Mass spectrometry

1. Overview

A mass spectrometer turns a sample into positive ions and sorts them by mass-to-charge ratio. Its output, a mass spectrum, shows which masses are present and how common each one is. This section shows how to read one: the relative atomic mass of an element from its isotope peaks and, for an organic compound, its relative molecular mass, its number of carbon atoms, any chlorine or bromine, and the groups it is built from.

Key Definitions

• Mass spectrum: a chart of the positive ions detected from a sample, showing the relative abundance of each ion (vertical axis) against its mass-to-charge ratio, m/e (horizontal axis).
• Mass-to-charge ratio (m/e): the relative mass of an ion divided by its charge; for an ion with a 1+ charge, m/e is simply the relative mass of the ion.
• Relative abundance: the height of a peak compared with the other peaks, showing how many ions of that m/e were detected; the tallest peak is often set to 100.
• Base peak: the tallest peak in a mass spectrum, usually given a relative abundance of 100; it is often a fragment ion, not the molecular ion.
• Molecular ion (M⁺): the positive ion formed when a whole molecule loses one electron; for a compound without chlorine or bromine, the m/e of the molecular ion peak equals the relative molecular mass.
• Fragmentation: the breaking up of a molecular ion into a smaller positive ion, which is detected, and a neutral species, which is not; in simple fragmentation only one bond breaks.
• [M + 1]⁺ peak: a small peak one m/e unit above the molecular ion peak, caused mainly by molecules that contain one carbon-13 atom; its height compared with the M⁺ peak gives the number of carbon atoms in the molecule.
• [M + 2]⁺ peak: a peak two m/e units above the molecular ion peak; a large one shows that the compound contains chlorine or bromine; the M⁺ and [M + 2]⁺ heights are in the ratio of about 3 : 1 for one chlorine atom and 1 : 1 for one bromine atom.
• Relative atomic mass (Ar): the weighted mean mass of an atom of an element, taking account of the abundance of each isotope, compared with one twelfth of the mass of one atom of carbon-12.

Content

What a mass spectrum shows

You do not need to know how the instrument works. Inside it, particles of the sample lose an electron to become positive ions, which are sorted by their mass-to-charge ratio, m/e, and counted. Only positive ions are detected; neutral atoms, molecules and radicals are not.

• Horizontal axis: m/e. Most ions have a 1+ charge, so their m/e is just their relative mass. An ion with a 2+ charge appears at half its mass: an ion of mass 56 with a 2+ charge gives a peak at m/e 28.
• Vertical axis: relative abundance (also labelled relative intensity or % abundance). The height of a peak shows how many ions of that m/e reached the detector. Often the tallest peak, the base peak, is set to 100 and the others are scaled to it.

Isotopes have different masses, so each isotope of an element gives its own peak, and a compound gives a small cluster of molecular ion peaks, one for each combination of isotopes its molecules can contain. The heights in a cluster follow the isotopic abundances.

The diagram is a simplified spectrum of an ester, E, used in Worked examples 1 and 2.

Isotope peaks of an element and its relative atomic mass

For an element made of single atoms, such as a metal or a noble gas, each 1+ peak sits at a relative isotopic mass (close to the mass number), and the peak heights are in the ratio of the isotopic abundances. That is all you need to calculate the relative atomic mass, $A_{\text{r}}$:

$$A_{\text{r}} = \frac{\sum (m/e \times \text{abundance})}{\sum \text{abundance}}$$

Divide by the total of the abundances. This total is 100 only when the abundances are percentages.

For example, the spectrum of lithium has peaks at m/e 6 (relative abundance 8.2) and m/e 7 (relative abundance 100):

$$A_{\text{r}} = \frac{(6 \times 8.2) + (7 \times 100)}{8.2 + 100}$$ $$= \frac{749.2}{108.2} = 6.92$$

This agrees with the Data Booklet value, 6.9, and lies much nearer 7 than 6 because lithium-7 is far more common. Topic 2.1 has more practice, including working back from $A_{\text{r}}$ to the abundances.

The molecular ion peak and Mr

When a molecule, M, loses one electron, it becomes the molecular ion, M⁺. The mass of the lost electron is negligible, so the m/e of the molecular ion peak equals the relative molecular mass, $M_{\text{r}}$, of the compound.

How to find the M⁺ peak:

• Look at the high-m/e end. M⁺ is the peak at the highest m/e, apart from its small [M + 1]⁺ peak (and the [M + 2]⁺ and [M + 4]⁺ peaks of chlorine and bromine compounds).
• Do not assume it is the tallest peak. Many molecular ions break up before they reach the detector, so the M⁺ peak is often small, and the base peak is often a fragment.

In the spectrum of E, the M⁺ peak is at m/e 88, so $M_{\text{r}}$ = 88. Chlorine and bromine compounds need more care (see the [M + 2]⁺ section).

