Mass Spectrometry: Ionization and the Molecular Ion

EI-MS coupled to gas chromatography (GC-MS) is the backbone of forensic toxicology and environmental analysis: a molecular ion peak at a specific $m/z$, cross-referenced against enormous spectral libraries (the NIST/EPA/NIH library holds hundreds of thousands of reference EI spectra), identifies unknown compounds in blood, urine, soil, or air in minutes. Pharmaceutical quality control leans on the same logic to confirm both the identity and the purity of a drug substance batch: an unexpected extra molecular ion peak signals a synthetic byproduct or degradation product before it ever reaches a patient. Doping-control laboratories detect anabolic steroids and their metabolites by exactly this kind of molecular-ion-plus-fragment-pattern fingerprinting, often after chemical derivatization designed to shift the molecular ion into a more diagnostic, more stable, more easily ionized mass region. Petroleum and flavor/fragrance chemistry both rely on EI-MS libraries built up over decades to instantly classify hydrocarbon mixtures and volatile natural-product components by their characteristic molecular ions and fragmentation fingerprints. In modern proteomics and metabolomics, softer ionization methods (electrospray, MALDI) have largely displaced classical EI for very large or fragile biomolecules, but the conceptual backbone — a single peak whose $m/z$ reveals mass, checked against parity and isotope logic before being trusted — remains identical; those softer methods are natural extensions once the reasoning built here for the molecular ion is secure.

Ask the tutor: How do I use the isotope pattern accompanying the molecular ion peak to figure out how many chlorine, bromine, or sulfur atoms a compound contains?

A mass spectrometer never weighs a molecule directly. It takes a molecule, strips off exactly one electron in the gas phase to make a charged species, accelerates that charged species through a known electric field, and then sorts the resulting ions by their mass-to-charge ratio as they fly (or curve, or oscillate, depending on the analyzer) through vacuum. Whatever the analyzer — magnetic sector, quadrupole, time-of-flight — the fundamental quantity extracted is always

where $m$ is the ion's mass (in daltons, atomic mass units) and $z$ is the number of elementary charges it carries. In electron-impact ionization the overwhelming majority of ions are singly charged, $z=1$, so for practical purposes in this tier,

This is why chemists speak casually of "the peak at $m/z = 86$" and mean, without further qualification, "an ion of mass 86 daltons."

How the ionization event happens. The sample is vaporized into a high-vacuum ion source (typically $10^{-6}\text{–}10^{-5}$ torr, so that molecules travel in isolation without colliding). A filament emits electrons that are accelerated through a potential of about 70 volts, giving each electron a kinetic energy of about 70 electron-volts (eV) — roughly $6700\ \text{kJ/mol}$, immensely larger than any bond dissociation energy (typical C–C or C–H bonds: $300$–$420\ \text{kJ/mol}$). This 70 eV beam is fired across the ionization chamber, and a molecule that happens to sit in its path is struck. The mechanism of ionization is not a bond-forming/bond-breaking curved-arrow event in the usual organic sense — it is an ejection. A single incident electron transfers enough energy to the molecule to knock one of the molecule's own electrons out of its orbital entirely, ejecting it into the vacuum as a free (low-energy) electron. This is written schematically as

The molecule $\mathrm{M}$, which had all electrons paired, loses one electron and becomes $\mathrm{M}^{+\bullet}$: a species bearing both a positive charge (one fewer electron than protons) and an unpaired electron (one electron short of a closed shell). This is a radical cation — simultaneously a radical and a cation, denoted with the dot-plus superscript. It is not the same species as a carbocation formed by heterolysis (which has no unpaired electron), nor a radical formed by homolysis (which has no charge). The molecular ion is chemically unique to this ionization method.

