Part 1 of 6How a Mass Spectrum Arises
A mass spectrometer separates ions by their mass-to-charge ratio, written m/z.
1
The sample is vaporised and bombarded with high-energy electrons, which knock an electron from a molecule.
2
This forms a positive molecular ion, M⁺, with the same mass as the molecule because the electron lost has negligible mass.
3
The ions are accelerated and separated according to their m/z by an electric or magnetic field.
4
A detector records the relative abundance of ions at each m/z to give the mass spectrum.
5
For ions with a single positive charge, the m/z value equals the relative mass of the ion.
Part 2 of 6Reading the Spectrum
Each peak in the spectrum corresponds to a particular ion.
Molecular ion peak
- The peak at the highest m/z (apart from the small M+1 peak) gives the relative molecular mass of the compound.
M+1 peak
- A small peak one unit above M⁺ arises because about 1% of carbon atoms are ¹³C, so larger molecules show a larger M+1 peak.
Fragmentation
- The molecular ion breaks into a positive fragment and a neutral radical, and only the positive fragments are detected.
Base peak
- The tallest peak in the spectrum belongs to the most abundant ion, often a stable fragment.
Fingerprint
- The pattern of fragments is characteristic of a compound, so spectra are matched against a database.
| m/z | Likely fragment |
|---|---|
| 15 | CH₃⁺ |
| 29 | C₂H₅⁺ |
| 43 | C₃H₇⁺ or CH₃CO⁺ |
| 57 | C₄H₉⁺ |
Part 3 of 6Low and High Resolution
Low and High Resolution
Low resolution
- Masses are given to the nearest whole number or one decimal place, enough to find the relative molecular mass.
High resolution
- Masses are measured to four or five decimal places using exact isotope masses, such as H = 1.00783 and O = 15.99491.
Distinguishing compounds
- Two molecules with the same whole-number mass usually have different exact masses, so high resolution can tell them apart.
Isomers
- Isomers have the same molecular formula and so the same exact mass, so they cannot be separated by their molecular ion alone.
Part 4 of 6Using Fragments
Using Fragments
Loss of small groups
- A drop of 15 from M⁺ suggests loss of CH₃, and a drop of 29 suggests loss of C₂H₅ or CHO.
Propanone
- It has M⁺ at 58 and a strong peak at 43 for CH₃CO⁺, formed by loss of a CH₃ radical.
Branching
- Branched alkanes tend to break at the branch because tertiary or secondary carbocations are more stable.
Combining techniques
- Mass spectra are used with infrared and elemental analysis to deduce a full structure.
Part 5 of 6Worked Example
Question
- Ethene (C₂H₄) and carbon monoxide (CO) both have a nominal mass of 28.
- Using Ar(H) = 1.00783, Ar(C) = 12.0000 and Ar(O) = 15.99491, show how high-resolution mass spectrometry distinguishes them.
- Mass of C₂H₄ = (2 × 12.0000) + (4 × 1.00783) = 24.0000 + 4.03132.
- This gives 28.03132 for ethene.
- Mass of CO = 12.0000 + 15.99491 = 27.99491.
- The two values differ by about 0.036, which a high-resolution instrument can detect.
Answer Ethene gives M⁺ at 28.03132 and carbon monoxide at 27.99491, so they can be told apart.
Part 6 of 6Exam Checklist & Tips
3.6.3 – Key Points for the Exam
- Molecular ion The ion formed when a molecule loses one electron, whose m/z gives the relative molecular mass.
- m/z The mass-to-charge ratio of an ion.
- Fragmentation Breakdown of the molecular ion into smaller ions and neutral species.
- Base peak The peak of greatest abundance in a mass spectrum.
Common Mistakes
- Reading the M+1 peak as the molecular ion rather than the peak of highest m/z in the main group.
- Forgetting that only positive fragments are seen, not neutral radicals.
- Claiming that high-resolution mass spectrometry can separate structural isomers, which share the same exact mass.
- Writing the molecular ion without the positive charge or giving fragments as neutral molecules.