What is mass spectrometry and how is it used in AS-Level Chemistry?

Mass spectrometry is an analytical technique used to measure the mass-to-charge ratio of ions. In AS-Level Chemistry, it is used to determine the relative atomic masses of elements, identify unknown compounds, and analyze the isotopic composition of elements. This technique is crucial for understanding molecular structures and chemical compositions.

How does a mass spectrometer work in the context of Cambridge International A Levels?

In a mass spectrometer, a sample is ionized, causing the molecules to break into charged particles. These ions are then accelerated through an electric field and deflected by a magnetic field. The degree of deflection depends on the mass-to-charge ratio of the ions. A detector measures the deflection, allowing for the determination of the mass of the ions. This process helps students understand the principles of ionization and detection, which are key components of the Cambridge International A Levels Chemistry curriculum.

What are the main components of a mass spectrometer that students need to know?

The main components of a mass spectrometer include the ionization source, the mass analyzer, and the detector. The ionization source converts the sample into ions, the mass analyzer separates these ions based on their mass-to-charge ratio, and the detector records the abundance of each ion. Understanding these components helps students grasp how mass spectrometry functions as an analytical technique in Chemistry.

Why is mass spectrometry important for analyzing isotopes in AS-Level Chemistry?

Mass spectrometry is crucial for analyzing isotopes because it can accurately measure the mass of different isotopes of an element. This allows students to determine the relative abundance of each isotope, which is essential for calculating the average atomic mass of elements. This understanding is vital for mastering concepts related to isotopic composition and atomic structure in AS-Level Chemistry.

What are some common applications of mass spectrometry in real-world scenarios?

Mass spectrometry has a wide range of applications, including drug testing, environmental analysis, and food safety. It is used to identify and quantify compounds in complex mixtures, detect pollutants, and ensure the authenticity and safety of food products. These applications demonstrate the practical importance of mass spectrometry beyond the classroom, highlighting its relevance in various industries.

Why is "Reading a fragment peak as the molecular ion." a common mistake in Mass spectrometry?

Why it happens: Picking a tall peak instead of the highest m/e. How to avoid it: The molecular ion is the highest m/e peak (not the tallest). Source: 9701 Examiner Reports 2022-2024

Why is "Swapping the Cl and Br [M+2] ratios." a common mistake in Mass spectrometry?

Why it happens: Not knowing the isotope abundances. How to avoid it: Cl ≈ 3:1; Br ≈ 1:1 (from their isotope abundances).

Why is "Confusing the molecular ion with the base peak." a common mistake in Mass spectrometry?

Why it happens: Assuming the tallest peak is M⁺. How to avoid it: The base peak is the tallest (most stable fragment); M⁺ is the highest m/e.

Analytical techniques(AS-Level Analysis)Cambridge International A Level Chemistry 9701

Mass spectrometry

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Mass Spectrometry — Cambridge International AS & A Level Chemistry 9701 (2025-2027 syllabus)

Reading a mass spectrum: the molecular ion peak and relative molecular mass, fragmentation patterns, calculating relative atomic mass from isotopic abundances, and using the [M+1] and [M+2] peaks to find carbon numbers and detect chlorine or bromine.

What you’ll learn

Mapped to the Cambridge International A Level 9701 syllabus (2025-2027).

22.2.1 — Analyse mass spectra in terms of m/e values and isotopic abundances.
22.2.2 — Calculate the relative atomic mass of an element from relative abundances or its mass spectrum.
22.2.3 — Deduce the molecular mass of an organic molecule from the molecular ion peak.
22.2.4 — Suggest the identity of molecules/fragments formed by simple fragmentation in a given mass spectrum.
22.2.5 — Deduce the number of carbon atoms, n, in a compound using the [M+1] peak and the formula n = (100 × abundance[M+1]) / (1.1 × abundance[M]).
22.2.6 — Deduce the presence of bromine and chlorine in a compound using the [M+2] peak.

The molecular ion peak and Mr

The peak at the highest m/e (the molecular ion) gives the relative molecular mass.

A mass spectrometer ionises a molecule to form the molecular ion (M⁺) and detects ions by their mass/charge ratio (m/e). Since most ions carry a 1+ charge, m/e ≈ the mass of the ion.

The molecular ion peak is the peak at the highest m/e (ignoring small isotope peaks like M+1). Its value equals the relative molecular mass ( $M_r$ ) of the compound.

