Why is "Reading the mass number as the number of neutrons" a common mistake in Key terms in atomic structure?

Why it happens: Students see the large top number and assume it is the neutrons. How to avoid it: The mass number A = protons + neutrons. Find neutrons by **subtracting**: neutrons = A − Z. Always write the subtraction line. Source: Edexcel Chemistry examiner reports — recurring error

Why is "Defining isotopes as having 'different numbers of protons'" a common mistake in Key terms in atomic structure?

Why it happens: Students remember that isotopes are 'different' but attach the difference to the wrong particle. How to avoid it: Isotopes have the **SAME protons** (same element) and **DIFFERENT neutrons**. Different protons would mean different elements. Source: Edexcel Chemistry examiner reports

Why is "Getting the electron count of an ion the wrong way round" a common mistake in Key terms in atomic structure?

Why it happens: The + and − signs feel like they should add or remove the obvious way, so students add electrons for a positive ion. How to avoid it: A **positive** ion has **lost** electrons (subtract); a **negative** ion has **gained** electrons (add). So O²⁻ has 8 + 2 = 10 electrons. Source: Edexcel Chemistry examiner reports

Why is "Changing the number of protons or neutrons when forming an ion" a common mistake in Key terms in atomic structure?

Why it happens: Students think the charge comes from the nucleus. How to avoid it: Ions form by gaining or losing **electrons only**. The protons and neutrons (the nucleus) **never change** when an ion forms. Source: Edexcel Chemistry examiner reports

Why is "Taking a simple average of isotope masses instead of weighting by abundance" a common mistake in Key terms in atomic structure?

Why it happens: A plain mid-point average is the familiar way to 'average' two numbers. How to avoid it: Use Ar = Σ(mass × % abundance) ÷ 100. For chlorine this gives 35.5, not (35 + 37) ÷ 2 = 36, because ³⁵Cl is far more abundant. Source: Edexcel Chemistry examiner reports

Why is "Forgetting to divide the weighted sum by 100" a common mistake in Key terms in atomic structure?

Why it happens: Students compute mass × % for each isotope and add, then stop. How to avoid it: After summing (mass × %), **divide by 100** (the total percentage). A chlorine answer of 3550 instead of 35.5 is a tell-tale sign of this error. Source: Edexcel Chemistry examiner reports

Atomic structureEdexcel IGCSE Science(Double Award)-Chemistry (4WSD0-1C)

Key terms in atomic structure

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Key Terms in Atomic Structure — Edexcel International GCSE Science (Double Award) Chemistry (4WSD0-1C) Study Notes

Atomic number (Z) and mass number (A), how to read ²³₁₁Na nuclide notation, counting protons/neutrons/electrons in atoms and ions, isotopes, and the weighted-average relative atomic mass calculation. Covers Double Award Chemistry spec 1.16–1.20, assessed on Chemistry Paper 1 (4WSD0/1C).

What you’ll learn

Mapped to the Edexcel IGCSE 4WSD0-1C syllabus (2026 onwards).

1.16 — Know that the atomic number is the number of protons and the mass number is the number of protons plus neutrons.
1.17 — Calculate numbers of protons, neutrons and electrons in atoms and ions.
1.19 — Understand isotopes as atoms with the same number of protons but different numbers of neutrons.
1.20 — Calculate the relative atomic mass of an element from the relative abundances of its isotopes.

Atomic number and mass number (spec 1.16)

Z = protons (defines the element); A = protons + neutrons.

Every atom has a tiny central nucleus containing protons and neutrons, with electrons in shells around it. Two numbers describe the make-up of that atom.

Atomic number, Z = the number of protons in the nucleus. Z is what makes an element that element — every carbon atom has 6 protons, every sodium atom has 11 protons.
Mass number, A = the total number of protons + neutrons (the heavy particles in the nucleus). Electrons are far too light to count here.

Because A counts the protons and the neutrons, you can always find the number of neutrons by subtracting:

number of neutrons = A − Z

Top number is the mass number A; bottom number is the atomic number Z; neutrons = A − Z.

Nuclide (isotope) notation. Atoms are written with the mass number on top and the atomic number on the bottom, in front of the chemical symbol — for example ²³₁₁Na (sodium-23). Both numbers go on the left of the symbol.

Mark-scheme tip. If you write the line "neutrons = A − Z" before you do the arithmetic, you secure the method mark even if the final number is wrong (error-carried-forward). Always show the subtraction.

