Mass Spectrometry

Mass spectrometry is a powerful analytical technique used to determine the relative atomic mass of elements, identify isotopes, and determine the molecular mass of compounds. At A-Level, you need to understand the time-of-flight (TOF) mass spectrometer in detail.

Spec mapping (AQA 7405): This lesson maps to §3.1.1.2 (mass spectrometry of elements and the operation of a time-of-flight mass spectrometer) and §3.3.6 / §3.3.15 (mass spectrometry of organic compounds, including the interpretation of molecular-ion and fragment-ion peaks). The mathematical link between kinetic energy, mass and time of flight is examined in the M0 mathematical-requirements section of the spec. Refer to the official AQA specification document for the exact wording of each section.

Assessment objectives in this lesson: The four-stage description of the TOF instrument is straight AO1; reading a spectrum to calculate Aᵣ or to identify an organic compound is AO2; questions that require you to distinguish ESI from EI for a specific sample, or to interpret an unusual fragmentation pattern, test AO3. The derivation of m ∝ t² from KE = ½mv² and v = d / t is a recurring AO2 maths-skills item.

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Principles of Mass Spectrometry

A mass spectrometer separates particles based on their mass-to-charge ratio (m/z). The key stages in a time-of-flight mass spectrometer are:

1. Ionisation
2. Acceleration
3. Ion drift (flight tube)
4. Detection

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Stage 1: Ionisation

The sample must be converted into gaseous ions. There are two main methods at A-Level:

Electrospray Ionisation (ESI)

• The sample is dissolved in a volatile solvent.
• The solution is passed through a fine hypodermic needle at high voltage.
• As the solvent evaporates, the sample molecules gain a proton (H⁺) from the solvent to form positive ions: M + H⁺ → MH⁺
• This is a "soft" ionisation technique — it causes minimal fragmentation.
• The ions formed typically have a charge of +1, so m/z ≈ M + 1.

Electron Impact Ionisation (EI)

• The sample is vaporised and then bombarded with high-energy electrons from an electron gun.
• An electron is knocked out of the sample molecule: M(g) → M⁺(g) + e⁻
• This is a "hard" ionisation technique — it can cause the molecule to fragment.
• The ion M⁺ is called the molecular ion (also called the parent ion).

Key Point: Electrospray ionisation produces [M+H]⁺ ions (molecular mass + 1), while electron impact produces M⁺ ions (molecular mass). You must account for this difference when reading spectra.

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Stage 2: Acceleration

The positive ions are accelerated by an electric field. All ions are given the same kinetic energy (KE).

The kinetic energy equation is:

KE = ½mv²

Since all ions have the same KE, lighter ions will have a greater velocity:

v = √(2KE / m)

This means ions with a smaller mass travel faster through the flight tube.

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Stage 3: Ion Drift (Flight Tube)

The ions enter a field-free region (the flight tube or drift region). There is no electric or magnetic field here, so the ions travel at constant velocity. Lighter ions reach the detector first because they are moving faster.

The time of flight is given by:

t = d / v

where d is the length of the flight tube and v is the velocity of the ion.

Substituting v = √(2KE / m):

t = d / √(2KE / m) = d × √(m / 2KE)

Therefore: t² = d²m / (2KE)

Rearranging: m = 2KE × t² / d²

Since KE and d are constants for a given experiment, the mass is proportional to the square of the time of flight: m ∝ t².

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Stage 4: Detection

When ions hit the detector, they generate an electrical signal. The size of the signal is proportional to the number of ions (abundance). The detector records both the time of flight (which gives m/z) and the abundance of each ion.

The data is displayed as a mass spectrum: a bar chart with m/z on the x-axis and relative abundance (or relative intensity) on the y-axis.

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Interpreting Mass Spectra of Elements

Reading a Mass Spectrum

Each peak in the mass spectrum of an element represents a different isotope. The position on the x-axis gives the mass-to-charge ratio (m/z), and the height gives the relative abundance.

For singly charged ions (z = 1), the m/z value equals the isotopic mass.

Worked Example 1: Calculating Aᵣ from a Mass Spectrum

A mass spectrum of neon shows:

• m/z = 20, relative abundance = 90.5
• m/z = 21, relative abundance = 0.3
• m/z = 22, relative abundance = 9.2

Calculate the relative atomic mass of neon.

