22.2 Mass Spectrometry
By the end of this topic, you should be able to:
- Analyse mass spectra in terms of m/e values and isotopic abundances (knowledge of the working of the mass spectrometer is not required).
- Calculate the relative atomic mass of an element given the relative abundances of its isotopes, or its mass spectrum.
- Deduce the molecular mass of an organic molecule from the molecular ion peak in a mass spectrum.
- Suggest the identity of molecules formed by simple fragmentation in a given mass spectrum.
- Deduce the number of carbon atoms, n, in a compound using the M+1 peak and the formula n = 100 × abundance of M+1 ion ÷ (1.1 × abundance of M+ ion).
- Deduce the presence of bromine and chlorine atoms in a compound using the M+2 peak.
1 How a Mass Spectrometer Works
A mass spectrometer measures the mass of each isotope present in a sample. It also compares how much of each isotope is present — the relative isotopic abundance.
Relative isotopic abundance is the proportion (usually given as a %) of a particular isotope present in a sample of an element, relative to the other isotopes present.
A mass spectrometer is a large, complex instrument, but for A level purposes you only need to be able to read and interpret the spectrum it produces — not explain how the instrument itself works.
The spectrum produced plots:
- y-axis — relative abundance (%)
- x-axis — mass-to-charge ratio, m/e
Almost all ions formed in a mass spectrometer carry a single positive charge (+1). This means e = 1, so the m/e value read off the x-axis is simply equal to the mass (nucleon number) of the ion.
Worked example: the mass spectrum of lead
The spectrum below shows four peaks for a sample of lead:
| Isotopic mass | Relative abundance / % |
|---|---|
| 204 | 2 |
| 206 | 24 |
| 207 | 22 |
| 208 | 52 |
| Total | 100 |
So 52% of this lead sample is lead-208, with the remainder made up of lead-204 (2%), lead-206 (24%) and lead-207 (22%).
2 Calculating Ar from a Mass Spectrum
Mass spectrometry lets us calculate the relative atomic mass of an element very accurately. The method is:
- Multiply each isotopic mass by its percentage abundance.
- Add the resulting figures together.
- Divide the total by 100.
Relative atomic mass, Ar, is the weighted mean mass of an atom of an element, measured relative to 1/12th the mass of an atom of carbon-12.
Worked example: relative atomic mass of neon
Neon has three peaks in its mass spectrum: 20Ne (90.9%), 21Ne (0.3%) and 22Ne (8.8%).
The answer is given to 3 significant figures, consistent with the precision of the data provided.
A high-resolution mass spectrometer can measure isotopic masses extremely precisely — for example, 16O = 15.995 and 32S = 31.972. This precision lets chemists distinguish between molecules that appear to share the same relative molecular mass, such as SO2 and S2 (both ≈ 64).
3 Identifying Organic Compounds: the Molecular Ion Peak
The main use of mass spectrometry is identifying organic compounds. A substance can be identified by matching its spectrum against spectra of known substances stored in a database — a technique known as fingerprinting.
Inside the spectrometer, high-energy electrons knock electrons out of the sample molecules and break covalent bonds, fragmenting the molecule. The diagram below shows the mass spectrum of propanone, CH3COCH3.
The peak at the highest m/e value is the molecular ion peak, M+. It’s formed when a sample molecule simply has one electron knocked out — no atoms are lost — so its mass equals the relative molecular mass of the compound.
e.g. CH4 → CH4+ + e−
For propanone, the molecular ion peak appears at m/e = 58, corresponding to CH3COCH3+:
Fragment ions
The smaller peaks at m/e 15 and 43 come from fragments produced when propanone molecules break apart under electron bombardment. You can deduce a fragment’s identity by adding up the atomic masses of the atoms it contains:
m/e = 15: C + 3H = 12 + (3 × 1) = 15 → +CH3
m/e = 43: could be +C3H7 or CH3CO+ (both give 43)
Since propanone’s M+ ion (m/e 58) breaks in two different ways, each cleavage produces a detected cation and an undetected neutral radical:
The breaking of single bonds (C–C, C–O or C–N) is the most common cause of fragmentation. Only the charged fragment shows up as a peak — the neutral radical formed alongside it is invisible to the detector.
Some fragments recur so often in organic mass spectra that it’s worth recognising them on sight:
| m/e | Fragment |
|---|---|
| 15 | +CH3 |
| 28 | +CO or C2H4+ |
| 29 | CH3CH2+ |
| 43 | C3H7+ or CH3CO+ |
4 The M+1 Peak: Counting Carbon Atoms
Carbon has two stable isotopes: 12C (98.9%) and 13C (1.1%). Out of every 100 methane molecules, about 99 are 12CH4 and just 1 is 13CH4 — giving a small M+1 peak alongside the main M peak.
The more carbon atoms a molecule contains, the more likely it is that one of them is 13C, so the M+1 peak grows relative to M as the number of carbon atoms increases. This lets us work backwards from the ratio of M+1 to M to find the number of carbon atoms, n:
Abundances can be quoted as percentages or as raw peak heights in arbitrary units — it doesn’t matter, because only the ratio between them is used.
