What are isotopes and how do they differ from each other?

Isotopes are variants of a particular chemical element that have the same number of protons but different numbers of neutrons. This means they have the same atomic number but different mass numbers. For example, carbon-12 and carbon-14 are isotopes of carbon, with mass numbers 12 and 14 respectively.

Why are isotopes important in the Cambridge IGCSE Chemistry curriculum?

Isotopes are important in the Cambridge IGCSE Chemistry curriculum because they help students understand the concept of atomic structure and the diversity of elements. They also play a crucial role in various applications such as radiocarbon dating, medical diagnostics, and nuclear energy, which are relevant to both theoretical and practical chemistry.

How do isotopes affect the average atomic mass of an element?

The average atomic mass of an element is calculated based on the relative abundance of its isotopes. Each isotope contributes to the average atomic mass in proportion to its abundance. For instance, if an element has two isotopes, one more abundant than the other, the average atomic mass will be closer to the mass of the more abundant isotope.

Can you explain the concept of stable and radioactive isotopes?

Stable isotopes are those that do not undergo radioactive decay and remain unchanged over time. Radioactive isotopes, on the other hand, are unstable and decay over time, emitting radiation. This property makes radioactive isotopes useful in various applications such as medical imaging and treatment, as well as in determining the age of archaeological finds through radiometric dating.

How are isotopes used in real-world applications?

Isotopes have a wide range of applications in the real world. In medicine, radioactive isotopes are used in diagnostic imaging and cancer treatment. In archaeology, isotopes like carbon-14 are used for dating ancient artifacts. In environmental science, isotopes help trace the origins of pollutants. These applications highlight the significance of isotopes beyond the classroom, making them an essential topic in the Cambridge IGCSE Chemistry curriculum.

Why is "Saying isotopes are different elements" a common mistake in Isotopes (New)?

Why it happens: Their different nucleon (mass) numbers make them look like different substances. How to avoid it: The proton number decides the element. Same proton number = SAME element — isotopes just differ in neutron number and mass. Source: 0620 Examiner Reports — recurring

Why is "Saying isotopes have different chemical properties" a common mistake in Isotopes (New)?

Why it happens: Students assume different masses must mean different behaviour. How to avoid it: Chemistry is determined by the outer-shell electrons, which are identical for all isotopes of an element. Chemical properties are therefore the same; only physical properties (e.g. density) differ slightly.

Why is "Forgetting to divide by the total percentage abundance" a common mistake in Isotopes (New)?

Why it happens: Rushing, or only adding the (mass × abundance) products. How to avoid it: Always divide the sum of (mass × abundance) by the total abundance (usually 100). The final A_r must lie between the smallest and largest isotope masses — if it is larger, you forgot to divide.

Why is "Taking a simple average of the isotope masses" a common mistake in Isotopes (New)?

Why it happens: Students average (35 + 37) ÷ 2 instead of weighting by abundance. How to avoid it: A_r is a WEIGHTED mean — each isotope mass must be multiplied by its own abundance before adding. A simple average only works when the abundances happen to be equal.

Cambridge IGCSE Chemistry (0620)

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Short Notes - Isotopes (New)

Detailed Study Notes

Detailed notes on Atoms, Elements and Compounds for Cambridge IGCSE Chemistry, covering key concepts, explanations, examples, and exam-focused revision points.

Isotopes — Cambridge IGCSE 0620 Chemistry Extended (2026)

Different forms of the same element with the same proton number but different nucleon numbers. Same chemistry, different mass — the basis for relative atomic masses.

At a glance

Isotopes: same proton number ( $Z$ ), different nucleon number ( $A$ ).
Same number of PROTONS and ELECTRONS → same chemistry.
Different number of NEUTRONS → different mass.
Some isotopes radioactive (e.g. $^{14}\text{C}$ , $^{60}\text{Co}$ ).
Relative atomic mass ( $A_r$ ): weighted average of all natural isotopes.
Cl: $A_r = 35.5$ because $^{35}\text{Cl}$ (75%) and $^{37}\text{Cl}$ (25%).

What you’ll learn

Mapped to the Cambridge IGCSE 0620 syllabus (2026-2028).