Worked. If the molecular ion peak is at m/e = 46, the compound has $M_r = 46$ (e.g. ethanol, C₂H₅OH, or methanoic acid, HCOOH).

m/e = mass/charge (≈ ion mass for 1+ ions).
Molecular ion peak (highest m/e) = $M_r$ .
Read $M_r$ straight off the spectrum.

Fragmentation patterns

M⁺ breaks into smaller ions; the mass lost or the fragment m/e suggests the structure.

The molecular ion is unstable and fragments into smaller ions and neutral pieces. The detected fragment ions appear at lower m/e, and the gaps between peaks reveal what was lost:

m/e of fragment	Likely fragment	Mass lost from M⁺
15	CH₃⁺	15 (loss of CH₃)
29	C₂H₅⁺ or CHO⁺	29
43	C₃H₇⁺ or CH₃CO⁺	43
17 (loss)	—	17 (loss of OH)
45	COOH⁺ / C₂H₅O⁺	45

Worked. A compound with M⁺ at 58 (propanone, CH₃COCH₃) shows a strong peak at 43 (loss of CH₃ → CH₃CO⁺) and at 15 (CH₃⁺) — consistent with the structure.

M⁺ fragments into smaller ions.
Common: 15 = CH₃, 29 = C₂H₅/CHO, 43 = C₃H₇/CH₃CO.
Use gaps (mass lost) to identify groups.

Relative atomic mass from isotopes

For an element, the spectrum gives isotope masses and abundances; Ar is their weighted mean.

For an element, the mass spectrum shows a peak for each isotope at its mass, with peak height proportional to relative abundance. The relative atomic mass is the weighted mean (as in Topic 1.2): $A_r = \frac{\sum(\text{isotopic mass} \times \text{abundance})}{\sum \text{abundance}}$

Worked. Neon's spectrum: ²⁰Ne (90.5%), ²¹Ne (0.3%), ²²Ne (9.2%): $A_r = \frac{(20 \times 90.5)+(21 \times 0.3)+(22 \times 9.2)}{100} = 20.2$

Each isotope = a peak; height ∝ abundance.
$A_r$ = weighted mean of isotopic masses.
Same calculation as Topic 1.2.

The [M+1] and [M+2] peaks

[M+1] from ¹³C gives the carbon count; [M+2] reveals chlorine (3:1) or bromine (1:1).

[M+1] peak — number of carbons. A small peak one unit above M⁺ arises because ~1.1% of carbon atoms are ¹³C. The more carbons, the bigger this peak. The number of carbons is: $n = \frac{100 \times \text{abundance of }[M{+}1]}{1.1 \times \text{abundance of }M^+}$ Worked. If [M+1] is 4.4% of the M⁺ height: $n = \dfrac{100 \times 4.4}{1.1 \times 100} = 4$ carbons.

[M+2] peak — chlorine and bromine. Their two isotopes give a characteristic peak two units above M⁺:

Chlorine (³⁵Cl : ³⁷Cl ≈ 3 : 1) → M and M+2 peaks in roughly a 3 : 1 ratio.
Bromine (⁷⁹Br : ⁸¹Br ≈ 1 : 1) → M and M+2 peaks in roughly a 1 : 1 ratio.

So a strong [M+2] peak about equal to M⁺ indicates bromine; one about one-third of M⁺ indicates chlorine.

[M+1] from ¹³C → carbon count: $n = (100 \times [M{+}1])/(1.1 \times M)$ .
[M+2] ≈ ⅓ of M → chlorine (3:1).
[M+2] ≈ equal to M → bromine (1:1).

How it’s examined

Reading $M_r$ from the molecular ion peak, identifying fragments, and using the [M+1]/[M+2] peaks are reliable Paper 2 marks; the Ar-from-isotopes calculation links to Topic 1.2. Examiner reports flag misreading the molecular ion peak (using a fragment instead), and confusing the Cl (3:1) and Br (1:1) [M+2] patterns.

Sources: Cambridge International AS & A Level Chemistry 9701 syllabus (2025-2027), section 22.2; 9701 Examiner Reports 2022-2024; 9701/22 & 9701/42 May/Jun 2024 question papers and mark schemes. Last reviewed 2026-05-20.

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Step-by-step worked examples — Mass spectrometry

Step-by-step solutions to past-paper-style questions on mass spectrometry, written exactly the way a tutor would explain them at the board.

