A molecular formula tells us which atoms are present, but it does not tell us immediately how those atoms are connected. Mass spectrometry helps bridge that gap. Instead of looking directly at a molecule, we convert it into ions and examine the masses of the ions that survive.
A mass spectrum therefore works rather like a molecular fingerprint. The molecular ion can reveal the mass of the original molecule, isotope peaks can reveal particular atoms, and fragments can reveal which parts of the molecule are especially stable. The real skill is learning to connect all these clues rather than treating individual peaks in isolation.
Try these five questions before reading the chapter. Focus on what each peak actually represents rather than simply memorising common mass-spectrum patterns.
- 5 diagnostic questions
- Mass spectra and ion equations where useful
- Instant explanation for every answer option
What mass spectrometry actually measures
A mass spectrometer does not directly measure the mass of an intact neutral molecule. The sample must first be converted into ions. Those ions are then separated according to their mass-to-charge ratio, m/z, before reaching a detector.
The resulting mass spectrum plots m/z on the horizontal axis and relative abundance on the vertical axis. Each line therefore tells us that an ion with a particular m/z value reached the detector, while the height of the line indicates how abundant that ion was relative to the others.
A peak in a mass spectrum represents an ion. It does not directly represent the neutral fragment that may have been lost when that ion formed.
Why molecules must first be ionised
Neutral molecules do not respond sufficiently to the electric or magnetic processes used to control ions inside a mass spectrometer. We therefore convert them into charged particles before analysing them.
One important ionisation method is electron impact ionisation, often abbreviated EI. A gaseous sample is bombarded by energetic electrons. When an incoming electron transfers enough energy to a molecule, one of the molecule's electrons is removed.
The product M⁺• contains one fewer electron than the original molecule. It therefore has a +1 charge. It also contains an unpaired electron, which is represented by the radical dot.
M⁺• is known as the molecular ion, or sometimes the parent ion.
M: the original neutral molecule.
M⁺•: the molecular radical cation formed after removal of one electron.
Why M⁺• is both an ion and a radical
It is easy to focus only on the positive charge, but the radical symbol is chemically important. Consider an ordinary closed-shell molecule in which the electrons are paired. Removing just one electron leaves behind an odd number of electrons.
The resulting molecular ion therefore simultaneously carries a positive charge and contains an unpaired electron:
For a molecule with relative molecular mass 58, electron impact can produce a molecular ion whose m/z is 58 if that ion carries a single positive charge. Seeing a peak at 58 is therefore potential evidence for the intact molecular ion — although isotope peaks and fragmentation must also be considered.
Understanding the mass-to-charge ratio, m/z
The position of a peak is determined not by mass alone but by the mass-to-charge ratio:
In many mass spectra encountered at this level, the important molecular and fragment ions carry a single positive charge. When z = +1, dividing by one does not change the numerical value.
| Mass of ion | Charge | m/z |
|---|---|---|
| 72 | +1 | 72 |
| 100 | +1 | 100 |
| 100 | +2 | 50 |
The final example shows why it is unsafe to say that m/z is simply the mass of an ion. That shortcut works only when the ion carries a single charge.
For a singly charged ion, z = 1, so the numerical value of m/z equals the mass of the ion.
Do all ions have a charge of +1?
No. Multiply charged ions are possible. If an ion of mass 120 carries a charge of +2, it appears at:
However, unless there is evidence for a multiply charged species, the molecular and fragment ions in the spectra normally encountered in these problems are interpreted as singly charged.
How to read a mass spectrum
Once the ions have been separated and detected, the instrument produces a pattern of peaks. Three ideas are especially important when you first inspect a spectrum: the molecular ion peak, the base peak and isotope peaks.
This diagram is schematic rather than an experimental spectrum. Its purpose is to show how different kinds of peaks are identified.
The molecular ion peak
The molecular ion is produced when the intact molecule loses an electron without the nuclei of the molecule yet being separated:
For a singly charged molecular ion, the m/z value of M⁺• therefore gives the mass of that particular molecular isotopic composition. In many introductory interpretations, this allows the molecule's relative molecular mass, Mᵣ, to be inferred from the molecular-ion region.
