A mass spectrum can look intimidating at first: a forest of vertical lines rising to different heights, each labelled by an m/z value. But there is much more order in the spectrum than the first glance suggests. Every important peak represents an ion, and each ion tells us something about the molecule from which it came.
The most useful way to interpret a spectrum is therefore not to ask “What does every peak mean?” Instead, begin with the whole molecule, examine its isotope pattern, and then use the major fragments to test possible structures. Once these clues are combined, mass spectrometry becomes a process of molecular deduction rather than peak memorisation.
Try these five questions before reading the chapter. Focus on what each peak actually represents and how isotope patterns can reveal information about molecular composition.
- 5 diagnostic questions
- Mass-spectrum diagrams where useful
- Instant explanation for every answer option
1. How to read a mass spectrum
A mass spectrum is a plot of relative ion abundance against mass-to-charge ratio, m/z. Each vertical line represents ions having a particular value of m/z.
For most of the molecular ions and fragment ions considered here, the charge is +1. Consequently, an ion of relative isotopic mass 43 has m/z = 43. If an ion had a charge of +2, the same mass would appear at half the m/z value, but such ions are not usually the main focus when interpreting simple organic spectra at this level.
Every peak represents an ion. The position of the peak tells you m/z; the height tells you how abundant that ion is relative to the other detected ions.
The base peak
The tallest peak in a mass spectrum is called the base peak. Its relative intensity is defined as 100%, and the heights of all other peaks are measured relative to it.
The base peak is simply the most abundant detected ion. It may be a particularly stable fragment ion, so it is not necessarily the molecular ion.
Base peak: the most intense ion peak in the spectrum. It is assigned 100% relative abundance.
Molecular ion peak: the peak corresponding to the intact molecule after ionisation. It can be much smaller than the base peak and may occasionally be very weak.
2. The molecular ion peak: finding the whole molecule
Under electron-impact ionisation, a gaseous molecule is bombarded by high-energy electrons. An electron is removed from the molecule, producing a positively charged radical cation.
The ion M⁺• is called the molecular ion. The superscript “+” shows that the species carries a positive charge, while the dot shows that it contains an unpaired electron.
Because only one electron has been removed, the loss of electron mass is negligible on the scale used in ordinary mass spectra. Therefore, for a singly charged molecular ion, its m/z value gives the relative isotopic mass of that particular molecular isotopologue. For the molecule containing the most abundant isotopes, this is normally the nominal molecular mass used when interpreting the spectrum.
Do not simply choose the highest-m/z peak
A common shortcut is to say that the molecular ion is the peak with the highest m/z. That is not a safe rule. Isotope-containing versions of the same molecular ion give peaks at M + 1, M + 2 and sometimes higher values, so peaks can legitimately occur to the right of the molecular ion peak.
Instead, identify the molecular-ion cluster. The monoisotopic molecular ion is followed by smaller isotope peaks whose spacings and relative heights should make chemical sense.
Suppose a spectrum has a significant peak at m/z 72, a much smaller peak at 73, and no meaningful molecular-size peaks beyond this region. A sensible interpretation is that 72 is M and 73 is M + 1. The compound therefore has a nominal relative molecular mass of 72.
Why the molecular ion may be weak
Formation of M⁺• gives the molecule a considerable amount of energy. Some molecular ions survive long enough to reach the detector, but others fragment readily. A molecule that fragments very easily may therefore show only a weak molecular ion peak.
This is another reason why the absence of an obvious tall high-mass peak should not automatically be interpreted as evidence that no larger molecule was present.
3. Isotopic peaks: why M + 1, M + 2 and M + 4 appear
Atoms of the same element can occur naturally as different isotopes. Molecules containing different combinations of these isotopes have slightly different masses but almost identical chemical structures. Mass spectrometry separates them, giving a characteristic group of neighbouring peaks.
The lightest important member of a molecular-ion isotope cluster is normally labelled M. Replacing one atom with an isotope that is one mass unit heavier gives an M + 1 peak. Replacing an atom with an isotope two mass units heavier gives an M + 2 peak.
The M + 1 peak and ¹³C
Natural carbon consists mainly of ¹²C, but about 1.1% is ¹³C. A molecule in which one ¹²C atom has been replaced by ¹³C is one mass unit heavier, so it contributes to the M + 1 peak.