1Mr from the molecular ion peak (2 marks)
Getting started• molecular ion
Question
A compound shows its highest m/z (molecular ion) peak at 46. Suggest its Mr and a possible identity.
Step-by-step solution
1. Step 1
  M⁺ = highest m/z (ignoring isotope peaks) → Mr = 46.
2. Step 2
  Mr 46 fits C₂H₆O → ethanol (CH₃CH₂OH).
Answer
Mr = 46; e.g. ethanol, C₂H₆O.
Examiner tip
The molecular ion is the HIGHEST m/z peak (not the tallest) — it gives the Mr directly.
2Ar from a mass spectrum (3 marks)
Getting started• Ar
Question
An element shows peaks at m/e 63 (69.0%) and 65 (31.0%). Calculate its Ar.
Step-by-step solution
1. Step 1
  Weighted mean.
  $A_r = \frac{(63 \times 69.0)+(65 \times 31.0)}{100}$
2. Step 2
  Evaluate.
  $A_r = 63.6$
Answer
$A_r = 63.6$ (copper).
Examiner tip
Divide by the total abundance (100); the answer lies nearer the more abundant isotope.
3Mr and fragments (3 marks)
Building confidence• fragmentation
Question
A spectrum has a molecular ion peak at m/e = 58 and strong peaks at 43 and 15. Suggest the compound and the fragments.
Step-by-step solution
1. Step 1
  M⁺ = 58 → Mr = 58 (e.g. propanone, CH₃COCH₃).
2. Step 2
  Loss of 15 (CH₃) → m/e 43 = CH₃CO⁺.
3. Step 3
  m/e 15 = CH₃⁺.
Answer
Propanone (Mr 58); fragments CH₃CO⁺ (43) and CH₃⁺ (15).
Examiner tip
Use the gaps (loss of 15) to identify the fragments lost.
4Detecting Cl or Br from [M+2] (3 marks)
Building confidence• M+2
Question
Two compounds each show an [M+2] peak. In compound A, M : M+2 is 3 : 1; in compound B it is 1 : 1. Identify the halogen in each.
Step-by-step solution
1. Step 1
  A (3:1) matches ³⁵Cl:³⁷Cl → contains chlorine.
2. Step 2
  B (1:1) matches ⁷⁹Br:⁸¹Br → contains bromine.
Answer
A contains chlorine (3:1); B contains bromine (1:1).
Examiner tip
Cl gives a ~3:1 M:M+2 ratio; Br gives ~1:1.
5Number of carbons from [M+1] (3 marks)
Stretch• M+1
Question
The M⁺ peak has relative abundance 100 and the [M+1] peak 3.3. Calculate the number of carbon atoms.
Step-by-step solution
1. Step 1
  Use the [M+1] formula.
  $n = \frac{100 \times 3.3}{1.1 \times 100}$
2. Step 2
  Evaluate.
  $n = 3$
Answer
3 carbon atoms.
Examiner tip
The [M+1] peak arises from the ¹³C isotope (1.1% natural abundance per carbon).
6Deducing a structure from fragments (4 marks)
Stretch• fragmentation, deduction
Question
A compound has M⁺ at m/e 60 with fragment peaks at 45 and 43. Deduce a likely structure.
Step-by-step solution
1. Step 1
  Mr = 60 — fits ethanoic acid, CH₃COOH.
2. Step 2
  Loss of 15 (CH₃) → m/e 45 = COOH⁺ (CO₂H⁺).
3. Step 3
  Loss of 17 (OH) → m/e 43 = CH₃CO⁺ (acylium).
4. Step 4
  Both fragments fit a –COOH and a CH₃CO– → ethanoic acid.
Answer
Ethanoic acid, CH₃COOH (Mr 60): 45 = COOH⁺ (loss of CH₃), 43 = CH₃CO⁺ (loss of OH).
Examiner tip
Work out each fragment as a 'loss' from M⁺ (here −15 and −17) and match to the structure.

Key Formulae — Mass spectrometry

The formulae you need to memorise for mass spectrometry on the Cambridge International A Level 9701 paper, with every variable defined in plain English and a note on when to use it.

Number of carbons from [M+1]
$n = \frac{100 \times \text{abundance}[M{+}1]}{1.1 \times \text{abundance}[M]}$
$n$
number of carbon atoms
$1.1$
% natural abundance of ¹³C
When to use
Finding the carbon count from the [M+1] peak height.
Relative atomic mass (weighted mean)
$A_r = \frac{\sum(\text{isotopic mass} \times \text{abundance})}{\sum \text{abundance}}$
$isotopic mass$
m/z of each isotope peak
$abundance$
relative height of each peak
When to use
Calculating Ar from the isotope peaks of an element's mass spectrum.