Z = number of protons (the bottom, smaller number).
A = number of protons + neutrons (the top, larger number).
Neutrons = A − Z.
Z defines the element — same element always has the same Z.

See the full worked example for key terms in atomic structure →

Counting protons, neutrons and electrons in atoms and ions (spec 1.17)

Neutral atom: electrons = protons. Ion: adjust electrons by the charge.

For a neutral atom, the positive and negative charges balance, so:

number of electrons = number of protons = Z

So for ²³₁₁Na: 11 protons, 12 neutrons (23 − 11) and 11 electrons.

A few more atoms:

¹²₆C — carbon-12: 6 protons, 6 neutrons, 6 electrons.
¹⁶₈O — oxygen-16: 8 protons, 8 neutrons, 8 electrons.
²⁷₁₃Al — aluminium-27: 13 protons, 14 neutrons, 13 electrons.

Ions are different. An ion is an atom that has gained or lost electrons, giving it an overall charge. The nucleus is unchanged, so the number of protons and neutrons stays the same — only the electron count changes:

Positive ion (cation): the atom has lost electrons. Electrons = Z − charge.
- Na atom (11 p, 11 e) → Na⁺ (11 p, 10 e).
- Mg atom (12 p, 12 e) → Mg²⁺ (12 p, 10 e).
Negative ion (anion): the atom has gained electrons. Electrons = Z + charge.
- Cl atom (17 p, 17 e) → Cl⁻ (17 p, 18 e).
- O atom (8 p, 8 e) → O²⁻ (8 p, 10 e).

For example ¹⁶₈O²⁻ has 8 protons (it is still oxygen), 8 neutrons (16 − 8) and 10 electrons (8 + 2, because it has gained 2 electrons).

Golden rule. Protons and neutrons are fixed by the nuclide symbol; only the electrons respond to the charge — add electrons for a negative charge, subtract electrons for a positive charge.

Neutral atom: electrons = protons = Z.
Ion: protons and neutrons unchanged — only electrons change.
Positive ion (lost electrons): electrons = Z − charge.
Negative ion (gained electrons): electrons = Z + charge.

See the full worked example for key terms in atomic structure →

Isotopes (spec 1.19)

Same element (same protons), different number of neutrons.

Isotopes are atoms of the same element — so they have the same number of protons (same atomic number Z) — but with different numbers of neutrons, and therefore different mass numbers A.

Common examples:

Chlorine: ³⁵Cl and ³⁷Cl — both have 17 protons; they differ by having 18 or 20 neutrons.
Carbon: ¹²C, ¹³C and ¹⁴C — all have 6 protons; differ by 6, 7 or 8 neutrons.
Hydrogen: ¹H, ²H and ³H — all have 1 proton; differ by 0, 1 or 2 neutrons.

Isotopes of an element have the same number of protons but different numbers of neutrons.

Same chemistry, slightly different physics.

Isotopes have identical chemical properties. Chemical reactions depend only on the electron arrangement, and isotopes have the same number of protons → same number of electrons → same electron arrangement → react in exactly the same way.
Isotopes have slightly different physical properties (anything that depends on mass), because the extra neutrons make some isotopes heavier — for example a higher mass and slightly higher density.

Watch the definition wording. A common exam trap is to say isotopes have "different numbers of protons" — that would make them different elements. The protons are the same; it is the neutrons that differ.

Isotopes: same number of protons (same Z), different number of neutrons (different A).
Same chemical properties — identical electron arrangement.
Slightly different physical properties (e.g. mass, density) — different masses.
Examples: ³⁵Cl / ³⁷Cl, ¹²C / ¹³C / ¹⁴C, ¹H / ²H / ³H.

See the full worked example for key terms in atomic structure →

Relative atomic mass from isotopic abundance (spec 1.20)

Weighted average: multiply each mass by its %, add up, divide by 100.

Because most elements are natural mixtures of isotopes, we use a weighted average called the relative atomic mass (Ar):

Ar = Σ(isotope mass × % abundance) ÷ 100

It is a ratio, so it has no units, and it is usually not a whole number because of the isotope mixture.

Worked example 1 — chlorine. Chlorine is 75% ³⁵Cl and 25% ³⁷Cl:

Ar(Cl) = (35 × 75 + 37 × 25) ÷ 100 = (2625 + 925) ÷ 100 = 3550 ÷ 100 = 35.5

This is exactly the value of chlorine on the periodic table.