The more carbon atoms a molecule contains, the greater the probability that one of them is ¹³C. This allows the number of carbon atoms to be estimated from the relative sizes of the M and M + 1 peaks.
Rearranging gives:
If the M peak has been set to 100, the calculation becomes even simpler:
Example: A molecular ion has relative intensity 100 and its M + 1 peak has intensity 5.5. Therefore, n(C) ≈ 5.5 ÷ 1.1 = 5. The molecule probably contains about five carbon atoms.
The ¹³C method is an approximation, not an exact carbon counter. Other naturally occurring isotopes can also contribute to the M + 1 peak, and the approximation becomes less exact when their contributions are significant.
The M + 2 peak: chlorine and bromine
Chlorine and bromine are especially easy to recognise because each has two abundant isotopes separated by two mass units.
| Element | Important isotopes | Approximate natural abundances | One-atom molecular-ion pattern |
|---|---|---|---|
| Chlorine | ³⁵Cl and ³⁷Cl | About 75.8% : 24.2% | M : M + 2 ≈ 3 : 1 |
| Bromine | ⁷⁹Br and ⁸¹Br | About 50.7% : 49.3% | M : M + 2 ≈ 1 : 1 |
What happens when there is more than one halogen atom?
With two halogen atoms, more isotope combinations become possible. For example, a molecule containing two chlorine atoms may contain:
- two ³⁵Cl atoms → M,
- one ³⁵Cl and one ³⁷Cl atom → M + 2,
- two ³⁷Cl atoms → M + 4.
Because there are two different ways to choose which chlorine atom is ³⁷Cl in the middle case, the central peak receives an additional statistical contribution.
| Halogen composition | Main isotope pattern | Interpretation |
|---|---|---|
| One Cl | M : M + 2 ≈ 3 : 1 | Characteristic chlorine doublet |
| One Br | M : M + 2 ≈ 1 : 1 | Characteristic bromine doublet |
| Two Cl | M : M + 2 : M + 4 ≈ 9 : 6 : 1 | Three-peak chlorine cluster |
| Two Br | M : M + 2 : M + 4 ≈ 1 : 2 : 1 | Three-peak bromine cluster |
| One Cl and one Br | M : M + 2 : M + 4 ≈ 3 : 4 : 1 | Mixed chlorine–bromine cluster |
A quick H3 way to predict halogen patterns
The isotope ratios can be handled almost like algebra. Represent the lighter isotope by the constant term and the heavier isotope by x, where each power of x corresponds to another +2 mass shift.
For chlorine, the approximate isotope ratio is 3:1, so one chlorine behaves like:
Two chlorine atoms therefore give:
This directly predicts:
Similarly, bromine behaves approximately like 1 + x. Hence two bromine atoms give:
and therefore a 1:2:1 pattern.
Do not assume that every M + 2 peak proves the presence of chlorine or bromine. Other isotopes, such as ³⁴S and ¹⁸O, can also contribute at M + 2. What makes Cl and Br especially useful is the size and characteristic ratio of their isotope peaks.
4. Fragment ions: using molecular breakage as structural evidence
The molecular ion formed during electron-impact ionisation has considerable internal energy. Bonds may break before the ion reaches the detector. This process is called fragmentation.
A simple cleavage often produces a positively charged fragment and a neutral radical:
Only the charged fragment gives the corresponding peak in the mass spectrum. The neutral radical is not detected as a normal ion peak.
Example: fragmentation of propanone
Propanone has a molecular ion at m/z 58. Cleavage next to the carbonyl group can form the relatively stable acylium ion CH₃CO⁺.
The CH₃CO⁺ fragment has:
A strong peak at m/z 43 is therefore consistent with this fragmentation pathway.
Notice the direction of the reasoning. We do not say “m/z 43 always means CH₃CO⁺.” Instead, we ask whether CH₃CO⁺ is a chemically sensible fragment of the proposed molecule. The same nominal m/z can sometimes correspond to different ionic formulae or structures.
