Module 1 · L14 of 2125 min⚡ +50 XP in Learn · +25 to completeYear 11 · Module 1 · IQ2
Isotopes and Relative Atomic Mass
Today's hook, The periodic table gives chlorine a mass of 35.5. No chlorine atom anywhere weighs 35.5. Every single one is either 35 or 37.
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Warm up and recall
Warm up first
Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.
Worksheets
Practise this lesson
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?
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What you'll master, and the words for it
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What you'll master
Know
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.
Understand
Concepts
Isotopes share virtually 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.
Can do
Skills
Calculate relative atomic mass from isotopic masses and percentage abundances.
Interpret mass spectra to identify isotopes and determine their relative abundances.
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Isotopes: Same Element, Different Mass
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Isotopes: Same Element, Different Mass
core concept
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 virtually 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%
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 ≈ 35.5. This uses the whole-number mass numbers, which is why it comes out near 35.5; the precise periodic-table value, 35.45, is obtained when you use the exact isotopic masses (34.97 and 36.97) instead of the mass numbers. Either way it is not a whole number, because it reflects the weighted average of two isotopes, not a single mass.
Isotopes are atoms of the same element with the same proton number (Z) but different neutron numbers. They have virtually 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.
protonsneutronselectronschemical
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.
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Structural Basis of Isotope Instability
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Structural Basis of Isotope Instability
core concept
We just saw that isotopes of the same element can be stable or radioactive. That raises a question: structurally, what actually makes one isotope stable and another unstable? This card answers it → the neutron-to-proton (N:Z) ratio, covered by the syllabus dot point on investigating the basic structure of stable and unstable isotopes (CH11-8 dot point 5).
Inside the nucleus, two forces compete. The strong nuclear force attracts all nucleons (protons and neutrons) to each other over very short range, holding the nucleus together. The electrostatic force repels protons from each other (like charges repel) and acts over a longer range. Whether a nucleus is stable depends on the balance between these two forces, which in turn depends on the ratio of neutrons to protons (N:Z).
The pattern: For light elements (up to about Z = 20, calcium), stable nuclei have N:Z ≈ 1:1, roughly equal numbers of protons and neutrons (e.g. ¹²C has 6 protons, 6 neutrons). For heavier elements, stable nuclei need progressively more neutrons than protons, rising toward N:Z ≈ 1.5:1 for the heaviest stable nuclei (e.g. ²⁰⁸Pb has 82 protons, 126 neutrons). Extra neutrons add strong-force attraction without adding extra proton-proton repulsion, this is why heavier stable nuclei need a growing neutron surplus to stay bound together.
Plotting the number of neutrons (N) against the number of protons (Z) for every stable isotope traces out a narrow curve called the band of stability. Isotopes that fall on this band are stable. Isotopes with too many or too few neutrons for their proton number fall outside the band, their combination of forces cannot hold the nucleus together indefinitely, and they are radioactive (unstable), they spontaneously decay, releasing particles and energy, in a process that moves them back toward the band of stability.
Looking ahead: Exactly what an unstable nucleus emits when it decays (alpha, beta or gamma radiation), how to write a balanced nuclear equation for that decay, and how quickly a radioactive sample decays (half-life), are covered in the next lesson, nuclear chemistry.
Nuclear stability depends on the neutron-to-proton (N:Z) ratio, the balance between the short-range strong nuclear force (attracts all nucleons) and the longer-range electrostatic force (repels protons from each other). Light stable nuclei have N:Z ≈ 1:1; heavy stable nuclei need progressively more neutrons, up to N:Z ≈ 1.5:1. Plotting N against Z for all stable isotopes traces the band of stability; isotopes outside this band are unstable (radioactive) and decay to move back toward it.
Pause, copy the highlighted N:Z ratio rule and band-of-stability definition into your book before moving on.
Quick check: why do heavier stable nuclei need progressively more neutrons than protons?
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Calculating Relative Atomic Mass
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Calculating Relative Atomic Mass
core concept
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.
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, on the scale where a carbon-12 atom is defined as exactly 12 (that is, relative to 1/12 the mass of a carbon-12 atom). 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?
