Mass Spectrometry

Spec Mapping — OCR H432 Module 2.1.2 — Relative masses and time-of-flight mass spectrometry, content statements covering the principles of time-of-flight (TOF) mass spectrometry — ionisation, acceleration, ion drift and detection — the interpretation of mass spectra to determine relative isotopic masses and abundances, the calculation of $A_r$Ar​ from spectra, and the identification of molecular ion and fragmentation peaks (refer to the official OCR H432 specification document for exact wording). This is the experimental anchor for the relative-mass framework of the previous lesson and the foundational analytical technique you will revisit in Module 6 (organic spectroscopy combined techniques).

Mass spectrometry is the experimental engine of A-Level chemistry's relative-mass framework. The previous lesson defined $A_r$Ar​ as a weighted mean across isotopes; mass spectrometry is the device that measures the very abundances on which that weighted mean is computed. The technique developed in the early twentieth century through the work of F. W. Aston at Cambridge (who built the first usable mass spectrograph in 1919 and discovered most of the non-radioactive isotopes between 1919 and 1940) and reached its modern instrumental form with the time-of-flight (TOF) variant pioneered in the 1940s. At A-Level, OCR expects you to describe the four stages of a TOF instrument qualitatively, derive the time-of-flight relation from $E_k = \tfrac{1}{2}mv^2$Ek​=21​mv2, and apply both to isotope spectra (giving $A_r$Ar​) and molecular spectra (giving $M_r$Mr​ and structural information through fragmentation). The wider analytical horizon — forensic toxicology, anti-doping, pharmaceutical purity testing, space probes such as the Cassini mission — illustrates the technique's reach but stays beyond the examination boundary.

Key Equation: For an ion of mass $m$m and charge $z = +1$z=+1 accelerated to kinetic energy $E_k$Ek​, the time of flight $t$t across a field-free drift region of length $L$L is $t = L\sqrt{\frac{m}{2E_k}}$t=L2Ek​m​​

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What is Mass Spectrometry?

Mass spectrometry is an analytical technique that measures the mass-to-charge ratio ($m/z$m/z) of ions. It allows chemists to identify isotopes and calculate $A_r$Ar​; determine the relative molecular mass of a compound directly from the molecular-ion peak; deduce structural information from fragmentation patterns; and detect trace amounts of unknown substances at sensitivities below $10^{-15}\text{ g}$10−15 g.

A mass spectrometer produces a mass spectrum: a plot of relative abundance ($y$y-axis) against ${m/z}$m/z ($x$x-axis). Because almost all ions formed in the conventional electron-impact source carry a single $+1$+1 charge, ${m/z}$m/z is effectively the ion's mass in atomic mass units.

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The Four Stages of Time-of-Flight (TOF) Mass Spectrometry

OCR expects you to be able to describe the four stages and connect them to the underlying physics.

flowchart LR
    A[Sample] --> B[1. Ionisation]
    B --> C[2. Acceleration]
    C --> D[3. Ion drift]
    D --> E[4. Detection]
    E --> F[Mass spectrum]

Inline TOF schematic

1. Ionisation

The sample is vaporised and ionised. Two methods are routinely seen at A-Level:

Electron impact (EI) ionisation — a high-energy electron beam (around 70 eV) bombards the vaporised sample, knocking an electron from each molecule:

$\text{X(g)} + e^- \rightarrow \text{X}^+\text{(g)} + 2e^-$X(g)+e−→X+(g)+2e−

EI is energetic and often causes fragmentation — useful for structural information but the molecular-ion peak can be weak.

Electrospray ionisation (ESI) — the sample is dissolved in a volatile polar solvent and forced through a fine needle at high voltage. Each droplet evaporates, leaving protonated molecules:

$\text{X(g)} + \text{H}^+ \rightarrow \text{XH}^+\text{(g)}$X(g)+H+→XH+(g)

ESI is a "soft" ionisation method causing little fragmentation, so it gives a clear molecular-ion peak. Note that the resulting ion is at $M + 1$M+1 because of the added proton — a discriminator move for top-band candidates is to flag this $+1$+1 offset when comparing EI and ESI spectra.

