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Predicting Diatomic Mass Spectra

A concise revision guide to predicting the molecular ion peaks for diatomic molecules from isotope abundance data, using chlorine, Cl2, as the key example.

Paper 1
AQA
3.1.1 Atomic Structure
7405/1
Dr. Mohammed Al-Fatah

Written by: Dr. Mohammed Al-Fatah

Chemistry specialist revision notes for A Level Chemistry.

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Before you start

GCSE Recap: Isotopes and Diatomic Molecules

Before you start, check what you remember about diatomic elements and isotopes.

1

Why Diatomic Molecules Give Several Peaks

A diatomic molecule contains two atoms joined together. If the element has more than one isotope, different isotope pairings are possible in the molecule.

For chlorine, the two main isotopes are 35Cl and 37Cl. This means a molecule of Cl2 can contain two 35Cl atoms, one of each isotope, or two 37Cl atoms.

Key idea: each possible isotope pairing gives a different molecular mass, so the molecular ion region of the mass spectrum shows more than one peak.

2

List Every Possible Isotope Pair

Start by listing the possible combinations. This prevents you from missing the mixed molecule, which is the most common exam error in this calculation.

Combination Molecule Molecular mass Expected m/z for 1+ ion
Light + light 35Cl-35Cl 35 + 35 = 70 70
Light + heavy 35Cl-37Cl 35 + 37 = 72 72
Heavy + light 37Cl-35Cl 37 + 35 = 72 72
Heavy + heavy 37Cl-37Cl 37 + 37 = 74 74

Exam focus: the two mixed arrangements have the same molecular mass, so they contribute to the same m/z peak.

Check your understanding

Quick Check: List the Pairs for a New Element

Write out every isotope pair on paper, then type the m/z values.

3

Multiply the Isotopic Abundances

Convert the percentage abundances into decimals, then multiply the two isotope abundances for each possible molecule.

For chlorine: 35Cl = 75% = 0.75 and 37Cl = 25% = 0.25. These are the rounded values used in exam questions: a 3:1 ratio, which gives Ar = 35.5.

Molecule Relative abundance calculation Relative abundance
35Cl-35Cl 0.75 × 0.75 0.5625
35Cl-37Cl 0.75 × 0.25 0.1875
37Cl-35Cl 0.25 × 0.75 0.1875
37Cl-37Cl 0.25 × 0.25 0.0625

Remember: do not add the isotope abundances for a molecule. Multiply them, because both atoms must be chosen together.

Check your understanding

Quick Check: Half the Atoms, Half the Molecules?

Decide whether the statement about bromine molecules is true or false.

4

Combine Equivalent Mixed Isotope Molecules

The molecules 35Cl-37Cl and 37Cl-35Cl have the same molecular mass. Both have molecular mass 72, so their abundances are added together.

0.1875 + 0.1875 = 0.375

This gives the total relative abundance of the m/z 72 peak.
Molecular ion m/z Total relative abundance
35Cl-35Cl+ 70 0.5625
35Cl-37Cl+ and 37Cl-35Cl+ 72 0.375
37Cl-37Cl+ 74 0.0625
Check your understanding

Quick Check: The Mixed Isotope Peak

Work out the total abundance of the mixed molecules for an element with 60% and 40% isotopes.

5

Convert Relative Abundances into a Peak Ratio

To find the simplest whole number ratio, divide all of the relative abundances by the smallest value.

m/z 70

0.5625 ÷ 0.0625 = 9

m/z 72

0.375 ÷ 0.0625 = 6

m/z 74

0.0625 ÷ 0.0625 = 1

Cl2 molecular ion peaks: m/z 70 : 72 : 74 = 9 : 6 : 1

The middle peak is larger than the m/z 74 peak because there are two ways to make a mixed isotope molecule.

A real spectrum of Cl2 shows two more peaks at lower m/z. Some Cl2+ ions break apart in the instrument and give atomic ions: 35Cl+ at m/z 35 and 37Cl+ at m/z 37. Their heights are in the ratio 3:1, the same as the isotope abundances.

Exam focus: the full spectrum of chlorine has five peaks. m/z 35 and 37 (3:1) are atomic ions, Cl+. m/z 70, 72 and 74 (9:6:1) are molecular ions, Cl2+.

Check your understanding

Quick Check: Predict the Whole Pattern

Predict the m/z values and the simplest peak ratio on paper before you flip each card.

Worked example showing how chlorine isotope abundances predict the mass spectrum of Cl2

The m/z 70, 72 and 74 peaks arise from the three possible molecular masses of Cl2+.

6

Use the Pattern to Check Your Answer

The 9:6:1 peak ratio is not random. It comes from the isotope abundance ratio of chlorine, which is approximately 3:1 for 35Cl:37Cl.

For a diatomic molecule, the combinations follow the pattern:

3 × 3 : 2 × 3 × 1 : 1 × 1 = 9 : 6 : 1

The factor of 2 appears because the mixed molecule can form in two equivalent orders.

Exam focus: if the chlorine molecular ion peaks are in a 9:6:1 ratio, this supports the presence of two chlorine atoms in the molecule.

Check your understanding

Quick Check: Patterns Beyond Chlorine

In each round, pick the one statement that is accurate.

7

Common Exam Points

These points help avoid the most frequent mistakes when predicting molecular mass spectra for diatomic molecules.

Use molecular mass, not atomic mass

For Cl2, the molecular ion peaks are at m/z 70, 72 and 74. The peaks at m/z 35 and 37 come from Cl+ atomic ions, not from molecules.

Remember the mixed molecule twice

35Cl-37Cl and 37Cl-35Cl are equivalent but both contribute to the m/z 72 peak.

Multiply probabilities

To find the relative abundance of a molecule, multiply the abundances of the two isotopes used to form it.

Use 1+ ions carefully

When the molecular ion has a single positive charge, the m/z value is equal to the relative molecular mass of that ion.

QuickSnap Summary

Predicting a diatomic mass spectrum means combining isotope masses and isotope abundances to find the expected molecular ion peaks.

1

List every possible isotope pair in the diatomic molecule.

2

Add the isotope mass numbers to find each molecular ion m/z value.

3

Multiply isotope abundances to calculate the relative abundance of each pairing.

4

Add equivalent mixed pair abundances, then divide by the smallest value to get the simplest ratio.

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FAQs

These common questions summarise the key calculation ideas for predicting diatomic molecular mass spectra.

Why does Cl2 give peaks at m/z 70, 72 and 74?

Chlorine has two isotopes, 35Cl and 37Cl. The possible Cl2 molecules have masses 35 + 35 = 70, 35 + 37 = 72, and 37 + 37 = 74.

Why is the middle Cl2 peak larger than expected?

The middle peak includes two equivalent mixed isotope arrangements: 35Cl-37Cl and 37Cl-35Cl. Their abundances must be added together.

How do you calculate the relative abundance of a diatomic molecule?

Convert the isotope abundances into decimals, then multiply the two isotope abundances for the atoms in that molecule.

What is the predicted Cl2 molecular ion peak ratio?

Using 75% 35Cl and 25% 37Cl, the predicted molecular ion peaks at m/z 70, 72 and 74 have a ratio of 9:6:1.

Copyright notice: This OLS revision content, including the explanations, layout, diagrams, tables and embedded learning structure, is authored for Online Learning System by Dr. Mohammed Al-Fatah. It may not be copied, reproduced, redistributed or adapted without written permission.