Which electron leaves? By an argument closely related to Koopmans' theorem, the electron most easily removed is the one in the highest-occupied molecular orbital (HOMO), because that is the electron least tightly bound (smallest ionization energy). Qualitatively, for organic molecules the ranking of "easiest to remove" is

This ordering explains a huge amount of EI-MS behavior at a glance. An amine or ether ionizes preferentially by removing a nitrogen or oxygen lone-pair electron; the positive charge and radical then sit substantially localized on that heteroatom, resonance-stabilized into the adjacent framework. An alkene or arene ionizes by removing a π electron, leaving the charge/radical delocalized over the π system — which is part of why aromatic and conjugated molecules tend to give strong, long-lived molecular ions. A simple saturated alkane has only σ electrons available, and ionizing a C–C or C–H σ bond gives a much less stabilized radical cation that fragments extensively, often leaving little or no observable $\mathrm{M}^{+\bullet}$.

Excess energy and fragmentation. The 70 eV beam energy vastly exceeds the roughly 7–15 eV needed to ionize almost any organic molecule. The leftover energy — many electron-volts — is deposited as vibrational excitation in the newly formed $\mathrm{M}^{+\bullet}$. A vibrationally hot radical cation, sitting in a high vacuum with no solvent or collision partner to carry away that energy, has only one way to relax: fall apart. It does so on a timescale of roughly $10^{-10}$ to $10^{-6}$ seconds inside the source and flight tube, breaking a bond (usually the weakest one, or the one whose cleavage gives the most stable fragment cation) to produce a smaller cation plus a neutral fragment, or occasionally two radicals plus a neutral. These lighter cations are what populate the rest of the spectrum below $M$; their systematic interpretation is the subject of fragmentation mechanisms elsewhere. Here, the essential point is binary: some fraction of the $\mathrm{M}^{+\bullet}$ population survives all the way to the detector without fragmenting, and it is that surviving population that produces the molecular ion peak.

Recognizing the molecular ion peak. In a real spectrum, the peak of highest $m/z$ is usually $\mathrm{M}^{+\bullet}$, but three complications require care. First, natural isotopes ($^{13}\mathrm{C}$, $^2\mathrm{H}$, $^{15}\mathrm{N}$, $^{18}\mathrm{O}$, and heavier halogen isotopes) put small satellite peaks one or two mass units above the all-light-isotope peak; the true $\mathrm{M}^{+\bullet}$ is the lowest-mass member of this isotope cluster, not necessarily the tallest peak in the cluster. Second, ion–molecule collisions in a poorly tuned or overloaded source can generate a spurious $[\mathrm{M}+1]^+$ peak by proton transfer, which must not be mistaken for the true molecular ion. Third — and most important for this tier — the molecular ion itself may be so unstable that it fragments essentially completely, leaving no detectable peak at $M$ at all; tertiary alcohols and highly branched alkanes are classic offenders. In such cases the chemist must use milder ionization (chemical ionization, CI) to coax a detectable $[\mathrm{M}+1]^+$ or $[\mathrm{M}-1]^+$ pseudo-molecular-ion peak, or must infer $M$ indirectly from a self-consistent fragmentation pattern. The chemical sanity checks for "is this peak really $\mathrm{M}^{+\bullet}$?" are: 1. It must be an odd-electron species. Losing a closed-shell neutral fragment (H₂O, CO, C₂H₄, HCl, CH₃•, …) from $\mathrm{M}^{+\bullet}$ gives another odd-electron or even-electron cation consistent with radical/molecule bookkeeping; a proposed $M$ peak that cannot lose any chemically sensible neutral fragment to explain the very next peak down is suspect. 2. Reasonable mass losses. The gap between the candidate $M$ peak and the next peak down should correspond to a plausible neutral loss. Losses of 3–14 mass units (except 12, 13 CH-type oddities) are chemically implausible and usually signal that the candidate peak is not $\mathrm{M}^{+\bullet}$ but a fragment or noise. 3. The nitrogen rule (derived below) — a parity check that either confirms or rules out a candidate $M$ instantly.

