Reading Mass Spectra for Elements and Isotopes

The mass spectrum of an element is one of the most direct pieces of evidence that isotopes exist. Instead of one peak at the element’s average mass, you see several peaks, one for each isotope, at whole-number masses. From their heights you can calculate the relative atomic mass that appears on the periodic table.

This guide works through the types of calculation that appear in exams, from straightforward to tricky. For how the instrument works, see mass spectrometry.

The key ideas

• Each peak in an element’s mass spectrum corresponds to an isotope: atoms with the same number of protons but different numbers of neutrons.
• The m/z value of the peak (for a +1 ion) is the isotope’s mass number (approximately its relative isotopic mass).
• The height (abundance or intensity) of each peak shows how common that isotope is.
• Relative atomic mass (Ar) is the weighted mean mass of all the isotopes, relative to 1/12 of the mass of a carbon-12 atom.

The formula:

Ar = Σ(m/z × abundance) ÷ Σ(abundance)

If the abundances are percentages that add to 100, you just divide by 100.

Worked example 1: percentages

Boron’s mass spectrum shows peaks at m/z 10 (19.9%) and m/z 11 (80.1%). Calculate Ar.

Ar = (10 × 19.9 + 11 × 80.1) ÷ 100 = (199 + 881.1) ÷ 100 = 10.8

Check: the answer must lie between the lightest and heaviest isotope (10 and 11) and closer to the more abundant one (11). ✔

Worked example 2: relative intensities

Neon shows peaks at m/z 20, 21 and 22 with relative intensities 114.0, 0.3 and 11.2. Calculate Ar.

Sum of intensities = 114.0 + 0.3 + 11.2 = 125.5

Ar = (20 × 114.0 + 21 × 0.3 + 22 × 11.2) ÷ 125.5 = (2280 + 6.3 + 246.4) ÷ 125.5 = 2532.7 ÷ 125.5 = 20.2

The key point: when intensities don’t add up to 100, divide by their total, not by 100.

Worked example 3: four isotopes

Chromium’s mass spectrum: m/z 50 (4.35%), 52 (83.79%), 53 (9.50%), 54 (2.36%). Calculate Ar to 2 decimal places.

Ar = (50 × 4.35 + 52 × 83.79 + 53 × 9.50 + 54 × 2.36) ÷ 100 = (217.5 + 4357.08 + 503.5 + 127.44) ÷ 100 = 5205.52 ÷ 100 = 52.06

The accepted value is 52.00. The small difference arises because real isotopic masses aren’t exact whole numbers (for example, ⁵²Cr has a mass of 51.9405). Using whole-number mass numbers is standard at school level and gives answers within about 0.1.

Worked example 4: finding an unknown abundance

Copper has two isotopes, ⁶³Cu and ⁶⁵Cu. Its Ar is 63.55. Calculate the percentage abundance of each.

Let the abundance of ⁶³Cu be x%. Then ⁶⁵Cu is (100 − x)%.

63.55 = [63x + 65(100 − x)] ÷ 100 6355 = 63x + 6500 − 65x 6355 = 6500 − 2x 2x = 145 x = 72.5% ⁶³Cu, so 27.5% ⁶⁵Cu

(The accepted values are 69.2% and 30.8%; the difference again comes from using whole-number masses rather than exact isotopic masses.)

This “let x equal the abundance” method works for any element with two isotopes.

Worked example 5: identifying the element

An element gives peaks at m/z 107 (51.8%) and 109 (48.2%). Identify it.

Ar = (107 × 51.8 + 109 × 48.2) ÷ 100 = (5542.6 + 5253.8) ÷ 100 = 107.96 ≈ 108

The element with Ar ≈ 107.9 is silver. You can check any element’s atomic mass on the interactive periodic table.

Worked example 6: diatomic molecules (Cl₂)

Chlorine exists as Cl₂ molecules. Its mass spectrum shows peaks for atoms (Cl⁺, formed when molecules break apart) and for whole molecules (Cl₂⁺).

Chlorine’s isotopes: ³⁵Cl (75%) and ³⁷Cl (25%), a ratio of 3 : 1.

