IB DP Chemistry Structure 1.2: The nuclear atom

This study guide teaches Structure 1.2, The nuclear atom, as set out in the IB’s Diploma Programme Chemistry guide (first assessment in 2025; published February 2023). Its scope runs from Structure 1.2.1 to 1.2.3. The first two are for SL and HL students; Structure 1.2.3, on mass spectra, sits in the guide’s additional higher level material and is HL only. It applies to the DP Chemistry course first assessed in 2025.

When you reach the end, check what you remember using the nuclear atom revision notes and then the nuclear atom practice questions. For the whole of Structure 1 in one place, see the Structure 1 overview. Other DP Chemistry topics are on the IB DP Chemistry hub, and you can tick off each understanding on the printable checklist. The free diagnostics are a quick way to see where your gaps are before you start.

What this unit covers

Understanding	Skills you need	Level
Structure 1.2.1	Describe the nuclear atom; recall the relative masses and charges of the three subatomic particles; work out the particle make-up of any atom or ion from its nuclear symbol	SL and HL
Structure 1.2.2	Define isotopes; calculate a non-integer relative atomic mass from isotope abundances, and an abundance from a relative atomic mass; explain why isotopes differ in physical properties	SL and HL
Structure 1.2.3	Interpret the mass spectrum of an element: identify isotopes, read relative abundances, calculate relative atomic mass and identify the element	HL only

Structure 1.2.1: Inside the atom

The nuclear model

An atom has a tiny central nucleus that carries all of its positive charge and almost all of its mass. The nucleus is made of protons and neutrons, which are together called nucleons. Electrons, which are negatively charged, occupy the space outside the nucleus.

The nucleus is extremely small compared with the whole atom: its diameter is tens of thousands of times smaller than the atom’s. Because nearly all the mass is packed into that small volume, the nucleus is very dense. Most of an atom’s volume is the region the electrons occupy.

Relative masses and charges

You need to know these relative values. The guide notes that the actual masses and charges are given in the data booklet.

Particle	Relative charge	Relative mass	Found
Proton	+1	1	in the nucleus
Neutron	0	1	in the nucleus
Electron	−1	about 5 × 10⁻⁴ (negligible)	outside the nucleus

Worked example 1: how small is the electron? The proton’s mass is 1.672622 × 10⁻²⁷ kg and the electron’s is 9.109384 × 10⁻³¹ kg. How many times heavier is the proton?

ratio = (1.672622 × 10⁻²⁷) / (9.109384 × 10⁻³¹)
      = 1836 (to 4 s.f.)

So an electron has about 1/1836 of a proton’s mass. That is why you can treat the electron’s mass as negligible: an atom’s mass is set by its nucleons. The neutron is very slightly heavier than the proton (by about 0.14%), but both are taken as relative mass 1.

The nuclear symbol

The nuclear symbol, written ᴬ_Z X in the guide’s notation, packs the composition of a nucleus into one line:

• A, the mass number, is written top left. It is the number of protons plus neutrons.
• Z, the atomic number, is written bottom left. It is the number of protons, and it fixes which element X is.
• An ion’s charge is written top right, for example ³⁺ or ²⁻.

From the symbol:

• protons = Z
• neutrons = A − Z
• electrons = Z − (charge), so a positive ion has fewer electrons than protons and a negative ion has more.

Worked example 2: counting particles in ions. Find the numbers of protons, neutrons and electrons in ⁵⁹₂₇Co³⁺ and ¹²⁸₅₂Te²⁻.

For ⁵⁹₂₇Co³⁺:

• protons = 27
• neutrons = 59 − 27 = 32
• electrons = 27 − 3 = 24 (three electrons lost)

For ¹²⁸₅₂Te²⁻:

• protons = 52
• neutrons = 128 − 52 = 76
• electrons = 52 + 2 = 54 (two electrons gained)

Notice that forming an ion changes only the electron count. The nucleus, and so A and Z, stays the same.

Worked example 3: from particles to symbol. A species contains 38 protons, 50 neutrons and 36 electrons. Write its nuclear symbol.

• Z = 38, so the element is strontium, Sr.
• A = 38 + 50 = 88.
• Charge = 38 − 36 = +2.
• Symbol: ⁸⁸₃₈Sr²⁺

Structure 1.2.2: Isotopes and relative atomic mass

What isotopes are

Isotopes of an element share an atomic number Z but have different mass numbers A, because their nuclei hold different numbers of neutrons. For example, ¹⁹¹Ir and ¹⁹³Ir are both iridium: each has 77 protons, but one has 114 neutrons and the other 116.

Same chemistry, different physical properties

Two isotopes of one element match in proton count and, as neutral atoms, in the number and arrangement of their electrons. Chemical reactions involve electrons, so isotopes react in the same way and form the same compounds.

Their nuclei differ in mass, so properties that depend on mass differ:

• density (a heavier isotope packs more mass into the same volume)
• melting and boiling points, which differ slightly
• the rate at which gaseous samples diffuse: lighter particles move faster on average at the same temperature

Worked example 4: separating isotopes by mass. Uranium hexafluoride, UF₆, is a gas. Compare the relative molecular masses of ²³⁵UF₆ and ²³⁸UF₆, using mass numbers and taking fluorine as ¹⁹F.