Common fragment masses
Certain fragment ions appear frequently because they are readily formed from common structural units. These values are useful clues, but they should always be interpreted in the context of the entire spectrum and the proposed molecular structure.
| m/z | Possible fragment | Where it may arise |
|---|---|---|
| 15 | CH₃⁺ | Methyl-containing structures |
| 29 | C₂H₅⁺ | Ethyl-containing structures |
| 31 | CH₂OH⁺ / CH₃O⁺ composition | Common in suitable oxygen-containing compounds |
| 43 | C₃H₇⁺ or CH₃CO⁺ | Alkyl fragments or carbonyl compounds; context is essential |
| 57 | C₄H₉⁺ | Larger alkyl fragments |
| 91 | C₇H₇⁺ | Common stable fragment from many alkylbenzenes |
Using mass differences correctly
The difference between the molecular ion and a fragment peak can help identify the neutral species that has been lost.
For example, if M occurs at 74 and a major fragment appears at 59, the mass difference is:
A loss of 15 is consistent with loss of a methyl radical, •CH₃, during simple cleavage.
However, a mass difference alone is not proof. The proposed neutral loss must also be chemically possible for the molecular structure being tested.
Fragment interpretation has two tests: the proposed ion must have the correct m/z, and it must be obtainable by a chemically reasonable fragmentation of the proposed molecular structure.
Why some fragment ions are much more intense
Fragmentation is not random. Pathways that form comparatively stable ions are often favoured, so those ions may produce strong peaks. This is why the base peak frequently corresponds to a particularly favourable fragment rather than to the molecular ion itself.
When several fragmentation pathways are possible, focus first on the major peaks. Minor peaks can contain useful information, but trying to explain every tiny signal often creates more ambiguity than insight.
5. Putting the molecular ion, isotope pattern and fragments together
The strongest interpretations come from combining independent pieces of evidence. A reliable sequence is:
Step 1 — Find the molecular-ion region
Look near the high-mass end for a plausible molecular-ion cluster rather than automatically choosing the final visible line.
Step 2 — Identify M
Distinguish the principal molecular ion from its M + 1 and M + 2 isotope peaks.
Step 3 — Inspect M + 1
Use the ¹³C contribution to estimate the number of carbon atoms where appropriate.
Step 4 — Inspect M + 2 and M + 4
Look for diagnostic chlorine or bromine isotope patterns.
Step 5 — Examine major fragments
Calculate plausible ionic formulae and consider which bonds could have broken.
Step 6 — Cross-check everything
The molecular mass, isotope evidence and fragment pattern should all be compatible with the same proposed structure.
6. Worked interpretation examples
Example 1: estimating the number of carbon atoms
A molecular ion peak has relative intensity 80.0 and the corresponding M + 1 peak has intensity 4.4.
First calculate the intensity of M + 1 relative to M:
(4.4 ÷ 80.0) × 100 = 5.5%
Therefore:
n(C) ≈ 5.5 ÷ 1.1 = 5
The molecule is therefore estimated to contain five carbon atoms.
Example 2: recognising a halogen
Suppose a molecular-ion cluster contains peaks at m/z 108 and 110 of almost equal intensity.
The peaks are separated by two mass units and have approximately a 1:1 ratio. This strongly indicates the presence of one bromine atom, corresponding to molecules containing ⁷⁹Br and ⁸¹Br.
Example 3: two chlorine atoms
An unknown compound gives molecular-ion peaks at M, M + 2 and M + 4 with approximate intensities 90, 60 and 10.
Dividing by 10 gives:
9 : 6 : 1
This is the characteristic pattern for a molecule containing two chlorine atoms.
Example 4: using a fragment together with M
A compound gives a molecular ion at m/z 58 and a very strong fragment at m/z 43.
A possible fragment formula at 43 is CH₃CO⁺:
(2 × 12) + (3 × 1) + 16 = 43
If other evidence suggests a carbonyl compound such as propanone, the strong 43 peak is readily explained by cleavage producing the stable CH₃CO⁺ fragment. The fragment evidence therefore supports, rather than independently proves, that structural assignment.
