A LevelChemistryEdexcelRevision Notes3. Organic ChemistryModern Analytical Techniques IMass Spectrometry

Mass Spectrometry (Edexcel A Level Chemistry): Revision Note

Exam code: 9CHO

Written by: Philippa Platt

Reviewed by: Caroline Carroll

Updated on 22 October 2026

Mass Spectrometry

Mass spectrometry is an analytical technique used to identify unknown compounds
The molecules in the small sample are bombarded with high energy electrons which can cause the molecule to lose an electron
This results in the formation of a positively charged molecular ion with one unpaired electron
- One of the electrons in the pair has been removed by the beam of electrons

MOLECULE $\overset{Electron bombardment}{\to}$ MOLECULE⁺• + e^–

The [M+1] peak is a smaller peak which is due to the natural abundance of the isotope carbon-13
The amount of naturally occurring C-13 is a little over 1%, so the [M+1] peak is very small
The height of the [M+1] peak for a particular ion depends on how many carbon atoms are present in that molecule; the more carbon atoms, the larger the [M+1] peak is
- For example, the height of the [M+1] peak for a hexane (containing six carbon atoms) ion will be greater than the height of the [M+1] peak of an ethane (containing two carbon atoms) ion

Worked Example

Analysing mass spectra

Determine whether the following mass spectrum corresponds to but-1-ene or pent-1-ene:

Mass spectrum for the worked example, plotted as relative intensity against m/z, with the molecular ion peak at m/z 70, a small M+1 peak at 71 and the base peak at 42, to be matched to but-1-ene (M = 56) or pent-1-ene (M = 70)

Answer

The mass spectrum corresponds to pent-1-ene as the molecular ion peak is at m/z = 70
- The small peak at m/z = 71 is a C-13 peak, which does not count as the molecular ion peak
- But-1-ene arises from the C₄H₈⁺ ion which has a molecular mass of 56
- Pent-1-ene arises from the C₅H₁₀⁺ ion which has a molecular mass of 70

The molecular ion can further fragment to form new ions, molecules, and radicals

Fragmentation of a molecular ion in three ways: the molecular ion splits into a positively charged fragment and a radical or molecule; the spectrometer detects only the positively charged ion

Fragmentation of a molecule in mass spectrometry

These fragmented ions are accelerated by an electric field
Based on their mass (m) to charge (z) ratio, the ion fragments are then separated by deflecting them into the detector
- Most ions will only gain a charge of 1+ and therefore an ion with mass 12 and charge 1+ will have an m/z value of 12
- It is, however, possible for a greater charge to occur. For example, an ion with mass 16 and charge 2+ will have a m/z value of 8
The smaller and more positively charged fragment ions will be detected first as they will get deflected the most and are the ions with the smallest m/z values
Each fragment corresponds to a specific peak with a particular m/z value in the mass spectrum
The base peak is the peak corresponding to the most abundant ion
The m/z is sometimes referred to as the m/e ratio and it is almost always 1:1

Isotopes

Isotopes are different atoms of the same element that contain the same number of protons and electrons but a different number of neutrons.
- These are atoms of the same elements but with different mass number
- For example, Cl-35 and Cl-37 are isotopes as they are both atoms of the same element (chlorine, Cl) but have a different mass number (35 and 37 respectively)
Mass spectrometry can be used to find the relative abundance of the isotopes experimentally
The relative abundance of an isotope is the proportion of one particular isotope in a mixture of isotopes found in nature
- For example, the relative abundance of Cl-35 and Cl-37 is 75% and 25% respectively
- This means that in nature, 75% of the chlorine atoms is the Cl-35 isotope and 25% is the Cl-37 isotope
The heights of the peaks in mass spectrometry show the proportion of each isotope present

Mass spectrum of boron: two peaks, boron-10 at m/e 10 with 19.9 percent abundance and boron-11 at m/e 11 with 80.1 percent abundance

The peak heights show the relative abundance of the boron isotopes: boron-10 has a relative abundance of 19.9% and boron-11 has a relative abundance of 80.1%

Worked Example

Calculating m/z ratio

In a sample of iron, the ions ⁵⁴Fe²⁺ and ⁵⁶Fe³⁺ are detected. Calculate their m/z ratio and determine which ion is deflected more inside the spectrometer.

Answer

Calculate m/z for Fe²⁺:

m/z (Fe²⁺) = $\frac{54}{2}$ = 27

Calculate m/z for Fe³⁺:

m/z (Fe³⁺) = $\frac{56}{3}$ = 18.7

⁵⁶Fe³⁺ has a smaller m/z ratio and will therefore be deflected more.
- It also has the largest positive charge and will be deflected more by the magnetic field within the mass spectrometer.