²³⁵UF₆: 235 + (6 × 19) = 349
²³⁸UF₆: 238 + (6 × 19) = 352
difference = 3/349 × 100 = 0.86 %

The two molecules are chemically identical, but the lighter one diffuses slightly faster. A small mass difference like this is what physical separation methods exploit. You do not need to learn specific isotopes; the guide says specific examples need not be learned. You do need to explain why the physical properties differ.

Relative atomic mass

Relative atomic mass, Ar, is the average mass of an element’s atoms, weighted by isotope abundance, on the scale that sets one ¹²C atom at exactly 12. Because most elements are mixtures of isotopes, Ar is usually not a whole number. It has no units.

Ar = Σ (relative isotopic mass × abundance) / Σ (abundances)

If the abundances are percentages, the bottom line is 100.

Worked example 5: Ar from abundances. A sample of iridium contains 37.3% ¹⁹¹Ir (relative isotopic mass 190.96) and 62.7% ¹⁹³Ir (relative isotopic mass 192.96). Calculate Ar to two decimal places.

Ar = (190.96 × 37.3 + 192.96 × 62.7) / 100
   = (7122.81 + 12098.59) / 100
   = 192.21

Check it makes sense: the answer lies between the two isotopic masses and nearer 192.96, the more abundant isotope. If a question gives only mass numbers, use them as the isotopic masses; the answer is then an approximation.

Worked example 6: abundance from Ar. Antimony has two isotopes, ¹²¹Sb (relative isotopic mass 120.90) and ¹²³Sb (relative isotopic mass 122.90). Its Ar is 121.76. Calculate the percentage abundance of each isotope.

Let the fraction of ¹²¹Sb be x, so the fraction of ¹²³Sb is (1 − x).

120.90x + 122.90(1 − x) = 121.76
122.90 − 2.00x = 121.76
2.00x = 1.14
x = 0.570

So antimony is 57.0% ¹²¹Sb and 43.0% ¹²³Sb. The two percentages must add to 100.

Structure 1.2.3: Mass spectra (HL only)

A mass spectrometer turns atoms of an element into positive ions and sorts them by their mass-to-charge ratio, m/z. The guide says the operational details of the instrument will not be assessed. What you must do is read the output.

Reading the spectrum of an element

• The x-axis is m/z. When every ion carries a single positive charge, the m/z of a peak is simply the isotope’s mass, so each peak marks one isotope.
• The y-axis shows relative abundance: either percentages, or peak heights measured against the tallest peak.
• The number of peaks tells you how many isotopes are present in the sample.
• Ar is the abundance-weighted mean of the m/z values. You can then identify the element by matching Ar with the relative atomic masses on the periodic table.

If the heights are not percentages, divide each height by the total of all heights to turn it into a fraction.

Worked example 7: identifying an element from its spectrum. The mass spectrum of a metal shows four peaks, all from 1+ ions:

m/z	84	86	87	88
Relative abundance / %	0.6	9.9	7.0	82.5

(a) State how many isotopes the sample contains. Four, one for each peak.

(b) Calculate Ar and identify the metal, given Ar values of 85.47 for Rb, 87.62 for Sr and 88.91 for Y.

total abundance = 0.6 + 9.9 + 7.0 + 82.5 = 100.0
Ar = (84 × 0.6 + 86 × 9.9 + 87 × 7.0 + 88 × 82.5) / 100
   = (50.4 + 851.4 + 609.0 + 7260.0) / 100
   = 87.71

87.71 is closest to 87.62, so the metal is strontium. The small gap comes from using whole-number m/z values rather than exact isotopic masses.

(c) Which isotope is most abundant? The tallest peak is at m/z 88, so ⁸⁸Sr.

Common errors

• Writing neutrons as A + Z or as A alone. Neutrons = A − Z.
• Changing the number of protons when an ion forms. Only electrons change.
• Getting the sign wrong for anions: a 2− ion has two more electrons than protons.
• Taking Ar as the mass of the most common isotope instead of the weighted mean.
• (HL only) Forgetting to divide by the total when peak heights do not add up to 100.
• Giving abundances in a reverse calculation that do not add up to 100%.
• Saying isotopes have “different chemical properties because they have different masses”. The electrons, not the nucleus, decide chemistry.
• Adding units to Ar. Relative atomic mass has none.
• (HL only) Describing how the mass spectrometer works when the question asks you to interpret its output.

Where to go next

Revise the key facts with the revision notes, then try the practice questions, which include HL-only mass spectrum problems. For how this unit fits into the whole course, read the IB DP Chemistry syllabus guide. Structure 1.3, on electron configurations, builds on the electron counts you worked out here.

Official syllabus

Structure 1.2, The nuclear atom: International Baccalaureate Organization, Chemistry guide, Diploma Programme, first assessment 2025, published February 2023.