Atomic Structure: Mass Number, Isotopes, and Calculations Revision Notes | A Level AQA Chemistry

1. Why Atomic Structure Underpins All of Chemistry

Every chemical property you will ever study — reactivity, bonding, spectroscopy — traces back to one question: what is inside the atom? Get this foundation right and the rest of A Level Chemistry starts to make sense.

At A Level, you need more than a vague picture of protons, neutrons, and electrons. You need to be able to count particles precisely, explain why atoms of the same element can have different masses, and calculate a weighted average mass from real isotopic data. These skills appear in calculation questions, extended-response questions, and as the assumed knowledge behind topics like mass spectrometry and nuclear chemistry.

2. Mass Number, Atomic Number, and the Nucleus

Picture the nucleus as a tightly packed core. It contains two types of particle: protons, each carrying a charge of +1, and neutrons, which are electrically neutral. Surrounding the nucleus are electrons, each carrying a charge of −1, arranged in shells (or, at A Level, in atomic orbitals).

The atomic number tells you two things at once: how many protons are present, and — for a neutral atom — how many electrons are present, because a neutral atom has equal numbers of protons and electrons.

From these two numbers you can derive everything you need. The number of neutrons is $A - Z$. For a neutral atom, the number of electrons equals $Z$. The standard notation places A as a superscript and Z as a subscript to the left of the element symbol: $^{A}_{Z}\text{X}$.

3. Extending to Ions: Adjusting the Electron Count

When an atom gains or loses electrons it becomes an ion. The nucleus does not change — the number of protons and neutrons stays exactly the same. Only the electron count shifts.

For a positive ion (cation), electrons have been removed. For a negative ion (anion), electrons have been added. The rule is: electrons = $Z$ − (charge on the ion), where the charge is treated as a signed integer (e.g. +2 gives −2 electrons relative to the neutral atom, and −1 gives +1 electron).

Worked example: Counting Particles in an Ion

Determine the number of protons, neutrons, and electrons in the ion $^{56}_{26}\text{Fe}^{3+}$.

1. $Z = 26$, $A = 56$ — Z is the subscript (atomic number); A is the superscript (mass number).
2. Protons $= Z = 26$ — The proton count never changes when forming an ion — only electrons are gained or lost.
3. Neutrons $= A - Z = 56 - 26 = 30$ — Neutrons are also unaffected by ionisation.
4. Electrons $= Z - \text{charge} = 26 - 3 = 23$ — A 3+ ion has lost 3 electrons relative to the neutral atom, so 26 − 3 = 23 electrons remain.

Answer: Protons = 26, Neutrons = 30, Electrons = 23

The same logic applies to negative ions. For $^{35}_{17}\text{Cl}^{-}$: protons = 17, neutrons = 35 − 17 = 18, electrons = 17 − (−1) = 18. The extra electron gives chloride one more electron than protons.

4. Isotopes: Same Element, Different Mass

Here is where it gets interesting. Chlorine has an atomic number of 17, so every chlorine atom contains 17 protons. Yet if you measure the masses of individual chlorine atoms, you find two distinct values: some have a mass number of 35 and others have a mass number of 37. These are isotopes.

Isotopes of an element have identical atomic numbers (same number of protons) but different mass numbers (different numbers of neutrons). Because the number of protons — and therefore the number of electrons in the neutral atom — is unchanged, isotopes have identical chemical properties. The extra neutrons affect nuclear stability and physical properties such as density, but they do not alter how the atom bonds.

Carbon-14 is a well-known example: it is radioactive and used in radiocarbon dating, yet chemically it behaves identically to carbon-12 because both have 6 protons and 6 electrons. The difference in neutron number is what makes carbon-14 unstable.

Isotopes exist because the nucleus can accommodate varying numbers of neutrons while remaining stable (up to a point). The strong nuclear force holds the nucleus together, and a range of neutron-to-proton ratios can produce stable nuclei for lighter elements. This is why most elements in the periodic table have more than one naturally occurring isotope.

5. Relative Atomic Mass: Calculating the Weighted Mean

Because most elements exist as a mixture of isotopes, the mass you see on the periodic table is not a whole number — it is a weighted mean that accounts for both the mass of each isotope and how abundant each one is in nature. This value is the relative atomic mass, $A_r$.

Think of it like a weighted average of exam scores: if 75% of students scored 80 and 25% scored 60, the class average is not (80 + 60) ÷ 2 = 70. You weight each score by how many students achieved it: (0.75 × 80) + (0.25 × 60) = 60 + 15 = 75. Relative atomic mass works on exactly the same principle, with isotopic abundances playing the role of the weightings.

Worked example: Calculating Relative Atomic Mass for Chlorine

Chlorine has two naturally occurring isotopes: $^{35}\text{Cl}$ with an abundance of 75.0% and $^{37}\text{Cl}$ with an abundance of 25.0%. Calculate the relative atomic mass of chlorine.

1. $^{35}\text{Cl}$: mass = 35, abundance = 75.0% $^{37}\text{Cl}$: mass = 37, abundance = 25.0% — Read these directly from the question. Check that the abundances sum to 100%: 75.0 + 25.0 = 100.0%. [M1]
2. $35 \times 75.0 = 2625$ $37 \times 25.0 = 925$ — Each product represents the contribution of that isotope to the total. [M1]
3. $2625 + 925 = 3550$ — This is the total of (mass × abundance) across all isotopes.
4. $A_r = \frac{3550}{100} = 35.5$ — Dividing by 100 converts the percentage-weighted sum into a weighted mean. [A1]
5. $A_r(\text{Cl}) = 35.5$ — No units — relative atomic mass is dimensionless (it is a ratio relative to $\frac{1}{12}$ of carbon-12). This matches the value on the periodic table.

Answer: $A_r(\text{Cl}) = 35.5$

Notice that 35.5 is not a whole number — no single chlorine atom has a mass of 35.5. The value is a statistical average reflecting the natural mixture of isotopes. This is why relative atomic masses on the periodic table are rarely whole numbers.

Worked example: Calculating Relative Atomic Mass for Boron

Boron has two isotopes: $^{10}\text{B}$ with an abundance of 19.9% and $^{11}\text{B}$ with an abundance of 80.1%. Calculate the relative atomic mass of boron to one decimal place.

1. $^{10}\text{B}$: mass = 10, abundance = 19.9% $^{11}\text{B}$: mass = 11, abundance = 80.1% — Check: 19.9 + 80.1 = 100.0%. [M1]
2. $10 \times 19.9 = 199$ $11 \times 80.1 = 881.1$ — Work through each multiplication carefully. [M1]
3. $199 + 881.1 = 1080.1$ — Add the two contributions together.
4. $A_r = \frac{1080.1}{100} = 10.801$ — Converting the percentage-weighted sum to a mean.
5. $A_r(\text{B}) = 10.8$ — The question specifies one decimal place. The periodic table value is approximately 10.81, confirming this is correct. [A1]

Answer: $A_r(\text{B}) = 10.8$

6. Exam Technique and Common Pitfalls

These four specification points appear across multiple question types — short-answer, calculation, and extended-response. The marks are very accessible if you apply precise language and method, but they are easy to drop through careless errors.

Bringing this all together: mass number and atomic number give you the complete particle inventory of any atom or ion. Isotopes arise when the neutron count varies while the proton count stays fixed. And relative atomic mass is the weighted mean of all isotopic masses, calculated using the percentage abundances found in nature. These three ideas are the bedrock of quantitative chemistry at A Level — you will use them again in moles calculations, mass spectrometry, and nuclear chemistry.