The molecular ion's mass, read straight off the spectrum, gives the molecular mass:

where:

• $M$ — the nominal molecular mass of the neutral molecule, in daltons (sum of the most abundant isotope masses: $^{12}\mathrm{C}=12$, $^1\mathrm{H}=1$, $^{16}\mathrm{O}=16$, $^{14}\mathrm{N}=14$, $^{32}\mathrm{S}=32$, $^{35}\mathrm{Cl}=35$, etc.)
• $(m/z)_{\mathrm{M}^{+\bullet}}$ — the mass-to-charge ratio of the lowest-mass peak in the molecular-ion isotope cluster, read directly from the spectrum, with $z=1$ assumed for EI.

Reading: because ionization only removes an electron of negligible mass ($m_e \approx 0.00055\ \mathrm{u}$, invisible at nominal-mass resolution), the mass of $\mathrm{M}^{+\bullet}$ equals the mass of the neutral $\mathrm{M}$ to within the precision used at this level. The molecular ion peak is therefore a direct readout of molecular weight — no calibration curve, no comparison standard needed, just the position of one peak. The second core relationship is the nitrogen rule, a parity statement:

Equivalently, $M$ even $\iff$ nitrogen count even (including zero). Reading: this is not an empirical curiosity but a direct consequence of nitrogen's valence arithmetic, derived rigorously below.

Derivation 1 — why $m/z$ of the molecular ion gives molecular mass

Goal: show explicitly that the mass measured for $\mathrm{M}^{+\bullet}$ differs from the neutral molecular mass $M$ by an amount too small to matter at the resolution used to read nominal masses, and quantify that difference for a real molecule.

Step 1 — mass balance on ionization. The ionization event is

Conservation of mass (nonrelativistic; kinetic energies here are far too small to invoke $E=mc^2$ corrections) requires

so that

Step 2 — size of the correction. The electron mass is

For essentially any organic molecule of interest (mass tens to thousands of daltons), this is a correction of order $10^{-4}$ to $10^{-6}$ relative to $M$ — utterly negligible at the nominal (integer) mass resolution used throughout this tier, and still negligible at typical unit-resolution or even most high-resolution work at the level of a few parts per million unless one is doing exact-mass formula assignment (a job for high-resolution MS, not treated here).

Step 3 — conclude the working equation. To the precision relevant for reading integer $m/z$ values off a spectrum,

Hence the $m/z$ value at which the molecular ion peak sits is the molecular weight of the compound. This is why "find $M$" and "find the molecular ion peak" are, for practical purposes, the same instruction.

Step 4 — worked numeric check with an asymmetric, real molecule. Take butan-2-one (methyl ethyl ketone), $\mathrm{CH_3COCH_2CH_3}$, molecular formula $\mathrm{C_4H_8O}$. Using nominal atomic masses $\mathrm{C}=12.000$, $\mathrm{H}=1.008$, $\mathrm{O}=15.999$ (or nominal integers $12,1,16$ for the mass a low-resolution EI spectrum reports):

An EI spectrum of butan-2-one indeed shows its molecular ion at $m/z = 72$. Correcting for the ejected electron,

a difference of about six parts in a hundred thousand — invisible on any unit-resolution instrument and irrelevant to the integer value $72$ that the chemist actually reads and uses.

Step 5 — limiting case. As the molecule gets arbitrarily large ($M \to \infty$, e.g. a small protein at $M \sim 15{,}000$), the absolute correction from the missing electron stays fixed at $\sim 0.00055$ u while the relative error shrinks further, confirming that the neglect of electron mass is always safe at nominal-mass resolution and becomes progressively safer, not less safe, for larger molecules.

Derivation 2 — the nitrogen rule

Goal: derive, from valence and mass-parity bookkeeping alone, why the parity of the nitrogen count in a neutral organic molecule fixes the parity of its nominal molecular mass.

Step 1 — set up the counting problem. Consider a neutral, closed-shell organic molecule built from atoms of C, H, N, O, S, and halogens (X = F, Cl, Br, I) — the common elements of undergraduate organic structure work. Each atom contributes an integer nominal mass and has a fixed valence (number of bonds it forms, ignoring charge-bearing exceptions for the moment):

Notice the pattern that drives the whole rule: for every element except nitrogen, valence and nominal mass have the same parity — H (valence 1, mass 1: both odd), C (valence 4, mass 12: both even), O (valence 2, mass 16: both even), S (valence 2, mass 32: both even), halogens (valence 1, mass odd: both odd). Nitrogen is the exception: valence 3 (odd) but mass 14 (even) — an odd valence paired with an even mass. This single mismatch is the entire origin of the nitrogen rule.