1.5 — Define isotope.
1.5 — Explain why isotopes of the same element have the same chemistry.
1.5 — Calculate relative atomic mass from isotopic abundance (Extended).

What is an isotope?

Same proton number, different number of neutrons → same element, different mass.

Isotope. A different form of the same element with the same number of PROTONS (so same $Z$ ) but a different number of NEUTRONS (so different $A$ ).

Examples.

Hydrogen: $^{1}_{1}\text{H}$ (protium, no neutrons), $^{2}_{1}\text{H}$ (deuterium, 1 neutron), $^{3}_{1}\text{H}$ (tritium, 2 neutrons).
Carbon: $^{12}_{6}\text{C}$ , $^{13}_{6}\text{C}$ , $^{14}_{6}\text{C}$ (radioactive).
Chlorine: $^{35}_{17}\text{Cl}$ and $^{37}_{17}\text{Cl}$ .

All three have one proton and one electron — same chemistry — but different neutron counts, so different masses.

Worked. $^{35}\text{Cl}$ has 17 protons (chlorine) and $35 - 17 = 18$ neutrons. $^{37}\text{Cl}$ has 17 protons and 20 neutrons.

Top number is the nucleon number A; bottom number is the proton number Z. Isotopes share Z but differ in neutron count.

Same proton number → same number of electrons → same chemistry.

Different mass. Heavier isotopes are slightly less mobile (e.g. $^{2}\text{H}_{2}\text{O}$ — heavy water — is denser and less volatile than ordinary water), but their chemical reactions are identical in type.

Same $Z$ , different $A$ .
Same protons + electrons.
Different neutrons.
Same chemistry, different mass.

Relative atomic mass from isotopes

$A_r$ = weighted average of the natural isotope masses, weighted by abundance.

Why is chlorine's $A_r$ shown as 35.5 on the Periodic Table? Because in nature, chlorine occurs as a mixture of two stable isotopes — neither has mass 35.5. The "35.5" is the WEIGHTED AVERAGE.

Calculation. $A_r = \dfrac{\sum (\text{isotope mass} \times \text{abundance \%})}{100}.$

Worked. Chlorine: $^{35}\text{Cl}$ at 75% abundance, $^{37}\text{Cl}$ at 25%.

$A_r = (35 \times 75 + 37 \times 25)/100 = (2625 + 925)/100 = 35.5$ .

Worked. Boron: $^{10}\text{B}$ at 20%, $^{11}\text{B}$ at 80%.

$A_r = (10 \times 20 + 11 \times 80)/100 = (200 + 880)/100 = 10.8$ .

Setting the chlorine calculation out as a table makes the weighting clear:

Isotope	Isotope mass	Abundance	mass × abundance
³⁵Cl	35	75%	35 × 75 = 2625
³⁷Cl	37	25%	37 × 25 = 925
Total	—	100%	3550

$A_r = 3550 \div 100 = 35.5$ .

The relative atomic mass sits between the isotope masses, pulled toward the more abundant one.

Tip. $A_r$ is rarely a whole number — that's a sign of multiple isotopes. The Periodic Table values (e.g. Cl 35.5, Cu 63.5) are the standardised weighted averages.

$A_r = \sum(m \times \%) / 100$ .
Weighted by ABUNDANCE.
Cl: 35.5 from 75% $^{35}$ Cl + 25% $^{37}$ Cl.
Non-whole $A_r$ → multiple isotopes.

Radioactive isotopes (briefly)

Some isotopes are unstable — used as tracers, dating, medical treatment.

Radioactive isotopes. Have unstable nuclei that emit radiation (alpha, beta, or gamma) over time.

Common uses.

$^{60}\text{Co}$ (cobalt-60): gamma source; cancer treatment, sterilising medical equipment.
$^{14}\text{C}$ (carbon-14): beta emitter; radiocarbon dating of organic remains (half-life $5730$ years).
$^{131}\text{I}$ (iodine-131): medical tracer; thyroid imaging.
$^{235}\text{U}$ (uranium-235): nuclear power.

Isotope	Use
⁶⁰Co (cobalt-60)	Gamma source — cancer treatment, sterilising medical equipment
¹⁴C (carbon-14)	Beta emitter — radiocarbon dating of organic remains
¹³¹I (iodine-131)	Medical tracer — thyroid imaging
²³⁵U (uranium-235)	Nuclear power

Worked qualitative. Why use carbon-14 for dating but not for cancer treatment? Half-life 5730 years means activity is too low for a significant medical effect; useful for archaeology because the long timescale matches sample ages.

