundance ~75% 35 ~25% 37 Two peaks = two isotopes
1

Worked Example 1 — Annotated: Calculate Ar from mass spectrum data

A mass spectrum of boron shows two peaks: mass 10 with relative abundance 19.9% and mass 11 with relative abundance 80.1%. Calculate the relative atomic mass of boron.
Step 1 — Convert percentages to fractionsFractional abundance of ¹⁰B = 19.9 ÷ 100 = 0.199Fractional abundance of ¹¹B = 80.1 ÷ 100 = 0.801Check: 0.199 + 0.801 = 1.000 ✓ (must sum to 1)
Always check that fractional abundances sum to 1.000 (or percentages sum to 100%). If they don't, there's an error in the data or you've missed an isotope.
Answer
Ar(B) = (10 × 0.199) + (11 × 0.801) = 1.990 + 8.811 = 10.8 (to 3 s.f.).
2

Worked Example 2 — Annotated: Find isotopic abundance from Ar

Silicon has two main isotopes: ²⁸Si (mass 28) and ²⁹Si (mass 29). If the Ar of silicon is 28.22, calculate the percentage abundance of each isotope. (Assume only these two isotopes for this calculation.)
Step 1 — Set up algebraLet x = fractional abundance of ²⁸SiThen (1 − x) = fractional abundance of ²⁹Si(abundances must sum to 1)
This is a two-variable problem with a constraint: abundances sum to 1. Using one variable is the most efficient approach. If there were three isotopes, you'd need two equations.
Answer
²⁸Si: 78%, ²⁹Si: 22% (for the simplified two-isotope model). Verify: (28 × 0.78) + (29 × 0.22) = 28.22 ✓
⚠️

Common Mistakes

Using percentage instead of fractional abundance in the formula. You must divide by 100: fractional abundance = % ÷ 100. Using 75.77 instead of 0.7577 gives Ar = 2646 — an obvious impossibility. Always check your answer is close to the mass number of the element.
Claiming isotopes have different chemical properties. Isotopes of the same element have the same electron configuration → identical chemical properties. Physical properties (mass, density, reaction rate) differ slightly, but chemical behaviour (reactivity, bond formation, valence) is the same.
Confusing Ar with mass number. Mass number (A) is a whole number for a specific nuclide. Ar is a decimal (weighted average of all isotopes). Chlorine A = 35 for ³⁵Cl, but Ar = 35.5. Never use a decimal as a mass number.

📓 Copy Into Your Books

⚛️ Isotopes

  • Same Z (protons), different A (neutrons)
  • Same chemical properties (same electrons)
  • Different physical properties (different mass)
  • Notation: mass A top, atomic Z bottom
  • Example: ³⁵Cl and ³⁷Cl — both chlorine

⚖️ Relative Atomic Mass

  • Ar = Σ(isotopic mass × fractional abundance)
  • Fractional abundance = % ÷ 100
  • All fractions must sum to 1.000
  • Reference: ¹²C = exactly 12
  • Ar is a decimal, not a whole number

📊 Mass Spectrometry

  • Separates ions by mass-to-charge ratio (m/z)
  • Peak mass = isotope mass number
  • Peak height ∝ relative abundance
  • Used to determine Ar and identify isotopes

⚠️ Exam Traps

  • Use fractions (not %) in Ar calculation
  • Ar pulled toward most abundant isotope
  • Isotopes ≠ different chemical properties
  • Mass number (A) is always a whole number
⚖️ Activity 1 — Ar Calculation Drill

1 Magnesium has three isotopes: ²⁴Mg (78.99%), ²⁵Mg (10.00%), ²⁶Mg (11.01%). Calculate the relative atomic mass of magnesium. Show full working.

✏️ Answer in your book

2 Copper has two stable isotopes: ⁶³Cu and ⁶⁵Cu. If Ar(Cu) = 63.55, calculate the percentage abundance of each isotope. Show working.

✏️ Answer in your book
📊 Activity 2 — Interpreting Mass Spectra

A mass spectrum of an unknown element X shows three peaks:

Relative abundance (%)
Column B

A Calculate the relative atomic mass of element X. Show full working.

✏️ Answer in your book

B Use the calculated Ar to identify the element (refer to the periodic table). State how many isotopes this element has.

✏️ Answer in your book
Interactive: Isotope Calculator Interactive
Revisit Your Initial Thinking

Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?