Mass Spectrum
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Mass Spectrum
core concept
Beyond the syllabus. Calculating relative atomic mass from isotopic masses and abundances is Core, and it will be examined. The mass spectrometry instrument itself, how ions are formed, accelerated and deflected by a magnetic field, is supporting: Module 1 does not set the mechanism as an endpoint. What you do need from this card is how to read a spectrum, peak position is an isotope's mass and peak height is its abundance, and how to feed those two numbers into the Ar 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."
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Short Answer Questions
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Short Answer Questions
core concept
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 in essentially the same way 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
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 virtually 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.
35370.757735.5
For chlorine, Ar = (___ × ___) + (___ × 0.2423) ≈ ___.
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Worked examples
Worked examples · reveal as you go
Worked example+5 XP on full reveal
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.
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Convert percentages to fractional abundances: 19.9% ÷ 100 = 0.199 and 80.1% ÷ 100 = 0.801
All percentage values must be divided by 100 to get fractions. Always check: 0.199 + 0.801 = 1.000 ✓
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Apply the Ar formula: Ar = (mass₁ × fraction₁) + (mass₂ × fraction₂)
This is the weighted average formula. Each isotope's mass is "weighted" by how common it is in nature.
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Ar(B) = (10 × 0.199) + (11 × 0.801)
Substitute the values. Keep careful track of significant figures during multiplication.
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Ar(B) = 1.99 + 8.811 = 10.801
Add the two weighted contributions together to get the final result.
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Ar(B) ≈ 10.8 (to 3 significant figures)
Round your answer. Note: 10.8 is between 10 and 11, pulled toward 11 because the heavier isotope (¹¹B) is more abundant.
Worked example+5 XP on full reveal
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.)
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Let x = fractional abundance of ²⁸Si. Then (1 − x) = fractional abundance of ²⁹Si
Since only two isotopes exist, their abundances must sum to 1. Using one variable makes the algebra cleaner.
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Set up the equation: 28x + 29(1 − x) = 28.22
The Ar formula gives us one equation with one unknown (x). This is solvable.
Always check your answer matches the given Ar. If it doesn't, you've made an error in algebra.
Put the method in order
Sort the steps+7 XP
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.
Misconception to fix
Common errors · the 3 traps that cost marks
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Misconception to fix
Wrong: Isotopes have different chemical properties because they have different masses.
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Misconception to fix
Right: Isotopes have virtually 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. (Strictly, very small isotope effects can make reaction rates differ slightly, most noticeably between hydrogen and deuterium; for HSC purposes the chemistry of isotopes is treated as the same.)
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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.
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Drill, then revisit
Quick-fire practice · 5 reps +2 XP per reveal
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Define an isotope and state two ways isotopes of the same element differ.
An isotope is an atom of the same element with the same proton number but a different neutron number. Isotopes differ in mass number and nuclear stability (some are radioactive).
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Why do chlorine-35 and chlorine-37 have virtually identical chemical properties?
Both isotopes have 17 protons and 17 electrons, giving them the same electron configuration. Chemical behaviour is determined by electrons, not neutrons, so isotopes react in essentially the same way.
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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?
Bromine (Br), which has two stable isotopes ⁷⁹Br (~50.7%) and ⁸¹Br (~49.3%) with nearly equal natural abundances.
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Explain why the relative atomic mass of chlorine (35.45) is closer to 35 than to 37.
The lighter isotope ³⁵Cl is more abundant (~75.77%) than ³⁷Cl (~24.23%). The weighted average is pulled toward the more abundant isotope's mass, giving Ar ≈ 35.45.
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Revisit your thinking
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
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 in essentially the same way with oxygen to form CO₂.
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ApplyApply3 MARKS
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.
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
Isotopes: same Z (protons), different A (neutrons count differs). Same element means same Z always.
Ar = (79 × 0.5069) + (81 × 0.4931) = 40.045 + 39.941 = 79.986 ≈ 79.99. Almost exactly 80 because the abundances are nearly equal.
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).
Same protons → same electrons → same electron configuration → identical chemical behaviour. Neutrons don't participate in bonding.
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 in essentially the same way 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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Check what actually stuck
Check what actually stuck
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Take the full module quiz
quiz
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.