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.
Core Content
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:
| Isotope | Protons (Z) | Neutrons (A−Z) | Mass number (A) | Stability | Natural abundance |
|---|---|---|---|---|---|
| ¹H (protium) | 1 | 0 | 1 | Stable | 99.985% |
| ²H (deuterium) | 1 | 1 | 2 | Stable | 0.015% |
| ³H (tritium) | 1 | 2 | 3 | Radioactive | Trace (artificial) |
| ¹²C | 6 | 6 | 12 | Stable | 98.89% |
| ¹³C | 6 | 7 | 13 | Stable | 1.11% |
| ¹⁴C | 6 | 8 | 14 | Radioactive (t½ = 5730 yr) | Trace |
| ³⁵Cl | 17 | 18 | 35 | Stable | 75.77% |
| ³⁷Cl | 17 | 20 | 37 | Stable | 24.23% |
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.
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.
Insert bar chart: x-axis = mass number (35 and 37); y-axis = relative abundance (%). Bar at 35 height ≈ 75.77%; bar at 37 height ≈ 24.23%. Label each bar with mass number and % abundance. Title: "Mass Spectrum of Chlorine".
Worked Examples
| Step 1 — Convert percentages to fractions Fractional abundance of ¹⁰B = 19.9 ÷ 100 = 0.199 Fractional abundance of ¹¹B = 80.1 ÷ 100 = 0.801 Check: 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. |
| Step 2 — Apply weighted average formula Ar = (10 × 0.199) + (11 × 0.801) Ar = 1.990 + 8.811 Ar = 10.801 | Each term = (isotopic mass) × (fractional abundance). The isotopic mass used here is the mass number (integer) — in more precise calculations, exact isotopic masses are given (e.g. ¹¹B = 11.009). At HSC level, use the mass number provided. |
| Step 3 — Round appropriately and report Ar(B) = 10.8 (rounded to 3 significant figures, consistent with the data provided) Compare to periodic table value: Ar(B) = 10.81 ✓ | The answer is closer to 11 than to 10 because ¹¹B is the more abundant isotope (80.1%). Weighted averages are always "pulled toward" the more abundant isotope. This is a useful cross-check: if 80% is one value and 20% is another, the average should be about 80% of the way from the 20% value to the 80% value. |
| Step 1 — Set up algebra Let x = fractional abundance of ²⁸Si Then (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. |
| Step 2 — Write the Ar equation Ar = (28 × x) + (29 × (1 − x)) = 28.22 28x + 29 − 29x = 28.22 −x = 28.22 − 29 = −0.78 x = 0.78 | Always expand the brackets before solving. A common error is distributing the wrong sign. Check: does the answer make sense? Ar = 28.22 is very close to 28, so ²⁸Si should be the overwhelmingly dominant isotope — x = 0.78 = 78% makes sense. |
| Step 3 — Convert and verify ²⁸Si: x = 0.78 → 78% ²⁹Si: 1 − 0.78 = 0.22 → 22% Verify: (28 × 0.78) + (29 × 0.22) = 21.84 + 6.38 = 28.22 ✓ | Note: real silicon has three stable isotopes (²⁸Si 92.2%, ²⁹Si 4.7%, ³⁰Si 3.1%), so this two-isotope example is simplified. In real mass spectrometry data, all peaks would be shown. |
Activities
1 Magnesium has three isotopes: ²⁴Mg (78.99%), ²⁵Mg (10.00%), ²⁶Mg (11.01%). Calculate the relative atomic mass of magnesium. Show full working.
2 Copper has two stable isotopes: ⁶³Cu and ⁶⁵Cu. If Ar(Cu) = 63.55, calculate the percentage abundance of each isotope. Show working.
A mass spectrum of an unknown element X shows three peaks:
| Mass number | Relative abundance (%) |
|---|---|
| 204 | 1.4 |
| 206 | 24.1 |
| 207 | 22.1 |
| 208 | 52.4 |
A Calculate the relative atomic mass of element X. Show full working.
B Use the calculated Ar to identify the element (refer to the periodic table). State how many isotopes this element has.
Multiple Choice
1. Which correctly describes isotopes of the same element?
2. Bromine has two isotopes: ⁷⁹Br (50.69%) and ⁸¹Br (49.31%). Its relative atomic mass is approximately:
3. Two isotopes of an element have masses of 6 and 7. The relative atomic mass of the element is 6.94. This means:
4. Why do isotopes of the same element have identical chemical properties?
5. A mass spectrum shows an element with peaks at mass 107 (52%) and 109 (48%). What is the relative atomic mass?
Short Answer
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
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
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 ✓.
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.
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.)
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.
Tick when you've finished all activities and checked your answers.