The big idea: Most elements exist as a mixture of isotopes — atoms of the same element with different masses (different numbers of neutrons).
So the relative atomic mass, Ar, quoted on the periodic table is not the mass of a single atom. It is the weighted average mass of all the isotopes, taking into account how common each one is.
Ar has no units — it is a ratio, comparing the average atom's mass with one-twelfth of a carbon-12 atom.
Mass spectrum of chlorine: two isotope peaks — ³⁵Cl (75%) and ³⁷Cl (25%). The taller peak is the more abundant isotope.
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Why chlorine's A_{r} is 35.5: Chlorine is 75% ³⁵Cl and 25% ³⁷Cl. No chlorine atom actually has a mass of 35.5 — that value is just the average, pulled close to 35 because the lighter isotope is three times more common.
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A mass spectrometer sorts the atoms of a sample by mass. It gives a mass spectrum: a bar chart with the mass-to-charge ratio (m/z) on the x-axis and the relative abundance of each isotope on the y-axis.
Reading the chart: For singly-charged ions the m/z value is the isotope's mass. The height of each peak gives the relative abundance.
- The number of peaks = the number of isotopes. - The position of a peak (m/z) = the mass of that isotope. - The height of a peak = how abundant that isotope is.
Mass spectrum of magnesium: three isotopes — ²⁴Mg (79%), ²⁵Mg (10%), ²⁶Mg (11%). Read each m/z value and its relative abundance.
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Abundances may not be given as %: Sometimes the heights are given as relative values (e.g. 7.9 : 1.0 : 1.1) instead of percentages. Convert to percentages by dividing each by the total and multiplying by 100 — or just divide the weighted sum by the total of the abundances instead of by 100.
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To find Ar, multiply each isotope mass by its percentage abundance, add the results, then divide by 100. Each isotope is weighted by how common it is.
- relative atomic mass (no units — it is a ratio)
- the m/z value of each isotope peak
- the relative abundance of that isotope, as a percentage
- divides by the total percentage, giving a weighted mean
Worked example — Ar of chlorine (two isotopes)
Chlorine has two isotopes: ³⁵Cl with 75.0% abundance and ³⁷Cl with 25.0% abundance. Calculate the relative atomic mass of chlorine.
Solution
- Formula first — weight each isotope mass by its abundance:
- Substitute the two isotopes:
- Work out the top line:
- Divide by 100:
Final answer
Ar(Cl) = 35.5 (no units) — closer to 35 because the lighter isotope is more abundant.
Worked example — Ar of magnesium (three isotopes)
Magnesium has three isotopes: ²⁴Mg (78.6%), ²⁵Mg (10.1%) and ²⁶Mg (11.3%). Calculate the relative atomic mass of magnesium to three significant figures.
Solution
- Formula first — extend the weighted sum to all three isotopes:
- Substitute each isotope:
- Work out the top line:
- Divide and round to 3 s.f.:
Final answer
Ar(Mg) = 24.3 — dominated by the most abundant isotope, ²⁴Mg.
How this is tested: This skill turns up as a short Paper 2 calculation ('calculate the relative atomic mass from the two isotope abundances') and as a Paper 1A MCQ that asks you to read peaks off a mass spectrum.
You are given the isotope masses and their abundances (as a chart or in words) and must return the weighted-average Ar — usually to a stated number of significant figures.
Where the marks go: Mark 1 is the correct set-up — masses multiplied by abundances and divided by the total. Mark 2 is the correct final value to the right significant figures. Always sanity-check: Ar must lie between the lightest and heaviest isotope, closest to the most abundant one.
IB-style question — Ar of element X
A sample of element X gives two peaks in its mass spectrum: m/z = 63 at 69.2% abundance and m/z = 65 at 30.8% abundance. Calculate the relative atomic mass of X to one decimal place. [2]
Solution
- Formula first — weight each isotope mass by its abundance:
- Read the two peaks off the spectrum and substitute:
- Work out the top line:
- Divide and round to 1 d.p.:
Final answer
Ar(X) = 63.6 (the element is copper — closer to 63 because ⁶³Cu is the more abundant isotope).