Mass Spectrometry

AS · 12 min

A mass spectrometer turns molecules into positive ions and sorts them by mass. The resulting mass spectrum gives the relative atomic mass of an element from its isotopes, the relative molecular mass of a compound from its molecular ion, clues to its structure from the fragments it breaks into, the number of carbon atoms from the M+1 peak, and the presence of chlorine or bromine from the M+2 peak. You do not need to know how the instrument works, but you must be able to read every feature of a spectrum. Mass spectra appear in Paper 1 and Paper 2, often alongside infrared spectra.

Reading a mass spectrum

A mass spectrum is a bar chart:

the horizontal axis is $m/e$ , the mass-to-charge ratio of each ion. Almost all the ions have a charge of $+1$ , so $m/e$ is simply the mass of the ion;
the vertical axis is the relative abundance (or relative intensity) of each ion, often scaled so that the tallest peak is 100%.

Only positive ions reach the detector. Neutral fragments and radicals are not detected.

Relative atomic mass from isotopes

For an element, each peak is one isotope, and its height is that isotope's relative abundance. The relative atomic mass is the weighted mean:

Key result

A_r = \frac{\sum (\text{isotopic mass} \times \text{relative abundance})}{\sum \text{relative abundances}}

For chlorine, peaks at $m/e$ 35 (75.8%) and 37 (24.2%):

A_r(\ce{Cl}) = \frac{35 \times 75.8 + 37 \times 24.2}{100} = 35.5

The same method is covered in Relative masses and the mole.

The molecular ion

When an organic molecule M enters the spectrometer, a high-energy electron knocks an electron out of it, forming a positive ion with an unpaired electron, the molecular ion:

\ce{M + e- -> M+ + 2e-}

Key result

The molecular ion peak, $\ce{M+}$ , is the peak with the highest $m/e$ value (ignoring the small M+1 and M+2 peaks next to it). Its $m/e$ value equals the relative molecular mass of the compound.

The base peak is the tallest peak in the spectrum, given an abundance of 100%. It is often not the molecular ion: it is the most stable (most abundant) fragment.

Fragmentation

Molecular ions are unstable and many break apart. When a molecular ion fragments, it splits into one positive ion and one radical:

\ce{M+ -> X+ + Y^.}

Only $\ce{X+}$ is detected. The fragment peaks are at lower $m/e$ than $\ce{M+}$ , and the difference between $\ce{M+}$ and a fragment peak tells you what was lost.

Simplified mass spectra. Ethanol: molecular ion at m/e 46, base peak at 31 (CH₂OH⁺). Chloroethane: molecular ion peaks at 64 and 66 in the ratio 3 : 1, showing one chlorine atom; the pair at 49 and 51 (CH₂Cl⁺) is also 3 : 1, while the peak at 29 (C₂H₅⁺) contains no chlorine.

For ethanol ( $M_r$ 46), the main peaks are:

$m/e$	ion	formed by
46	$\ce{CH3CH2OH+}$	molecular ion
45	$\ce{C2H5O+}$	loss of H (46 − 1)
31	$\ce{CH2OH+}$ (base peak)	loss of $\ce{CH3}$ (46 − 15): $\ce{CH3CH2OH+ -> CH2OH+ + CH3^.}$
29	$\ce{C2H5+}$	loss of OH (46 − 17)
15	$\ce{CH3+}$

Common fragments to recognise

Key result

$m/e$	likely ion	$m/e$	likely ion
15	$\ce{CH3+}$	43	$\ce{C3H7+}$ or $\ce{CH3CO+}$
17	$\ce{OH+}$ (or loss of 17 = OH)	45	$\ce{COOH+}$ or $\ce{C2H5O+}$
29	$\ce{C2H5+}$ or $\ce{CHO+}$	57	$\ce{C4H9+}$ or $\ce{C2H5CO+}$
31	$\ce{CH2OH+}$ or $\ce{CH3O+}$

Several masses can be more than one ion (29 is $\ce{C2H5+}$ or $\ce{CHO+}$ ; 43 is $\ce{C3H7+}$ or $\ce{CH3CO+}$ ), so use the molecular formula and other evidence to decide.