Fragmentation

Many molecular ions are unstable and break apart. In simple fragmentation, one covalent bond in M⁺ breaks. One piece keeps the positive charge and is detected as a fragment ion. The other piece is a neutral radical, which is not detected:

[CH₃COCH₂CH₃]⁺ → CH₃CO⁺ + •CH₂CH₃

This is butanone, $M_{\text{r}}$ = 72. Either piece can keep the charge, so one bond can give two different peaks: the same C-C bond can also break to give C₂H₅⁺ (m/e 29) and a neutral CH₃CO• radical.

To identify a fragment ion:

1. Work out the mass lost: m/e of M⁺ minus m/e of the fragment. For butanone, 72 − 43 = 29, the mass of CH₂CH₃.
2. Find a group of atoms with the fragment's mass that is joined together in the molecule and comes away when one bond breaks.
3. Write it with a positive charge: CH₃CO⁺, not CH₃CO.

These fragments are worth knowing:

m/e	Possible ion	Often from
15	CH₃⁺	any CH₃ group
29	C₂H₅⁺ or CHO⁺	ethyl groups; aldehydes
31	CH₂OH⁺ or CH₃O⁺	primary alcohols; methyl esters
43	C₃H₇⁺ or CH₃CO⁺	propyl groups; ethanoates, methyl ketones
45	COOH⁺ or C₂H₅O⁺	carboxylic acids; ethyl esters
57	C₄H₉⁺ or C₂H₅CO⁺	butyl groups; propanoates, ethyl ketones

Different ions can share an m/e value (43 can be C₃H₇⁺ or CH₃CO⁺), so use the other clues, such as the molecular formula, to choose. Some peaks come from rearrangements or from fragments breaking up further; at this level, suggest ions formed by breaking one bond.

The [M + 1]⁺ peak: counting carbon atoms

About 1.1% of carbon atoms are carbon-13 (¹³C); the rest are carbon-12. A molecule containing one ¹³C atom is one mass unit heavier, so it gives a small peak one m/e unit above M⁺: the [M + 1]⁺ peak.

The more carbon atoms a molecule has, the more likely it is to contain a ¹³C atom. Each carbon atom adds about 1.1% of the height of the M⁺ peak to the height of the [M + 1]⁺ peak. So for a molecule with $n$ carbon atoms:

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

• Use the abundances (peak heights), never the m/e values.
• Heights in mm, percentages or relative abundances all work if both are on the same scale: only their ratio matters.
• Round $n$ to the nearest whole number (3.9 or 4.1 means 4): small peaks are hard to measure exactly, and other heavy isotopes, such as hydrogen-2, add a little to the [M + 1]⁺ peak.

Working backwards, a molecule with $n$ carbon atoms has an [M + 1]⁺ peak about $1.1n$% of the height of its M⁺ peak (3.3% for 3 carbon atoms).

To find the molecular formula, combine $n$ with $M_{\text{r}}$ from the M⁺ peak. Subtract the mass of the carbon atoms, $12.0n$, and share what is left between the other atoms. Then check that the number of hydrogen atoms is possible: a compound of C, H and O has at most $2n + 2$ hydrogen atoms.

Worked example 1: molecular formula of ester E

Question: Compound E contains carbon, hydrogen and oxygen only. Its spectrum is in the diagram above, and an expanded view of the high-m/e end gives:

m/e	Relative abundance
88	8.0
89	0.35

Calculate the number of carbon atoms in a molecule of E and deduce its molecular formula. ($A_{\text{r}}$: H 1.0, C 12.0, O 16.0)

Step 1: identify the peaks. The peak at 88 is M⁺, so $M_{\text{r}}$ = 88. The peak at 89 is [M + 1]⁺.

Step 2: substitute the abundances. $$n = \frac{100 \times 0.35}{1.1 \times 8.0} = \frac{35}{8.8} = 3.98$$

The nearest whole number is 4.

Step 3: share out the remaining mass. Four carbon atoms have a mass of 4 × 12.0 = 48.0, leaving 88 − 48 = 40 for H and O.

• One O atom: 40 − 16 = 24 H atoms, more than the maximum of 2(4) + 2 = 10. Impossible.
• Two O atoms: 40 − 32 = 8 H atoms. Possible.
• Three O atoms: 3 × 16 = 48, more than 40. Impossible.

Answer: E has 4 carbon atoms per molecule, and its molecular formula is C₄H₈O₂ ($M_{\text{r}}$ = 88.0).

Worked example 2: which ester?

Question: E is an ester: either ethyl ethanoate, CH₃COOCH₂CH₃, or methyl propanoate, CH₃CH₂COOCH₃. Use its peaks at m/e 43 and 45 to decide which, and identify those ions.

