GATE Mass Spectrometry — Isotope Patterns, Fragments and Resolution

By Aniket Bhardwaj · 22 September 2026 · GATE Chemistry

Mass spectrometry rewards a very small amount of memorised arithmetic. If you know the natural isotope ratios of chlorine and bromine, the nitrogen rule, the 1.1 % carbon-13 abundance and the degrees-of-unsaturation formula, most GATE Chemistry questions on the topic become a two-line calculation. What follows is each recurring type, worked out in full — including exact-mass resolution and time-of-flight scaling, which are the two that students skip and then meet in the paper.

The quantities and what they mean

m/z = mass of the ion ÷ its charge number · Resolving power R = m / Δm

Term	Meaning
Molecular ion M^+•	Intact molecule minus one electron; a radical cation with an odd number of electrons
Base peak	The most intense peak, scaled to 100 % — not necessarily the molecular ion
Nominal mass	Sum of integer masses of the most abundant isotopes (C = 12, H = 1, N = 14, O = 16)
Exact (monoisotopic) mass	Sum of accurate isotope masses: ¹H = 1.00783, ¹²C = 12.00000, ¹⁴N = 14.00307, ¹⁶O = 15.99491
Average molar mass	Weighted over all isotopes — the number a molar-mass calculator gives, and it is not what the spectrum shows

That last row causes real confusion. For chlorobenzene the average molar mass is about 112.56 g mol⁻¹, but the spectrum shows discrete peaks at m/z 112 and 114 in a 3 : 1 ratio and nothing at all at 112.56. A mass spectrum resolves individual isotopologues; a balance weighs their average.

Type 1 — the nitrogen rule and degrees of unsaturation

DBE = C − H/2 + N/2 + 1 (halogens count as H; oxygen and sulfur are ignored)

The nitrogen rule: for a compound of C, H, N, O and halogens, an odd nominal molecular mass means an odd number of nitrogen atoms; an even nominal mass means zero or an even number.

An unknown gives M^+• at m/z 135. Elemental analysis suggests C₈H₉NO. Check both, and find the DBE.

Nominal mass = 8(12) + 9(1) + 1(14) + 1(16) = 96 + 9 + 14 + 16 = 135 ✓

135 is odd, and the formula has one nitrogen — odd. The nitrogen rule is satisfied.

DBE = 8 − 9/2 + 1/2 + 1 = 8 − 4.5 + 0.5 + 1 = 5

Five degrees: a benzene ring accounts for four (three C=C plus the ring), leaving one for a C=O. Acetanilide, CH₃CONHC₆H₅, fits exactly. Its spectrum shows a strong fragment at m/z 93 from loss of ketene (CH₂=C=O, 42 mass units): 135 − 42 = 93, the aniline radical cation C₆H₇N (72 + 7 + 14 = 93 ✓), and an acylium ion at m/z 43, CH₃CO⁺ (24 + 3 + 16 = 43 ✓).

Type 2 — halogen isotope patterns

Chlorine is roughly 75.8 % ³⁵Cl and 24.2 % ³⁷Cl, close enough to 3 : 1. Bromine is roughly 50.7 % ⁷⁹Br and 49.3 % ⁸¹Br, i.e. 1 : 1. Multiply the contributions of each halogen to get the pattern.

Halogens present	M : M+2 : M+4 : M+6	How it is obtained
1 Cl	3 : 1	Direct from abundances
2 Cl	9 : 6 : 1	(3 + 1)² expansion
3 Cl	27 : 27 : 9 : 1	(3 + 1)³ expansion
1 Br	1 : 1	Direct from abundances
2 Br	1 : 2 : 1	(1 + 1)² expansion
1 Cl + 1 Br	3 : 4 : 1	(3 + 1)(1 + 1) expansion

(a) A spectrum shows peaks at m/z 94 and 96 of almost equal height. What halogen is present, and what is the compound if the formula is CH₃X?