MC

Multiple Choice

5 random questions from a replayable lesson bank — feedback shown immediately

02

Short Answer Questions

6. Carbon-12 (¹²C) and carbon-14 (¹⁴C) are isotopes of carbon. (a) State one similarity and two differences between these isotopes. (b) Explain why both isotopes react identically with oxygen to form CO₂. 4 MARKS

✏️ Answer in your book

7. The Ar of neon is 20.18. Neon has three isotopes: ²⁰Ne (90.48%), ²¹Ne (0.27%), and ²²Ne (9.25%). Using the Ar formula, verify that these abundances are consistent with Ar = 20.18. Show full working. 3 MARKS

✏️ Answer in your book

✅ Comprehensive Answers

⚖️ Activity 1

1. Fractions: ²⁴Mg = 78.99÷100 = 0.7899, ²⁵Mg = 0.1000, ²⁶Mg = 0.1101. Sum = 1.0000 ✓. Ar = (24 × 0.7899) + (25 × 0.1000) + (26 × 0.1101) = 18.958 + 2.500 + 2.863 = 24.321 ≈ 24.32. (Compare to periodic table: Ar(Mg) = 24.31 ✓).

2. Let x = fractional abundance of ⁶³Cu. Then (1−x) = fractional abundance of ⁶⁵Cu. Ar = (63×x) + (65×(1−x)) = 63.55. 63x + 65 − 65x = 63.55. −2x = 63.55 − 65 = −1.45. x = 0.725. ⁶³Cu = 72.5%; ⁶⁵Cu = 27.5%. Verify: (63×0.725) + (65×0.275) = 45.675 + 17.875 = 63.55 ✓.

📊 Activity 2

A: Check: 1.4 + 24.1 + 22.1 + 52.4 = 100.0% ✓. Fractions: 204→0.014, 206→0.241, 207→0.221, 208→0.524. Ar = (204×0.014) + (206×0.241) + (207×0.221) + (208×0.524) = 2.856 + 49.646 + 45.747 + 108.992 = 207.24 ≈ 207.2.

B: Element X is Lead (Pb), Ar ≈ 207.2 (matches periodic table value of 207.2). Lead has 4 naturally occurring isotopes: ²⁰⁴Pb, ²⁰⁶Pb, ²⁰⁷Pb, ²⁰⁸Pb.

❓ Multiple Choice

1. C — Isotopes: same Z (protons), different A (neutrons count differs). Same element means same Z always.

2. B — Ar = (79 × 0.5069) + (81 × 0.4931) = 40.045 + 39.941 = 79.986 ≈ 79.99. Almost exactly 80 because the abundances are nearly equal.

3. D — Ar = 6.94, very close to 6. The lighter isotope dominates. If x = fraction of mass 6: 6x + 7(1−x) = 6.94 → −x = −0.06 → x = 0.94 = 94% mass-6 (this is lithium: ⁶Li 7.5%, ⁷Li 92.5% — the question uses hypothetical values).

4. A — Same protons → same electrons → same electron configuration → identical chemical behaviour. Neutrons don't participate in bonding.

5. C — Ar = (107 × 0.52) + (109 × 0.48) = 55.64 + 52.32 = 107.96. (This is silver, Ag.)

📝 Short Answer Model Answers

Q6 (4 marks): (a) Similarity: both have 6 protons (atomic number Z = 6) and 6 electrons (1 mark). Differences: (1) ¹²C has 6 neutrons, ¹⁴C has 8 neutrons (1 mark). (2) ¹²C is stable; ¹⁴C is radioactive (unstable nucleus) (1 mark). (b) Both react identically with oxygen because chemical behaviour is determined by electron configuration. ¹²C and ¹⁴C both have 6 protons → 6 electrons in a neutral atom → identical electron arrangement → same bonding behaviour → both form 2 C=O bonds with 2 oxygen atoms to give CO₂ (1 mark).

Q7 (3 marks): Fractions: ²⁰Ne = 90.48÷100 = 0.9048, ²¹Ne = 0.0027, ²²Ne = 0.0925. Sum = 0.9048 + 0.0027 + 0.0925 = 1.0000 ✓ (1 mark). Ar = (20 × 0.9048) + (21 × 0.0027) + (22 × 0.0925) (1 mark) = 18.096 + 0.0567 + 2.035 = 20.188 ≈ 20.18 ✓ (1 mark). The calculated value matches the stated Ar, confirming the abundances are consistent.

03

Revisit Your Thinking

Return to your Think First response. You should now be able to explain the 35.45 value precisely:

Consolidation Game

Isotopes and Relative Atomic Mass

Mark lesson as complete

Tick when you've finished all activities and checked your answers.