Mass spectrometry
How chemists weigh molecules, and the pieces they break into, to determine molecular formulas and structures. Mass spectrometry answers a different question from IR: how heavy is this molecule, and what is its molecular formula?
Why MSThe instrumentAnatomyMolecular formulaFragmentationStep by stepExamplesPractice
01Why do we use mass spectrometry?
IR tells you which functional groups are present; NMR tells you how the carbon and hydrogen skeleton is arranged. Mass spectrometry supplies the piece neither can: the molecular mass, and from it the molecular formula. It is also extraordinarily sensitive; nanograms of sample are enough, which is why it is the method of choice for trace analysis in forensics, drug testing, and environmental work. As a bonus, the way a molecule breaks apart inside the instrument is itself structural evidence.
02How the instrument works
Every mass spectrometer, whatever its design, performs the same three operations in sequence. The classic teaching version uses electron ionization (EI), and the whole process happens under high vacuum so that ions travel without colliding with air.
Why m/z is effectively just mass. Almost every ion produced by EI carries a single positive charge, so z = 1 and the m/z value is numerically equal to the ion’s mass. Neutral molecules and neutral fragments are invisible; only charged species are detected, which is worth remembering when you interpret fragmentation.
03Anatomy of a mass spectrum
A mass spectrum is a line (bar) graph rather than a curve. Three terms describe everything you need:
- The x-axis is m/z, increasing left to right; each line is one ion of a particular mass.
- The y-axis is relative abundance. The tallest line is called the base peak and is assigned 100%; every other peak is scaled against it. The base peak is simply the most stable, most easily formed ion; it is not necessarily the molecular ion.
- The molecular ion peak (M⁺) is normally the line at the highest m/z value (ignoring small isotope peaks to its right). Its position gives the molecular mass of the compound.
In the mass spectrum of butan-2-one, CH₃COCH₂CH₃ (molecular mass 72), the molecular ion appears at m/z 72; the base peak at m/z 43 is the acetyl cation CH₃CO⁺ formed by cleaving the bond next to the carbonyl. A small line at 73 is the M+1 isotope peak.
04The molecular ion and the molecular formula
Finding M⁺ gives you the molecular mass; two simple rules then narrow the formula dramatically.
| Isotope clue | What you see | Interpretation |
|---|---|---|
| Chlorine (³⁵Cl / ³⁷Cl) | M and M+2 in a 3 : 1 ratio | One chlorine atom present |
| Bromine (⁷⁹Br / ⁸¹Br) | M and M+2 of nearly equal height | One bromine atom present |
| Carbon (¹³C) | M+1 about 1.1% of M per carbon | (M+1 ÷ M) × 100 ÷ 1.1 estimates the number of carbons |
05Fragmentation: the common losses
The energy used to ionize a molecule is far more than needed, so many molecular ions shatter into smaller fragments before reaching the detector. Fragmentation is not random: bonds break to leave the most stable cation. A useful technique is to look at the difference between M⁺ and a fragment peak, since that difference is the mass of the neutral piece lost.
| Fragment m/z | Likely ion | Suggests |
|---|---|---|
| 15 | CH₃⁺ | A methyl group; M–15 is loss of CH₃ |
| 29 | C₂H₅⁺ or CHO⁺ | Ethyl group, or an aldehyde |
| 43 | CH₃CO⁺ or C₃H₇⁺ | A methyl ketone (acetyl), or a propyl group |
| 57 | C₄H₉⁺ | A butyl group; very strong for tert-butyl |
| 77 | C₆H₅⁺ | A phenyl group (monosubstituted benzene) |
| 91 | C₇H₇⁺ (tropylium) | A benzylic CH₂ next to a ring; unusually stable |
| 105 | C₆H₅CO⁺ | A benzoyl group (aryl ketone) |
| M – 18 | loss of H₂O | An alcohol (dehydration in the instrument) |
| M – 28 | loss of CO or C₂H₄ | A ketone/aldehyde, or an alkyl chain |
The pattern behind the numbers. Three preferences explain most spectra: cleavage happens next to a carbonyl (giving 43, 57, 105 type ions), next to a benzene ring (giving the very stable tropylium ion at 91), and at branch points (giving the most substituted, most stable carbocation). Alcohols additionally like to eject a neutral water molecule, so M–18 is common and M⁺ itself can be weak.
06How to read a spectrum, step by step
- Find M⁺ at the far right. The highest significant m/z (excluding isotope peaks) is the molecular mass. If the value is odd, apply the nitrogen rule.
