# Mass spectrometry

> CIE A-Level Chemistry · 9701
> Source: https://www.owlsprep.com/study/cie-9701-u16-mass-spectrometry/

This sub-topic covers the working principles of mass spectrometry, its use to calculate relative atomic masses of elements, and interpretation of organic mass spectra to find molecular mass and structural fragments.

**Prerequisites:** [Relative atomic mass and isotopic abundance](https://www.owlsprep.com/study/cie-9701-u1-isotopes-relative-mass/); [Basic organic functional group structure](https://www.owlsprep.com/study/cie-9701-u10-introduction-organic-chemistry/)

## Learning objectives

- Describe the four main stages of operation of a mass spectrometer
- Calculate relative atomic mass from isotopic abundance data from mass spectra
- Interpret organic mass spectra to identify molecular mass and common fragment ions
- Relate fragment peaks to functional groups and structure of organic molecules

## 1. Key Stages of Mass Spectrometer Operation

Mass spectrometry separates positive ions based on their mass-to-charge ratio ($m/z$) to produce a spectrum that gives information about the mass and relative abundance of particles.

**Mass-to-charge ratio (m/z)** — The ratio of the mass of an ion (in atomic mass units, u) to its charge (in number of elementary charges, e). Most detected ions are +1, so m/z equals the mass of the ion.

*Example:* A singly charged ion of mass 44 u has m/z = 44

1. Ionisation: The sample is bombarded with high-energy electrons to form positive ions.
2. Acceleration: Positive ions are accelerated by an electric field to constant kinetic energy.
3. Deflection: Ions are deflected by a magnetic field; lighter/ higher charge ions are deflected more.
4. Detection: Ions reach a detector, and their abundance and m/z are recorded.

> **tip**
>
> For CIE 9701, you may be asked to describe all four stages in order, so always remember the correct sequence.

## 2. Calculating Relative Atomic Mass from Isotopic Spectra

For an element with multiple isotopes, mass spectrometry measures the mass and relative abundance of each isotope, which we use to calculate the weighted average relative atomic mass ($A_r$).

**Worked example:** A sample of neon has three isotopes with the following mass and percentage abundance: $^{20}$Ne (19.99, 90.5%), $^{21}$Ne (20.99, 0.3%), $^{22}$Ne (21.99, 9.2%). Calculate the relative atomic mass of neon.

1. Relative atomic mass is calculated as the weighted average of isotopic masses, using the formula:
2. $$A_r = \frac{(m_1 \times a_1) + (m_2 \times a_2) + ... + (m_n \times a_n)}{100}$$
3. Substitute the values from the question into the formula:
4. $$A_r = \frac{(19.99 \times 90.5) + (20.99 \times 0.3) + (21.99 \times 9.2)}{100}$$
5. Calculate the numerator, then divide by 100 to get the final value:
6. $$A_r = \frac{2017.7}{100} = 20.18 \text{ (4 s.f.)}$$

**Check your understanding**

Check your understanding

1. Chlorine has two isotopes: $^{35}$Cl (75% abundance) and $^{37}$Cl (25% abundance). What is the $A_r$ of chlorine?

   - 35.0
   - 35.5
   - 36.0
   - 36.5

   *Why:* Correct: $(35 \times 75 + 37 \times 25)/100 = 35.5$

## 3. Interpreting Organic Mass Spectra

For organic compounds, mass spectrometry provides information about the molecular mass and structure of the compound, via the molecular ion and fragment peaks.

**Molecular ion ($M^+$)** — The whole organic molecule ionised after losing one electron. The m/z of this peak equals the relative molecular mass ($M_r$) of the compound.