Be careful with the common statement that the molecular ion is simply the peak furthest to the right. Small isotope peaks may occur at M + 1 or M + 2, so these can lie to the right of the principal molecular-ion peak.
Furthermore, some molecules fragment so readily under electron impact that their molecular-ion peak is weak or may be difficult to observe.
Look for the molecular-ion region, not merely the rightmost line in the entire spectrum.
The base peak
The base peak is the most intense peak in the mass spectrum. Its relative abundance is defined as 100.
A base peak does not necessarily correspond to the intact molecular ion. Frequently it belongs to a particularly stable fragment ion that is formed readily during fragmentation.
Molecular ion peak: evidence from the intact ionised molecule and therefore useful for identifying molecular mass.
Base peak: the most intense detected ion; it may be either a molecular ion or a fragment ion.
Isotope peaks reveal information about composition
Natural samples contain isotopes. If otherwise identical molecules contain different isotopes, their molecular ions have different masses and therefore appear at slightly different m/z values.
This gives isotope peaks such as M + 1, M + 2 and, when several isotope-containing atoms are present, sometimes further peaks such as M + 4.
The M + 1 peak and ¹³C
Carbon occurs mainly as ¹²C, but about 1.1% of naturally occurring carbon is ¹³C. Replacing one ¹²C atom by one ¹³C atom increases the molecular mass by one unit.
A molecule containing carbon therefore normally has a small M + 1 peak next to its molecular-ion peak.
For a molecule containing n carbon atoms, the approximate relationship is:
Rearranging:
Suppose the M peak has relative intensity 100 and the M + 1 peak has relative intensity 4.4. Then n ≈ (4.4 / 100) × (100 / 1.1) = 4. The spectrum is therefore consistent with a molecule containing about four carbon atoms.
This treatment is an approximation because isotopes of atoms other than carbon can also contribute slightly to the M + 1 region. Nevertheless, the relationship is extremely useful when carbon dominates the M + 1 contribution.
Chlorine and bromine give distinctive M + 2 patterns
Chlorine and bromine are especially useful because each possesses two major naturally occurring isotopes differing by two mass units.
| Element | Important isotopes | Approximate natural ratio |
|---|---|---|
| Chlorine | ³⁵Cl and ³⁷Cl | 3 : 1 |
| Bromine | ⁷⁹Br and ⁸¹Br | 1 : 1 |
One chlorine atom
Consider a molecule containing one chlorine atom. Some molecules contain ³⁵Cl and others contain ³⁷Cl. Because ³⁷Cl is two mass units heavier, two molecular-ion peaks appear separated by two m/z units.
This approximate 3 : 1 pair is therefore strong evidence for one chlorine atom.
One bromine atom
For bromine, ⁷⁹Br and ⁸¹Br occur in approximately equal abundance. A molecule containing one bromine atom therefore produces:
Two neighbouring molecular-ion-region peaks separated by two units and of almost equal height are therefore a classic signature of bromine.
What happens when there are two halogen atoms?
With two halogen atoms, there are three possible isotope combinations. This produces three principal molecular-ion-region peaks:
Their intensities follow the statistical combinations of the isotopes.
| Atoms present | Main peak pattern | Useful clue |
|---|---|---|
| 1 Cl | M : M + 2 ≈ 3 : 1 | Two peaks separated by 2 |
| 1 Br | M : M + 2 ≈ 1 : 1 | Two nearly equal peaks |
| 2 Cl | M : M + 2 : M + 4 ≈ 9 : 6 : 1 | Three-peak chlorine pattern |
| 2 Br | M : M + 2 : M + 4 ≈ 1 : 2 : 1 | Symmetrical three-peak pattern |
Do not identify chlorine or bromine simply because an M + 2 peak exists. Use the relative intensities of the isotope peaks.
Fragmentation turns a molecular ion into structural clues
Electron impact does more than remove an electron. The molecular ion has considerable internal energy and may break apart before reaching the detector. This process is called fragmentation.
A simple cleavage of a molecular radical cation can be represented schematically as:
Charge must be conserved. If the original molecular ion has a +1 charge and one product is neutral, the other fragment retains the positive charge.
Only the charged fragment is recorded as the corresponding peak in an ordinary mass spectrum. The neutral radical does not itself appear as that peak.