Step 1: break the same bond in each ester. The C-O single bond next to the C=O group gives these ions:

Ester	Ions formed	m/e
ethyl ethanoate	CH₃CO⁺ or C₂H₅O⁺	43 or 45
methyl propanoate	C₂H₅CO⁺ or CH₃O⁺	57 or 31

Step 2: compare with the spectrum. E has peaks at 43 and 45, and none at 57 or 31. Methyl propanoate cannot give an ion of mass 43 or 45 by breaking one bond: neither of its CH₃ groups is bonded to the C=O carbon, and its ethyl group is bonded to carbon, not to oxygen.

Step 3: check the other peaks. The peaks at 29 (C₂H₅⁺), 15 (CH₃⁺) and 73 (M⁺ minus CH₃, 88 − 15) could come from either ester, so they do not decide.

Answer: E is ethyl ethanoate. The peak at m/e 43 is CH₃CO⁺ and the peak at m/e 45 is C₂H₅O⁺ (CH₃CH₂O⁺).

The [M + 2]⁺ peak: chlorine and bromine

Chlorine and bromine each have two common isotopes, 2 mass units apart:

• Chlorine: chlorine-35 (³⁵Cl) and chlorine-37 (³⁷Cl), in the ratio of about 3 : 1. This is why $A_{\text{r}}$(Cl), 35.5, is much nearer 35 than 37.
• Bromine: bromine-79 (⁷⁹Br) and bromine-81 (⁸¹Br), in the ratio of about 1 : 1. This is why $A_{\text{r}}$(Br), 79.9, is close to halfway between 79 and 81.

So a compound with one chlorine atom per molecule gives two molecular ion peaks, M⁺ (with ³⁵Cl) and [M + 2]⁺ (with ³⁷Cl), with heights in the ratio of the isotopes. A large [M + 2]⁺ peak therefore signals chlorine or bromine; for compounds of only C, H, N and O it is tiny.

With more halogen atoms there are more combinations of isotopes, so there are more peaks:

Halogen atoms	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, one Br	M, M + 2, M + 4	3 : 4 : 1

For two chlorine atoms, the chance that both are ³⁵Cl is $\frac{3}{4} \times \frac{3}{4} = \frac{9}{16}$. The chance of one of each is $2 \times \frac{3}{4} \times \frac{1}{4} = \frac{6}{16}$, and the chance that both are ³⁷Cl is $\frac{1}{16}$. Hence 9 : 6 : 1.

Fragment ions follow the same rule. A fragment that still contains a Cl or Br atom shows as a pair of peaks 2 units apart, in the ratio 3 : 1 or 1 : 1. A fragment that has lost the halogen gives a single peak.

Because M⁺ contains only the lighter isotope, the $M_{\text{r}}$ of a halogen compound (calculated with $A_{\text{r}}$ = 35.5 or 79.9) lies between the m/e values of M⁺ and [M + 2]⁺.

Worked example 3: a bromine compound

Question: Compound J contains carbon, hydrogen and one halogen atom only. The main peaks at the high-m/e end of its mass spectrum are:

m/e	Relative abundance
136	20.0
137	0.9
138	19.5

There are also fragment peaks at m/e 57 (the base peak) and at m/e 93 and 95 (of almost equal height). Identify the halogen, deduce the molecular formula and $M_{\text{r}}$ of J, identify the fragment ions and suggest structures for J. ($A_{\text{r}}$: H 1.0, C 12.0, Cl 35.5, Br 79.9)

Step 1: the halogen. The peaks at 136 and 138 are 2 units apart and almost equal in height (20.0 : 19.5, about 1 : 1). This is the pattern for one bromine atom; with one chlorine atom, the second peak would be about a third of the first. So M⁺ (with ⁷⁹Br) is at 136 and [M + 2]⁺ (with ⁸¹Br) is at 138.

Step 2: the number of carbon atoms. The peak at 137 is [M + 1]⁺. $$n = \frac{100 \times 0.9}{1.1 \times 20.0} = 4.09$$

The nearest whole number is 4.

Step 3: the molecular formula. 136 − 79 (one ⁷⁹Br) − 48 (four C) = 9, so there are 9 hydrogen atoms: C₄H₉Br.

Step 4: the relative molecular mass. $$M_{\text{r}} = 48.0 + 9.0 + 79.9 = 136.9$$

This is neither 136 nor 138: each of those is one isotopic form of J, while $M_{\text{r}}$ is a weighted mean.

Step 5: the fragments.

• m/e 57 is a single peak, so the ion has no bromine. 136 − 57 = 79, the mass of ⁷⁹Br, so the ion is C₄H₉⁺, formed when the C-Br bond breaks. Molecules containing ⁸¹Br lose 81 and give the same ion at 57.
• m/e 93 and 95 are a 1 : 1 pair, so this ion contains bromine. 93 − 79 = 14, the mass of CH₂, so the ion is CH₂Br⁺. The neutral piece lost is C₃H₇ (136 − 93 = 43).