- Check for M+2. A 3 : 1 M/M+2 pair means chlorine; a roughly 1 : 1 pair means bromine. Both are unmistakable once you have seen them.
- Identify the base peak. The most abundant ion is the most stable fragment, and it usually names the key structural feature (43 for a methyl ketone, 91 for a benzyl group, 57 for tert-butyl).
- Subtract to find the neutral losses. Compute M minus each major fragment. Differences of 15, 18, 28, 29, and 45 correspond to CH₃, H₂O, CO, CHO, and COOH.
- Assemble the pieces. Add the fragments back up to the molecular mass, then cross-check the result against your IR and NMR data; MS rarely settles a structure alone.
07Worked interpretation examples
Example 1. Butan-2-one
The highest significant line is at m/z 72, so the molecular mass is 72; an even number, so no nitrogen. The base peak at 43 is 72 – 29, meaning a neutral ethyl radical (mass 29) was lost, leaving CH₃CO⁺. The peak at 29 is the complementary ion, C₂H₅⁺, and 57 is 72 – 15, the loss of a methyl group. A molecule of mass 72 that cleaves into an acetyl piece and an ethyl piece is butan-2-one, CH₃COCH₂CH₃.
Example 2. An unknown with peaks at m/z 108 and 110
A spectrum shows two peaks of almost equal height at m/z 108 and 110, and one strong fragment at m/z 29. Two peaks of nearly equal intensity separated by two mass units are the fingerprint of a single bromine atom, so M⁺ is 108 (the ⁷⁹Br form) and 110 is its ⁸¹Br partner. Subtracting bromine gives 108 – 79 = 29 for the rest of the molecule, which matches the observed fragment: C₂H₅⁺. The compound is bromoethane, CH₃CH₂Br. Note how the fragment and the halogen mass add back up to M⁺; that arithmetic check is always worth doing.
Example 3. Distinguishing three isomers of C₄H₁₀O
All three have M⁺ at m/z 74, so the molecular ion alone cannot separate them; the fragmentation does.
| Compound | Key fragments | Why |
|---|---|---|
| Butan-1-ol | 56 (M–18), 31, weak 74 | Primary alcohols readily lose water; 31 is CH₂OH⁺ |
| tert-Butanol | 59 (M–15) dominant | Losing a methyl leaves a resonance-stabilised oxocarbenium ion |
| Diethyl ether | 59 (M–15), 45, 31, 29 | Cleavage beside oxygen gives 45 (CH₃CH₂O⁺) and 29 |
08Practice questions
Try each before opening the answer.
Q1Why is the base peak not always the molecular ion peak?
The base peak is simply the most abundant ion. Many molecular ions fragment readily before reaching the detector, so a stable fragment ion can be formed in greater numbers than M⁺ survives.
Q2A spectrum has M⁺ at m/z 122 and a peak of nearly the same height at 124. What element is present, and what is the mass of the rest of the molecule?
Bromine (a 1 : 1 M/M+2 pair). The rest of the molecule is 122 – 79 = 43, which fits C₃H₇, so the compound is a bromopropane.
Q3An unknown has M⁺ at m/z 87. What does that odd mass tell you immediately?
By the nitrogen rule, the molecule contains an odd number of nitrogen atoms, so at least one.
Q4Ethylbenzene (M = 106) gives an intense peak at m/z 91 rather than at 15. Explain.
Benzylic cleavage breaks the CH₂–CH₃ bond. The charge stays on the ring fragment, giving the highly stable benzyl/tropylium cation C₇H₇⁺ at 106 – 15 = 91. The methyl leaves as a neutral radical, which is not detected; a CH₃⁺ cation would be far less stable.
Q5A compound with M⁺ at 60 loses 18 to give a peak at 42, and also shows m/z 31. What functional group is likely?
An alcohol, most likely a primary one. Loss of 18 is dehydration, and m/z 31 is CH₂OH⁺; together with M = 60 this points to propan-1-ol.
Key takeaways
- MS ionizes molecules, sorts the ions by m/z, and counts them; only charged species are detected.
- The molecular ion at the highest m/z gives the molecular mass; the base peak is the most abundant fragment and is set to 100%.
- Interpret in this order: find M⁺ → check M+2 for Cl or Br → identify the base peak → subtract to find neutral losses → assemble.
- Fragmentation favours the most stable cation, so cleavage happens beside carbonyls and benzene rings and at branch points.