Mass spectrometry
| English | Chinese | Pinyin |
|---|---|---|
| mass spectrometry | 质谱 | zhì pǔ |
| mass-to-charge ratio | 质荷比 | zhì hé bǐ |
| abundance | 丰度 | fēng dù |
| molecular ion | 分子离子 | fèn zǐ lí zi |
| fragmentation | 碎裂 | suì liè |
Weighing a single molecule
- In mass spectrometry 质谱, a molecule is ionised and sorted by mass-to-charge ratio 质荷比 ($m/e$).
- The spectrum is a set of peaks.
- From it we get relative masses and structural clues.
Mass spectrometry route
Follow a molecule through ionisation, separation and detection.
In mass spectrometry, the peak at the highest m/z (the molecular ion) gives the relative molecular ______.
The M⁺ peak equals the Mr.
Relative atomic mass from isotopes
- An element's isotopes give several peaks. From their abundances 丰度, the relative atomic mass is a weighted average:
- Chlorine (75% $^{35}\text{Cl}$, 25% $^{37}\text{Cl}$): $A_r = \dfrac{35 \times 75 + 37 \times 25}{100} = 35.5$.
A mass spectrum: the molecular ion 分子离子 gives the Mr, fragments show what was lost
Chlorine is 75% ³⁵Cl and 25% ³⁷Cl. What is its relative atomic mass? (use (35×75 + 37×25)/100)
Ar = (35×75 + 37×25)/100 = (2625 + 925)/100 = 35.5.
Molecular ion and fragmentation 碎裂
- The highest $m/e$ peak is the molecular ion ($M^{+}$) — it gives the relative molecular mass.
- The molecule also breaks into pieces (fragmentation). The gap between peaks = the mass lost: a loss of 15 = a $\text{CH}_3$ group; a loss of 29 = $\text{CHO}$ or $\text{C}_2\text{H}_5$.
A modern mass spectrometer in a laboratory; the sample is loaded at the front and the machine sorts its ions by mass-to-charge ratio
The molecular ion peak (the highest m/e) gives the:
The highest-m/e peak (M⁺) is the whole molecule, so it gives the Mr.
A gap of 15 between two peaks means the molecule has lost:
A loss of 15 mass units corresponds to a CH₃ fragment; a loss of 29 is CHO or C₂H₅.
The [M+1] and [M+2] peaks
- a small [M+1] peak (from $^{13}\text{C}$) tells you the number of carbons.
- an [M+2] peak (two units up) shows a halogen: chlorine gives a 3:1 ratio, bromine a 1:1 ratio.
An [M+2] peak about the same height as M⁺ (a 1:1 ratio) shows the molecule contains:
One bromine gives a 1:1 [M+2]:M⁺ ratio; one chlorine gives about 3:1.
You've got it
- mass spec sorts ions by $m/e$; $A_r = \dfrac{\sum(\text{mass} \times \text{abundance})}{\sum \text{abundance}}$ (Cl = 35.5)
- the molecular ion ($M^{+}$, highest $m/e$) = the $M_r$; fragmentation gaps give lost pieces (loss 15 = $\text{CH}_3$)
- [M+1] ($^{13}\text{C}$) → number of carbons; [M+2] → halogen (Cl 3:1, Br 1:1)