Method

Using fragmentation to distinguish isomers

Draw each possible isomer.
For each, list the fragments formed by breaking each C–C bond (and C–O, C–X bonds), and their $m/e$ values.
Find a fragment that one isomer can give but the other cannot.
Check the spectrum for that peak.

For example, propan-1-ol ( $\ce{CH3CH2CH2OH}$ ) gives a strong peak at 31 ( $\ce{CH2OH+}$ , from breaking C1–C2), but propan-2-ol ( $\ce{CH3CH(OH)CH3}$ ) cannot form $\ce{CH2OH+}$ ; instead it gives a strong peak at 45 ( $\ce{CH3CHOH+}$ , from losing a $\ce{CH3}$ ).

The M+1 peak: counting carbon atoms

About 1.1% of all carbon atoms are carbon-13. A molecule containing one $\ce{^{13}C}$ atom is one mass unit heavier than the same molecule containing only $\ce{^{12}C}$ , so it appears as a small peak at $m/e$ one unit above the molecular ion: the M+1 peak.

The more carbon atoms in the molecule, the more likely it is that one of them is $\ce{^{13}C}$ , so the bigger the M+1 peak relative to M.

Key result

The number of carbon atoms, $n$ , in the molecule:

n = \frac{100 \times \text{abundance of } [\ce{M + 1}]^+ \text{ ion}}{1.1 \times \text{abundance of } \ce{M+} \text{ ion}}

For example, if $\ce{M+}$ has abundance 31.0 and the M+1 peak has abundance 1.37:

n = \frac{100 \times 1.37}{1.1 \times 31.0} = 4.02 \approx 4

Round to the nearest whole number. The abundances can be in any units (percent, peak heights in mm), as long as both are in the same units.

The M+2 peak: chlorine and bromine

Chlorine and bromine each have two common isotopes, two mass units apart, in roughly fixed proportions:

element	isotopes	approximate ratio
chlorine	$\ce{^{35}Cl}$ : $\ce{^{37}Cl}$	$3 : 1$
bromine	$\ce{^{79}Br}$ : $\ce{^{81}Br}$	$1 : 1$

A molecule containing one chlorine atom therefore produces two molecular ion peaks: M (containing $\ce{^{35}Cl}$ ) and M+2 (containing $\ce{^{37}Cl}$ ), with heights in the ratio $3 : 1$ . One bromine atom gives M and M+2 of equal height.

Key result

halogen atoms in the molecule	peaks	height ratio
one Cl	M, M+2	$3 : 1$
one Br	M, M+2	$1 : 1$
two Cl	M, M+2, M+4	$9 : 6 : 1$
two Br	M, M+2, M+4	$1 : 2 : 1$
one Cl and one Br	M, M+2, M+4	$3 : 4 : 1$

The two-atom ratios follow from probability. For two chlorine atoms, each is $\ce{^{35}Cl}$ with probability $\tfrac{3}{4}$ and $\ce{^{37}Cl}$ with probability $\tfrac{1}{4}$ :

both $\ce{^{35}Cl}$ (M): $\tfrac{3}{4} \times \tfrac{3}{4} = \tfrac{9}{16}$ ;
one of each (M+2): $2 \times \tfrac{3}{4} \times \tfrac{1}{4} = \tfrac{6}{16}$ (the factor 2 because either atom can be the heavier one);
both $\ce{^{37}Cl}$ (M+4): $\tfrac{1}{4} \times \tfrac{1}{4} = \tfrac{1}{16}$ .

Fragment ions that still contain the halogen also appear as pairs (for chloroethane, $\ce{CH2Cl+}$ at 49 and 51 in a $3 : 1$ ratio). Fragments that have lost the halogen appear as single peaks.

Worked examples

Routine: relative atomic mass

The mass spectrum of bromine atoms shows peaks at $m/e$ 79 and 81 with relative abundances 50.7 and 49.3. Calculate $A_r$ of bromine to one decimal place.

Solution

A_r = \frac{79 \times 50.7 + 81 \times 49.3}{50.7 + 49.3} = \frac{4005.3 + 3993.3}{100} = 79.99 \approx 80.0

(The Periodic Table value, 79.9, is slightly lower because the exact isotopic masses are 78.92 and 80.92, not whole numbers.)

