Revision Notes
IB DP Chemistry Structure 1.2: The nuclear atom – Revision Notes
Condensed IB DP Chemistry Structure 1.2 notes: particle counts from nuclear symbols, isotopes, Ar calculations, HL mass spectra and a quick self-test.
- Subject
- Chemistry
- Level
- IB
- Topic
- The nuclear atom
- Author
- Marlbridge Academic Team
- Updated
Aligned to International Baccalaureate IB Diploma Programme Chemistry (DP Chemistry), First assessment 2025. Official specification .
Syllabus page (what it covers and how it is assessed): IB Diploma Programme Chemistry.
Syllabus points this page covers
DP Chemistry
- 1.2 The nuclear atom
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The nuclear atom study guide explains everything here at length. These notes compress Structure 1.2, The nuclear atom, from the IB’s Diploma Programme Chemistry guide (first assessment in 2025; published February 2023) into a form for the last weeks before your exams. They span Structure 1.2.1 and 1.2.2, which are for SL and HL, and Structure 1.2.3, which is HL only. They apply to the DP Chemistry course first assessed in 2025.
When you have finished, try the nuclear atom practice questions. The IB DP Chemistry hub lists the other topics, and the printable checklist lets you mark each understanding as done. For a quick check of your gaps across subjects, use the free diagnostics. A broader summary of all five Structure 1 sub-topics is in the Structure 1 revision notes.
Key definitions
- Nucleus: the small, dense, positively charged centre of an atom. It holds the protons and neutrons.
- Nucleons: protons and neutrons, the particles in the nucleus.
- Atomic number, Z: the number of protons. It identifies the element.
- Mass number, A: the total count of nucleons (protons and neutrons together).
- Isotopes: forms of one element whose nuclei differ in neutron count (same Z, different A).
- Relative atomic mass, Ar: the abundance-weighted average mass of an element’s atoms, measured on the scale where ¹²C counts as exactly 12. No units.
- m/z (HL only): mass-to-charge ratio, the quantity on the x-axis of a mass spectrum.
Structure 1.2.1: Subatomic particles
| Particle | Relative charge | Relative mass | Found |
|---|---|---|---|
| Proton | +1 | 1 | in the nucleus |
| Neutron | 0 | 1 | in the nucleus |
| Electron | −1 | negligible (about 1/1836) | outside the nucleus |
Key points:
- The nucleus holds nearly the whole mass of the atom in a tiny volume, so it is very dense.
- Electrons occupy the space outside the nucleus, and that space makes up almost all of the atom’s volume.
- In a neutral atom the proton count and the electron count are equal.
Method in steps: counting particles from ᴬ_Z X
- Protons = Z (bottom left).
- Neutrons = A − Z (top left minus bottom left).
- Electrons = Z − charge. For n+ subtract n; for n− add n.
Worked reminders
| Species | Protons | Neutrons | Electrons |
|---|---|---|---|
| ¹¹⁹₅₀Sn²⁺ | 50 | 119 − 50 = 69 | 50 − 2 = 48 |
| ⁷⁵₃₃As³⁻ | 33 | 75 − 33 = 42 | 33 + 3 = 36 |
To go the other way, Z comes from the proton count, A = protons + neutrons, and charge = protons − electrons.
Structure 1.2.2: Isotopes and Ar
Why isotopes behave alike chemically
Same proton number → same number and arrangement of electrons in the neutral atom → same chemical reactions.
Why isotopes differ physically
Different neutron number → different nuclear mass → different mass-dependent properties: density, melting point, boiling point, and rate of diffusion of gases (lighter diffuses faster). You do not need to learn named examples of isotopes.
Worked reminder: compare hydrogen gas made of ¹H atoms with hydrogen gas made of ²H atoms, both at the same temperature and pressure. Using mass numbers, the molecules have relative masses of 2 and 4. Equal volumes of gas under the same conditions hold the same number of molecules, so the ²H₂ sample has twice the mass in the same volume: its density is twice as large. Both gases burn in oxygen to form water in the same way, because each molecule has the same two electrons.
How to write the explanation in an answer:
- State what is the same: protons and electrons.
- Link that to chemistry: reactions depend on electrons, so reactions are the same.
- State what is different: neutrons, so the mass of each atom.
- Link that to the physical property asked about.
Formulas
| Situation | Formula |
|---|---|
| Ar from percentage abundances | Ar = Σ(mass × %) / 100 |
| Ar from relative heights or other amounts | Ar = Σ(mass × amount) / Σ(amounts) |
| Two isotopes, find abundance | m₁x + m₂(1 − x) = Ar, then solve for x |
Method in steps: Ar from data
- For each isotope, find mass × abundance.
- Sum these values.
- Divide that sum by the sum of all abundances (100 for percentages).
- Check the answer lies between the lightest and heaviest isotope and nearest the most abundant one.
