NSWChemistrySyllabus dot point

Inquiry Question 2: How is information about the reactivity and structure of organic compounds obtained?

Investigate the processes used to analyse the structure of simple organic compounds, including mass spectroscopy

Q: 2020 HSC HSC: A mass spectrum shows a molecular ion at m/z = 78 with a peak at m/z = 80 of roughly one-third the height of the m/z = 78 peak. Explain what this isotope pattern tells you about the compound.

The signature of a 3:1 ratio between M and M+2 is **one chlorine atom**. Natural chlorine is 75% $^{35}Cl$ and 25% $^{37}Cl$ (a 3:1 ratio). Any compound with one chlorine therefore shows two molecular ion peaks, the lighter one with $^{35}Cl$ at M and the heavier one with $^{37}Cl$ at M+2, in a 3:1 ratio. For $m/z = 78$: a compound containing one chlorine has a non-chlorine part of mass $78 - 35 = 43$, which is $C_3H_7$. The unknown is therefore $C_3H_7Cl$, chloropropane (either 1-chloropropane or 2-chloropropane, indistinguishable from this evidence alone). A 1:1 M to M+2 ratio would indicate **bromine** ($^{79}Br$ and $^{81}Br$ are roughly equal). Two chlorines give a 9:6:1 pattern at M : M+2 : M+4. Markers reward (1) identifying chlorine from the 3:1 ratio, (2) calculating the rest of the molecular formula, (3) noting the contrast with bromine.

A focused answer to the HSC Chemistry Module 8 dot point on mass spectrometry. The five stages of a mass spectrometer (ionisation, acceleration, deflection, detection, recording), how to read a mass spectrum, identifying the molecular ion and the base peak, recognising fragment loss of 15, 17, 29, 45, the M+2 isotope pattern of chlorine and bromine, and worked HSC past exam questions.

Generated by Claude OpusReviewed by Better Tuition Academy9 min answerUpdated 2026-05-18

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What this dot point is asking

NESA wants you to describe how a mass spectrometer ionises and separates molecules by mass-to-charge ratio, read a mass spectrum to find the molecular ion (M+), the base peak (tallest, set to 100%) and fragment ions, recognise common fragment losses, and identify chlorine and bromine from M and M+2 patterns.

The answer

How a mass spectrometer works

The instrument has five stages:

Vaporisation. The sample is heated and admitted to a high-vacuum chamber as a gas.
Ionisation. A beam of high-energy electrons (about 70 eV) collides with each molecule, knocking out one electron to form a radical cation:

$M + e^- \rightarrow M^{+ \bullet} + 2e^-$

This is the molecular ion. Some molecular ions break apart into smaller cations and neutral radicals; these are the fragment ions.
Acceleration. Positive ions are accelerated through a potential difference, gaining the same kinetic energy. Lighter ions reach higher velocities.
Deflection. A magnetic field bends each ion's path. Lighter (and more highly charged) ions deflect more. The radius of deflection depends on the mass-to-charge ratio, $m/z$ .
Detection. Ions hit a detector; their abundance at each $m/z$ is recorded.

The result is a mass spectrum: a bar chart of relative abundance (0 to 100%) against $m/z$ . The base peak is set to 100%.

Reading a mass spectrum

Molecular ion (M+). The peak at the highest $m/z$ (excluding the small isotope satellites) is the molecular ion. Its $m/z$ is the molecular mass of the compound.

Base peak. The tallest peak, set to 100% relative abundance. Often a fragment, not the molecular ion. The base peak indicates the most stable cation formed in fragmentation.

Fragment peaks. Other peaks. The difference between $M^+$ and a fragment is the mass of the neutral radical lost.