Cambridge tip. Cambridge IGCSE Chemistry doesn't go deep into radioactive decay — that's more Physics 0625. Just know: some isotopes are radioactive, used in dating, medicine and energy.

Some isotopes unstable → radioactive.
$^{14}\text{C}$ for archaeological dating.
$^{60}\text{Co}$ for cancer treatment / sterilisation.
$^{131}\text{I}$ for thyroid scans.

Quick recap

Isotopes: same $Z$ , different $A$ .
Same chemistry; different mass.
$A_r = \sum(m \times \%)/100$ .
Cl: $A_r = 35.5$ from 75/25 mix of $^{35,37}$ Cl.
Some isotopes radioactive: dating, medicine, energy.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Isotope — different forms of the same element with the same proton number but different nucleon numbers.
Relative atomic mass ( $A_r$ ) — weighted average mass of an element's isotopes (relative to $\frac{1}{12}$ of $^{12}\text{C}$ ).
Radioactive isotope — unstable isotope that emits radiation over time.

How it’s examined

Isotopes appear most years on Paper 2 (3-4 marks: definition, identify isotope pairs) and Paper 4 (4-5 marks: $A_r$ calculation from abundances). Examiner reports flag students saying 'isotopes have different chemistries' — they don't, they have the same chemistry but different mass.

Sources: Cambridge IGCSE Chemistry 0620 syllabus 2026-2028 (1.5); 0620/42 Oct/Nov 2024 — Q4 ($A_r$ from isotopic abundances); 0620 Examiner Reports 2022-2024. Last reviewed 2026-05-06.

Take this whole topic with you

Step-by-step worked examples — Isotopes (New)

Step-by-step solutions to past-paper-style questions on isotopes (new), written exactly the way a tutor would explain them at the board.