Key Definitions and Keywords — Mass spectrometry

Definitions to memorise and the exact keywords mark schemes credit for mass spectrometry answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9701 sitting.

Mass spectrometry
Examiner keyword
A technique that ionises a sample and separates the ions by mass/charge (m/z), giving Mr and structural information.
Molecular ion peak (M⁺)
Examiner keyword
The peak at the highest m/e (ignoring isotope peaks), equal to the relative molecular mass.
Fragmentation
Examiner keyword
The breakdown of the molecular ion into smaller ions, whose m/e values help deduce the structure.
[M+1] peak
Examiner keyword
A small peak one unit above M⁺ caused by ¹³C; its height gives the number of carbon atoms.
[M+2] peak
Examiner keyword
A peak two units above M⁺ indicating chlorine (M:M+2 ≈ 3:1) or bromine (M:M+2 ≈ 1:1).

Common Mistakes and Misconceptions — Mass spectrometry

The traps other students keep falling into on mass spectrometry questions — taken from recent Cambridge International A Level 9701 examiner reports and mark schemes — and how to avoid them.

✕Reading a fragment peak as the molecular ion.
9701 Examiner Reports 2022-2024
Why it happens
Picking a tall peak instead of the highest m/e.
How to avoid it
The molecular ion is the highest m/e peak (not the tallest).
✕Swapping the Cl and Br [M+2] ratios.
Why it happens
Not knowing the isotope abundances.
How to avoid it
Cl ≈ 3:1; Br ≈ 1:1 (from their isotope abundances).
✕Confusing the molecular ion with the base peak.
Why it happens
Assuming the tallest peak is M⁺.
How to avoid it
The base peak is the tallest (most stable fragment); M⁺ is the highest m/e.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

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Mass spectrometry — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

m/e of fragment

Likely fragment

Mass lost from M⁺

CH₃⁺

15 (loss of CH₃)

C₂H₅⁺ or CHO⁺

C₃H₇⁺ or CH₃CO⁺

17 (loss)

—

17 (loss of OH)

COOH⁺ / C₂H₅O⁺

[M+2] peak — chlorine and bromine. Their two isotopes give a characteristic peak two units above M⁺:

Chlorine (³⁵Cl : ³⁷Cl ≈ 3 : 1) → M and M+2 peaks in roughly a 3 : 1 ratio.
Bromine (⁷⁹Br : ⁸¹Br ≈ 1 : 1) → M and M+2 peaks in roughly a 1 : 1 ratio.

So a strong [M+2] peak about equal to M⁺ indicates bromine; one about one-third of M⁺ indicates chlorine.

Mass spectrometry

Short Study Notes in the form of Slides

Short Study Notes — Mass spectrometry

Detailed Notes

What you’ll learn

How it’s examined

1Mr from the molecular ion peak (2 marks)

2Ar from a mass spectrum (3 marks)

3Mr and fragments (3 marks)

4Detecting Cl or Br from [M+2] (3 marks)

5Number of carbons from [M+1] (3 marks)

6Deducing a structure from fragments (4 marks)

Number of carbons from [M+1]

Relative atomic mass (weighted mean)

Mass spectrometry

Molecular ion peak (M⁺)

Fragmentation

[M+1] peak

[M+2] peak

✕Reading a fragment peak as the molecular ion.

✕Swapping the Cl and Br [M+2] ratios.

✕Confusing the molecular ion with the base peak.

Practice questions

Past paper style quiz

Assess your understanding

Mass spectrometry — frequently asked questions

Mass spectrometry

Short Study Notes in the form of Slides

Short Study Notes — Mass spectrometry

Detailed Notes

What you’ll learn

How it’s examined

1Mr from the molecular ion peak (2 marks)

2Ar from a mass spectrum (3 marks)

3Mr and fragments (3 marks)

4Detecting Cl or Br from [M+2] (3 marks)

5Number of carbons from [M+1] (3 marks)

6Deducing a structure from fragments (4 marks)

Number of carbons from [M+1]

Relative atomic mass (weighted mean)

Mass spectrometry

Molecular ion peak (M⁺)

Fragmentation

[M+1] peak

[M+2] peak

✕Reading a fragment peak as the molecular ion.

✕Swapping the Cl and Br [M+2] ratios.

✕Confusing the molecular ion with the base peak.

Practice questions

Past paper style quiz

Assess your understanding

Mass spectrometry — frequently asked questions