A 1 : 1 doublet separated by 2 mass units is the bromine signature. CH₃⁷⁹Br = 12 + 3 + 79 = 94, and CH₃⁸¹Br = 96 ✓. The compound is bromomethane.

(b) Peaks appear at m/z 162, 164 and 166 in the ratio 9 : 6 : 1. Interpret them.

9 : 6 : 1 is the two-chlorine pattern, so the molecular ion carrying two ³⁵Cl is the lowest of the three, m/z 162. Test the formula C₆H₄Cl₂O (2,4-dichlorophenol): 72 + 4 + 70 + 16 = 162 ✓. Even nominal mass, no nitrogen ✓. Counting halogens as hydrogens, DBE = 6 − (4 + 2)/2 + 1 = 6 − 3 + 1 = 4, exactly one benzene ring ✓.

Note what the isotope pattern did and did not tell you: it fixed the number of chlorines, and nothing else. Everything after that came from the mass balance.

Type 3 — counting carbons from the M+1 peak

Carbon-13 has an abundance of about 1.1 %. Each carbon in the molecule adds roughly 1.1 % to the M+1 intensity relative to M.

number of carbons ≈ [ intensity(M+1) / intensity(M) ] × 100 ÷ 1.1

M^+• has an intensity of 100 units and M+1 has 6.6 units. How many carbons?

Ratio = 6.6 / 100 = 6.6 %. Carbons = 6.6 / 1.1 = 6.

And if M+1 read 13.2 units? 13.2 / 1.1 = 12 carbons.

Treat this as an estimate, not a measurement: nitrogen (¹⁵N, about 0.37 %) and sulfur also contribute to M+1, so the raw number can come out slightly high. If the calculation gives 6.4, read it as 6.

Type 4 — the standard fragmentation routes

Route	What breaks	Diagnostic ion
α-cleavage (ketones)	C–C bond next to C=O	Acylium R–CO⁺
McLafferty rearrangement	γ-H transfers to the carbonyl oxygen, then β-cleavage	Even-mass enol radical cation
Benzylic cleavage	Bond β to the ring	m/z 91 (tropylium, C₇H₇⁺)
Retro Diels–Alder	Cyclohexene ring opens	Diene + dienophile fragments
Loss of small neutrals	—	−18 (H₂O), −28 (CO or C₂H₄), −44 (CO₂)

Pentan-2-one, CH₃COCH₂CH₂CH₃. Predict the main fragments.

Molecular ion: C₅H₁₀O = 60 + 10 + 16 = m/z 86 (even mass, no nitrogen ✓).

α-cleavage losing CH₃ (15): 86 − 15 = 71, the acylium CH₃CH₂CH₂CO⁺ = C₄H₇O = 48 + 7 + 16 = 71 ✓

α-cleavage losing C₃H₇ (43): 86 − 43 = 43, the acylium CH₃CO⁺ = 24 + 3 + 16 = 43 ✓ — usually the base peak.

McLafferty: pentan-2-one has a γ-hydrogen on C5, so the rearrangement can run. Identify the two pieces before doing any arithmetic. The charge is retained on the enol CH₂=C(OH)CH₃, C₃H₆O = 36 + 6 + 16 = 58; the neutral expelled is ethene, C₂H₄ = 28. Balance: 58 + 28 = 86 ✓, which is the molecular ion. Always close that loop — fragment plus neutral must return M exactly, and this single check kills most wrong options.

Note the parity: 71 and 43 are odd-mass even-electron cations from simple bond cleavage, while 58 is an even-mass odd-electron radical cation from a rearrangement. That parity check alone often identifies a McLafferty product in a multiple-choice list.

Type 5 — resolving power

What resolving power is needed to separate CO^+• from N₂^+•, both nominally m/z 28?