*Notation:* M^+ peak at m/z = M_r

**Base peak** — The most abundant (tallest) peak in the mass spectrum, always assigned an abundance of 100%.

*Example:* Often the most stable fragment ion, e.g. the acylium ion at m/z 43 in aliphatic ketones

**Worked example:** A straight chain alkane has a molecular ion peak at m/z = 72. What is its molecular formula?

1. For +1 ions, the m/z of the molecular ion equals the relative molecular mass, so $M_r = 72$. The general formula for an alkane is $C_nH_{2n+2}$.
2. Substitute atomic masses (C = 12, H = 1) to solve for n:
3. $$12n + (2n + 2) = 72 \\ 14n = 70 \\ n = 5$$
4. The molecular formula is confirmed to be $C_5H_{12}$, as $12(5) + 12(1) = 72$.

> **info**
>
> The small M+1 peak (1 m/z higher than M+) comes from $^{13}$C, but it is rarely tested in CIE 9701 questions.

## 4. Identifying Common Fragment Ions

When the molecular ion breaks apart during ionisation, it forms stable fragment ions that give clues about the structure of the original molecule. Common fragments correspond to common functional groups or alkyl groups.

| Fragment | m/z value | Common origin |
| --- | --- | --- |
| CH$_3^+$ | 15 | Alkyl chains |
| C$_2$H$_5^+$ | 29 | Ethyl groups/alkanes |
| CO^+$ | 28 | Aldehydes/ketones |
| CH$_3$CO^+$ | 43 | Ethanoyl groups/ketones |
| C$_6$H$_5^+$ | 77 | Aromatic benzene rings |
| COOH^+$ | 45 | Carboxylic acids |

> **tip**
>
> Always check for peaks at $M_r - X$, where X is the mass of a common lost group. For example, $M_r - 15$ means loss of a CH$_3$ group.

## Common pitfalls

- **Wrong:** Writing the order of stages as Ionisation → Deflection → Acceleration → Detection
  - Why it fails: Ions must be accelerated to uniform kinetic energy before deflection can separate them by mass
  - Correct: Memorise the order: Ionisation → Acceleration → Deflection → Detection
- **Wrong:** Calculating $A_r$ as the simple average of isotopic masses, ignoring abundance
  - Why it fails: This does not account for different proportions of each isotope, leading to incorrect results
  - Correct: Always calculate the weighted average: multiply each mass by its percentage abundance, sum, then divide by 100
- **Wrong:** Assuming the tallest (base) peak is the molecular ion peak
  - Why it fails: The base peak is the most abundant fragment, not the whole molecular ion, so this gives the wrong $M_r$
  - Correct: The molecular ion is the highest significant m/z peak (excluding the small M+1 peak)
- **Wrong:** Dividing m/z by 2 for all ions to get mass
  - Why it fails: Almost all ions detected in standard mass spectrometry are singly charged (+1)
  - Correct: Only adjust for charge if the question explicitly states the ion has a +2 or higher charge

## Cheatsheet

| Concept | Key Fact |
| --- | --- |
| Stage order | Ionisation → Acceleration → Deflection → Detection |
| m/z for +1 ions | Equals the mass of the ion |
| Relative atomic mass | $A_r = \sum (m_i \times a_i) / 100$ |
| Molecular ion peak | Gives the relative molecular mass $M_r$ |
| Base peak | Most abundant peak = 100% abundance |
| CH$_3^+$ fragment | m/z 15, M-15 = CH$_3$ loss |
| C$_6$H$_5^+$ fragment | m/z 77, indicates benzene ring |

## What's next

Mass spectrometry is a core analytical technique that forms the foundation for advanced structural problem-solving in organic chemistry, a common extended question in CIE 9701 papers. You will combine mass spectrometry data with data from other spectroscopic techniques to identify unknown organic compounds, a key skill for Paper 2 and Paper 4. Understanding mass spectrometry also supports the study of isotopic labelling used to investigate reaction mechanisms. Mastering the skills here will make more advanced analytical topics much easier to tackle.

- [Infra-red Spectroscopy](https://www.owlsprep.com/study/cie-9701-u16-infra-red-spectroscopy/)
- [Chromatography basics](https://www.owlsprep.com/study/cie-9701-u16-chromatography-basics/)
- [Further chemical energetics](https://www.owlsprep.com/study/cie-9701-u17-overview/)

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