2. Acceleration

The positive ions are accelerated through a known potential difference towards the field-free drift region. Because the accelerating potential is the same for every ion, all $+1$+1 ions gain the same kinetic energy regardless of mass:

$E_k = \tfrac{1}{2}mv^2$Ek​=21​mv2

Rearranging for velocity:

$v = \sqrt{\frac{2E_k}{m}}$v=m2Ek​​​

Velocity is inversely proportional to $\sqrt{m}$m​ — lighter ions move faster than heavier ions carrying the same kinetic energy.

3. Ion drift

The accelerated ions pass through a field-free region of known length $L$L. With no electric field, each ion travels at the constant velocity gained during acceleration. The time of flight is:

$t = \frac{L}{v} = L\sqrt{\frac{m}{2E_k}}$t=vL​=L2Ek​m​​

Time of flight is proportional to $\sqrt{m}$m​. Measuring $t$t in nanoseconds determines $m$m with very high precision.

4. Detection

When an ion strikes the detector (typically an electron multiplier), it gains an electron, producing a tiny current that is amplified by a factor of $10^6$106 or more. The size of the current at each arrival time gives the relative abundance; the time of flight gives the $m/z$m/z value. A computer assembles this into the mass spectrum.

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Reading a Mass Spectrum

Each peak represents an ion of a particular ${m/z}$m/z. The $x$x-axis is the mass-to-charge ratio — effectively mass for $+1$+1 ions — and the $y$y-axis is relative abundance, either as a percentage of the base peak (tallest peak, conventionally assigned 100 %) or in arbitrary peak-height units.

Example — Chlorine (Cl₂)

Chlorine gas shows peaks at ${m/z} = 35,\ 37,\ 70,\ 72,\ 74$m/z=35, 37, 70, 72, 74. The 35 and 37 peaks are atomic ions $^{35}\text{Cl}^+$35Cl+ and $^{37}\text{Cl}^+$37Cl+ from fragmentation. The diatomic Cl₂⁺ peaks split by isotopologue combination:

• $70$70 = $^{35}\text{Cl}{-}^{35}\text{Cl}^+$35Cl−35Cl+
• $72$72 = $^{35}\text{Cl}{-}^{37}\text{Cl}^+$35Cl−37Cl+ (and the equivalent $^{37}{-}^{35}$37−35)
• $74$74 = $^{37}\text{Cl}{-}^{37}\text{Cl}^+$37Cl−37Cl+

With $^{35}\text{Cl}:{}^{37}\text{Cl} \approx 0.75 : 0.25$35Cl:37Cl≈0.75:0.25:

Combination	Probability	Relative
$^{35}{-}^{35}$35−35	$0.75 \times 0.75 = 0.5625$0.75×0.75=0.5625	9
$^{35}{-}^{37}$35−37 (× 2 orderings)	$2 \times 0.75 \times 0.25 = 0.375$2×0.75×0.25=0.375	6
$^{37}{-}^{37}$37−37	$0.25 \times 0.25 = 0.0625$0.25×0.25=0.0625	1

So Cl₂ shows a classic 9 : 6 : 1 triplet at 70 : 72 : 74.

Example — Bromine (Br₂)

Bromine has $^{79}\text{Br}:{}^{81}\text{Br} \approx 50 : 50$79Br:81Br≈50:50, so Br₂ shows peaks at $158, 160, 162$158,160,162 in the ratio 1 : 2 : 1. This 1:2:1 triplet is a diagnostic fingerprint for bromine-containing compounds and is the discriminator examiners use to test for "did the candidate recognise the bromine signature?".

