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Analytical Techniques: Infrared Spectroscopy and Mass Spectrometry

Reading an infrared spectrum for functional groups, and interpreting a mass spectrum for relative atomic mass, molecular mass, fragmentation and halogen content, for Cambridge International AS & A Level Chemistry 9701.

Subject
Chemistry
Level
AS LEVEL
Topic
Analytical techniques
Updated

Aligned to Cambridge A Level Chemistry (9701), 2025-2027. Official specification .

Syllabus page (what it covers and how it is assessed): Cambridge A Level Chemistry.

Syllabus points this page covers

9701 (AS Level)

  • 22.1 Infrared spectroscopy
  • 22.2 Mass spectrometry

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This guide covers subtopics 22.1, Infrared spectroscopy, and 22.2, Mass spectrometry, from Topic 22 of Cambridge International AS & A Level Chemistry 9701, 2025–2027 series. Both are AS Level content, grouped together because they’re the two interpretation skills examiners routinely combine — a question may give both spectra for the same unknown compound and ask you to use them together. 22.1 is a single, narrow outcome; the substance of this resource is mostly in 22.2.

Before studying this

This resource assumes isotopes from Atomic Structure: Particles, Radius and Isotopes — mass spectrometry is, at its core, isotope data read as a spectrum — and functional group recognition from the organic resources it references below. Neither the working of the mass spectrometer itself, nor NMR spectroscopy, is required at AS — this page stays strictly within the six mass-spectrometry outcomes and the one infrared outcome the syllabus lists.

Syllabus coverage

CAMBRIDGE INTERNATIONAL AS & A LEVEL CHEMISTRY 9701 — AS Level, Topic 22

22.1 Infrared spectroscopy — analysing an infrared spectrum of a simple molecule to identify functional groups, using reference absorption data.

22.2 Mass spectrometry — analysing mass spectra in terms of m/e values and isotopic abundances (the instrument’s working is not required); calculating relative atomic mass from isotopic abundances or a mass spectrum; deducing molecular mass from the molecular ion peak; suggesting the identity of fragments from simple fragmentation; deducing the number of carbon atoms using the [M+1]⁺ peak; deducing the presence of bromine or chlorine using the [M+2]⁺ peak.

Infrared spectroscopy

An infrared spectrum plots absorption against wavenumber (cm⁻¹). Different bonds absorb IR radiation at characteristic wavenumbers, so identifying which absorptions are present in a spectrum tells you which functional groups the molecule contains. These exact values are printed in the Data section supplied in the examination (syllabus outcome 22.1.1 directs you to it), so they do not need to be memorised precisely — but it’s worth knowing the shape of the table:

Bond Wavenumber range (cm⁻¹) Found in
C–O 1040–1300 alcohols, esters
C=C 1500–1680 alkenes
C=O (amide) 1640–1690 amides
C=O (carbonyl/carboxyl) 1670–1740 aldehydes, ketones, carboxylic acids
C=O (ester) 1710–1750 esters
C≡N 2200–2250 nitriles
C–H (alkane) 2850–2950 almost all organic compounds
O–H (carboxylic acid) 2500–3000, very broad carboxylic acids
N–H 3300–3500 amines
O–H (hydroxy/alcohol) 3200–3600, broad alcohols

Reading a spectrum: a strong, sharp peak around 1700 cm⁻¹ with no broad O–H absorption suggests an aldehyde or ketone rather than a carboxylic acid; the same C=O peak together with a very broad absorption spanning roughly 2500–3000 cm⁻¹ indicates a carboxylic acid specifically, since its O–H absorption is unusually broad and sits below (not overlapping) the alcohol O–H and N–H regions.

Mass spectrometry: the basics

A mass spectrum plots relative abundance against m/e (mass-to-charge ratio). You are not required to know how the instrument itself works — only how to interpret the output.

Relative atomic mass from a spectrum

Worked example. A mass spectrum of copper shows two peaks, at m/e = 63 with relative abundance 69, and m/e = 65 with relative abundance 31. Calculate the Ar of copper.

Ar = (63 × 69 + 65 × 31) / 100 = (4347 + 2015) / 100 = 6362 / 100 = 63.6 (consistent with copper’s accepted Ar of 63.5).

