How to Identify Fragment Ions from Their Mass-to-Charge Ratios

A mass spectrum shows a strong peak at \(m/z=43\). What ion produced it? It could be \(\ce{C3H7+}\), \(\ce{C2H3O+}\), or another ion with the same nominal mass. The number alone doesn’t tell you.

To identify a fragment ion from its mass-to-charge ratio, work through four pieces of evidence: its \(m/z\), the molecular ion’s mass, the possible neutral loss, and the fragment’s isotope pattern. Then check whether your proposed ion is chemically possible.

For a singly charged ion, the numerical value of \(m/z\) is approximately its nominal mass in atomic mass units. That’s your starting point. The rest of the spectrum helps you decide which ion is actually present.

01What does a mass spectrum tell you?

Before assigning fragments, understand what the instrument has measured.

In a typical electron-impact (EI) mass spectrometry experiment, the sample is vaporised and enters an ionisation chamber. A beam of high-energy electrons passes through the sample and removes electrons from some molecules.

This produces positively charged molecular ions. Some remain intact, while others break into smaller charged fragments and neutral particles.

The positive ions then enter a mass analyser, which separates them according to their mass-to-charge ratios. A detector records the ions that arrive, and the instrument produces a mass spectrum.

The experimental sequence is:

1. The sample enters the ion source.
2. Electron impact produces molecular ions.
3. Some molecular ions fragment.
4. The mass analyser separates the resulting ions by \(m/z\).
5. The detector records their relative abundance.

The detector produces a graph of relative abundance against mass-to-charge ratio.

If you need to understand the physics of how the analyser separates ions, see How a Mass Spectrometer Separates Ions by Mass-to-Charge Ratio.

Molecular ion versus fragment ion

Two types of peaks are particularly important.

The molecular ion provides information about the mass of the original molecule.

The fragment ions provide information about which parts of that molecule can remain as charged particles.

Here’s a common trap. Is the tallest peak necessarily the molecular ion?

No. The tallest peak is the base peak. A particularly stable fragment can be more abundant than the original molecular ion.

The molecular ion is often found at a higher \(m/z\) than its fragments, but it can be quite weak or even absent if the molecule fragments readily.

For the calculations below, assume positive-ion EI spectra and singly charged ions unless stated otherwise.

01Convert the mass-to-charge ratio into an ion mass

The mass-to-charge ratio is written as:

\[ \frac{m}{z}=\frac{m_{\text{ion}}}{|z|} \]

Where:

• \(m_{\text{ion}}\) is the mass of the ion, measured in unified atomic mass units (u).
• \(z\) is the charge number of the ion.
• \(|z|\) is the magnitude of that charge number.

For a singly charged positive ion, \(z=+1\). Therefore:

\[ \frac{m}{z}=\frac{m_{\text{ion}}}{1} \]

So the numerical \(m/z\) value is approximately equal to the ion’s nominal mass in u.

For example, a singly charged ion at \(m/z=43\) has a nominal mass of 43 u.

However, this does not automatically tell you its chemical formula.

Calculating possible ion formulas

Use the following approximate atomic masses when working with nominal masses:

Suppose you observe a fragment at \(m/z=43\).

Two possible ion formulas are:

\[ \ce{C3H7+} \]

Its nominal mass is:

\[ 3(12)+7(1)=43 \]

Alternatively:

\[ \ce{C2H3O+} \]

Its nominal mass is:

\[ 2(12)+3(1)+16=43 \]

Both ions have the same nominal mass.

This is why identifying a fragment ion requires more than matching a number to a formula. You must also consider the atoms available in the original molecule.

An important qualification: the examples above use singly charged ions. A doubly charged ion with a mass of 86 u also produces a peak at \(m/z=43\).

Always check the charge state before treating \(m/z\) as the ion’s actual mass.

02Use the molecular ion to calculate the neutral loss

Once you have identified the molecular ion, you can compare its mass with the fragment ion.

The difference gives the nominal mass lost during fragmentation, provided both ions have the same charge state.

For singly charged molecular and fragment ions:

\[ \Delta m = (m/z)_M-(m/z)_F \]

Here:

• \((m/z)_M\) is the molecular ion’s mass-to-charge ratio.
• \((m/z)_F\) is the fragment ion’s mass-to-charge ratio.
• \(\Delta m\) is the nominal mass of the lost neutral species.

The neutral species might be a radical or a stable molecule.

Remember that the detected fragment retains the charge. The other part of the original molecule leaves as a neutral species.

03Use isotope patterns to identify which atoms remain

Mass differences are useful, but isotope patterns provide another important clue.

Some elements naturally occur as different isotopes. These isotopes have different masses, so ions containing them produce additional peaks in a mass spectrum.

Two particularly useful elements are chlorine and bromine.

Chlorine isotope pattern

Chlorine has two naturally occurring stable isotopes:

• Chlorine-35, with a nominal mass of 35 u.
• Chlorine-37, with a nominal mass of 37 u.