When interpreting a fragment peak, draw or identify the species that remains positively charged — not merely the group that has been lost.
Why some fragments are much more intense than others
Fragmentation is not equally favourable at every bond. A fragmentation pathway becomes important when cleavage can generate a comparatively stable positive ion.
This means peak intensity can often be understood by considering the stability of the fragment ion. Resonance-stabilised ions, for example, may form especially readily.
However, avoid turning this into the oversimplified rule that a molecule always breaks at its single “weakest bond”. Fragmentation depends on the complete ion structure and the stability of the products formed.
Using mass differences to analyse fragmentation
One of the quickest ways to interpret a fragment peak is to compare its m/z value with the molecular-ion peak.
If the ions are singly charged:
The difference suggests the mass of the neutral species that has been lost. You can then ask whether that loss makes chemical sense for the proposed structure.
Example: propanone
Propanone has molecular formula CH₃COCH₃ and relative molecular mass 58. Its molecular ion can therefore appear at m/z 58.
Cleavage adjacent to the carbonyl group can produce the acylium ion CH₃CO⁺:
The fragment CH₃CO⁺ has mass:
It therefore produces a peak at m/z 43.
We can also arrive at the same conclusion from the mass difference:
A mass loss of 15 is consistent with loss of a CH₃• radical.
Notice the direction of the reasoning. The peak at m/z 43 belongs to CH₃CO⁺. The CH₃• radical explains the mass that disappeared from the molecular ion, but CH₃• itself is neutral and is not the m/z 43 peak.
Common neutral losses are clues, not automatic answers
Certain mass differences appear frequently, but they should be used as starting clues rather than as rules that identify a structure by themselves.
| Mass difference | Possible neutral loss | Important caution |
|---|---|---|
| 15 | CH₃• | The proposed molecule must actually contain a plausible methyl-loss pathway. |
| 17 | OH• | This is structure-dependent and should not be assigned automatically. |
| 18 | H₂O | Common for suitable oxygen-containing compounds where dehydration is possible. |
| 28 | CO or C₂H₄ | The mass difference alone cannot distinguish species of the same nominal mass. |
| 29 | C₂H₅• or CHO• | The original structure determines which interpretation is chemically reasonable. |
A mass difference tells you how much mass was lost. It does not, by itself, uniquely identify what was lost.
A systematic method for interpreting a mass spectrum
A complicated spectrum becomes much easier to analyse when you approach it in a consistent order.
Step 1 — Locate the molecular-ion region.
Identify a plausible M peak while remembering that M + 1 and M + 2 isotope
peaks may lie immediately to its right.
Step 2 — Examine isotope patterns.
Look for an M + 1 contribution from ¹³C and for characteristic chlorine
or bromine M + 2 patterns.
Step 3 — Identify important fragment peaks.
Give particular attention to intense peaks because they often correspond
to relatively stable fragment ions.
Step 4 — Calculate mass differences.
Compare fragment m/z values with the proposed molecular ion and ask what
neutral loss is chemically possible.
Step 5 — Check the complete structure.
Every assignment should be compatible with the molecular formula,
functional groups and possible fragmentation pathways.
Do not interpret one peak in isolation
An m/z value rarely gives a unique structure. For example, different ions can possess the same nominal mass. A good structural assignment therefore uses several pieces of evidence simultaneously.
Mass spectrometry becomes especially powerful when combined with other spectroscopic techniques. A molecular ion can constrain molecular mass, isotope patterns can reveal elemental composition, fragmentation can reveal structural units, while IR and NMR spectroscopy can provide complementary information about functional groups and molecular environments.