Molecular mass from the molecular ion peak

The molecular ion peak, M⁺, is (barring isotope peaks at higher m/e) the peak at the highest m/e value in the spectrum, formed when the whole molecule loses one electron. Its m/e value gives the Mr of the compound directly, since losing one electron changes the mass by a negligible amount.

Fragmentation

Molecular ions often break apart in the spectrometer, producing smaller fragment ions at lower m/e values. Common, recognisable losses include 15 (loss of •CH₃), 17 (loss of •OH), 29 (loss of •CHO or •C₂H₅), and 45 (loss of •COOH) — recognising these differences between peaks is often more useful than trying to identify an isolated fragment peak on its own.

Worked example. Propan-1-ol, CH₃CH₂CH₂OH (Mr = 60), gives a molecular ion peak at m/e = 60. Suggest the m/e value and identity of a fragment ion formed by loss of the –OH group.

Loss of •OH (mass 17) from the molecular ion: 60 − 17 = m/e = 43, corresponding to the propyl cation, CH₃CH₂CH₂⁺.

Counting carbon atoms with the [M+1]⁺ peak

A small proportion of carbon atoms in any sample are the heavier isotope ¹³C (natural abundance about 1.1%), so every molecular ion peak is accompanied by a small [M+1]⁺ peak — and the more carbon atoms a molecule has, the larger that [M+1]⁺ peak is relative to M⁺. This is quantified by:

n = (100 × abundance of [M+1]⁺ ion) / (1.1 × abundance of M⁺ ion)

Worked example. A compound’s mass spectrum shows the M⁺ peak with relative abundance 100 and the [M+1]⁺ peak with relative abundance 4.4. Calculate the number of carbon atoms in the molecule.

n = (100 × 4.4) / (1.1 × 100) = 440 / 110 = 4 carbon atoms

Detecting bromine or chlorine with the [M+2]⁺ peak

Chlorine and bromine both occur naturally as two isotopes in characteristic ratios, which shows up directly as an [M+2]⁺ peak alongside M⁺:

  • Chlorine (³⁵Cl : ³⁷Cl ≈ 3 : 1) gives an M⁺ : [M+2]⁺ peak height ratio of roughly 3 : 1.
  • Bromine (⁷⁹Br : ⁸¹Br ≈ 1 : 1) gives an M⁺ : [M+2]⁺ peak height ratio of roughly 1 : 1.

Worked example. A compound’s mass spectrum shows two peaks close together at high m/e, in a height ratio of approximately 1 : 1. What does this indicate about the compound’s composition?

A roughly 1 : 1 ratio between the M⁺ and [M+2]⁺ peaks is the signature of one bromine atom in the molecule (a 3 : 1 ratio would indicate chlorine instead; no significant [M+2]⁺ peak at all would indicate neither is present).

Common mistakes

  • Trying to recall exact IR wavenumbers from memory instead of using given reference data. The syllabus expects you to interpret a spectrum using supplied data, not to memorise precise cut-off values.
  • Assuming the tallest peak in a mass spectrum is the molecular ion. The tallest peak (the base peak) is the most abundant fragment, which is often not the molecular ion — M⁺ is identified by being at the highest m/e value, not the greatest height.
  • Mixing up the 3:1 (chlorine) and 1:1 (bromine) ratios. They’re easy to swap under exam pressure — it helps to remember chlorine, the lighter and more common halogen, has the more lopsided (3:1) ratio.
  • Forgetting the [M+1]⁺ formula has a fixed 1.1 in the denominator. This comes directly from ¹³C’s natural abundance (about 1.1%) and is not something to derive from scratch — just apply the formula as given.

Quick revision checklist

  • IR: characteristic wavenumber ranges for O–H, N–H, C=O, C≡N, C=C, C–O, using reference data rather than memorisation
  • Mass spectrometry basics: m/e axis, relative abundance axis, no need to know the instrument’s mechanism
  • Ar from isotopic abundances (weighted mean)
  • Mr from the molecular ion peak (highest m/e, not necessarily tallest)
  • Recognising common fragment losses (15, 17, 29, 45)
  • The [M+1]⁺ formula for counting carbon atoms
  • The 3:1 (Cl) vs 1:1 (Br) ratio for the [M+2]⁺ peak

Written against Cambridge International AS & A Level Chemistry 9701, 2025–2027 series. Always check the current syllabus for your examination year.

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