Worked example 2 — boron. Boron is 20% ¹⁰B and 80% ¹¹B:

Ar(B) = (10 × 20 + 11 × 80) ÷ 100 = (200 + 880) ÷ 100 = 1080 ÷ 100 = 10.8

Worked example 3 — three isotopes (magnesium). Mg is 79% ²⁴Mg, 10% ²⁵Mg and 11% ²⁶Mg:

Ar(Mg) = (24 × 79 + 25 × 10 + 26 × 11) ÷ 100 = (1896 + 250 + 286) ÷ 100 = 2432 ÷ 100 = 24.3 (to 3 s.f.)

Why you must weight. Taking a simple average of the masses is wrong. For chlorine a simple average would give (35 + 37) ÷ 2 = 36, but because ³⁵Cl is three times more common than ³⁷Cl, the true Ar is pulled down to 35.5. Always multiply each mass by its abundance.

Ar = Σ(mass × % abundance) ÷ 100 — a weighted average.
Ar has no units and is usually not a whole number.
Multiply each isotope mass by its %, add, then divide by 100.
Never take a simple (mid-point) average — weight by abundance.

See the full worked example for key terms in atomic structure →

How it’s examined

Double Award Chemistry is assessed on one untiered paper — Chemistry Paper 1 (4WSD0/1C), 2 hours, 110 marks, 33.3% of the Double Award (calculator allowed). Key terms in atomic structure appear most often as: 3–4 mark counting of protons / neutrons / electrons from a nuclide symbol or for an ion; the isotope definition (same protons, different neutrons) plus a same-chemistry / different-physics comparison; and a 4-mark relative atomic mass calculation from isotopic abundances. The biggest mark-losers are forgetting neutrons = A − Z, mishandling the electron count for ions, and taking a simple average instead of weighting by abundance.

Sources: Pearson Edexcel International GCSE Science (Double Award) 4SD0 specification — Issue 4 (valid for 2026 onwards); Pearson Edexcel International GCSE Chemistry 4CH1 specification — Issue 4 (shared Chemistry Paper 1 content); Pearson Edexcel 4SD0 / 4CH1 examiner reports 2022–2024. Last reviewed 2026-06-07.

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Step-by-step worked examples — Key terms in atomic structure

Step-by-step solutions to past-paper-style questions on key terms in atomic structure, written exactly the way a tutor would explain them at the board.

