Mass spectrometry: WACE Year 12 Chemistry

Syllabus dot point

“Interpret mass spectra to determine molar mass and identify fragments of organic molecules”

8 min readUpdated 2 Jun 2026

Quick answer

In a mass spectrometer a molecule is ionised and broken into fragments, and the mass-to-charge ratio of each ion is measured. The molecular ion peak (the peak at the highest mass for the parent ion) gives the molar mass. The fragment peaks below it, and the mass differences between peaks, reveal which groups have broken off and so help deduce the structure.

What this dot point is asking

Mass spectrometry is an instrumental technique that measures the mass of molecules and their fragments. It is the most direct way to find the molar mass of an organic compound and provides clues to its structure.

How it works in outline

A sample is vaporised and ionised (usually by knocking off an electron to form a positive ion). The ions are accelerated and separated according to their mass-to-charge ratio (m/z). A detector records the abundance of each, producing a spectrum of peaks. Most ions carry a single positive charge, so m/z effectively reads as the mass.

Molecular ion peak

The peak produced by the intact molecule after losing one electron (the parent ion, $\text{M}^+$ ). Its m/z value equals the molar mass of the molecule, so it is read at the highest mass peak (ignoring small isotope peaks).

Finding the molar mass

The molecular ion peak appears at the highest m/z (apart from minor isotope peaks). Its value is the molar mass of the compound. For example, a molecular ion peak at m/z 46 is consistent with ethanol, $\text{C}_2\text{H}_5\text{OH}$ ( $M = 46$ ).

The molecular ion gives the molar mass

Read the molar mass directly from the m/z of the molecular ion peak (the parent peak at the highest mass). Small peaks just above it usually come from heavier isotopes such as carbon-13, not a larger molecule.

Fragmentation and structure

Inside the spectrometer the molecular ion breaks into smaller fragment ions. Each fragment gives a peak, and the difference in mass between peaks tells you what was lost. Common losses include 15 (a $\text{CH}_3$ group), 17 (an -OH group), 29 (a $\text{CHO}$ or $\text{C}_2\text{H}_5$ group) and 45 (a $\text{COOH}$ group). Identifying these losses lets you piece together the structure.

The base peak

The tallest peak in the spectrum is the base peak, corresponding to the most stable (most abundant) fragment ion. Its abundance is set to 100 percent and other peaks are scaled relative to it. A particularly stable fragment dominates the spectrum.

Interpreting fragment losses

Question

A compound has a molecular ion peak at m/z 60 and a strong fragment at m/z 45. Suggest what has been lost and a possible identity.

Solution

The molar mass is 60, read from the molecular ion peak. The fragment at m/z 45 is 15 mass units lower, so a fragment of mass 15 (a $\text{CH}_3$ group) has been lost from the molecular ion.

A molar mass of 60 with loss of $\text{CH}_3$ to give a fragment of 45 is consistent with ethanoic acid, $\text{CH}_3\text{COOH}$ ( $M = 60$ ), losing the methyl group to leave the $\text{COOH}^+$ -related fragment at 45. The data support ethanoic acid (further evidence would come from other techniques).

Isotope patterns as a structural clue

Beyond the molar mass and fragments, the pattern of peaks around the molecular ion can reveal which elements are present. A small peak one unit above the molecular ion (the $\text{M}+1$ peak) comes from molecules containing a carbon-13 atom, and its relative size hints at how many carbon atoms the molecule has. More strikingly, elements with two abundant isotopes leave a characteristic doublet: chlorine gives molecular-ion peaks two units apart in roughly a $3:1$ ratio (chlorine-35 to chlorine-37), and bromine gives a near $1:1$ doublet two units apart (bromine-79 and bromine-81). Spotting one of these patterns immediately tells you the compound contains chlorine or bromine, which is a quick, high-value deduction in an unknown-identification question.

Why this matters

Mass spectrometry is usually the first technique applied when identifying an unknown, because the molar mass narrows the possibilities dramatically. Combined with infrared and NMR spectroscopy, it lets chemists determine a full structure, which is exactly the skill the analysis section of Unit 4 assesses.

Misreading the molecular ion or the loss

A common error is reading the molar mass from the base peak (the tallest peak) rather than the molecular ion peak (the highest-mass parent peak); they are usually different. Another is miscalculating the lost fragment by subtracting peaks the wrong way; the loss is the molecular ion mass minus the fragment mass. Always identify the molecular ion peak first, then work down.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2022-style6 marks

The mass spectrum of an organic compound shows a molecular ion peak at

m/z = 58

, a base peak at

m/z = 43

, and a peak at

m/z = 15

. (a) Determine the molar mass. (b) Identify the fragments lost and the fragment ions, and suggest a structure. (c) Explain why the base peak is the most abundant.

Show worked answer →

A 6 mark question rewards the molar mass, the fragment analysis, and the base-peak reasoning.

(a) The molar mass equals the m/z of the molecular ion peak (the highest-mass parent peak), so $M = 58\ \text{g mol}^{-1}$ .

(b) Loss from $58 \rightarrow 43$ is $15$ , a $\text{CH}_3$ group, so the $m/z = 43$ ion is $\text{CH}_3\text{CO}^+$ (an acylium ion) and the $m/z = 15$ peak is $\text{CH}_3^+$ . A molar mass of 58 with these fragments is consistent with propanone, $\text{CH}_3\text{COCH}_3$ ( $M = 58$ ), which loses a methyl group to give the stable $\text{CH}_3\text{CO}^+$ at 43.

(c) The base peak (at $m/z = 43$ ) is most abundant because $\text{CH}_3\text{CO}^+$ is a particularly stable fragment ion (the positive charge is stabilised on the carbonyl carbon), so this fragmentation is favoured and produces the largest number of ions.

Markers reward $M = 58$ , the loss of $\text{CH}_3$ giving $\text{CH}_3\text{CO}^+$ at 43 with propanone as the structure, and the stability reason for the base peak.

WACE 2020-style4 marks

Chloromethane

\text{CH}_3\text{Cl}

shows two molecular ion peaks at

m/z = 50

and

m/z = 52

in a ratio of about

3:1

. Explain this observation, and explain what additional information the molecular ion peak alone provides about an unknown compound.

Show worked answer →

A 4 mark answer needs the isotope explanation and the molar-mass role.

Chlorine has two stable isotopes, chlorine-35 and chlorine-37, in an approximate $3:1$ natural abundance. Molecules containing chlorine-35 give a molecular ion at $m/z = 50$ ( $12 + 3 + 35$ ) and those containing chlorine-37 give a peak at $m/z = 52$ ( $12 + 3 + 37$ ), so two peaks appear two units apart in the same $3:1$ ratio as the isotopes.

For an unknown compound, the molecular ion peak gives the molar mass directly (the m/z of the parent ion), which dramatically narrows the possible molecular formulae before any fragmentation is considered.

Markers reward the chlorine-35/chlorine-37 isotope explanation with the $3:1$ ratio, and the molecular ion giving the molar mass.

Practise this

Sources & how we know this

Chemistry ATAR course: syllabus and support materials — SCSA
Chemistry: past ATAR course examinations — SCSA