1Identify a pair of isotopes
Getting started• definition, nuclide notation
Question
Are $^{35}_{17}\text{Cl}$ and $^{37}_{17}\text{Cl}$ isotopes of each other? Explain your answer in terms of subatomic particles.
Step-by-step solution
1. Step 1
  Read the nuclide notation. The bottom number is the proton number ( $Z$ ); the top number is the nucleon (mass) number ( $A$ ). Both atoms have $Z = 17$ , so both are chlorine.
2. Step 2
  Find the neutrons using $\text{neutrons} = A - Z$ . For $^{35}\text{Cl}$ : $35 - 17 = 18$ neutrons. For $^{37}\text{Cl}$ : $37 - 17 = 20$ neutrons.
3. Step 3
  Same proton number, different number of neutrons (hence different nucleon number) — this is exactly the definition of isotopes.
Answer
Yes. Both have 17 protons (so both are chlorine), but $^{35}\text{Cl}$ has 18 neutrons and $^{37}\text{Cl}$ has 20 neutrons. Same proton number, different nucleon number = isotopes.
Examiner tip
Cambridge rewards the link to particles: same protons (and electrons), different neutrons. Naming them 'the same element' alone is not enough.
2Count protons, neutrons and electrons in an isotope
Getting started• nuclide notation
Question
An atom of the isotope $^{40}_{19}\text{K}$ is neutral. State the number of protons, neutrons and electrons it contains.
Step-by-step solution
1. Step 1
  Protons = proton number = 19.
2. Step 2
  Neutrons = nucleon number − proton number.
  $\text{neutrons} = 40 - 19 = 21$
3. Step 3
  The atom is neutral, so electrons = protons = 19. (Changing the isotope changes neutrons, never the electrons of a neutral atom.)
Answer
19 protons, 21 neutrons, 19 electrons.
Examiner tip
Every isotope of an element has the SAME number of protons and (when neutral) electrons — only the neutron count varies between isotopes.
3Calculate relative atomic mass from two isotopes
Building confidence• Adapted from 0620/42 May/Jun 2024 Q4• A_r, weighted mean
Question
A sample of chlorine contains $75\%$ $^{35}\text{Cl}$ and $25\%$ $^{37}\text{Cl}$ . Calculate the relative atomic mass of chlorine. Give your answer to one decimal place.
Step-by-step solution
1. Step 1
  $A_r$ is a weighted mean — multiply each isotope's mass by its percentage abundance, add them, then divide by the total abundance (100).
  $A_r = \dfrac{(35 \times 75) + (37 \times 25)}{100}$
2. Step 2
  Work out the top line first.
  $(35 \times 75) + (37 \times 25) = 2625 + 925 = 3550$
3. Step 3
  Divide by 100.
  $A_r = \dfrac{3550}{100} = 35.5$
Answer
$A_r(\text{Cl}) = 35.5$
Examiner tip
The answer must lie between the two isotope masses (35 and 37) and be closer to the more abundant one (35) — a quick sanity check. Always divide by the total abundance, here 100.
4Why isotopes share chemical but not physical properties
Building confidence• properties
Question
Explain why the two isotopes of chlorine, $^{35}\text{Cl}$ and $^{37}\text{Cl}$ , have identical chemical properties but slightly different physical properties.
Step-by-step solution
1. Step 1
  Chemical properties depend on the electronic configuration — specifically the number of outer-shell electrons. Both isotopes have 17 electrons arranged $2,8,7$ , so they react identically.
2. Step 2
  Isotopes differ only in their number of neutrons, which adds mass but does not affect the electrons, so chemistry is unchanged.
3. Step 3
  Physical properties such as density depend on mass. $^{37}\text{Cl}$ has two more neutrons, so an atom is heavier — giving a slightly higher density (and small differences in melting/boiling point and rate of diffusion).
Answer
Both isotopes have the same number and arrangement of electrons, so their chemical reactions are identical. They differ in neutron number and therefore mass, which gives small differences in physical properties such as density.
Examiner tip
Mark schemes want the cause stated, not just the effect: 'same electron configuration → same chemistry' and 'different mass → different physical properties (e.g. density)'.
5Find an unknown abundance from the relative atomic mass
Stretch• 0620/42-style structured question• A_r, weighted mean, algebra
Question
Boron exists as two isotopes, $^{10}\text{B}$ and $^{11}\text{B}$ . The relative atomic mass of boron is $10.8$ . Calculate the percentage abundance of each isotope.
Step-by-step solution
1. Step 1
  Let the abundance of $^{11}\text{B}$ be $x\%$ . Then the abundance of $^{10}\text{B}$ is $(100 - x)\%$ , because the two must total 100%.
2. Step 2
  Write the weighted-mean equation for $A_r = 10.8$ and multiply through by 100.
  $\dfrac{10(100 - x) + 11x}{100} = 10.8 \;\Rightarrow\; 10(100 - x) + 11x = 1080$
3. Step 3
  Expand and simplify.
  $1000 - 10x + 11x = 1080 \;\Rightarrow\; 1000 + x = 1080$
4. Step 4
  Solve for $x$ (the $^{11}\text{B}$ abundance), then subtract from 100 for $^{10}\text{B}$ .
  $x = 80 \;\Rightarrow\; {}^{11}\text{B} = 80\%,\; {}^{10}\text{B} = 20\%$
Answer
$^{11}\text{B} = 80\%$ and $^{10}\text{B} = 20\%$ .
Examiner tip
Setting the unknown to $x$ and the other to $(100 - x)$ is the standard reverse- $A_r$ method. Check: $\frac{(10 \times 20) + (11 \times 80)}{100} = \frac{200 + 880}{100} = 10.8$ ✓.
6Relative atomic mass from three isotopes
Stretch• A_r, weighted mean
Question
Magnesium has three isotopes: $^{24}\text{Mg}$ ( $79\%$ ), $^{25}\text{Mg}$ ( $10\%$ ) and $^{26}\text{Mg}$ ( $11\%$ ). Calculate the relative atomic mass of magnesium to one decimal place.
Step-by-step solution
1. Step 1
  Extend the weighted-mean formula to three terms — one product per isotope — and divide by the total abundance (100).
  $A_r = \dfrac{(24 \times 79) + (25 \times 10) + (26 \times 11)}{100}$
2. Step 2
  Evaluate each product.
  $(24 \times 79) = 1896,\;\; (25 \times 10) = 250,\;\; (26 \times 11) = 286$
3. Step 3
  Add the products and divide by 100.
  $A_r = \dfrac{1896 + 250 + 286}{100} = \dfrac{2432}{100} = 24.3$
Answer
$A_r(\text{Mg}) = 24.3$
Examiner tip
First check the abundances sum to 100% ( $79 + 10 + 11 = 100$ ). The result should sit nearest $^{24}\text{Mg}$ because it is by far the most abundant — and $24.3$ does.