Worked reminder: rhenium is 37.4% ¹⁸⁵Re (184.95) and 62.6% ¹⁸⁷Re (186.96).
Ar = (184.95 × 37.4 + 186.96 × 62.6) / 100 = 186.21
Method in steps: abundance from Ar (two isotopes)
- Let the fraction of the lighter isotope be x; the heavier is (1 − x).
- Write m₁x + m₂(1 − x) = Ar.
- Solve for x, then convert both fractions to percentages.
- Check that they add to 100%.
Worked reminder: element X has isotopes of relative isotopic masses 64.00 and 66.00 and Ar = 64.70.
64.00x + 66.00(1 − x) = 64.70
66.00 − 2.00x = 64.70 → x = 0.650
So 65.0% of mass 64.00 and 35.0% of mass 66.00.
Structure 1.2.3: Mass spectra (HL only)
- Each peak = one isotope (assuming 1+ ions, m/z = isotope mass).
- Peak height = relative abundance.
- Number of peaks = number of isotopes in the sample.
- Ar = weighted mean of m/z values; match it to the periodic table to identify the element.
- The guide does not assess how the instrument works. Interpret the spectrum; don’t describe the machine.
Worked reminder: a spectrum shows peaks at m/z 182, 183, 184 and 186 with abundances 26.5%, 14.3%, 30.7% and 28.5%.
Ar = (182 × 26.5 + 183 × 14.3 + 184 × 30.7 + 186 × 28.5) / 100
= 183.90
That is close to the Ar of tungsten, 183.84. Whole-number m/z values cause the small difference.
Must-know distinctions
- Atomic number vs mass number: Z counts protons only; A counts protons and neutrons.
- Mass number vs relative isotopic mass: A is a whole-number count; the relative isotopic mass is a measured mass and is close to, but not exactly, A.
- Relative isotopic mass vs relative atomic mass: one isotope vs the weighted mean of all isotopes of the element.
- Atom vs ion: an ion has the same nucleus but a different number of electrons.
- Chemical vs physical properties of isotopes: chemical the same (electrons), physical different (mass).
- Percentage abundance vs relative peak height (HL only): heights may need converting to fractions of the total before use.
Quick self-test
- How many neutrons are in ⁵⁵₂₅Mn?
- How many electrons are in ⁵⁶₂₆Fe³⁺?
- A species has 16 protons, 18 neutrons and 18 electrons. Write its nuclear symbol.
- State the relative mass and relative charge of a neutron.
- Why do isotopes of an element have the same chemical properties?
- Name two physical properties that can differ between isotopes.
- Element J is 25.0% of isotopic mass 52.00 and 75.0% of isotopic mass 54.00. Calculate Ar.
- An element has isotopes of masses 98.00 and 100.00 and Ar = 98.40. What percentage is the lighter isotope?
- Why does relative atomic mass have no units?
- (HL only) Peaks at m/z 120 and 122 have relative heights 40 and 60. Calculate Ar.
- (HL only) Two peaks have heights 50 and 100. Give their percentage abundances.
- (HL only) What does the number of peaks in an element’s mass spectrum tell you?
Answers
- 55 − 25 = 30
- 26 − 3 = 23
- ³⁴₁₆S²⁻ (Z = 16 is sulfur; A = 16 + 18 = 34; charge = 16 − 18 = −2)
- Relative mass 1, relative charge 0
- Equal proton numbers give equal electron numbers and arrangements, and electrons are what take part in reactions.
- Any two of: density, melting point, boiling point, rate of diffusion.
- (52.00 × 25.0 + 54.00 × 75.0) / 100 = 53.50
- 98.00x + 100.00(1 − x) = 98.40 gives x = 0.800, so 80.0%
- It compares one mass with another (a twelfth of a ¹²C atom’s mass), so the units cancel out.
- (120 × 40 + 122 × 60) / 100 = 121.20
- 50/150 = 33.3% and 100/150 = 66.7%
- The number of isotopes present in the sample.
Where marks are usually lost
- Subtracting the charge the wrong way for negative ions, giving too few electrons.
- Changing the proton or neutron number when an atom becomes an ion.
- Swapping A and Z positions in a nuclear symbol, or leaving the charge off an ion.
- Using the mass of the most abundant isotope as Ar.
- (HL only) Dividing by 100 when the abundances given are peak heights that do not total 100.
- Rounding part-way through an Ar calculation, so the final decimals are wrong.
- Reverse calculations where the two percentages do not add to 100%.
- Saying isotopes differ chemically, or giving “different number of protons” as the difference.
- Explaining physical differences without linking them to the extra mass of the neutrons.
- (HL only) Describing the inside of a mass spectrometer when the question asks what the spectrum shows.
Official syllabus
Structure 1.2, The nuclear atom: International Baccalaureate Organization, Chemistry guide, Diploma Programme, first assessment 2025, published February 2023.
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