Mass Spectrometry Curriculum Scope
Mass spectrometry is an Additional Higher Level (AHL) topic. The IB syllabus states that the operational details of the mass spectrometer instrument will not be assessed, but you must be able to interpret mass spectra data, calculate relative atomic masses, and deduce molecular fragmentations.
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.
Focus on Spectra Interpretation
The IB will not test you on the hardware components or operational mechanisms of the mass spectrometer. Focus your revision strictly on reading and interpreting mass spectra peaks, \(m/z\) ratios, and fragmentation losses.
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:
\[A_r = \frac{\sum (\text{isotope mass} \times \text{% abundance})}{100}\]
Calculating Ar of Chlorine
From the mass spectrum: ³⁵Cl = 75%, ³⁷Cl = 25%
\[A_r = \frac{(35 \times 75) + (37 \times 25)}{100}\]
\[A_r = \frac{2625 + 925}{100} = \frac{3550}{100} = \mathbf{35.50}\]
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).
Empirical to Molecular Formula via M⁺ Peak
If given an empirical formula and a mass spectrum:
- Calculate the empirical formula mass (EFM)
- Read the \(M^+\) peak to find \(M_r\)
- Divide: \(n = \frac{M_r}{\text{Empirical Formula Mass}}\)
- Multiply the empirical formula subscripts 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.
| Loss of Mass | Fragment Lost (Radical) | Common Identity |
|---|---|---|
| 15 | \(\bullet\text{CH}_3\) | Methyl group |
| 17 | \(\bullet\text{OH}\) | Hydroxyl group |
| 29 | \(\bullet\text{C}_2\text{H}_5\) or \(\bullet\text{CHO}\) | Ethyl group or aldehyde group |
| 31 | \(\bullet\text{OCH}_3\) | Methoxy group |
| 45 | \(\bullet\text{COOH}\) | Carboxyl group |
Deducing Compound Structure from Fragmentation
A mass spectrum shows: \(M^+ = 46\), major fragment peak at \(m/z = 29\)
Mass lost = \(46 - 29 = \mathbf{17}\)
A mass loss of 17 corresponds to an \(\bullet\text{OH}\) radical group being detached.
The fragment at \(29 = \text{C}_2\text{H}_5^+\) (ethyl cation) or \(\text{CHO}^+\) (formyl cation).
This fragmentation pattern confirms ethanol (\(\text{C}_2\text{H}_5\text{OH}\), \(M_r = 46\)).
Distinguishing Structural Isomers by Fragmentation
Two structural isomers have the identical \(M^+\) peak (same molecular formula and molar mass) but different fragmentation patterns because of differing bond connectivities. This is a very common IB structured question.
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