Model Answers — Isotopes (New)

High-scoring sample answers for isotopes (new) on the Cambridge IGCSE 0620 paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
Paper 4 short-answer style1 mark
Define the term isotope. (1 mark)
Model answer
Isotopes are atoms of the same element that have the same proton number but a different nucleon (mass) number — that is, the same number of protons but a different number of neutrons.
Why this scores
One mark for 'same proton number, different nucleon number' (or 'same protons, different neutrons'). 'Same element' on its own is not enough.
Question 2
Paper 4 short-answer style2 marks
$^{63}_{29}\text{Cu}$ and $^{65}_{29}\text{Cu}$ are isotopes of copper. State the difference in the number of neutrons between one atom of each. (2 marks)
Model answer
Neutrons $=$ nucleon number $-$ proton number. For $^{63}\text{Cu}$ : $63 - 29 = 34$ neutrons. For $^{65}\text{Cu}$ : $65 - 29 = 36$ neutrons. The difference is $36 - 34 = 2$ neutrons.
Why this scores
One mark for correctly finding the neutron numbers (34 and 36), one for the difference (2). Protons and electrons are unchanged.
Question 3
Paper 4 (Extended) structured style3 marks
Explain why the isotopes of an element have the same chemical properties. (3 marks)
Model answer
Isotopes of an element have the same number of electrons because they have the same proton number. This means they have the same electronic configuration / the same number of outer-shell electrons. Chemical properties are determined by the outer-shell electrons, so the isotopes react in exactly the same way. They differ only in their number of neutrons, which affects mass but not the electrons and therefore not the chemistry.
Why this scores
Three marks: (1) same number of electrons; (2) same electronic configuration / same outer electrons; (3) chemistry is decided by outer electrons, so reactions are identical. Mentioning that neutrons do not affect chemistry secures the link.
Question 4
Paper 4 (Extended) structured style4 marks
Bromine consists of $50\%$ $^{79}\text{Br}$ and $50\%$ $^{81}\text{Br}$ . Calculate the relative atomic mass of bromine. Show your working. (4 marks)
Model answer
Relative atomic mass is the weighted mean of the isotope masses:

$A_r = \dfrac{(79 \times 50) + (81 \times 50)}{100}$

Working out the top line: $(79 \times 50) + (81 \times 50) = 3950 + 4050 = 8000$ .

Dividing by the total abundance: $A_r = \dfrac{8000}{100} = 80$ .

So the relative atomic mass of bromine is 80. (Because the two abundances are equal, the answer is simply the average of 79 and 81.)
Why this scores
Four marks: (1) correct method — multiply mass by abundance; (2) correct products summed (8000); (3) divide by 100; (4) correct answer (80) with working shown. A bald answer with no working can lose method marks.
Question 5
0620/42-style structured question5 marks
Neon has three isotopes: $^{20}\text{Ne}$ ( $90.5\%$ ), $^{21}\text{Ne}$ ( $0.3\%$ ) and $^{22}\text{Ne}$ ( $9.2\%$ ). Calculate the relative atomic mass of neon to one decimal place. (5 marks)
Model answer
Use the weighted-mean formula with one term per isotope, dividing by the total abundance (100):

$A_r = \dfrac{(20 \times 90.5) + (21 \times 0.3) + (22 \times 9.2)}{100}$

Evaluate each product: $(20 \times 90.5) = 1810$ ; $(21 \times 0.3) = 6.3$ ; $(22 \times 9.2) = 202.4$ .

Add them: $1810 + 6.3 + 202.4 = 2018.7$ .

Divide by 100: $A_r = \dfrac{2018.7}{100} = 20.187 \approx 20.2$ .