Their natural abundances produce an approximate 3:1 ratio.

If a molecular ion contains one chlorine atom, you expect two peaks separated by two mass units.

For example:

The same pattern can appear in fragment ions that retain chlorine.

If a fragment contains chlorine, you should look for its own M and M+2 pair.

If the fragment has lost chlorine, that characteristic pair should disappear.

Bromine isotope pattern

Bromine also has two naturally occurring stable isotopes:

• Bromine-79, with a nominal mass of 79 u.
• Bromine-81, with a nominal mass of 81 u.

These occur in approximately equal abundances.

Consequently, a bromine-containing ion typically produces two peaks separated by two mass units, with an approximate 1:1 intensity ratio.

Comparing the isotope patterns

For two chlorine atoms, the three possible isotope combinations are:

• Two \({}^{35}\mathrm{Cl}\) atoms.
• One \({}^{35}\mathrm{Cl}\) and one \({}^{37}\mathrm{Cl}\).
• Two \({}^{37}\mathrm{Cl}\) atoms.

Their approximate relative proportions are 9:6:1.

These ratios are approximate. Other isotopes, overlapping peaks, and instrument resolution can affect the measured intensities.

Also remember that isotope peaks are not necessarily different fragments. They can be different isotopic versions of the same ion.

04A common trap: the same m/z can represent different ions

Let’s return to the original problem of identifying a peak at \(m/z=43\).

Imagine a compound with molecular formula \(\ce{C3H6O}\). Its molecular ion appears at:

\[ m/z=58 \]

A prominent fragment appears at:

\[ m/z=43 \]

You might immediately suggest \(\ce{C3H7+}\), because:

\[ 3(12)+7=43 \]

But check the molecular formula.

The original molecule contains only six hydrogen atoms.

A fragment containing seven hydrogen atoms cannot be produced from that molecule through ordinary fragmentation without an external hydrogen source.

The proposed formula is therefore inconsistent with the parent molecule.

Consider the alternative:

\[ \ce{C2H3O+} \]

Its nominal mass is:

\[ 2(12)+3+16=43 \]

This formula is consistent with the atoms available in the original molecule.

The neutral loss is:

\[ 58-43=15\ \mathrm{u} \]

A methyl radical has a mass of 15 u, so the fragmentation is consistent with:

\[ \ce{C3H6O^{+\bullet} -> C2H3O+ + CH3^.} \]

The ion \(\ce{C2H3O+}\) can be represented as an acetyl cation.

This example demonstrates an important distinction:

A matching mass is not the same as a valid fragment formula.

You must check whether the proposed ion can actually be formed from the original molecule.

There is another limitation worth remembering. Even if you correctly assign a fragment formula, you have not necessarily identified the original molecule’s structure.

Different structural isomers can produce ions with the same elemental composition. Rearrangements during fragmentation can also produce ions that are not explained by simply breaking one bond.

Mass spectrometry provides structural evidence, not an automatic identification of every bond in the original molecule.

A practical decision process

When an exam question asks you to identify a fragment ion, use this sequence:

1. Check the charge state. Establish whether the ion is singly or multiply charged.
2. Find the molecular ion. Identify \(M^{+\bullet}\), if it is present.
3. Calculate the fragment’s nominal mass. Use its \(m/z\) and charge state.
4. Generate possible formulas. Use the elements and atom counts available in the parent molecule.
5. Calculate the neutral loss. Compare the molecular ion and fragment masses, accounting for their charge states.
6. Check isotope patterns. Look for chlorine, bromine, or carbon isotope evidence.
7. Check chemical plausibility. Make sure your proposed fragmentation is consistent with the original molecule.

If more than one formula fits, do not claim that the peak uniquely proves one structure.

A useful exam response identifies the most plausible assignment and explains the evidence supporting it.

What happens when one mass spectrum is not enough?

A conventional mass spectrum records many ions produced from the sample. It does not directly establish which precursor ion produced every observed fragment.

Tandem mass spectrometry, or MS/MS, improves this experimental control.

The instrument first selects a particular precursor ion. That ion enters a collision region, where collisions with a gas can cause fragmentation. A second mass analyser then measures the resulting product ions.

This provides a more direct connection between a chosen precursor and its fragments.

However, even tandem mass spectrometry does not guarantee that every structural isomer can be distinguished.

06Questions and solutions

The following questions move from straightforward fragment calculations to isotope interpretation and charge-state reasoning. Assume nominal masses and positive ions unless a question states otherwise.

Question 1

An EI mass spectrum of hexane, \(\ce{C6H14}\), contains a molecular ion peak at \(m/z=86\) and a fragment ion peak at \(m/z=57\).

Assuming both ions are singly charged, identify a plausible fragment-ion formula and calculate the nominal mass and formula of the neutral species lost.