Step 6: possible structures. Breaking one bond gives CH₂Br⁺ only if the Br atom is on a CH₂ group. Of the four structural isomers of C₄H₉Br:

• CH₃CH₂CH₂CH₂Br (1-bromobutane) and (CH₃)₂CHCH₂Br (1-bromo-2-methylpropane) contain CH₂Br, so both fit.
• In CH₃CHBrCH₂CH₃ and (CH₃)₃CBr, the Br atom is on a carbon with fewer than two hydrogen atoms, so neither fits.

Answer: J contains one bromine atom and is C₄H₉Br, $M_{\text{r}}$ = 136.9. The peak at 57 is C₄H₉⁺ and the pair at 93 and 95 is CH₂Br⁺. J is 1-bromobutane or 1-bromo-2-methylpropane; these peaks cannot tell those two apart.

Key Equations

Relative atomic mass from a mass spectrum (sum over all the isotope peaks; the abundances can be percentages or relative heights): $$A_{\text{r}} = \frac{\sum (m/e \times \text{abundance})}{\sum \text{abundance}}$$

Number of carbon atoms, $n$ (both abundances on the same scale; round to the nearest whole number): $$n = \frac{100 \times \text{abundance of [M + 1]}^+}{1.1 \times \text{abundance of M}^+}$$

Mass-to-charge ratio (no units; the charge is the number of positive charges): $$m/e = \frac{\text{relative mass of ion}}{\text{charge on ion}}$$

Mass lost in a fragmentation: m/e of M⁺ − m/e of the fragment ion = mass of the neutral piece lost.

Common Mistakes to Avoid

• Mistake: putting m/e values into the carbon-number formula. Fix: use the abundances. In Worked example 1, $n = \frac{100 \times 0.35}{1.1 \times 8.0} = 3.98$, so 4; using the m/e values, $\frac{100 \times 89}{1.1 \times 88}$, gives 91.9, which is meaningless.
• Mistake: taking the tallest peak as the molecular ion peak. Fix: M⁺ is the peak at the highest m/e, apart from its small isotope peaks. In the spectrum of E, the base peak is at 43 but M⁺ is at 88.
• Mistake: writing a fragment with no charge or with a negative charge. Fix: only positive ions are detected, so write CH₃CO⁺, not CH₃CO or CH₃CO⁻.
• Mistake: suggesting a fragment whose atoms are not joined together in the molecule. Fix: check that the fragment comes away when one bond breaks. Butanone, CH₃COCH₂CH₃, has no C₃H₇ group, so its peak at m/e 43 is CH₃CO⁺, not C₃H₇⁺.
• Mistake: linking chlorine to the [M + 1]⁺ peak. Fix: ³⁷Cl is 2 units heavier than ³⁵Cl, so chlorine shows up at [M + 2]⁺. The [M + 1]⁺ peak comes from ¹³C and is present for every organic compound.
• Mistake: swapping the halogen ratios. Fix: about 3 : 1 for one Cl and 1 : 1 for one Br, as the $A_{\text{r}}$ values 35.5 and 79.9 remind you. In Worked example 3, peaks of 20.0 and 19.5 mean bromine.
• Mistake: giving the m/e of M⁺ as the $M_{\text{r}}$ of a chlorine or bromine compound. Fix: use $A_{\text{r}}$ values. For C₄H₉Br, M⁺ is at 136 and [M + 2]⁺ at 138, but $M_{\text{r}}$ = 136.9.

Exam Tips

• Calculate or show that for the number of carbon atoms: write the formula, substitute the two abundances, show the unrounded value (such as 3.98) and then state the whole number.
• Identify or suggest a fragment: give its formula with a + charge (C₂H₅⁺) and check that its atoms add up to the m/e value. If asked to draw it, draw the structure and put the + on it.
• Explain the heights of the M⁺ and [M + 2]⁺ peaks: name both isotopes and link their abundance ratio to the peak ratio, for example "M⁺ contains ³⁵Cl and [M + 2]⁺ contains ³⁷Cl; ³⁵Cl is about three times as abundant, so M⁺ is about three times as tall."
• Deduce $M_{\text{r}}$ or a structure: say what each peak shows, not just where it is ("the M⁺ peak at m/e 88 shows that $M_{\text{r}}$ = 88"; [M + 1]⁺ gives the number of carbon atoms, [M + 2]⁺ any Cl or Br, fragments the groups of atoms), then check your structure against every peak given.
• Calculate $A_{\text{r}}$ from a spectrum: multiply each m/e by its abundance, add, divide by the total abundance, and round as the question asks.