Isotopes and Relative Atomic Mass
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Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
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?
Key facts
- Isotopes of an element have the same proton number (Z) but different neutron numbers.
- Relative atomic mass is a weighted average of all naturally occurring isotopes, using the ¹²C = 12 standard.
Concepts
- Isotopes share identical chemical properties because chemistry is determined by electron configuration, but differ in physical properties and nuclear stability.
- Ar values on the periodic table are rarely whole numbers because they reflect the weighted average of isotopes with different natural abundances.
Skills
- Calculate relative atomic mass from isotopic masses and percentage abundances.
- Interpret mass spectra to identify isotopes and determine their relative abundances.
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
| 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% |
Isotopes are atoms of the same element with the same proton number (Z) but different neutron numbers. They have identical chemical properties (chemistry depends on electron configuration, which is the same) but differ in mass, density, BP, and nuclear stability (some are radioactive). Examples: ¹H, ²H, ³H; ¹²C, ¹³C, ¹⁴C; ³⁵Cl, ³⁷Cl.
Pause — copy the highlighted isotope definition into your book before moving on.
Fill the blanks: drag each word into the right gap.
Isotopes of an element share the same number of ___ but have different numbers of ___. Because they share the same number and arrangement of ___, isotopes have identical ___ properties.
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.
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.
We just saw that isotopes have the same chemical behaviour but different masses. That raises a question: since elements have multiple isotopes with different masses, how is the single "atomic mass" value on the periodic table calculated? This card answers it → it is the weighted average of all naturally occurring isotopic masses, using fractional abundance as the weighting.
Relative atomic mass (Ar) = weighted average of all naturally occurring isotopic masses, using ¹²C = 12 as the standard. Formula: Ar = Σ(isotopic mass × fractional abundance). Always convert percentage abundances to fractions (÷ 100). Example: for neon, Ar = (20 × 0.9048) + (21 × 0.0027) + (22 × 0.0925) ≈ 20.18.
Add the highlighted Ar formula to your notes before the check below.
Odd one out: which step does NOT belong in a relative atomic mass calculation?
We just saw how Ar is calculated from fractional abundances. That raises a question: how do chemists actually measure isotopic composition experimentally? This card answers it → mass spectrometry separates ions by mass and measures their relative abundance, producing a spectrum where peak height equals abundance.
A mass spectrum shows peaks at each isotope's mass-to-charge ratio (m/z), with peak height proportional to relative abundance. More abundant isotopes produce taller peaks. Mass spectrometry is the primary method for determining isotopic composition and calculating Ar. A spectrum with two peaks at m/z 35 and 37 (ratio 3:1) gives Ar = (35 × 0.75) + (37 × 0.25) = 35.5.
Pause — write the highlighted mass spectrum rule into your book.
True or false: "On a mass spectrum, the height of each peak is proportional to the relative abundance of that isotope."
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
We just saw how mass spectra reveal isotopic composition. That raises a question: how do you write full-mark exam answers on isotopes, Ar calculations, and mass spectra? This card answers it → for each question type, follow the structured method: define, calculate (showing working), and interpret from the data given.
For exam answers on isotopes: state that isotopes have identical chemical properties (same Z, same electron configuration) but different physical properties (different mass, density). For Ar calculations: convert % to fractions, multiply each isotopic mass by its fraction, then sum — always show the calculation. For mass spectra: peak position = mass number, peak height ∝ relative abundance.
Pause — copy the highlighted exam strategy into your book before moving on.
Fill the blanks: drag each value into the chlorine Ar calculation.
For chlorine, Ar = (___ × ___) + (___ × 0.2423) ≈ ___.
Worked examples · reveal as you go
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.
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.)
Click two steps to swap them. Put the relative atomic mass calculation for chlorine (³⁵Cl 75.77%, ³⁷Cl 24.23%) in the correct order.
- Multiply each isotopic mass by its fractional abundance: 35 × 0.7577 = 26.52 and 37 × 0.2423 = 8.97.
- Identify the two isotopes and their percentage abundances from the mass spectrum.
- Convert each percentage to a fractional abundance: 75.77% → 0.7577 and 24.23% → 0.2423.
- Add the weighted contributions: 26.52 + 8.97 = 35.49.
- Round and report Ar(Cl) ≈ 35.5, matching the periodic table value.
Common errors · the 3 traps that cost marks
Misconception to fix
Wrong: Isotopes have different chemical properties because they have different masses.
Misconception to fix
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.
Ar is the simple average of isotopic masses
Wrong: To find relative atomic mass, just add up the isotopic masses and divide by how many isotopes there are.
Fix: Always use the weighted average: multiply each isotopic mass by its fractional abundance (percentage ÷ 100) and sum the results. The simple average only works when all isotopes are equally abundant, which never happens in nature. Check: chlorine's simple average would be (35 + 37) ÷ 2 = 36, but its actual Ar is 35.45 because ³⁵Cl is more abundant.
Quick-fire practice · 5 reps +2 XP per reveal
Define an isotope and state two ways isotopes of the same element differ.
Why do chlorine-35 and chlorine-37 have identical chemical properties?
Boron has isotopes ¹⁰B (19.9%) and ¹¹B (80.1%). Calculate Ar(B) to 3 significant figures.
A mass spectrum shows two peaks at m/z = 79 and m/z = 81 with roughly equal heights. What element is this?
Explain why the relative atomic mass of chlorine (35.45) is closer to 35 than to 37.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
Pick your answer, then rate your confidence — that tells the system what to drill next.
Complete the Ar calculation for neon: ²⁰Ne 90.48%, ²¹Ne 0.27%, ²²Ne 9.25%. Type each missing value, then click Check.
Apply Ar = Σ(mass × fraction):
Ar(Ne) = (20 × ) + (21 × 0.0027) + ( × 0.0925)
Ar(Ne) = 18.096 + 0.0567 + 2.035 =
Q1. 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.
Q2. 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.
📖 Comprehensive answers (click to reveal)
️ 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.
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