Step 2 — total mass versus total valence, summed over the whole molecule. For a valid closed-shell neutral molecule with $c$ carbons, $h$ hydrogens, $n$ nitrogens, $o$ oxygens, $s$ sulfurs, and $x$ halogens, the nominal mass is

Every coefficient except the ones multiplying $h$ and $n$ is even (12, 16, 32, and each halogen mass — 19, 35, 79, 127 — is odd, matching that halogens are odd-valence too; but each halogen appears alongside exactly one bond, so we track halogens together with $h$ below). Group the odd-valence atoms (H, N, and monovalent halogens X) as the ones whose parity needs tracking, since even-valence atoms (C, O, S) always contribute an even mass regardless of how many there are:

Since $14$ is even, $14n \equiv 0 \pmod 2$ regardless of $n$ — mass-wise nitrogen looks "invisible" mod 2. But halogens are odd in mass, so each halogen contributes $1 \pmod 2$; write $x$ for the total halogen count. So

Step 3 — bring in the valence (degree) sum rule. In any neutral, fully-bonded organic structure (no radicals, no formal charges), the sum of all valences over all atoms must be even, because every bond is shared between two atoms and contributes $2$ to the total valence count (one unit to each partner):

Write out that sum using the valences from the table: each C contributes 4 (even), each O contributes 2 (even), each S contributes 2 (even) — all irrelevant mod 2 — while each H contributes 1, each halogen contributes 1, and each N contributes 3 (odd). So

(since $3n \equiv n \pmod 2$). Rearranged,

Step 4 — combine Steps 2 and 3. Step 2 gave $M \equiv h+x \pmod 2$; Step 3 gave $h + x \equiv n \pmod 2$. Chaining these,

This is the nitrogen rule in its entirety: the nominal molecular mass has the same parity as the number of nitrogen atoms. If $n$ is even (0, 2, 4, …), $M$ is even; if $n$ is odd (1, 3, 5, …), $M$ is odd.

Step 5 — why this is really a "hidden hydrogen" bookkeeping trick in disguise. The clean way to see it physically: every bond you could imaginatively "cut" to isolate one atom's valence requires a partner electron/atom on the other end, and only odd-valence atoms (H, N, halogens) can appear an "un-paired" number of times without violating the closed-shell requirement — but N does so while contributing an even mass number, silently absorbing an odd count of bonding partners (H atoms, typically, since amines and amides load extra H's around N) without disturbing the mass parity the way a genuinely odd-mass odd-valence atom (H, halogen) would. This is exactly the "nitrogen anomaly:" the only common element whose valence parity and mass parity disagree.

Step 6 — worked numeric check with an asymmetric molecule containing more than one nitrogen and a halogen, to stress-test the rule. Take procaine hydrochloride's free base, procaine, $\mathrm{C_{13}H_{20}N_2O_2}$ (a local anesthetic; two nitrogens — one aromatic amine, one tertiary amine in the side chain). Compute:

$M = 236$ is even, and $n=2$ is even — consistent with the rule. Now perturb it to a genuinely asymmetric odd-nitrogen, halogen-bearing test case: chlorpheniramine, $\mathrm{C_{16}H_{19}ClN_2}$ — wait, that has two nitrogens (even), so instead take a single-nitrogen halogenated amine, 4-chloroaniline, $\mathrm{C_6H_6ClN}$:

$M = 127$ is odd, and $n=1$ is odd — again consistent. As a genuinely different-flavored check with three nitrogens, take melamine, $\mathrm{C_3H_6N_6}$:

even, with $n=6$ (even) — consistent. And with one nitrogen and no halogen, pyridine, $\mathrm{C_5H_5N}$:

odd, with $n=1$ (odd) — consistent. Every one of these independent, structurally unrelated, asymmetric test molecules obeys $M \equiv n \pmod 2$ exactly as Step 4 predicts.