Routine: molecular ion and fragments

The mass spectrum of propanone, $\ce{CH3COCH3}$ , has peaks at $m/e$ 58, 43 (base peak) and 15. (a) Identify the ion responsible for each peak. (b) Write an equation for the formation of the ion at $m/e$ 43 from the molecular ion.

Solution

(a) 58: the molecular ion, $\ce{CH3COCH3+}$ . 43: $\ce{CH3CO+}$ (58 − 15, loss of $\ce{CH3}$ ). 15: $\ce{CH3+}$ .

(b) $\ce{CH3COCH3+ -> CH3CO+ + CH3^.}$

Standard: the M+1 peak

A hydrocarbon has a molecular ion peak at $m/e$ 72 with relative abundance 40.0, and an M+1 peak with relative abundance 2.20. Find the number of carbon atoms and the molecular formula, and suggest the structure if the spectrum has a large peak at $m/e$ 57.

Solution

n = \frac{100 \times 2.20}{1.1 \times 40.0} = 5.0

Five carbons: $5 \times 12 = 60$ , leaving $72 - 60 = 12$ for hydrogen, so $\ce{C5H12}$ (pentane or an isomer).

$m/e$ 57 is $72 - 15$ : loss of a $\ce{CH3}$ group to give $\ce{C4H9+}$ . All three isomers can lose a methyl group, but 2,2-dimethylpropane, $\ce{C(CH3)4}$ , does so especially readily (giving the stable tertiary $\ce{(CH3)3C+}$ ion), so a very large 57 peak points to it. Further fragments would confirm the choice.

Standard: the M+2 peak

A halogenoalkane shows molecular ion peaks at $m/e$ 122 and 124 of equal height, and a large peak at $m/e$ 43. (a) Which halogen is present, and how many atoms? (b) Deduce the molecular formula. (c) Identify the ion at $m/e$ 43.

Solution

(a) M and M+2 in a $1 : 1$ ratio: one bromine atom.

(b) With $\ce{^{79}Br}$ : $122 - 79 = 43$ for the hydrocarbon part, $\ce{C3H7}$ . Formula $\ce{C3H7Br}$ (1-bromopropane or 2-bromopropane).

(c) $\ce{C3H7+}$ (122 − 79): the molecule has lost a bromine atom, $\ce{C3H7Br+ -> C3H7+ + Br^.}$ . Because the fragment contains no bromine, it is a single peak.

Exam-hard: distinguishing positional isomers

1-bromopropane and 2-bromopropane both give molecular ion peaks at $m/e$ 122 and 124. One of them has a pair of peaks at $m/e$ 93 and 95 (equal height) and a peak at 29; the other has a pair at 107 and 109 and a peak at 15. Assign the spectra.

Solution

1-bromopropane, $\ce{CH3CH2CH2Br}$ : breaking the C1–C2 bond gives $\ce{CH2Br+}$ ( $14 + 79 = 93$ and $14 + 81 = 95$ , equal height because the ion still contains Br) and $\ce{C2H5}$ ; the alternative split gives $\ce{C2H5+}$ at 29. So the spectrum with 93/95 and 29 is 1-bromopropane.

2-bromopropane, $\ce{CH3CHBrCH3}$ : breaking a C–C bond loses a $\ce{CH3}$ group, giving $\ce{CH3CHBr+}$ ( $28 + 79 = 107$ , and 109) or $\ce{CH3+}$ at 15. It cannot form $\ce{CH2Br+}$ . So the spectrum with 107/109 and 15 is 2-bromopropane.

Exam-hard: combining mass spectrometry and IR

Compound E contains C, H and O only. Its mass spectrum shows $\ce{M+}$ at $m/e$ 88 (abundance 40.0) and an M+1 peak (abundance 1.76), with the base peak at $m/e$ 43 and other peaks at 29 and 45. Its IR spectrum shows strong absorptions at $1740$ and $1240\ \text{cm}^{-1}$ and no O–H absorption. Identify E.

Solution

Carbon atoms: $n = \dfrac{100 \times 1.76}{1.1 \times 40.0} = 4.0$ , so four carbons.

$M_r = 88$ : $4 \times 12 = 48$ , leaving 40 for H and O. With two oxygens (32), 8 hydrogens: $\ce{C4H8O2}$ .

IR: C=O at $1740\ \text{cm}^{-1}$ (ester range) and C–O at $1240\ \text{cm}^{-1}$ , no O–H: an ester.