The relative atomic mass of neon is 20.2.
Why this scores
Five marks: (1) correct method extended to three isotopes; (2) each product evaluated correctly; (3) products summed (2018.7); (4) divided by 100; (5) answer rounded correctly to 20.2. Check abundances sum to 100% first.
Question 6
0620/42-style structured question6 marks
Copper has two isotopes, $^{63}\text{Cu}$ and $^{65}\text{Cu}$ . The relative atomic mass of copper is $63.5$ . Calculate the percentage abundance of each isotope. Show all your working. (6 marks)
Model answer
Let the abundance of $^{65}\text{Cu}$ be $x\%$ ; then $^{63}\text{Cu}$ must be $(100 - x)\%$ because the two abundances add up to 100%.

Write the weighted-mean equation and set it equal to 63.5:

$\dfrac{63(100 - x) + 65x}{100} = 63.5$

Multiply both sides by 100:

$63(100 - x) + 65x = 6350$

Expand the bracket: $6300 - 63x + 65x = 6350$ .

Simplify the $x$ terms: $6300 + 2x = 6350$ .

Solve: $2x = 50$ , so $x = 25$ .

Therefore $^{65}\text{Cu} = 25\%$ and $^{63}\text{Cu} = 100 - 25 = 75\%$ .

Check: $\dfrac{(63 \times 75) + (65 \times 25)}{100} = \dfrac{4725 + 1625}{100} = \dfrac{6350}{100} = 63.5$ ✓.
Why this scores
Six marks: (1) define unknowns as $x$ and $(100 - x)$ ; (2) correct weighted-mean equation; (3) clear the fraction (×100); (4) expand correctly; (5) solve to $x = 25$ ; (6) state both abundances (25% and 75%). A verification line confirms the answer and guards against algebra slips.

Key Formulae — Isotopes (New)

The formulae you need to memorise for isotopes (new) on the Cambridge IGCSE 0620 paper, with every variable defined in plain English and a note on when to use it.

Relative atomic mass from isotopic abundances
$A_r = \dfrac{\Sigma\,(\text{isotope mass} \times \%\,\text{abundance})}{\text{total }\%\,\text{abundance}}$
$A_r$
Relative atomic mass (weighted mean)
$\text{isotope mass}$
Nucleon (mass) number of each isotope
$\%\,\text{abundance}$
Percentage of each isotope in the sample
When to use
Whenever an element has two or more isotopes with known percentage abundances. The total abundance is almost always 100.
Example
Chlorine: $A_r = \frac{(35 \times 75) + (37 \times 25)}{100} = 35.5$ .

Key Definitions and Keywords — Isotopes (New)

Definitions to memorise and the exact keywords mark schemes credit for isotopes (new) answers — sharpened from recent examiner reports for the 2026 0620 sitting.

Isotope
Examiner keyword
Atoms of the same element with the same proton number ( $Z$ ) but different nucleon numbers ( $A$ ) — i.e. the same number of protons but different numbers of neutrons.
Nucleon (mass) number, $A$
Examiner keyword
The total number of protons and neutrons in the nucleus of an atom. Shown as the top number in nuclide notation, e.g. $^{35}_{17}\text{Cl}$ has $A = 35$ .
Proton number (atomic number), $Z$
Examiner keyword
The number of protons in the nucleus. It defines the element and is the bottom number in nuclide notation. All isotopes of an element share the same $Z$ .
Relative atomic mass ( $A_r$ )
Examiner keyword
The average mass of the atoms of an element, taking into account the masses and abundances of all its isotopes, on a scale where an atom of $^{12}\text{C}$ has a mass of exactly 12.

Common Mistakes and Misconceptions — Isotopes (New)

The traps other students keep falling into on isotopes (new) questions — taken from recent Cambridge IGCSE 0620 examiner reports and mark schemes — and how to avoid them.