Common fragment losses

Loss (mass)	Fragment lost	Group
15	IMATH_7	Methyl
17	IMATH_8	Hydroxyl
18	IMATH_9	Water (alcohol dehydration)
28	IMATH_10 or IMATH_11	Carbonyl or ethene
29	IMATH_12 or IMATH_13	Aldehyde or ethyl
31	IMATH_14	Methoxy
35, 37	IMATH_15 (35 or 37)	Chlorine
43	IMATH_16 or IMATH_17	Propyl or acetyl
45	IMATH_18	Carboxyl
77	IMATH_19	Phenyl

So a loss of 17 ( $OH$ ) suggests an alcohol; a loss of 45 ( $COOH$ ) suggests a carboxylic acid; a loss of 29 ( $CHO$ ) suggests an aldehyde.

Common diagnostic fragments

IMATH_23	Cation	Hint
15	IMATH_24	Methyl group present
17	IMATH_25	Rare; usually appears as loss not as cation
29	IMATH_26 or IMATH_27	Aldehyde or ethyl
31	IMATH_28	Primary alcohol
43	IMATH_29 or IMATH_30	Propyl or acetyl (ketone, ester)
45	IMATH_31 or IMATH_32	Carboxylic acid or ether
77	IMATH_33	Phenyl (aromatic)

Isotope patterns

Chlorine. $^{35}Cl$ : $^{37}Cl$ = 3 : 1. A compound with one $Cl$ shows M : M+2 = 3 : 1. Two chlorines give M : M+2 : M+4 = 9 : 6 : 1.

Bromine. $^{79}Br$ : $^{81}Br$ $\approx$ 1 : 1. A compound with one $Br$ shows M : M+2 of roughly equal height. Two bromines give M : M+2 : M+4 = 1 : 2 : 1.

Carbon. $^{12}C$ : $^{13}C$ = 98.9 : 1.1, so the M+1 peak is about 1.1% per carbon atom. A 10-carbon molecule shows an M+1 peak about 11% of M; this can be used to count carbons.

Worked logic for an unknown

Suppose a mass spectrum shows:

IMATH_43 molecular ion (small peak).
IMATH_44 (loss of 15, $CH_3$ ).
IMATH_46 (base peak, $CH_3CO^+$ ).
IMATH_48 ( $CH_3^+$ ).

Molecular formula at 88 is consistent with $C_4H_8O_2$ . Base peak at 43 ( $CH_3CO^+$ ) and loss of $CH_3$ from M+ both point to an acetyl group. Loss of 45 (88 to 43) is $C_2H_5O$ , i.e. an $OC_2H_5$ ethoxy group. The compound is ethyl ethanoate $CH_3COOC_2H_5$ .

Strengths and limits

Strengths. Gives the exact molecular mass and structural fragments. Picks out chlorine, bromine and sulfur by isotope pattern. Sensitive to nanograms of sample. Used routinely in forensics, drug testing, environmental analysis (often coupled to gas chromatography, GC-MS).

Limits. Destroys the sample during analysis. Cannot distinguish stereoisomers. Cannot always distinguish structural isomers if they fragment similarly (propan-1-ol and propan-2-ol have very similar spectra). Best combined with NMR and IR for unambiguous structural assignment.

Common traps

Calling the tallest peak the molecular ion. The molecular ion is the highest m/z, not necessarily the tallest. The tallest is the base peak.

Forgetting that the molecular ion can be weak or absent. Some compounds (alcohols, highly branched alkanes) fragment so readily that M+ is barely visible. Confirm M+ by the isotope pattern (M+1 from $^{13}C$ ).

Reading M+2 as a second compound. A 3:1 M : M+2 means chlorine. A 1:1 M : M+2 means bromine. These are isotopologues, not impurities.

Mistaking the m/z value for the mass of the ion. $m/z$ is mass divided by charge. For singly charged ions $z = 1$ and $m/z$ equals the mass; this is the usual HSC case.

Writing "loss of $OH^-$ " or "loss of $CH_3^+$ ". Fragmentation produces a cation and a neutral radical, not a cation and an anion. The cation is detected; the neutral radical is not.