Step 7 — limiting case. Set $n=0$ (no nitrogen at all): the rule then says $M$ must be even for any ordinary CHO(S)(halogen) organic molecule with no nitrogen — check against glucose, $\mathrm{C_6H_{12}O_6}$: $M = 72+12+96=180$, even, consistent, since $n=0$ is even. This is the "trivial" limiting case built into the rule and it recovers a fact organic chemists use constantly without naming it: ordinary hydrocarbons, alcohols, ketones, ethers, esters, and sugars — anything with zero nitrogen — always have even nominal molecular weight.

Example 1 — Foundational: reading M and a formula candidate from a clean EI spectrum. An EI-MS of a neutral, colorless liquid shows its highest chemically sensible peak at $m/z = 88$, with a very weak, chemically consistent $[\mathrm{M}+1]$ satellite at $m/z=89$ (about 4–5% the height of the $88$ peak — consistent with natural $^{13}\mathrm{C}$ abundance for a handful of carbons, not with a distinct compound). We assign

By the nitrogen rule, $M=88$ is even, so this compound contains an even number of nitrogens — most simply, zero. Suppose independent evidence (IR: strong C=O stretch near 1735 cm⁻¹, no N–H or O–H stretch; a companion technique) suggests an ester. A reasonable formula with $M=88$ and no nitrogen is $\mathrm{C_4H_8O_2}$:

consistent with, e.g., ethyl acetate or methyl propanoate — both common $C_4H_8O_2$ esters, distinguished later by fragmentation patterns, not by $M$ alone. The key skill exercised here is entirely arithmetic-plus-parity: one peak position, one parity check, and a short list of candidate formulas collapses dramatically.

Example 2 — Applied: catching a deceptively small molecular ion peak in a real drug fragment. An EI spectrum of a suspected metabolite shows a tall base peak at $m/z=58$, a moderate peak at $m/z=91$, and only a very faint peak at $m/z=135$, so faint a careless reader might miss it and instead (wrongly) call $91$ the molecular ion. Applying the sanity checks: is the gap from $135$ down to $91$ chemically plausible? $135-91 = 44$, a plausible loss of $\mathrm{CO_2}$ or of $\mathrm{C_3H_8O}$-type neutral, or — more diagnostically — is the gap down to the base peak plausible? $135 - 58 = 77$, not an obviously clean single neutral loss, but $91-58=33$ is unusual too, suggesting $91$ is itself already a fragment, not $M$. Apply the nitrogen rule as the tie-breaker: is $135$ or $91$ the more sensible candidate for $M$? If we hypothesize the molecule is an amine, expect an odd $M$ if it carries one nitrogen. $135$ is odd — consistent with one nitrogen. $91$ is also odd, so parity alone can't fully discriminate, but $91$ corresponds suspiciously exactly to the mass of the extremely common, extremely stable tropylium cation $\mathrm{C_7H_7^+}$ ($7\times12 + 7\times1 = 84+7=91$), a famous fragmentation product of essentially any compound with a benzylic C–C bond (arising from loss of a radical to give an aromatic, 6π-electron, fully conjugated cation — a fragmentation covered in full elsewhere but worth flagging here because its characteristic mass, 91, is one of the most notorious "fake molecular ion" traps in the whole subject). Recognizing $91$ as tropylium, not $M$, forces the correct assignment

odd, one nitrogen, consistent with an aromatic amine bearing a benzylic substituent — for instance a molecule like $N$-methyl-$\alpha$-methylbenzylamine-type fragment ($\mathrm{C_9H_{13}N}$: $9(12)+13(1)+14 = 108+13+14=135\checkmark$). This example is deliberately built to show why "highest visible peak" and "true $\mathrm{M}^{+\bullet}$" are not automatically the same peak, and why chemical reasoning about both neutral-loss plausibility and nitrogen parity must be combined, not used in isolation.

