🔬 HL Only
Mass spectrometry is an Additional Higher Level topic. The IB syllabus states that the operational details of the mass spectrometer will not be assessed, but you must be able to interpret mass spectra data.
How a Mass Spectrometer Works
A mass spectrometer determines the masses and relative abundances of atoms or molecules in a sample. While you do not need to memorise the details for the exam, understanding the process helps you interpret the output.
⚠️ Examiner Note
The IB will not test you on the inner workings of the mass spectrometer. Focus your revision on reading and interpreting mass spectra.
Reading a Mass Spectrum
A mass spectrum is a bar chart with:
- x-axis: mass-to-charge ratio (\(m/z\)). Since the charge is usually +1, this effectively shows the mass number.
- y-axis: relative abundance (%). The height of each peak shows how common that isotope or fragment is.
Example: Mass Spectrum of Chlorine
Calculating \(A_r\) from a Mass Spectrum
The relative atomic mass is the weighted average of all the isotopes shown in the spectrum:
\[A_r = \frac{\sum (\text{isotope mass} \times \text{% abundance})}{100}\]
Worked Example: Chlorine
From the mass spectrum: ³⁵Cl = 75%, ³⁷Cl = 25%
\(A_r = \frac{(35 \times 75) + (37 \times 25)}{100}\)
\(A_r = \frac{2625 + 925}{100}\)
\(A_r = \frac{3550}{100} = \) 35.5
The Molecular Ion Peak (\(M^+\))
When a molecule (rather than an element) is placed in the mass spectrometer, the entire molecule can lose a single electron to form a positively charged molecular ion, \([M]^+\).
- The \(m/z\) value of the \(M^+\) peak tells you the relative molecular mass (\(M_r\)) of the compound.
- The \(M^+\) peak is typically the highest m/z value peak on the spectrum (ignoring any small M+1 peak from ¹³C).
🎯 Key Skill: Empirical to Molecular Formula
If given an empirical formula and a mass spectrum:
- Calculate the empirical formula mass
- Read the \(M^+\) peak to find \(M_r\)
- Divide: \(n = \frac{M_r}{\text{Empirical formula mass}}\)
- Multiply the empirical formula by \(n\) to get the molecular formula
Fragmentation Patterns
The high-energy electron beam can cause bonds in the molecular ion to break, producing smaller fragment ions. Only positively charged fragments are detected; neutral fragments are invisible to the detector.
Fragmentation is like a molecular fingerprint. It lets you deduce the structure of an unknown compound by analysing which pieces break off.
Common Mass Losses (IB Data Booklet)
You can identify what neutral fragment was lost by looking at the difference between the \(M^+\) peak and a fragment peak:
| Mass Lost | Fragment Lost | Indicates |
|---|---|---|
| 15 | \(\bullet CH_3\) | Methyl group |
| 17 | \(\bullet OH\) | Hydroxyl group |
| 18 | \(H_2O\) | Water (alcohol or carboxylic acid) |
| 28 | \(CO\) or \(CH_2{=}CH_2\) | Carbonyl or ethene |
| 29 | \(\bullet C_2H_5\) or \(\bullet CHO\) | Ethyl group or formyl group |
| 31 | \(\bullet OCH_3\) | Methoxy group |
| 45 | \(\bullet COOH\) | Carboxyl group |
Worked Example: Identifying a Compound
A mass spectrum shows: \(M^+ = 46\), fragment at \(m/z = 29\)
Mass lost = 46 - 29 = 17
A loss of 17 corresponds to an \(\bullet OH\) group being removed.
The fragment at 29 = \(CHO^+\) (formyl cation)
This is consistent with ethanol (\(C_2H_5OH\), \(M_r\) = 46)
⚠️ Examiner Tip: Structural Isomers
Two structural isomers have the same \(M^+\) peak (same molecular formula) but different fragmentation patterns because they have different connectivity. This is a common exam question.