Atomic peaks: m/z 35 and 37, in a 3 : 1 ratio.

Molecular peaks: a Cl₂ molecule can contain:

Combination	m/z	Probability
³⁵Cl–³⁵Cl	70	0.75 × 0.75 = 0.5625
³⁵Cl–³⁷Cl (either order)	72	2 × 0.75 × 0.25 = 0.375
³⁷Cl–³⁷Cl	74	0.25 × 0.25 = 0.0625

Ratio 0.5625 : 0.375 : 0.0625 = 9 : 6 : 1

So the molecular ion region shows peaks at m/z 70, 72 and 74 in a 9 : 6 : 1 ratio. The factor of 2 for the mixed molecule comes from the two ways of arranging it (³⁵Cl–³⁷Cl and ³⁷Cl–³⁵Cl).

Worked example 7: bromine (Br₂)

Bromine’s isotopes, ⁷⁹Br and ⁸¹Br, are about 50 : 50.

Combination	m/z	Probability
⁷⁹Br–⁷⁹Br	158	0.25
⁷⁹Br–⁸¹Br	160	0.50
⁸¹Br–⁸¹Br	162	0.25

Ratio 1 : 2 : 1, a very recognisable pattern. See isotopes, ions, isomers and allotropes for more on isotopes.

Worked example 8: 2+ ions

Sometimes an atom loses two electrons, forming a 2+ ion. Its m/z is half its mass.

A peak appears at m/z 12 in the mass spectrum of magnesium, whose isotopes are 24, 25 and 26. Explain.

It’s the ²⁴Mg²⁺ ion: mass 24, charge 2+, so m/z = 24 ÷ 2 = 12.

2+ ions are usually much less abundant than 1+ ions, because removing a second electron needs more energy.

Why isotope abundances vary slightly

The isotope abundances in data tables are averages for material found on Earth. Real samples can differ slightly depending on where they came from, because physical, chemical and biological processes can favour lighter or heavier isotopes. That’s why IUPAC now gives the standard atomic weights of some elements, such as hydrogen, carbon and oxygen, as ranges rather than single numbers. Scientists exploit these tiny variations: isotope ratio mass spectrometry can reveal whether honey has been diluted with cheap sugar syrup, which region a wine came from, or what an ancient person ate, by measuring carbon, nitrogen and oxygen isotope ratios with great precision.

Common mistakes

• Dividing by 100 when intensities don’t add to 100. Divide by the total.
• Getting an Ar outside the range of the isotope masses. Always sense-check.
• Forgetting the factor of 2 for mixed isotope combinations in diatomic molecules.
• Confusing m/z with mass for 2+ ions.
• Rounding too early. Keep extra figures until the end.
• Mixing up mass number and relative isotopic mass. For most calculations at this level, whole-number mass numbers are fine, but data books give exact isotopic masses for precise work.

Practice questions

1. Lithium: m/z 6 (7.6%), m/z 7 (92.4%). Calculate Ar.
2. Gallium: m/z 69 (60.1%), m/z 71 (39.9%). Calculate Ar.
3. An element has two isotopes, mass 10 and 11, and Ar 10.8. Find the abundance of each.
4. Predict the m/z values and ratio of peaks for the HCl molecular ion (H is all ¹H; Cl is 3 : 1 ³⁵Cl : ³⁷Cl).

Answers:

1. (6 × 7.6 + 7 × 92.4) ÷ 100 = 6.92
2. (69 × 60.1 + 71 × 39.9) ÷ 100 = 69.80
3. 10x + 11(100 − x) = 1080 → x = 20% mass 10, 80% mass 11
4. m/z 36 and 38 in a 3 : 1 ratio

Key takeaways

• Each peak in an element’s mass spectrum is an isotope; its height is its abundance.
• Ar = Σ(m/z × abundance) ÷ Σ(abundance).
• For two isotopes with a known Ar, let one abundance be x and solve.
• Diatomic molecules give characteristic patterns: 9 : 6 : 1 for Cl₂, 1 : 2 : 1 for Br₂.
• 2+ ions appear at half their mass.