In one sentence

A mass spectrometer ionises, accelerates and deflects molecular and fragment cations by mass-to-charge ratio to produce a spectrum where the molecular ion gives the molecular mass, fragment losses (15 $CH_3$ , 17 $OH$ , 29 $CHO$ , 45 $COOH$ ) reveal functional groups, and characteristic M+2 isotope patterns (3:1 for $Cl$ , 1:1 for $Br$ ) identify halogens.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2022 HSC5 marksA mass spectrum of an unknown organic compound shows a molecular ion at m/z = 60, a base peak at m/z = 43, and a fragment at m/z = 15. The molecular formula contains only C, H and O. Deduce the structure and explain the origin of each peak. Show your working.

Show worked answer →

A 5 mark answer needs the molecular formula derived from m/z = 60, the structure deduced from the fragments, and the chemistry of each loss.

Step 1: Molecular formula from M+. $m/z = 60$ with only C, H, O. Trying $C_3H_8O$ : $3(12) + 8(1) + 16 = 60$ . Matches. Other options ( $C_2H_4O_2$ = 60 is also valid) need to be checked against the fragmentation.

Step 2: Identify the loss from M+ to base peak. $60 - 43 = 17$ , which is the loss of $OH$ . Loss of $OH$ is diagnostic of an alcohol losing a hydroxyl radical to give a carbocation.

Step 3: Identify the m/z = 15 fragment. This is $CH_3^+$ (methyl cation), formed by cleavage of a C-C bond.

Step 4: Deduce structure. Loss of OH plus presence of $CH_3^+$ is consistent with $CH_3CH_2CH_2-OH$ (propan-1-ol) or $CH_3CH(OH)CH_3$ (propan-2-ol). Both are $C_3H_8O$ , both can lose $OH$ , and both can lose $CH_3$ . Propan-2-ol is more likely the answer because the secondary carbocation $CH_3CH^+CH_3$ at $m/z = 43$ is more stable than the propan-1-ol fragment, so the base peak at 43 is expected to dominate. The unknown is most consistent with propan-2-ol.

Step 5: Origin of each peak.

IMATH_16 : molecular ion $[CH_3CH(OH)CH_3]^+$ , formed by loss of one electron.
IMATH_18 : $[CH_3CHCH_3]^+$ (loss of OH, 17). Stable secondary carbocation, so the base peak.
IMATH_20 : $[CH_3]^+$ (loss of $CH_3CHOH$ , 45). Methyl cation.

Markers reward (1) the molecular formula at 60, (2) loss of 17 identified as $OH$ and the alcohol implication, (3) loss of 15 ( $CH_3$ ) and identification of the methyl cation, (4) a self-consistent structure, (5) stability argument for the base peak.

2020 HSC3 marksA mass spectrum shows a molecular ion at m/z = 78 with a peak at m/z = 80 of roughly one-third the height of the m/z = 78 peak. Explain what this isotope pattern tells you about the compound.

Show worked answer →

The signature of a 3:1 ratio between M and M+2 is one chlorine atom.

Natural chlorine is 75% $^{35}Cl$ and 25% $^{37}Cl$ (a 3:1 ratio). Any compound with one chlorine therefore shows two molecular ion peaks, the lighter one with $^{35}Cl$ at M and the heavier one with $^{37}Cl$ at M+2, in a 3:1 ratio.

For $m/z = 78$ : a compound containing one chlorine has a non-chlorine part of mass $78 - 35 = 43$ , which is $C_3H_7$ . The unknown is therefore $C_3H_7Cl$ , chloropropane (either 1-chloropropane or 2-chloropropane, indistinguishable from this evidence alone).

A 1:1 M to M+2 ratio would indicate bromine ( $^{79}Br$ and $^{81}Br$ are roughly equal). Two chlorines give a 9:6:1 pattern at M : M+2 : M+4.

Markers reward (1) identifying chlorine from the 3:1 ratio, (2) calculating the rest of the molecular formula, (3) noting the contrast with bromine.