Step 1: Multiply each mass by its abundance.

20 × 90.5 = 1810.0 21 × 0.3 = 6.3 22 × 9.2 = 202.4

Step 2: Sum the products.

1810.0 + 6.3 + 202.4 = 2018.7

Step 3: Divide by the total abundance.

Total abundance = 90.5 + 0.3 + 9.2 = 100.0

Aᵣ = 2018.7 / 100.0

Aᵣ = 20.2 (to 1 decimal place)

Worked Example 2: Calculating Aᵣ of Magnesium

A mass spectrum of magnesium shows:

• m/z = 24, relative abundance = 78.6
• m/z = 25, relative abundance = 10.1
• m/z = 26, relative abundance = 11.3

Aᵣ = (24 × 78.6 + 25 × 10.1 + 26 × 11.3) / (78.6 + 10.1 + 11.3)

Aᵣ = (1886.4 + 252.5 + 293.8) / 100.0

Aᵣ = 2432.7 / 100.0

Aᵣ = 24.3 (to 1 decimal place)

Exam Tip: If the relative abundances do not add up to 100, you must still divide by the total of the abundances, not by 100. Always check the sum first.

Worked Example 3: Abundances Not Summing to 100

A mass spectrum of copper shows:

• m/z = 63, relative abundance = 69.2
• m/z = 65, relative abundance = 30.8

Total abundance = 69.2 + 30.8 = 100.0

Aᵣ = (63 × 69.2 + 65 × 30.8) / 100.0

Aᵣ = (4359.6 + 2002.0) / 100.0

Aᵣ = 6361.6 / 100.0

Aᵣ = 63.6 (to 1 decimal place)

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Mass Spectra of Molecules

When electron impact ionisation is used on molecules, the molecular ion M⁺ may fragment into smaller ions. The resulting mass spectrum shows:

• The molecular ion peak (M⁺): The peak at the highest m/z value (ignoring the M+1 peak from ¹³C). This gives the relative molecular mass of the compound.
• Fragment ion peaks: Peaks at lower m/z values corresponding to pieces of the molecule.

The M+1 Peak

A small peak is often seen one mass unit above the molecular ion peak. This is due to molecules containing ¹³C (natural abundance 1.1%). For a molecule with n carbon atoms, the height of the M+1 peak relative to the M peak is approximately n × 1.1%.

Worked Example 4: Identifying the Molecular Ion Peak

A mass spectrum of a compound shows peaks at m/z = 15, 29, 43, and 58. What is the relative molecular mass?

The molecular ion peak is at m/z = 58 (the highest value), so Mᵣ = 58.

The fragment peaks can be interpreted:

• m/z = 43: loss of 15 (a CH₃ group, since 58 − 43 = 15)
• m/z = 29: loss of 29 from the molecular ion (a CHO group, since 58 − 29 = 29)
• m/z = 15: a CH₃⁺ fragment

This pattern is consistent with propanone (CH₃COCH₃, Mᵣ = 58) or butanal (CH₃CH₂CH₂CHO, Mᵣ = 72... no, that would be 72). Actually, a compound with Mᵣ = 58 losing CHO (29) would leave 29, and losing CH₃ (15) would leave 43. The compound could be propanal (CH₃CH₂CHO, Mᵣ = 58).

Exam Tip: The molecular ion peak gives you the Mᵣ directly. Common fragment losses to remember: 15 = CH₃, 17 = OH, 29 = CHO or C₂H₅, 45 = OC₂H₅, 77 = C₆H₅ (phenyl).

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Using Mass Spectrometry to Determine Isotopic Abundance

You can also work backwards from a known Aᵣ to find the percentage abundance of isotopes.

Worked Example 5: Finding Percentage Abundance

Bromine has two isotopes: ⁷⁹Br and ⁸¹Br. The relative atomic mass of bromine is 79.9. Calculate the percentage abundance of each isotope.

Let the percentage abundance of ⁷⁹Br = x. Then the abundance of ⁸¹Br = (100 − x).

Aᵣ = (79x + 81(100 − x)) / 100 = 79.9

79x + 8100 − 81x = 7990

−2x = 7990 − 8100 = −110

x = 55

Therefore: ⁷⁹Br = 55% and ⁸¹Br = 45%.