Common Mistakes
| Common mistake | Why it is wrong | Better approach |
|---|---|---|
| Assuming the highest-m/z peak must be the molecular ion | Isotope peaks such as M + 1 and M + 2 occur at higher m/z than M. | Identify the molecular-ion isotope cluster and decide which member corresponds to M. |
| Assuming the base peak is the molecular ion | The base peak is simply the most intense detected ion and is often a fragment. | Treat base-peak intensity and molecular-ion identity as separate questions. |
| Treating every M + 2 peak as proof of Cl or Br | Other isotopes can also produce M + 2 contributions. | Use the characteristic ratios: about 3:1 for one Cl and about 1:1 for one Br. |
| Using M + 1 intensity without comparing it with M | The carbon estimate depends on the ratio I(M + 1) / I(M), not the absolute height of M + 1 alone. | First express M + 1 as a percentage of M, then divide by approximately 1.1%. |
| Saying that m/z 43 always corresponds to one particular fragment | Different ions can have the same nominal mass. For example, C₃H₇⁺ and CH₃CO⁺ both have m/z 43. | Use the molecular structure and other spectral evidence to decide which assignment is chemically plausible. |
| Identifying a neutral loss solely from a mass difference | Several species can sometimes have the same nominal mass, and not every imaginable loss is chemically reasonable. | Check both the mass difference and the proposed bond-cleavage pathway. |
What you should be able to do
Read a mass spectrum
Identify the m/z axis, relative abundance axis, base peak and important high-mass peaks.
Identify the molecular ion
Distinguish M from nearby isotope peaks and use it to obtain the nominal molecular mass.
Use the M + 1 peak
Estimate the number of carbon atoms using the approximate natural abundance of ¹³C.
Recognise halogen patterns
Distinguish one or more chlorine and bromine atoms from M, M + 2 and M + 4 peak ratios.
Interpret major fragments
Propose positively charged fragment ions with the correct m/z and a reasonable route of formation.
Combine independent clues
Test whether molecular mass, isotope pattern and fragmentation all support the same proposed molecular structure.
These ten questions test whether you can combine molecular-ion, isotope and fragmentation evidence rather than interpreting each peak in isolation.
- 10 mastery questions
- Isotope clusters and mass-spectrum diagrams
- Instant explanation for every answer option
Frequently asked questions
Questions students often ask about interpreting mass spectra
Most difficulties with mass spectrometry come from treating one visual feature as an absolute rule. These questions help separate the useful shortcuts from the actual chemistry behind them.
Is the molecular ion always the peak with the highest m/z?
No. The molecular ion is accompanied by isotope peaks such as M + 1 and M + 2, which necessarily have higher m/z values. There may also be small background or impurity peaks. It is safer to identify a coherent molecular-ion isotope cluster rather than simply choosing the final line in the spectrum.
Is the base peak the same as the molecular ion peak?
Not necessarily. The base peak is simply the most intense peak in the spectrum and is assigned 100% relative abundance. If a particular fragmentation pathway produces a very stable ion, that fragment may become the base peak while the molecular ion is much smaller.
Why does one chlorine atom give a 3:1 pattern but one bromine gives a 1:1 pattern?
The pattern reflects natural isotope abundance. About three quarters of natural chlorine is ³⁵Cl and about one quarter is ³⁷Cl, so molecules containing one chlorine produce approximately a 3:1 M:M + 2 pair. The two major bromine isotopes, ⁷⁹Br and ⁸¹Br, occur in almost equal abundance, so the corresponding peaks are approximately equal in height.
Is the M + 1 method an exact way to count carbon atoms?
No. The familiar 1.1% per carbon rule is an approximation based on the natural abundance of ¹³C. Other isotopes can also contribute to the M + 1 peak, and real experimental peak intensities have finite uncertainty. It is therefore best used as supporting evidence for a plausible carbon count.
Why can the same m/z value correspond to different fragment ions?
A mass spectrometer separates ions according to mass-to-charge ratio, not according to structural identity. Two different ions can therefore have the same nominal mass. For example, C₃H₇⁺ and CH₃CO⁺ both have nominal m/z 43. Deciding between them requires the proposed molecular structure and the rest of the spectral evidence.
A mass spectrum becomes much easier to interpret once you stop treating it as a collection of unrelated peaks. First identify the molecular-ion region, then read the isotope pattern, and finally use the major fragment ions to test possible structures. Each clue narrows the possibilities, and when mass spectrometry is combined with techniques such as IR and NMR spectroscopy, the structure of an unknown molecule can often be built up with remarkable confidence.