Example 3 — Exam/competition level: combining the nitrogen rule with a formula search under an isotope constraint. A compound of unknown structure shows $\mathrm{M}^{+\bullet}$ at $m/z = 151$ (confirmed as the true molecular ion by a clean, chemically sensible loss of 15 mass units — a methyl radical — down to a strong peak at $m/z=136$, i.e. $\mathrm{M-CH_3}$, a classic and diagnostic loss). Elemental analysis independently establishes the empirical composition contains carbon, hydrogen, one nitrogen, and one oxygen, with no halogens or sulfur. Determine the molecular formula. Step 1 — apply the nitrogen rule as a consistency check first. $M=151$ is odd. One nitrogen (odd count) is exactly what makes $M$ odd, per $M \equiv n \pmod 2$ with $n=1$. This is consistent and does not by itself pin the formula, but it rules out any candidate formula with an even nitrogen count (e.g., accidentally proposing 0 or 2 N would immediately fail the parity check and can be discarded without further arithmetic). Step 2 — set up the mass equation. Let the formula be $\mathrm{C_cH_hN_1O_1}$ (one N, one O, fixed by the elemental analysis):

Step 3 — apply the degree-of-unsaturation / valence constraint to narrow $(c,h)$. For a molecule $\mathrm{C_cH_hN_1O_1}$, the maximum possible hydrogen count (fully saturated, acyclic) is given by the standard valence formula

(the $+n$ correction because a trivalent N can carry one extra H relative to a tetravalent-C skeleton count; oxygen, being divalent, doesn't change the H count). We need $h \le 2c+3$ and $h\ge$ some small positive number, and $12c+h=121$ with $h>0$ forces

Try $c=9$: $h = 121-108=13$; check $h_{\max}=2(9)+3=21 \ge 13$ ✓, and the degree of unsaturation

is a whole, non-negative number — exactly right for, e.g., one aromatic ring (4 degrees: ring + 3 π bonds) which is also completely consistent with a facile loss of $\mathrm{CH_3}$ (methyl loss from an aromatic methyl ketone-oxime or a methoxy/methyl-substituted aromatic amine is common and stabilized by the resulting aromatic cation). So the formula $\mathrm{C_9H_{13}NO}$ is fully consistent:

Step 4 — sanity-check alternative $c$ values are rejected. Try $c=10$: $h=121-120=1$, giving $\mathrm{DoU}=\frac{20+2+1-1}{2}=11$ for only 10 carbons and 1 nitrogen — implausibly unsaturation-dense (would require more π-bonds/rings than there are heavy atoms to support sensibly) and is discarded as chemically unreasonable at this level. Try $c=8$: $h=121-96=25 > h_{\max}=2(8)+3=19$ — violates the valence bound outright, impossible. So $c=9$, $h=13$ is the unique sensible solution, giving $\mathrm{C_9H_{13}NO}$, $\mathrm{DoU}=4$ — a formula matching, for instance, a methoxy-substituted phenethylamine-type structure, whose facile loss of a methyl radical (from the aromatic methoxy group, stabilizing the resulting oxocarbenium-like aromatic cation) explains the clean $151\to136$ transition observed experimentally.

Aston's mass spectrograph, built at the Cavendish Laboratory under J. J. Thomson's influence (Thomson had already shown that "positive rays" — early ion beams — could be deflected by electric and magnetic fields to sort by charge-to-mass ratio), used combined electric and magnetic sector fields to focus ions of the same $m/z$ regardless of their initial velocity spread — the key instrumental innovation ("velocity focusing") that gave dramatically sharper mass resolution than Thomson's parabola method. With it, Aston identified the isotopes of neon (²⁰Ne and ²²Ne, resolving a decades-old puzzle about neon's non-integer average atomic weight), then went on to characterize the isotopic composition of most of the periodic table, work recognized with the 1922 Nobel Prize in Chemistry. The organic chemist's EI-MS instrument, though built for very different molecules and questions than Aston's atomic-mass work, is a direct descendant of his sector-field, velocity-focusing design philosophy — and the entire practice of trusting an $m/z$ peak position as a faithful readout of mass rests on the resolving power his instruments first achieved.