1Count particles from a nuclide symbol (Getting started)
Getting started• 4WSD0/1C style — short answer• atomic-number, mass-number, spec-1.16, spec-1.17
Question
An atom is written as ²³₁₁Na. State the number of protons, neutrons and electrons in this atom. (3 marks)
Step-by-step solution
1. Step 1
  Protons (1). The bottom number is the atomic number Z = number of protons = 11.
2. Step 2
  Neutrons (1). Neutrons = A − Z = 23 − 11 = 12. Always write the subtraction line.
3. Step 3
  Electrons (1). It is a neutral atom, so electrons = protons = 11.
Answer
11 protons, 12 neutrons, 11 electrons.
Examiner tip
Three independent marks. The classic slip is reading the top number (23) as the number of neutrons — that is the mass number (protons + neutrons), so you must subtract Z.
2Define an isotope (Getting started)
Getting started• 4WSD0/1C style — short answer• isotopes, spec-1.19
Question
What is meant by the term isotopes? (2 marks)
Step-by-step solution
1. Step 1
  Same protons (1). Isotopes are atoms of the same element — they have the same number of protons (same atomic number).
2. Step 2
  Different neutrons (1). They have different numbers of neutrons (and so different mass numbers).
3. Step 3
  Watch the trap. Saying 'different numbers of protons' scores zero — that would make them different elements.
Answer
Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons.
Examiner tip
Both marks hinge on the protons being the SAME and the neutrons being DIFFERENT. Mentioning 'same element' or 'same atomic number' helps secure the first mark.
3Count particles in an ion (Building confidence)
Building confidence• 4WSD0/1C style — structured• ions, spec-1.17, application
Question
An oxide ion is written as ¹⁶₈O²⁻. State the number of protons, neutrons and electrons in this ion. (3 marks)
Step-by-step solution
1. Step 1
  Protons (1). Z = 8, so there are 8 protons — the charge does not change the nucleus.
2. Step 2
  Neutrons (1). Neutrons = A − Z = 16 − 8 = 8.
3. Step 3
  Electrons (1). The 2− charge means the atom has gained 2 electrons, so electrons = 8 + 2 = 10.
Answer
8 protons, 8 neutrons, 10 electrons.
Examiner tip
The discriminator is the electron count. A 2− ion has gained 2 electrons (add), a 2+ ion has lost 2 (subtract). Protons and neutrons are unaffected by charge.
4Relative atomic mass of chlorine (Building confidence)
Building confidence• 4WSD0/1C style — structured• relative-atomic-mass, spec-1.20, calculation
Question
Chlorine has two isotopes: ³⁵Cl with 75% abundance and ³⁷Cl with 25% abundance. Calculate the relative atomic mass of chlorine. (2 marks)
Step-by-step solution
1. Step 1
  Set up the weighted sum (1). Ar = (35 × 75 + 37 × 25) ÷ 100.
2. Step 2
  Work it out. = (2625 + 925) ÷ 100 = 3550 ÷ 100.
3. Step 3
  Answer (1). Ar = 35.5 (matches the periodic table value).
Answer
Ar(Cl) = (35 × 75 + 37 × 25) ÷ 100 = 35.5
Examiner tip
Marks are for the correct weighted method and the correct final value. Do not take the simple average (35 + 37) ÷ 2 = 36 — you must weight by the 75:25 abundances.
5Relative atomic mass with three isotopes (Stretch)
Stretch• 4WSD0/1C style — extended• relative-atomic-mass, spec-1.20, data
Question
Silicon has three isotopes: ²⁸Si (92.0%), ²⁹Si (4.7%) and ³⁰Si (3.3%). Calculate the relative atomic mass of silicon, giving your answer to one decimal place. (3 marks)
Step-by-step solution
1. Step 1
  Weighted sum (1). Numerator = 28 × 92.0 + 29 × 4.7 + 30 × 3.3.
2. Step 2
  Evaluate each term (1). = 2576 + 136.3 + 99 = 2811.3.
3. Step 3
  Divide by 100 (1). Ar = 2811.3 ÷ 100 = 28.113 ≈ 28.1.
Answer
Ar(Si) = (28 × 92.0 + 29 × 4.7 + 30 × 3.3) ÷ 100 = 28.1 (1 d.p.)
Examiner tip
With three isotopes, keep every term; dropping the smallest one loses the answer mark. Check the abundances add to 100% (92.0 + 4.7 + 3.3 = 100) before dividing by 100.
6Find isotope abundances from Ar (Stretch)
Stretch• 4WSD0/1C style — extended• relative-atomic-mass, spec-1.20, reasoning
Question
Copper has two isotopes, ⁶³Cu and ⁶⁵Cu, and a relative atomic mass of 63.5. Calculate the percentage abundance of each isotope. (4 marks)
Step-by-step solution
1. Step 1
  Define a variable (1). Let the abundance of ⁶³Cu be x%, so ⁶⁵Cu = (100 − x)%.
2. Step 2
  Form the equation (1). (63x + 65(100 − x)) ÷ 100 = 63.5.
3. Step 3
  Expand (1). 63x + 6500 − 65x = 6350 → −2x = −150.
4. Step 4
  Solve (1). x = 75, so ⁶³Cu = 75% and ⁶⁵Cu = 25%.
Answer
⁶³Cu = 75%, ⁶⁵Cu = 25%.
Examiner tip
A reverse Ar question. Setting the second abundance as (100 − x) is the key step. A quick check: 75% of mass-63 plus 25% of mass-65 should land just below the mid-point — closer to 63 — which matches Ar = 63.5.

Model Answers — Key terms in atomic structure

High-scoring sample answers for key terms in atomic structure on the Pearson Edexcel IGCSE Chemistry Paper 1 (4WSD0/1C) paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
4WSD0/1C style3 marks
An atom of carbon is written as ¹²₆C. State the number of (a) protons, (b) neutrons and (c) electrons in this atom. (3 marks)
Model answer
(a) Protons = atomic number Z = 6.

(b) Neutrons = A − Z = 12 − 6 = 6.

