S1.2 The nuclear atom | IB DP Chemistry

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🎯What you need to be able to do

State the relative masses and charges of the proton, neutron and electron.
Use the nuclear symbol notation to deduce the number of protons, neutrons and electrons in an atom or an ion.
Define isotopes and explain how they differ in physical, but not chemical, properties.
Calculate a relative atomic mass from isotopic abundances, and work backwards from a relative atomic mass to an abundance.
AHL Interpret a mass spectrum in terms of the identity and relative abundance of isotopes, and calculate the relative atomic mass from it.

📚The chemistry

Inside the atom

An atom has a positively charged, dense nucleus containing protons and neutrons — collectively nucleons — with negatively charged electrons occupying the space outside it. The nucleus holds essentially all the mass in an almost vanishing fraction of the volume; the electrons decide all of the chemistry.

Proton
relative mass 1
relative charge +1
in the nucleus

Neutron
relative mass 1
relative charge 0
in the nucleus

Electron
relative mass negligible (about 1/1836)
relative charge −1
outside the nucleus

The actual values in kilograms and coulombs are in the data booklet, so do not memorise them — but do memorise the relative values, because they are what deductions are built on. Treating the electron mass as negligible is the reason an ion has essentially the same mass as its parent atom.

The nuclear symbol

An atom is specified by two numbers written on the left of its symbol:

\[ ^{A}_{Z}\mathrm{X}^{\,c} \]

\( Z \), the atomic number — the number of protons. This alone defines which element it is. Change \( Z \) and you have a different element.
\( A \), the mass number — the total number of nucleons, protons plus neutrons. It is a count, so it is always a whole number.
Neutrons \( = A - Z \).
Electrons \( = Z - c \), where \( c \) is the charge. A 2+ ion has lost two electrons; a 2− ion has gained two.

So \( ^{56}_{26}\mathrm{Fe}^{3+} \) has 26 protons, \( 56 - 26 = 30 \) neutrons and \( 26 - 3 = 23 \) electrons.

Adding electrons for a positive ion. A positive charge means electrons were lost, so subtract. The sign of the charge tells you the direction, and the arithmetic \( Z - c \) handles both cases automatically if you keep the sign: for \( \mathrm{Cl}^- \), \( 17 - (-1) = 18 \).

Isotopes

Isotopes are atoms of the same element — same number of protons — with different numbers of neutrons, and therefore different mass numbers.

Their chemical properties are identical, because chemical behaviour is decided by the electron configuration, which depends only on the number of protons. Their physical properties differ slightly, because those depend on mass: heavier isotopes have marginally higher melting and boiling points, higher density, and diffuse and effuse more slowly. Some isotopes are radioactive, which is a nuclear property rather than a chemical one.

You are not expected to learn specific examples, but the standard ones are worth recognising: \( ^{1}\mathrm{H} \), \( ^{2}\mathrm{H} \) (deuterium) and \( ^{3}\mathrm{H} \) (tritium); the three isotopes of carbon; the two of chlorine, \( ^{35}\mathrm{Cl} \) and \( ^{37}\mathrm{Cl} \), in roughly a 3 : 1 ratio, which is why the relative atomic mass of chlorine is 35.45.

Relative atomic mass from abundances

The relative atomic mass \( A_{\mathrm{r}} \) of an element is the weighted mean of the masses of its isotopes, weighted by their natural abundances:

\[ A_{\mathrm{r}} \;=\; \frac{\sum (\text{isotopic mass} \times \% \text{ abundance})}{100} \]

That is why relative atomic masses in the periodic table are not whole numbers, even though every individual mass number is. It is also why an \( A_{\mathrm{r}} \) close to one isotope’s mass tells you that isotope is the abundant one — chlorine’s 35.45 sits much nearer 35 than 37, so \( ^{35}\mathrm{Cl} \) dominates, and a quick estimate gives about 3 : 1.

Running the calculation backwards is standard. With two isotopes, let the abundance of the lighter be \( x \)% and the heavier \( (100-x) \)%, form the weighted-mean expression, set it equal to the known \( A_{\mathrm{r}} \) and solve the single linear equation.

AHL Mass spectrometry

A mass spectrometer ionises a sample, accelerates the ions, deflects them according to their mass-to-charge ratio and records how many arrive at each value. The output is a mass spectrum: a set of peaks whose positions give \( m/z \) and whose heights give relative abundance.

The syllabus is explicit that the operational details of the instrument are not assessed. You will not be asked how the magnetic sector works. What you will be asked is to read a spectrum: identify the isotopes present from the \( m/z \) values, read the relative abundances from the peak heights, and compute the relative atomic mass as a weighted mean.

