IB Chemistry HLTopic 1 — The Nuclear AtomHL onlyCore skill~12 min read
Interpreting Mass Spectra
A mass spectrum is a bar chart of an element’s isotopes, sorted by mass, with the height of each bar telling you how common that isotope is. Read those two axes properly and the relative atomic mass falls straight out.
📚 What you need to know
A mass spectrometer measures the relative abundance of the isotopes in a sample.
The stages are: vaporise, ionise, accelerate, detect.
Ions are detected by their mass-to-charge ratio, m/z.
Most ions carry a single positive charge, so m/z is effectively the isotopic mass.
On the spectrum, x-axis is m/z and y-axis is abundance.
If abundances are given as relative heights rather than percentages, convert them first.
Then apply the same weighted average as before to get Ar.
What the instrument does
The IB does not test the engineering, but knowing the purpose of each stage makes the output make sense — particularly the ionisation step, which is the one students find arbitrary.
Losing one electron changes the mass by about 0.03%, far below what the instrument resolves. So the m/z of a 1+ ion is, for every practical purpose, the mass of the atom.
Reading the spectrum
Each peak is one isotope. Its position on the x-axis gives the mass, and its height gives how much of the element is that isotope.
Magnesium’s Ar is close to 24 because almost four fifths of its atoms are magnesium-24. The two heavier isotopes barely shift the average.
Watch the y-axis label. Some spectra give percentage abundance, which adds to 100 and can be used directly. Others give relative abundance, where the tallest peak is simply set to 100 and the rest scaled against it. Those do not add to 100, and using them as percentages gives a wrong answer every time.
When the peaks are relative heights
The fix is one extra line of arithmetic: divide each height by the total of all the heights, rather than by 100. Or, equivalently, use the total of the heights as the denominator in the weighted average.
From relative heights
Ar = Σ(height × mass) ÷ Σ(heights)
A subtlety: the z in m/z
The detector sorts by mass divided by charge. Nearly every ion formed carries a 1+ charge, so dividing by 1 changes nothing and m/z equals the mass. Occasionally a 2+ ion forms, and it appears at half the expected value — a small peak that can look like a mystery isotope until you spot what it is.
A spectrum of a diatomic element shows the combinations, not the individual atoms. Three peaks two units apart is the fingerprint of an element with two isotopes.
WORKED EXAMPLE
The mass spectrum of magnesium shows peaks at m/z 24 (78.99%), 25 (10.00%) and 26 (11.01%). Calculate the relative atomic mass to 2 decimal places.
Check they are percentages78.99 + 10.00 + 11.01 = 100.00They total 100, so these are true percentage abundances and can be used directly.Weighted sum(78.99 × 24) + (10.00 × 25) + (11.01 × 26)= 1895.76 + 250.00 + 286.26 = 2432.02Divide by 1002432.02 ÷ 100 = 24.3202Ar = 24.32Close to 24, as it should be when nearly 80% of the atoms are the lightest isotope.
WORKED EXAMPLE
A copper spectrum shows two peaks: m/z 63 with a relative height of 100.0 and m/z 65 with a relative height of 44.6. Calculate the percentage abundance of each isotope and the relative atomic mass.
Spot the problem firstThese are relative heights, not percentages: 100.0 + 44.6 = 144.6, not 100. They must be converted.Convert to percentages63Cu: 100.0 ÷ 144.6 × 100 = 69.15%65Cu: 44.6 ÷ 144.6 × 100 = 30.85%Now the weighted average(69.15 × 63) + (30.85 × 65) = 4356.45 + 2005.25 = 6361.706361.70 ÷ 100 = 63.617Ar = 63.62The shortcut[(100.0 × 63) + (44.6 × 65)] ÷ 144.6 = 63.62Dividing by the total of the heights does the conversion and the averaging in one step. Same answer, fewer chances to slip.
WORKED EXAMPLE
(a) A sample of magnesium produces a small peak at m/z 12. Suggest what causes it. (b) A spectrum of chlorine gas shows peaks at m/z 70, 72 and 74. Explain why there are three, and predict which is tallest.
(a) the peak at 12Magnesium has no isotope of mass 12. But the detector sorts by mass divided by charge, so a doubly charged ion appears at half its mass.24 ÷ 2 = 12a 24Mg2+ ion(b) why three peaksChlorine gas is Cl2, and with two isotopes there are three possible molecules: both atoms 35Cl (70), one of each (72), or both 37Cl (74).Which is tallest35Cl is roughly three times as common as 37Cl, so the molecule made of two light atoms is by far the most likely combination.m/z 70 is tallest, in roughly a 9 : 6 : 1 patternThe middle peak is doubled because either atom in the pair can be the heavy one, which is why 6 rather than 3.
💡 Exam tip
Add the abundances up first. If they do not make 100, they are relative heights.
For relative heights, divide by the sum of the heights, not by 100.
Say ionisation is needed because only charged particles can be accelerated and deflected.
Remember m/z is a ratio, so a 2+ ion appears at half the mass.
Give Ar to the number of decimal places asked for, rounding only at the end.
Check the answer sits nearest the tallest peak.
⚠️ Common mix-up
Treating relative heights as percentages without converting.
Dividing by the number of isotopes instead of taking a weighted average.
Reading the tallest peak as the answer rather than calculating the average.
Forgetting that m/z is divided by charge, so a 2+ peak looks like a light isotope.
Expecting one peak per atom in a diatomic spectrum rather than one per combination.
Saying the sample is ionised “to make it detectable” without mentioning fields acting on charge.
Up next: The Electromagnetic Spectrum — the nucleus is done. The next section moves outward to the electrons, starting with the light that reveals where they sit.
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