(c) Electrons = protons (neutral atom) = 6.
Why this scores
1 mark each. Show the neutron subtraction (12 − 6) so the method is credited even if you slip on the arithmetic.
Question 2
4WSD0/1C style3 marks
Bromine exists as two isotopes, ⁷⁹Br and ⁸¹Br. Explain what is meant by the term isotopes and state how the two bromine atoms differ. (3 marks)
Model answer
Isotopes are atoms of the same element with the same number of protons (same atomic number). (1)

They have different numbers of neutrons (and so different mass numbers). (1)

⁷⁹Br and ⁸¹Br both have 35 protons, but ⁷⁹Br has 44 neutrons while ⁸¹Br has 46 neutrons. (1)
Why this scores
The definition marks need 'same protons' AND 'different neutrons'. The third mark applies the idea to the named example — both have 35 protons, differing only in neutrons.
Question 3
4WSD0/1C style4 marks
Complete the particle counts for the magnesium ion ²⁴₁₂Mg²⁺: number of protons, neutrons and electrons. Explain why the number of electrons is not equal to the number of protons. (4 marks)
Model answer
Protons = Z = 12. (1)

Neutrons = A − Z = 24 − 12 = 12. (1)

Electrons = 12 − 2 = 10 (because the 2+ ion has lost 2 electrons). (1)

The electrons do not equal the protons because the atom has lost two electrons to form a 2+ ion; the nucleus (protons + neutrons) is unchanged, so there are now more protons than electrons, giving the overall positive charge. (1)
Why this scores
Marks for the three counts plus the explanation. A 2+ ion means 2 electrons LOST (subtract); a common error is to add electrons or to change the proton number.
Question 4
4WSD0/1C style4 marks
Boron has two isotopes: ¹⁰B with 20% abundance and ¹¹B with 80% abundance. Calculate the relative atomic mass of boron, showing your working. (4 marks)
Model answer
Multiply each isotope mass by its abundance: 10 × 20 = 200 and 11 × 80 = 880. (1)

Add the products: 200 + 880 = 1080. (1)

Divide by 100 (the total abundance): 1080 ÷ 100. (1)

Ar(B) = 10.8. (1)
Why this scores
Full working is rewarded: weighted products, the sum, division by 100, and the final value. A simple average (10 + 11) ÷ 2 = 10.5 is wrong because the isotopes are not equally abundant.
Question 5
4WSD0/1C style — extended6 marks
Chlorine-35 and chlorine-37 are isotopes. (a) State, with a reason, whether they have the same or different chemical properties. (b) State, with a reason, one physical property in which they differ. (6 marks)
Model answer
(a) Chemical properties — the SAME. (3)

Both isotopes have the same number of protons (17), so the same number of electrons (17). (1)

They therefore have the same electron arrangement (2,8,7), with the same number of outer-shell electrons. (1)

Chemical reactions depend only on the (outer) electrons, so the two isotopes react in exactly the same way — identical chemical properties. (1)

(b) Physical property — DIFFERENT. (3)

³⁷Cl has two more neutrons than ³⁵Cl, so a ³⁷Cl atom has a greater mass. (1)

A greater mass means a slightly higher density for ³⁷Cl. (1)

Any physical property that depends on mass (such as density) therefore differs between the isotopes, even though their chemistry is identical. (1)
Why this scores
The top band needs the chemistry tied to 'same electrons / same electron arrangement' and the physics tied to 'different mass (more neutrons)'. Vague answers like 'they are a bit different' score nothing.
Question 6
4WSD0/1C style — extended6 marks
Neon has three isotopes: ²⁰Ne (90.5%), ²¹Ne (0.3%) and ²²Ne (9.2%). (a) Calculate the relative atomic mass of neon to one decimal place. (b) Explain why the relative atomic mass of neon is not a whole number. (6 marks)
Model answer
(a) Calculation. (4)

Weighted products: 20 × 90.5 = 1810; 21 × 0.3 = 6.3; 22 × 9.2 = 202.4. (1)

Sum of products: 1810 + 6.3 + 202.4 = 2018.7. (1)

Divide by 100: 2018.7 ÷ 100. (1)

Ar(Ne) = 20.2 (1 d.p.). (1)

(b) Explanation. (2)

Naturally occurring neon is a mixture of isotopes with different masses. (1)

The relative atomic mass is the weighted average of those isotope masses, which falls between the whole-number masses and so is not a whole number itself. (1)
Why this scores
Keep all three isotopes in the sum — dropping the tiny 0.3% term still gives ~20.2 here, but in general missing a term loses the answer mark. The explanation marks need 'mixture of isotopes' AND 'weighted average'.

Key Definitions and Keywords — Key terms in atomic structure

Definitions to memorise and the exact keywords mark schemes credit for key terms in atomic structure answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IGCSE Chemistry Paper 1 (4WSD0/1C) sitting.