Fragments: 43 = $\ce{CH3CO+}$ (base peak), 29 = $\ce{C2H5+}$ , 45 = $\ce{OC2H5+}$ . These fit ethyl ethanoate, $\ce{CH3COOCH2CH3}$ . (Methyl propanoate, $\ce{CH3CH2COOCH3}$ , would instead give $\ce{C2H5CO+}$ at 57 and $\ce{COOCH3+}$ at 59.)

Watch out

Taking the base peak as the molecular ion. The molecular ion is the peak at the highest $m/e$ (apart from M+1 and M+2), not the tallest peak.
Detecting neutral fragments. Only positive ions are detected. In $\ce{M+ -> X+ + Y^.}$ , the radical $\ce{Y^.}$ is invisible; give fragment ions with a + charge.
M+1 formula upside down. $n = 100 \times (\text{M+1}) / (1.1 \times \text{M})$ . A check: the answer must be a sensible whole number, close to an integer.
Ratios the wrong way round. One Cl: M : M+2 = 3 : 1. One Br: 1 : 1. Do not confuse with the abundance of the isotopes themselves.
Missing the charge on fragments. Write $\ce{CH3CO+}$ , not $\ce{CH3CO}$ . Equations for fragmentation must balance in mass and charge.

Exam tip

Typical questions: "Identify the species responsible for the peaks at $m/e$ X and Y" (give the formula with the + charge), "Use the M+1 peak to calculate the number of carbon atoms" (show the formula and substitution), "What does the M+2 peak show?".
Fragmentation equations: molecular ion → fragment ion + radical. Both mass and charge balance.
Combined spectroscopy questions reward a clear chain of reasoning: carbon count from M+1, $M_r$ from M, functional groups from IR, structure from fragments. State each conclusion and the evidence for it.
Remember that the working of the mass spectrometer is not examined at AS; spend your time on interpreting spectra.

Summary

$m/e$ on the horizontal axis; relative abundance on the vertical axis; only positive ions are detected.
$A_r$ = weighted mean of isotopic masses using the relative abundances.
The molecular ion $\ce{M+}$ (highest $m/e$ , apart from M+1 and M+2) gives $M_r$ . The base peak is the tallest peak.
Fragmentation: $\ce{M+ -> X+ + Y^.}$ ; only $\ce{X+}$ is detected. Common fragments: 15 $\ce{CH3+}$ , 29 $\ce{C2H5+}$ / $\ce{CHO+}$ , 31 $\ce{CH2OH+}$ , 43 $\ce{C3H7+}$ / $\ce{CH3CO+}$ , 45 $\ce{COOH+}$ .
M+1 peak (from $\ce{^{13}C}$ ): $n = 100 \times \text{abundance(M+1)} / (1.1 \times \text{abundance(M)})$ .
M+2 peak: one Cl gives M : M+2 = 3 : 1; one Br gives 1 : 1; two Cl 9 : 6 : 1; two Br 1 : 2 : 1.

Practice

Question

Copper has isotopes $\ce{^{63}Cu}$ (69.2%) and $\ce{^{65}Cu}$ (30.8%). Calculate $A_r$ of copper.
Explain the difference between the molecular ion peak and the base peak.
Identify the ions responsible for the peaks at $m/e$ 15, 29, 43 and 58 in the mass spectrum of butane.
Write an equation for the fragmentation of the molecular ion of ethanol that produces the peak at $m/e$ 31.
The molecular ion peak of a compound has an abundance of 12.5 and its M+1 peak an abundance of 0.69. How many carbon atoms does the molecule contain?
A compound shows molecular ion peaks at $m/e$ 64 and 66 in a $3 : 1$ ratio. Identify the halogen and deduce the molecular formula, assuming the rest of the molecule is $\ce{C_{n}H_{2n+1}}$ .
Bromoethane has molecular ion peaks at $m/e$ 108 and 110. Explain why there are two peaks and predict their relative heights.
Propan-1-ol and propan-2-ol both have $M_r = 60$ . Explain how their mass spectra could be used to tell them apart.
A compound shows peaks at $m/e$ 98, 100 and 102 in the ratio $9 : 6 : 1$ , and a pair of peaks at 49 and 51 in the ratio $3 : 1$ , but no peaks at 83 and 85. (a) How many chlorine atoms does it contain? (b) Deduce its molecular formula, given that it contains only C, H and Cl. (c) Identify the compound, explaining your reasoning.
Compound G contains C, H and O. Its mass spectrum shows $\ce{M+}$ at $m/e$ 74 with an M+1 peak whose abundance is 3.3% of that of $\ce{M+}$ , and fragment peaks at 29, 45 and 57. Its IR spectrum shows a very broad absorption from 2500 to $3300\ \text{cm}^{-1}$ and a strong absorption at $1710\ \text{cm}^{-1}$ . Identify G, explaining the evidence from each technique.