CO: 12.00000 + 15.99491 = 27.99491
N₂: 2 × 14.00307 = 28.00614
Δm = 28.00614 − 27.99491 = 0.01123

R = m / Δm = 27.995 / 0.01123 = 2.49 × 10³

So a modest high-resolution instrument distinguishes them, while a unit-resolution quadrupole reports a single peak at 28. This is exactly why exact-mass measurement can confirm a molecular formula that nominal mass alone leaves ambiguous.

Type 6 — analyser physics

Magnetic sector: m/z = B²r² / (2V) · Time of flight: t = L √( m / (2 z e V) ) , so t ∝ √(m/z)

(a) In a TOF analyser an ion of m/z 100 arrives after 20.0 µs. When does an ion of m/z 400 arrive, all else equal?

t ∝ √(m/z), so t = 20.0 × √(400/100) = 20.0 × √4 = 20.0 × 2 = 40.0 µs.

(b) In a magnetic sector at fixed r and V, the field B is doubled. Which m/z now reaches the detector?

m/z ∝ B², so doubling B multiplies the transmitted m/z by 2² = 4. Halving the accelerating voltage V instead would double it, since m/z ∝ 1/V.

Ionisation methods — which one the question is describing

Method	Character	Typical ion seen	Best for
Electron ionisation (EI)	Hard — rich fragmentation	M^+• plus many fragments; M^+• may be absent	Small volatile organics; library matching
Chemical ionisation (CI)	Soft	[M+H]⁺	Recovering a molecular mass EI destroyed
Electrospray (ESI)	Soft, from solution	[M+H]⁺, [M+Na]⁺, multiply charged	Polar and large biomolecules
MALDI	Soft, from a solid matrix	Mostly singly charged	Polymers and proteins

The soft methods are where the "+1" trap lives: an ESI peak at m/z 181 for a compound of nominal mass 180 is [M+H]⁺, not an M+1 isotope peak. Read the ionisation method before you read the spectrum.

Common mistakes

Assuming the base peak is the molecular ion. In EI the molecular ion is often weak or completely absent, especially for branched alkanes and alcohols.
Treating an ESI [M+H]⁺ as an isotope peak. Check the ionisation method; a protonated ion shifts everything by one unit.
Using average molar mass for m/z. Use nominal or exact isotope masses. Chlorine is 35 or 37 in a spectrum, never 35.45.
Confusing the Cl and Br patterns. 3 : 1 is chlorine; 1 : 1 is bromine. Getting this backwards changes the whole structure.
Forgetting the even-electron rule. Simple bond cleavage of an odd-electron M^+• gives an even-electron cation; only a rearrangement gives another odd-electron (even-mass, in the absence of N) species.
Not balancing fragment against neutral lost. Fragment mass + neutral mass must equal the molecular ion mass exactly. This one check catches most wrong answers.
Applying the nitrogen rule to a fragment. It is stated for molecular ions; fragments follow a related but different parity argument.
Quoting resolution as a mass. R = m/Δm is dimensionless.

Where this appears in GATE Chemistry

Question type	What is tested	Usual form
Molecular formula from M and DBE	Nitrogen rule + unsaturation count	Choose the formula
Halogen count from M/M+2	Isotope statistics	Multiple choice
Carbon count from M+1	1.1 % ¹³C abundance	Numerical answer
Fragment identification	α-cleavage, McLafferty, tropylium	Match m/z to a structure
Resolving power	Exact masses, R = m/Δm	Numerical answer
Analyser behaviour	TOF √m scaling, sector B and V dependence	Conceptual or numerical

The relative emphasis given to instrumental methods changes between syllabus revisions, so confirm the current scope from the official GATE notification for your exam year rather than assuming last year's pattern carries over.

Check your formula masses before you trust a fragment assignment. The Molar Mass & Composition tool sums a formula for you, so you can confirm that a proposed fragment and the neutral it lost really do add back to the molecular ion. One honest caveat, because it matters here: it returns the average molar mass over natural isotopes, so use it to verify formulas, and use the integer isotope masses in this article for the m/z values themselves.

Open the Molar Mass & Composition Calculator →

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