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Calculating $A_r$Ar​ from a Mass Spectrum

Worked Example 1 — Neon

${m/z}$m/z	Relative abundance
20	90.9
21	0.3
22	8.8

$A_r = \frac{(20 \times 90.9) + (21 \times 0.3) + (22 \times 8.8)}{100.0} = \frac{1818.0 + 6.3 + 193.6}{100.0} = 20.2$Ar​=100.0(20×90.9)+(21×0.3)+(22×8.8)​=100.01818.0+6.3+193.6​=20.2

Worked Example 2 — Strontium (peak heights total 100 by coincidence)

${m/z}$m/z	Peak height
84	0.56
86	9.86
87	7.00
88	82.58

$A_r = \frac{(84)(0.56) + (86)(9.86) + (87)(7.00) + (88)(82.58)}{0.56 + 9.86 + 7.00 + 82.58} = \frac{8771.04}{100.00} = 87.7$Ar​=0.56+9.86+7.00+82.58(84)(0.56)+(86)(9.86)+(87)(7.00)+(88)(82.58)​=100.008771.04​=87.7

Note that even though the total happens to be 100 here, the disciplined approach is to always divide by the actual sum of peak heights.

Worked Example 3 — Rubidium (peak heights, total = 100 by coincidence)

Peak heights $72.2$72.2 at ${m/z} = 85$m/z=85 and $27.8$27.8 at ${m/z} = 87$m/z=87.

$A_r = \frac{(85)(72.2) + (87)(27.8)}{72.2 + 27.8} = \frac{6137.0 + 2418.6}{100.0} = 85.6$Ar​=72.2+27.8(85)(72.2)+(87)(27.8)​=100.06137.0+2418.6​=85.6

Worked Example 4 — Germanium (five isotopes)

${m/z}$m/z	Abundance / %
70	20.84
72	27.54
73	7.73
74	36.28
76	7.61

$A_r = \frac{(70)(20.84) + (72)(27.54) + (73)(7.73) + (74)(36.28) + (76)(7.61)}{100} = 72.7$Ar​=100(70)(20.84)+(72)(27.54)+(73)(7.73)+(74)(36.28)+(76)(7.61)​=72.7

Worked Example 4b — Inverse problem (gallium from peak heights)

A mass spectrum of an unknown alkali-metal-like element X gives two peaks: $^{69}\text{X}$69X with peak height 36.1 and $^{71}\text{X}$71X with peak height 19.4. Calculate $A_r$Ar​ and identify the element.

• Total peak height $= 36.1 + 19.4 = 55.5$=36.1+19.4=55.5
• $A_r = [(69)(36.1) + (71)(19.4)] / 55.5 = [2490.9 + 1377.4] / 55.5 = 3868.3 / 55.5 = 69.7$Ar​=[(69)(36.1)+(71)(19.4)]/55.5=[2490.9+1377.4]/55.5=3868.3/55.5=69.7
• Match against the periodic table: gallium ($A_r = 69.7$Ar​=69.7).

The discipline of dividing by the actual sum of peak heights (here 55.5, not 100) is the key trap. A candidate who assumes peak heights sum to 100 would compute $3868.3 / 100 = 38.7$3868.3/100=38.7 and conclude the element is — wrongly — close to potassium or argon.

Worked Example 4c — Mixed-isotope diatomic molecule (HCl)

The mass spectrum of HCl shows peaks at \{m/z} = 35$and$and${m/z} = 37$(atomic Cl⁺ from fragmentation), and at$(atomicCl+fromfragmentation),andat${m/z} = 36$and$and${m/z} = 38$(molecular HCl⁺). Using$(molecularHCl+).Using^{35}\text{Cl}:{}^{37}\text{Cl} \approx 0.75:0.25$and assuming H is monoisotopic ($andassumingHismonoisotopic(^{1}\text{H}$), predict the intensity ratio of the 36 and 38 peaks.