A hydrocarbon’s mass spectrum shows an M peak (m/e 86) with abundance 100, and an M+1 peak with abundance 6.6.
The molecule contains 6 carbon atoms — consistent with hexane, C6H14 (Mr = 86).
5 The M+2 Peak: Detecting Bromine and Chlorine
Fluorine and iodine each have only one stable isotope, but chlorine and bromine have two:
| Element | Isotope | Natural abundance / % | Approx. ratio |
|---|---|---|---|
| Chlorine | 35Cl | 75.5 | 3 : 1 |
| 37Cl | 24.5 | ||
| Bromine | 79Br | 50.5 | 1 : 1 |
| 81Br | 49.5 |
A compound containing one chlorine or bromine atom therefore shows two molecular ion peaks, 2 mass units apart (M and M+2), because some molecules contain the lighter isotope and some the heavier one.
M : M+2 ≈ 1 : 1 → one bromine atom present.
M : M+2 ≈ 3 : 1 → one chlorine atom present.
Chloromethane, CH3Cl, gives two molecular ion peaks:
- m/e = 12 + 3 + 35 = 50 (CH335Cl+)
- m/e = 12 + 3 + 37 = 52 (CH337Cl+)
These peaks appear in a 3 : 1 ratio, matching the natural abundance ratio of 35Cl to 37Cl.
Two halogen atoms: the M+2 and M+4 peaks
Things get more interesting with two halogen atoms. Take 1,2-dibromoethane, C2H4Br2. Each carbon could be bonded to either a 79Br or 81Br atom, with roughly equal chance:
| Formula | m/e | Peak |
|---|---|---|
| 79BrCH2CH279Br | 186 | M |
| 79BrCH2CH281Br | 188 | M+2 |
| 81BrCH2CH279Br | 188 | M+2 |
| 81BrCH2CH281Br | 190 | M+4 |
Since the middle two combinations both give m/e 188, three molecular ion peaks appear (at 186, 188 and 190) in a 1 : 2 : 1 ratio.
Don’t assume two halogens always means two peaks — work out all the isotope combinations. Two identical halogens give three peaks (M, M+2, M+4); the peak ratio tells you which halogen and how many atoms.
6 Practice Questions
Chlorine exists as 35Cl (75.5%) and 37Cl (24.5%). Calculate the relative atomic mass of chlorine to 3 significant figures.
Show mark scheme
Ar = (35 × 75.5 + 37 × 24.5) ÷ 100 = (2642.5 + 906.5) ÷ 100 = 3549 ÷ 100 = 35.5
An alkane gives a molecular ion peak at m/e = 72 and a fragment peak at m/e = 57. (a) Suggest the molecular formula of the alkane. (b) Identify the fragment lost, and give the formula of the ion at m/e 57.
Show mark scheme
(a) Mr = 72 → C5H12 (5 × 12 + 12 × 1 = 72), pentane.
(b) 72 − 57 = 15, so a CH3 (methyl radical) is lost. The remaining ion is C4H9+.
A compound’s mass spectrum has a molecular ion peak (M) with abundance 100, and an M+1 peak with abundance 5.5. Calculate the number of carbon atoms in the molecule.
Show mark scheme
n = (100 × 5.5) ÷ (1.1 × 100) = 550 ÷ 110 = 5 carbon atoms
The mass spectrum of a halogenoalkane shows two peaks of equal height at m/e 108 and m/e 110, with no peak at 112. What does this tell you about the halogen present, and why is there no third peak?
Show mark scheme
Equal-height M and M+2 peaks (ratio 1:1) indicate one bromine atom (79Br and 81Br are almost equally abundant). There’s no M+4 peak because the molecule contains only one Br atom — a third peak only appears when two halogen atoms are present.
A compound shows three molecular ion peaks at m/e 164, 166 and 168, in the approximate ratio 9 : 6 : 1. What does this indicate about the halogen content of the molecule?
Show mark scheme
The molecule contains two chlorine atoms. Each Cl is independently 35Cl or 37Cl in a 3:1 ratio, so combining two of them gives peak ratios of (3×3) : (3×1 + 1×3) : (1×1) = 9 : 6 : 1.
Do remember
- M+ = molecule minus one electron only — no atoms lost.
- For singly-charged ions, m/e = the actual mass of the ion.
- Divide by 100 when calculating a weighted-mean Ar.
- 13C natural abundance (1.1%) is the basis of the M+1 formula.
- M : M+2 of 1:1 → Br; M : M+2 of 3:1 → Cl.
Do not confuse
- The M+1 peak (counts carbon atoms) with the M+2 peak (detects Cl/Br).
- A charged fragment (detected) with a neutral radical (not detected).
- Relative abundance (%) with the actual number of molecules.
- The 35Cl : 37Cl ratio (3:1) with the 79Br : 81Br ratio (≈1:1).
Recap in 20 seconds
A mass spectrum plots relative abundance against m/e. The M+ peak gives the relative molecular mass; smaller peaks are fragments formed by bond breaking. The size of the M+1 peak reveals the number of carbon atoms; the size and pattern of the M+2 peak reveals bromine or chlorine atoms.