Examiner Tips and Tricks

A small m/z value corresponds to fragments that are either small or have a high positive charge or a combination of both.

Fragmentation

The molecular ion peak can be used to identify the molecular mass of a compound
However, different compounds may have the same molecular mass
To further determine the structure of the unknown compound, fragmentation is used
Fragments may appear due to the formation of characteristic fragments or the loss of small molecules
- For example, a peak at 29 is due to the characteristic fragment C₂H₅⁺
- Loss of small molecules gives rise to peaks at M-18 (loss of H₂O), M-28 (loss of CO) and M-44 (loss of CO₂)

Alkanes

Simple alkanes are fragmented in mass spectrometry by breaking the C-C bonds
M/e values of some of the common alkane fragments are given in the table below

m/e Values of Fragments Table

Fragment	m/e
CH₃⁺	15
C₂H₅⁺	29
C₃H₇⁺	43
C₄H₉⁺	57
C₅H₁₁⁺	71
C₆H₁₃⁺	85

Mass spectrum of decane, C10H22: molecular ion at m/e 142 and fragment peaks at 43, 56, 57 (C4H9+), 71 (C5H11+) and 113 (C8H17+)

Mass spectrum showing the fragmentation of C₁₀H₂₂

Halogenoalkanes

Halogenoalkanes often have multiple peaks around the molecular ion peak
This is caused by the fact that there are different isotopes of the halogens

Mass spectrum of a bromoalkane with two molecular ion peaks of equal height at m/e 108 and 110, from molecules containing 79Br and 81Br

Mass spectrum showing different isotopes of bromine in the molecular ion

Alcohols

Alcohols often tend to lose a water molecule giving rise to a peak at 18 below the molecular ion
Another common peak is found at m/e value 31 which corresponds to the CH₂OH⁺ fragment
For example, the mass spectrum of propan-1-ol shows that the compound has fragmented in four different ways:
- Loss of H^• to form a C₃H₇O⁺ fragment with m/e = 59
- Loss of a water molecule to form a C₃H₆⁺ fragment with m/e = 42
- Loss of a ^•C₂H₅ to form a CH₂OH⁺ fragment with m/e = 31
- And the loss of ^•CH₂OH to form a C₂H₅⁺ fragment with m/e = 29

Mass spectrum of propan-1-ol with peaks at m/e 60 (molecular ion), 59 (C3H7O+), 42 (loss of water), 31 (CH2OH+, the base peak) and 29 (C2H5+), with the fragmentation equations below

Mass spectrum showing the fragmentation patterns in propan-1-ol (alcohol)

Worked Example

Ion fragmentation

Which of the following statements about the mass spectrum of CH₃Br is correct?

A. There is one peak for the molecular ion with an m/e value of 44

B. There is one peak for the molecular ion with an m/e value of 95

C. The last two peaks have abundances in the ratio 3:1 and occur at m/e values of 94 and 96

D. The last two peaks are of equal size and occur at m/e values of 94 and 96

Answer

The correct answer is option D

Bromomethane (CH₃Br) can produce 3 peaks
- CH₃⁸¹Br → [CH₃⁸¹Br]⁺ + e⁻ at m/e 96
- CH₃⁷⁹Br → [CH₃⁷⁹Br]⁺ + e⁻ at m/e 94
- CH₃Br → [CH₃]⁺ + ^•Br at m/e 15
- The last two peaks (which correspond to the molecular ion peak) therefore are equal in size and occur at m/e values of 94 and 96

Mass spectrum of bromomethane: peak at m/e 15 for CH3+ and two peaks of equal height at m/e 94 (CH3 79Br+) and 96 (CH3 81Br+)

Worked Example

Alcohol fragmentation

Which alcohol is not likely to have a fragment ion at m/e at 43 in its mass spectrum?

A. (CH₃)₂CHCH₂OH

B. CH₃CH(OH)CH₂CH₂CH₃

C. CH₃CH₂CH₂CH₂OH

D. CH₃CH₂CH(OH)CH₃

Answer

The correct answer is option D

Because a line at m/e = 43 corresponds to an ion with a mass of 43 for example:
- [CH₃CH₂CH₂]⁺
- [(CH₃)₂CH]⁺
- butan-2-ol is not likely to have a fragment at m/e = 43 as it does not have either of these fragments in its structure.

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Mass Spectrometry (Edexcel A Level Chemistry): Revision Note

Mass Spectrometry

Worked Example

Isotopes

Worked Example

Examiner Tips and Tricks

Fragmentation

Alkanes

Halogenoalkanes

Alcohols

Worked Example

Worked Example

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Author: Philippa Platt

Reviewer: Caroline Carroll