Molecular HCl⁺ is formed by direct ionisation of HCl without fragmentation. Each Cl atom in the molecule is either $^{35}\text{Cl}$35Cl (75 %) or $^{37}\text{Cl}$37Cl (25 %), and the H is always $^{1}\text{H}$1H. So:

• 36 = $^{1}\text{H}{-}^{35}\text{Cl}^+$1H−35Cl+, probability 0.75 → relative intensity 3
• 38 = $^{1}\text{H}{-}^{37}\text{Cl}^+$1H−37Cl+, probability 0.25 → relative intensity 1

The ratio is 3:1, mirroring the elemental abundance because H contributes no isotope splitting. This is the simplest possible diatomic-with-mixed-isotope test and a good warm-up before tackling Cl₂ (9:6:1) and Br₂ (1:2:1).

Worked Example 4d — Identifying an unknown bromine compound from spectrum

A compound contains C, H and Br. The mass spectrum shows a molecular-ion pair at \{m/z} = 108$and$and${m/z} = 110$ in ratio 1:1. What is the compound?

• The 1:1 doublet at $M^+$M+ and $M^+ + 2$M++2 is the diagnostic signature of one bromine atom in the molecule (because $^{79}\text{Br}$79Br and $^{81}\text{Br}$81Br are nearly 1:1).
• Removing the Br (80 average) from 108: $108 - 80 = 28$108−80=28 for the C/H residue. C₂H₄ has $M_r = 28$Mr​=28.
• The compound is therefore C₂H₄Br⁺ in the spectrum, suggesting bromoethane $\text{CH}_3\text{CH}_2\text{Br}$CH3​CH2​Br with $M_r = 108$Mr​=108 (using $^{79}\text{Br}$79Br) or 110 (using $^{81}\text{Br}$81Br).

Two-Br compounds give a 1:2:1 triplet; three-Br compounds give a 1:3:3:1 quartet (binomial expansion with $p = q = 0.5$p=q=0.5). The number of equally-spaced $M$M-related peaks reveals the number of bromine atoms.

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The Molecular Ion Peak

When an entire molecule is ionised without breaking, the resulting ion is the molecular ion $M^+$M+. The peak at the highest ${m/z}$m/z (excluding small $M+1$M+1 peaks from $^{13}\text{C}$13C) is the molecular-ion peak, and its ${m/z}$m/z value gives the relative molecular mass $M_r$Mr​ directly.

Worked Example 5 — Unknown alcohol

Molecular-ion peak at ${m/z} = 46$m/z=46. Ethanol $\text{CH}_3\text{CH}_2\text{OH}$CH3​CH2​OH has $M_r = 2(12) + 6(1) + 16 = 46$Mr​=2(12)+6(1)+16=46, so the compound is ethanol.

Worked Example 6 — Unknown hydrocarbon

A hydrocarbon shows $M^+$M+ at $72$72 and fragments at $57, 43, 29$57,43,29. For an alkane $\text{C}_n\text{H}_{2n+2}$Cn​H2n+2​: $14n + 2 = 72 \Rightarrow n = 5$14n+2=72⇒n=5. The compound is $\text{C}_5\text{H}_{12}$C5​H12​ (pentane or a branched isomer). The fragmentation losses $-15$−15 (CH₃) and $-29$−29 (C₂H₅) are diagnostic of alkane chain breaks.

Base peak vs molecular ion

The base peak is conventionally assigned 100 % abundance and serves as the reference for all other peaks in the spectrum. It is rarely the molecular ion in EI spectra because of fragmentation. The molecular ion is identified by:

• The highest \{m/z}$peak (ignoring small$peak(ignoringsmallM+1$contributions from$contributionsfrom^{13}\text{C}$)
• A small but distinct peak (usually <20 % of base peak in EI; often dominant in ESI)
• Confirmation via fragmentation pattern — the fragments should be consistent with loss of common neutrals (CH₃ = 15, OH = 17, H₂O = 18, etc.) from the proposed $M_r$Mr​