✕Saying isotopes are different elements
0620 Examiner Reports — recurring
Why it happens
Their different nucleon (mass) numbers make them look like different substances.
How to avoid it
The proton number decides the element. Same proton number = SAME element — isotopes just differ in neutron number and mass.
✕Saying isotopes have different chemical properties
Why it happens
Students assume different masses must mean different behaviour.
How to avoid it
Chemistry is determined by the outer-shell electrons, which are identical for all isotopes of an element. Chemical properties are therefore the same; only physical properties (e.g. density) differ slightly.
✕Forgetting to divide by the total percentage abundance
Why it happens
Rushing, or only adding the (mass × abundance) products.
How to avoid it
Always divide the sum of (mass × abundance) by the total abundance (usually 100). The final $A_r$ must lie between the smallest and largest isotope masses — if it is larger, you forgot to divide.
✕Taking a simple average of the isotope masses
Why it happens
Students average $(35 + 37) \div 2$ instead of weighting by abundance.
How to avoid it
$A_r$ is a WEIGHTED mean — each isotope mass must be multiplied by its own abundance before adding. A simple average only works when the abundances happen to be equal.

Isotopes (New) — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Isotope. A different form of the same element with the same number of PROTONS (so same $Z$ ) but a different number of NEUTRONS (so different $A$ ).

Examples.

Hydrogen: $^{1}_{1}\text{H}$ (protium, no neutrons), $^{2}_{1}\text{H}$ (deuterium, 1 neutron), $^{3}_{1}\text{H}$ (tritium, 2 neutrons).
Carbon: $^{12}_{6}\text{C}$ , $^{13}_{6}\text{C}$ , $^{14}_{6}\text{C}$ (radioactive).
Chlorine: $^{35}_{17}\text{Cl}$ and $^{37}_{17}\text{Cl}$ .

All three have one proton and one electron — same chemistry — but different neutron counts, so different masses.

Worked. $^{35}\text{Cl}$ has 17 protons (chlorine) and $35 - 17 = 18$ neutrons. $^{37}\text{Cl}$ has 17 protons and 20 neutrons.

Top number is the nucleon number A; bottom number is the proton number Z. Isotopes share Z but differ in neutron count.

Same proton number → same number of electrons → same chemistry.

Isotope

Isotope mass

Abundance

mass × abundance

³⁵Cl

75%

35 × 75 = 2625

³⁷Cl

25%

37 × 25 = 925

Total

—

100%

3550

Isotope

Use

⁶⁰Co (cobalt-60)

Gamma source — cancer treatment, sterilising medical equipment

¹⁴C (carbon-14)

Beta emitter — radiocarbon dating of organic remains

¹³¹I (iodine-131)

Medical tracer — thyroid imaging

²³⁵U (uranium-235)

Nuclear power

Isotopes (New)

Short Notes - Isotopes (New)

Detailed Study Notes

At a glance

What you’ll learn

What is an isotope?

Relative atomic mass from isotopes

Radioactive isotopes (briefly)

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Identify a pair of isotopes

2Count protons, neutrons and electrons in an isotope

3Calculate relative atomic mass from two isotopes

4Why isotopes share chemical but not physical properties

5Find an unknown abundance from the relative atomic mass

6Relative atomic mass from three isotopes

Relative atomic mass from isotopic abundances

Isotope

Nucleon (mass) number, AAA

Proton number (atomic number), ZZZ

Relative atomic mass (ArA_rAr​)

✕Saying isotopes are different elements

✕Saying isotopes have different chemical properties

✕Forgetting to divide by the total percentage abundance

✕Taking a simple average of the isotope masses

Isotopes (New) — frequently asked questions

Isotopes (New)

Short Notes - Isotopes (New)

Detailed Study Notes

At a glance

What you’ll learn

What is an isotope?

Relative atomic mass from isotopes

Radioactive isotopes (briefly)

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Identify a pair of isotopes

2Count protons, neutrons and electrons in an isotope

3Calculate relative atomic mass from two isotopes

4Why isotopes share chemical but not physical properties

5Find an unknown abundance from the relative atomic mass

6Relative atomic mass from three isotopes

Relative atomic mass from isotopic abundances

Isotope

Nucleon (mass) number, AAA

Proton number (atomic number), ZZZ

Relative atomic mass (ArA_rAr​)

✕Saying isotopes are different elements

✕Saying isotopes have different chemical properties

✕Forgetting to divide by the total percentage abundance

✕Taking a simple average of the isotope masses

Isotopes (New) — frequently asked questions

Nucleon (mass) number, $A$

Proton number (atomic number), $Z$

Relative atomic mass ( $A_r$ )

Nucleon (mass) number, $A$

Proton number (atomic number), $Z$

Relative atomic mass ( $A_r$ )