Answers

$A_r = (63 \times 69.2 + 65 \times 30.8) / 100 = 63.6$ .
The molecular ion peak is at the highest $m/e$ (apart from the small M+1/M+2 peaks); it is produced by the whole molecule losing one electron, and its $m/e$ gives the $M_r$ . The base peak is the tallest peak, the most abundant ion, which is often a stable fragment rather than the molecular ion.
15: $\ce{CH3+}$ . 29: $\ce{C2H5+}$ . 43: $\ce{C3H7+}$ . 58: $\ce{C4H10+}$ (molecular ion).
$\ce{CH3CH2OH+ -> CH2OH+ + CH3^.}$
$n = 100 \times 0.69 / (1.1 \times 12.5) = 5.0$ , so five carbon atoms.
M : M+2 = 3 : 1: one chlorine atom. With $\ce{^{35}Cl}$ : $64 - 35 = 29 = \ce{C2H5}$ . Formula $\ce{C2H5Cl}$ , chloroethane.
Bromine has two isotopes, $\ce{^{79}Br}$ and $\ce{^{81}Br}$ , in roughly equal amounts. Molecules containing $\ce{^{79}Br}$ give the peak at 108 ( $29 + 79$ ) and those containing $\ce{^{81}Br}$ give 110. The peaks are of approximately equal height ( $1 : 1$ ).
Propan-1-ol, $\ce{CH3CH2CH2OH}$ , can lose $\ce{C2H5}$ to form $\ce{CH2OH+}$ , giving a strong peak at $m/e$ 31. Propan-2-ol, $\ce{CH3CH(OH)CH3}$ , cannot form $\ce{CH2OH+}$ ; it loses $\ce{CH3}$ to give $\ce{CH3CHOH+}$ , a strong peak at 45. (The spectrum of propan-1-ol also shows a peak at 29, $\ce{C2H5+}$ .)
(a) M, M+2, M+4 in a $9 : 6 : 1$ ratio: two chlorine atoms. (b) With two $\ce{^{35}Cl}$ : $98 - 70 = 28 = \ce{C2H4}$ . Formula $\ce{C2H4Cl2}$ . (c) The isomers are 1,1-dichloroethane, $\ce{CH3CHCl2}$ , and 1,2-dichloroethane, $\ce{CH2ClCH2Cl}$ . The 49/51 pair ( $3 : 1$ ) is $\ce{CH2Cl+}$ , containing one Cl, formed by breaking the C–C bond of 1,2-dichloroethane. 1,1-dichloroethane would break to give $\ce{CHCl2+}$ at 83, 85 and 87, which are absent. The compound is 1,2-dichloroethane.
M+1 is 3.3% of M: $n = 100 \times 3.3 / (1.1 \times 100) = 3$ carbon atoms. $M_r = 74$ : $3 \times 12 = 36$ , leaving 38 for H and O; with two O (32), six H: $\ce{C3H6O2}$ . IR: very broad O–H at $2500$ – $3300\ \text{cm}^{-1}$ (carboxyl) and C=O at $1710\ \text{cm}^{-1}$ : a carboxylic acid. Fragments: 29 $\ce{C2H5+}$ , 45 $\ce{COOH+}$ , 57 $\ce{C2H5CO+}$ (74 − 17, loss of OH). G is propanoic acid, $\ce{CH3CH2COOH}$ .

Reading a mass spectrum

Relative atomic mass from isotopes

The molecular ion

Fragmentation

Common fragments to recognise

The M+1 peak: counting carbon atoms

The M+2 peak: chlorine and bromine

Worked examples

Practice

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