Common Mistakes
| Common mistake | Why it is wrong | Better approach |
|---|---|---|
| Calling the rightmost peak the molecular ion automatically | M + 1 and M + 2 isotope peaks can lie to the right of M. | Interpret the complete molecular-ion region and its isotope pattern. |
| Assuming the base peak must be M⁺• | A stable fragment ion may be much more abundant than the intact molecular ion. | Define the base peak only as the most intense detected ion. |
| Calling the neutral group that was lost the observed fragment | The detector records the charged species. | Identify the positive fragment ion first, then describe the corresponding neutral loss. |
| Treating every m/z value as the ion's mass | This is true numerically only for singly charged ions. | Remember m/z = mass ÷ charge. |
| Memorising mass losses without checking the structure | Several different species can have the same nominal mass. | Use the parent structure and chemically reasonable fragmentation pathways. |
| Identifying any M + 2 peak as chlorine | Bromine and other isotopic contributions can also produce M + 2 peaks. | Use the relative intensity pattern: about 3 : 1 for one Cl and 1 : 1 for one Br. |
What you should be able to do
Explain electron impact ionisation
Write and interpret M(g) + e⁻ → M⁺•(g) + 2e⁻ and explain why M⁺• is a radical cation.
Use m/z correctly
Calculate mass-to-charge ratios and recognise when m/z is numerically equal to ion mass.
Identify important peaks
Distinguish the molecular ion, base peak, fragment peaks and isotope peaks.
Interpret isotope patterns
Use M + 1 to estimate carbon number and recognise characteristic Cl and Br patterns.
Analyse fragmentation
Identify positive fragment ions and calculate the corresponding neutral mass loss.
Combine several clues
Use molecular mass, isotope information and fragment ions together to support a proposed structure.
These ten questions test whether you can combine ionisation, isotope patterns, m/z values and fragmentation rather than treating each idea separately.
- 10 mastery questions
- Mass spectra and isotope patterns where useful
- Instant explanation for every answer option
Frequently asked questions
Questions students often ask about mass spectra
Once the basic ideas are connected, many apparent complications in mass spectrometry reduce to the same questions: what ion produced this peak, where did its charge come from, and what happened to the rest of the molecule?
Why does electron impact produce two electrons on the right-hand side of the equation?
One electron entered as the incident electron. It collides with the molecule and removes one of the molecule's electrons. There are therefore two electrons after the collision: the incident electron and the electron that was ejected from M.
Why is the molecular ion written M⁺• rather than simply M⁺?
Electron impact removes one electron from an initially closed-shell molecule. The ion therefore has both a positive charge and an unpaired electron. M⁺• shows both pieces of information.
Is the molecular ion always the highest-mass peak in a mass spectrum?
Not quite. Isotope peaks such as M + 1 and M + 2 may occur to the right of the main molecular-ion peak. It is better to identify the molecular-ion region and interpret its isotope pattern rather than simply choosing the furthest-right line.
Is the base peak always the molecular ion?
No. The base peak is simply the most intense peak in the spectrum. It is assigned a relative abundance of 100. In many spectra it is produced by a particularly stable fragment ion rather than the intact molecular ion.
Why can an M + 1 peak be used to estimate the number of carbon atoms?
Approximately 1.1% of naturally occurring carbon is ¹³C. Each carbon atom therefore gives another opportunity for a molecule to contain one ¹³C atom instead of ¹²C. As the number of carbon atoms increases, the M + 1 peak becomes correspondingly larger relative to M.
How can I distinguish chlorine from bromine in a mass spectrum?
For a molecule containing one chlorine atom, the molecular-ion-region peaks M and M + 2 are approximately in a 3 : 1 ratio because ³⁵Cl is about three times as abundant as ³⁷Cl. For one bromine atom, M and M + 2 are approximately equal because ⁷⁹Br and ⁸¹Br occur in roughly equal abundances.
If I calculate a neutral loss from two peaks, does that uniquely identify the group lost?
No. It identifies the mass difference. Different species may possess the same nominal mass, so the parent structure and possible fragmentation pathways must also support the assignment. For example, a mass loss of 28 could correspond to CO or C₂H₄ depending on the molecule.
Why does the neutral radical not appear as a peak?
The mass analyser separates and detects charged particles. A neutral radical does not respond in the same way and therefore does not give the corresponding detected ion peak. The positive fragment remaining after cleavage is the species represented in the spectrum.
Mass spectrometry becomes much easier once every peak is treated as evidence about an ion. Ionisation explains where the charge originates, m/z tells us where the ion appears, isotope patterns reveal elemental information, and fragmentation shows how parts of a molecule can separate. Rather than memorising isolated peak values, build the proposed structure from all of these clues together.