Check: (79 × 55 + 81 × 45) / 100 = (4345 + 3645) / 100 = 7990 / 100 = 79.9 ✓

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Synoptic Links

Mass spectrometry sits at the centre of multiple A-Level Chemistry topics:

• atomic-models-subatomic-particles (previous lesson, §3.1.1.2): the relative atomic mass calculations you saw in Lesson 1 are derived directly from mass-spectrometric data. This lesson explains how the instrument produces those numbers.
• organic-foundations (§3.3.6) and analytical-techniques (§3.3.15): mass spectrometry of organic compounds — interpreting the molecular ion peak, deducing Mᵣ, and using fragmentation patterns to identify a compound — is examined alongside infrared and NMR spectroscopy in the analytical-techniques course.
• inorganic / transition-metals (§3.2.5): mass-spectrometric identification of transition-metal isotopes is the historical foundation of natural-abundance Aᵣ values used throughout inorganic chemistry.
• chromatography (§3.3.16): gas chromatography–mass spectrometry (GC-MS) couples chromatographic separation to mass-spectrometric identification — the workhorse technique in modern analytical labs. The same physical principles that operate in the TOF mass spectrometer here are extended in the GC-MS lessons.
• ionisation-energies (§3.1.1.3): the energy required to remove an electron in EI ionisation is conceptually identical to the first ionisation energy of an atom. The mass spectrum of a noble gas at high resolution can be used to estimate its first ionisation energy directly.

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Specimen Exam Question — modelled on the AQA A-Level Chemistry paper format

Question 1. [12 marks total]

(a) State the four main stages of a time-of-flight mass spectrometer and briefly describe what happens at each. [4 marks]

(b) A mass spectrum of an element Y shows two peaks:

m/z	Relative abundance / %
79	50.7
81	49.3

Calculate the relative atomic mass of Y to one decimal place and identify the element. [3 marks]

(c) A pharmaceutical compound is analysed by electrospray ionisation mass spectrometry. A single peak is observed at m/z = 187. Explain why this corresponds to a relative molecular mass of 186, not 187. [2 marks]

(d) A mass spectrum of an unknown organic compound shows peaks at m/z = 15, 29, 43, 58 (largest peak), and 59 (small peak). Identify the molecular ion peak, deduce the relative molecular mass, suggest the identity of the compound, and explain the origin of the m/z = 43 peak. [3 marks]

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Mark Scheme — Specimen Question 1 (modelled, not a real paper)

(a) Four stages of TOF mass spectrometry [4 marks, AO1]

Award 1 mark per stage with a correct accompanying description:

• Ionisation: atoms or molecules are converted to positive ions, by electrospray ionisation (sample dissolved, sprayed through high-voltage needle, gains H⁺) or by electron impact ionisation (sample vaporised, bombarded with high-energy electrons, an electron is knocked off).
• Acceleration: positive ions accelerated by an electric field; all given the same kinetic energy.
• Ion drift (flight tube): ions travel through a field-free region; lighter ions travel faster and reach the detector first.
• Detection: ions hit the detector producing an electrical signal; signal magnitude is proportional to ion abundance; data displayed as a bar chart of relative abundance against m/z.

(b) Relative atomic mass of Y and identification [3 marks, AO2]

• 1 mark: weighted-mean expression Aᵣ = (79 × 50.7 + 81 × 49.3) / 100.
• 1 mark: correct arithmetic — (4005.3 + 3993.3) / 100 = 7998.6 / 100 = 79.99 ≈ 80.0 (to 1 d.p.). Alternatively the answer 79.99 to 2 d.p. or rounded 80.0 to 1 d.p. should be accepted.
• 1 mark: identification — bromine (Br), with reference to the periodic-table Aᵣ of 79.9 or 80.0.

Common mistake: rounding 79.99 to 80 and identifying the wrong element. The periodic-table Aᵣ of bromine is 79.9 (standard value 79.904) — accept this with appropriate tolerance.

(c) ESI gives Mᵣ = 186 not 187 [2 marks, AO2 + AO3]

• 1 mark: in electrospray ionisation, the compound M gains a proton (H⁺) to form the [M+H]⁺ ion. The m/z value therefore equals Mᵣ + 1.
• 1 mark: subtracting 1 from m/z = 187 gives Mᵣ = 186.

Accept either order of reasoning. Do not award the second mark unless the calculation Mᵣ = m/z − 1 is explicit.