Two conventions to watch. First, abundances may be given as percentages, as relative heights summing to some arbitrary number, or with the tallest peak set to 100 — in the last two cases you must divide by the total of all the peaks, not by 100. Second, singly charged ions are assumed, so \( m/z \) is numerically equal to the mass. Mass spectra of compounds, where fragmentation occurs, belong to S3.2.

✏️Worked example

(a) Deduce the number of protons, neutrons and electrons in \( ^{81}_{35}\mathrm{Br}^{-} \).
(b) The mass spectrum of an element X shows two peaks: \( m/z = 63 \) with a relative height of 69.2, and \( m/z = 65 \) with a relative height of 30.8. Calculate the relative atomic mass of X to two decimal places and identify the element.
(c) Boron has two isotopes, \( ^{10}\mathrm{B} \) and \( ^{11}\mathrm{B} \), and a relative atomic mass of 10.81. Calculate the percentage abundance of \( ^{11}\mathrm{B} \).
(d) State one physical property in which \( ^{10}\mathrm{B} \) and \( ^{11}\mathrm{B} \) differ, and one property in which they do not.

(a) \( Z = 35 \), so 35 protons. Neutrons \( = A - Z = 81 - 35 = \) 46. The charge is \( -1 \), so the ion has gained one electron: \( 35 - (-1) = \) 36 electrons.

(b) The heights total \( 69.2 + 30.8 = 100.0 \), so they are already percentages.

\[ A_{\mathrm{r}} = \frac{(63 \times 69.2) + (65 \times 30.8)}{100} = \frac{4359.6 + 2002.0}{100} = \frac{6361.6}{100} = 63.62 \]

Relative atomic mass 63.62, which is copper (the data booklet gives 63.55; the small difference is because the question’s abundances are rounded).

(c) Let the abundance of \( ^{11}\mathrm{B} \) be \( x \)%, so \( ^{10}\mathrm{B} \) is \( (100 - x) \)%.

\[ \frac{11x + 10(100-x)}{100} = 10.81 \;\Longrightarrow\; 11x + 1000 - 10x = 1081 \;\Longrightarrow\; x = 81 \]

So \( ^{11}\mathrm{B} \) is 81% and \( ^{10}\mathrm{B} \) is 19%.

(d) They differ in density, and in melting and boiling point, and in rate of diffusion — all consequences of the different mass. They do not differ in chemical properties, because both have five protons and therefore the identical electron configuration \( 1s^2\,2s^2\,2p^1 \), and chemistry is decided by electrons.

Check it. Every weighted mean must lie between the two extreme values, and closer to the more abundant one. In (b), 63.62 lies between 63 and 65 and is much nearer 63 — consistent with 69.2% of the lighter isotope. In (c), 10.81 lies between 10 and 11 and is much nearer 11, so the abundance of \( ^{11}\mathrm{B} \) must be well over 50%, and 81% fits. If your answer ever falls outside the range of the isotopic masses, you have multiplied a mass by the wrong abundance.

Taking the arithmetic mean instead of the weighted mean. Averaging 63 and 65 to get 64 ignores the abundances entirely, and is the single most common error on this topic. The second most common is dividing by 100 when the peak heights do not sum to 100 — if a spectrum gives heights of 3.0 and 1.0, the divisor is 4.0, not 100, and the answer must still land between the two isotopic masses. Use the sanity check above every time.

📝Practise

Work through these on paper, then reveal the answer.

1. Complete the table for these species: \( ^{24}_{12}\mathrm{Mg}^{2+} \), \( ^{31}_{15}\mathrm{P}^{3-} \), \( ^{40}_{18}\mathrm{Ar} \).

\( ^{24}_{12}\mathrm{Mg}^{2+} \): 12 protons, \( 24-12 = 12 \) neutrons, \( 12-2 = \) 10 electrons. \( ^{31}_{15}\mathrm{P}^{3-} \): 15 protons, \( 31-15 = 16 \) neutrons, \( 15-(-3) = \) 18 electrons. \( ^{40}_{18}\mathrm{Ar} \): 18 protons, \( 40-18 = 22 \) neutrons, 18 electrons. Notice that the magnesium ion and neon both have 10 electrons, and the phosphide ion and argon both have 18 — species with the same electron count are isoelectronic, which matters for the ionic radius trends in S3.1.

2. Silicon has three isotopes: \( ^{28}\mathrm{Si} \) (92.2%), \( ^{29}\mathrm{Si} \) (4.7%) and \( ^{30}\mathrm{Si} \) (3.1%). Calculate its relative atomic mass to two decimal places.