Atomic number (Z)
Examiner keyword
The number of protons in the nucleus of an atom. It defines the element — all atoms of an element have the same atomic number. In a neutral atom it also equals the number of electrons.
Mass number (A)
Examiner keyword
The total number of protons plus neutrons in the nucleus of an atom.
Proton
A positively charged particle (relative charge +1) found in the nucleus; its number equals the atomic number.
Neutron
Examiner keyword
An uncharged particle (relative charge 0) found in the nucleus. Number of neutrons = mass number − atomic number (A − Z).
Electron
A negatively charged particle (relative charge −1) found in shells around the nucleus. In a neutral atom the number of electrons equals the number of protons.
Nuclide notation (ᴬ_Z X)
The way of writing an atom with the mass number A on top and the atomic number Z on the bottom, to the left of the chemical symbol, e.g. ²³₁₁Na.
Ion
Examiner keyword
A charged particle formed when an atom loses electrons (positive ion / cation) or gains electrons (negative ion / anion). The number of protons and neutrons is unchanged.
Isotopes
Examiner keyword
Atoms of the same element with the same number of protons (same atomic number) but different numbers of neutrons (different mass numbers).
Relative atomic mass (Ar)
Examiner keyword
The weighted-average mass of the atoms of an element, taking account of the relative abundances of its isotopes: Ar = Σ(isotope mass × % abundance) ÷ 100. It has no units.
Relative (isotopic) abundance
The percentage of each isotope present in a naturally occurring sample of an element. The abundances of all isotopes add up to 100%.

Common Mistakes and Misconceptions — Key terms in atomic structure

The traps other students keep falling into on key terms in atomic structure questions — taken from recent Pearson Edexcel IGCSE Chemistry Paper 1 (4WSD0/1C) examiner reports and mark schemes — and how to avoid them.

✕Reading the mass number as the number of neutrons
Edexcel Chemistry examiner reports — recurring error
Why it happens
Students see the large top number and assume it is the neutrons.
How to avoid it
The mass number A = protons + neutrons. Find neutrons by subtracting: neutrons = A − Z. Always write the subtraction line.
✕Defining isotopes as having 'different numbers of protons'
Edexcel Chemistry examiner reports
Why it happens
Students remember that isotopes are 'different' but attach the difference to the wrong particle.
How to avoid it
Isotopes have the SAME protons (same element) and DIFFERENT neutrons. Different protons would mean different elements.
✕Getting the electron count of an ion the wrong way round
Edexcel Chemistry examiner reports
Why it happens
The + and − signs feel like they should add or remove the obvious way, so students add electrons for a positive ion.
How to avoid it
A positive ion has lost electrons (subtract); a negative ion has gained electrons (add). So O²⁻ has 8 + 2 = 10 electrons.
✕Changing the number of protons or neutrons when forming an ion
Edexcel Chemistry examiner reports
Why it happens
Students think the charge comes from the nucleus.
How to avoid it
Ions form by gaining or losing electrons only. The protons and neutrons (the nucleus) never change when an ion forms.
✕Taking a simple average of isotope masses instead of weighting by abundance
Edexcel Chemistry examiner reports
Why it happens
A plain mid-point average is the familiar way to 'average' two numbers.
How to avoid it
Use Ar = Σ(mass × % abundance) ÷ 100. For chlorine this gives 35.5, not (35 + 37) ÷ 2 = 36, because ³⁵Cl is far more abundant.
✕Forgetting to divide the weighted sum by 100
Edexcel Chemistry examiner reports
Why it happens
Students compute mass × % for each isotope and add, then stop.
How to avoid it
After summing (mass × %), divide by 100 (the total percentage). A chlorine answer of 3550 instead of 35.5 is a tell-tale sign of this error.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Key terms in atomic structure

Detailed Notes

What you’ll learn

How it’s examined

1Count particles from a nuclide symbol (Getting started)

2Define an isotope (Getting started)

3Count particles in an ion (Building confidence)

4Relative atomic mass of chlorine (Building confidence)

5Relative atomic mass with three isotopes (Stretch)

6Find isotope abundances from Ar (Stretch)

Atomic number (Z)

Mass number (A)

Proton

Neutron

Electron

Nuclide notation (ᴬ_Z X)

Ion

Isotopes

Relative atomic mass (Ar)

Relative (isotopic) abundance

✕Reading the mass number as the number of neutrons

✕Defining isotopes as having 'different numbers of protons'

✕Getting the electron count of an ion the wrong way round

✕Changing the number of protons or neutrons when forming an ion

✕Taking a simple average of isotope masses instead of weighting by abundance

✕Forgetting to divide the weighted sum by 100

Past paper style quiz