\( A_{\mathrm{r}} = \dfrac{(28 \times 92.2)+(29 \times 4.7)+(30 \times 3.1)}{100} = \dfrac{2581.6 + 136.3 + 93.0}{100} = \dfrac{2810.9}{100} = \mathbf{28.11} \). Check: the abundances sum to 100.0, and the answer lies between 28 and 30, very close to 28, as it must when over 92% of the atoms are the lightest isotope. The data booklet value is 28.09.

3. Chlorine has relative atomic mass 35.45 and two isotopes of mass number 35 and 37. Calculate the percentage abundance of each, and predict the relative heights of the three peaks in the mass spectrum of Cl₂.

Let \( ^{35}\mathrm{Cl} \) be \( x \)%: \( \dfrac{35x + 37(100-x)}{100} = 35.45 \Rightarrow 35x + 3700 - 37x = 3545 \Rightarrow -2x = -155 \Rightarrow x = 77.5 \). So 77.5% \( ^{35}\mathrm{Cl} \) and 22.5% \( ^{37}\mathrm{Cl} \), roughly 3.4 : 1. For Cl₂ there are three possible molecular masses: 70 (both \( ^{35} \)), 72 (one of each) and 74 (both \( ^{37} \)). Their probabilities are \( 0.775^2 = 0.601 \), \( 2 \times 0.775 \times 0.225 = 0.349 \) and \( 0.225^2 = 0.051 \), giving peaks at \( m/z \) 70, 72 and 74 in the ratio 0.601 : 0.349 : 0.051, or about 12 : 7 : 1. (With the rougher 3 : 1 abundance approximation you get the familiar textbook ratio 9 : 6 : 1; both are acceptable if you state which abundances you used.) The factor of two on the middle term is the one people forget: there are two ways to make a mixed molecule.

4. Explain why the relative atomic masses in the periodic table are not whole numbers, even though the mass number of every individual atom is.

The mass number of a single atom counts nucleons, so it is necessarily an integer. The relative atomic mass of an element is not a property of one atom: it is the weighted mean of the masses of all its naturally occurring isotopes, weighted by abundance. A weighted mean of integers is an integer only in the special case that one isotope is 100% abundant, so for elements with more than one isotope the value is fractional. (A second, much smaller contributor is that isotopic masses are themselves not exactly integers, because of the mass defect associated with nuclear binding energy — but the syllabus explanation is the weighted mean.)

5. A student claims that \( ^{12}\mathrm{C} \) and \( ^{14}\mathrm{C} \) will react at different rates with oxygen because they have different masses, and that this makes them chemically different. Evaluate this claim.

The claim conflates two things. Chemically, the isotopes are the same: both have six protons and therefore the electron configuration \( 1s^2\,2s^2\,2p^2 \), and it is the electrons that form bonds. Both burn in oxygen to give carbon dioxide, with the same products, the same equations and the same enthalpy change to the precision the course works to. The student is right that mass affects physical behaviour — the heavier isotope diffuses more slowly and its bonds vibrate at lower frequency — and there is a small kinetic isotope effect on rate for exactly that reason, largest for hydrogen where the mass ratio is greatest. But a small difference in rate is not a difference in chemistry: the substances undergo the same reactions. Additionally, \( ^{14}\mathrm{C} \) is radioactive, which is a nuclear property, unrelated to its chemical behaviour.

6. AHL The mass spectrum of an element shows peaks at \( m/z \) 204, 206, 207 and 208 with relative heights 0.3, 5.2, 4.8 and 11.7 respectively. Calculate the relative atomic mass.

The heights do not total 100, so first find the total: \( 0.3 + 5.2 + 4.8 + 11.7 = 22.0 \). Then take the weighted mean, dividing by that total: \( A_{\mathrm{r}} = \dfrac{(204 \times 0.3)+(206 \times 5.2)+(207 \times 4.8)+(208 \times 11.7)}{22.0} \) \( = \dfrac{61.2 + 1071.2 + 993.6 + 2433.6}{22.0} = \dfrac{4559.6}{22.0} = \mathbf{207.3} \). The element is lead. Check: the answer lies between 204 and 208 and nearer 208, which is right since the 208 peak is over half the total signal.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

PhET — Build an Atom and Isotopes and Atomic Mass; the second lets you mix isotopes and watch the weighted mean move, which makes the sanity check above obvious.
The Royal Society of Chemistry periodic table — isotopic abundances for every element, useful for making up your own practice calculations.
Khan Academy — a slower walk through mass spectrometry than the syllabus needs, useful if the \( m/z \) axis is not yet intuitive.