Solution 1

The likely fragment is \(\ce{C4H9+}\), formed by loss of an ethyl radical, \(\ce{C2H5^.}\).

First, calculate the nominal mass of the proposed fragment:

\[ m(\ce{C4H9+})=4(12)+9(1) \]

\[ =57\ \mathrm{u} \]

This matches the observed peak at \(m/z=57\).

Next, calculate the mass difference:

\[ \Delta m=86-57 \]

\[ \boxed{\Delta m=29\ \mathrm{u}} \]

An ethyl radical has a nominal mass of:

\[ m(\ce{C2H5^.})=2(12)+5=29\ \mathrm{u} \]

The proposed fragmentation is therefore:

\[ \ce{C6H14^{+\bullet} -> C4H9+ + C2H5^.} \]

The atom balance is consistent with the original hexane molecule.

This is a plausible assignment, although the peak does not distinguish between different structural arrangements of the \(\ce{C4H9+}\) ion.

Question 2

An EI mass spectrum of pentane, \(\ce{C5H12}\), contains the following peaks:

A student claims that the molecular ion has a mass of 43 u because this is the tallest peak.

Identify the molecular ion and the base peak. Assign a plausible formula to the fragment at \(m/z=43\), and calculate the neutral loss.

Solution 2

The molecular ion is at \(m/z=72\), while the base peak is the fragment at \(m/z=43\).

The tallest peak does not necessarily represent the original molecule. It represents the most abundant detected ion.

Step 1: Identify the molecular ion.

The nominal mass of pentane is:

\[ m(\ce{C5H12})=5(12)+12 \]

\[ =72\ \mathrm{u} \]

Therefore, the peak at 72 corresponds to the molecular ion.

Step 2: Assign the fragment ion.

A suitable formula for the peak at 43 is:

\[ \ce{C3H7+} \]

Checking its nominal mass:

\[ m(\ce{C3H7+})=3(12)+7 \]

\[ =43\ \mathrm{u} \]

This matches the observed fragment peak.

Step 3: Calculate the neutral loss.

\[ \Delta m=72-43 \]

\[ \boxed{\Delta m=29\ \mathrm{u}} \]

The lost species can be an ethyl radical:

\[ \ce{C2H5^.} \]

The proposed fragmentation is:

\[ \ce{C5H12^{+\bullet} -> C3H7+ + C2H5^.} \]

The key misconception is confusing abundance with mass. The molecular ion can be weak because the molecule fragments readily, while a particularly stable fragment becomes the base peak.

Question 3

A compound has the molecular formula \(\ce{C4H9Cl}\).

Its illustrative EI mass spectrum contains the following peaks:

Assume all ions are singly charged and the spectrum contains no significant overlapping peaks.

Answer the following:

a) Explain what the peaks at 92 and 94 indicate.

b) Assign a plausible formula to the ions at 63 and 65.

c) Calculate the neutral loss associated with the fragment at 63.

d) Assign a plausible formula to the ion at 57 and explain why it does not have a corresponding chlorine isotope peak at 59.

Solution 3

The molecular ion contains one chlorine atom, the fragment at 63 retains chlorine, and the fragment at 57 does not.

Step 1: Interpret the molecular ion isotope pattern.

The molecular formula is \(\ce{C4H9Cl}\).

Using chlorine-35:

\[ m(\ce{C4H9^{35}Cl})=48+9+35 \]

\[ =92\ \mathrm{u} \]

Using chlorine-37:

\[ m(\ce{C4H9^{37}Cl})=48+9+37 \]

\[ =94\ \mathrm{u} \]

The two peaks are separated by two mass units and have an approximate 3:1 ratio.

This supports the presence of one chlorine atom.

Step 2: Identify the fragment at 63.

A suitable formula is:

\[ \ce{C2H4^{35}Cl+} \]

Its nominal mass is:

\[ 2(12)+4+35=63\ \mathrm{u} \]

The corresponding chlorine-37 ion has a mass of:

\[ 2(12)+4+37=65\ \mathrm{u} \]

This explains the pair at 63 and 65.

The approximate 3:1 ratio also supports the assignment.

Step 3: Calculate the neutral loss.

\[ \Delta m=92-63 \]

\[ \boxed{\Delta m=29\ \mathrm{u}} \]

A possible neutral species with this mass is an ethyl radical:

\[ \ce{C2H5^.} \]

The proposed fragmentation is:

\[ \ce{C4H9^{35}Cl^{+\bullet} -> C2H4^{35}Cl+ + C2H5^.} \]

The fragment retains chlorine, which explains its isotope pair.

Step 4: Identify the fragment at 57.

A suitable formula is:

\[ \ce{C4H9+} \]

Its nominal mass is:

\[ 4(12)+9=57\ \mathrm{u} \]

The difference from the molecular ion is:

\[ 92-57=35\ \mathrm{u} \]

This is consistent with the loss of a chlorine-35 radical.

The fragment contains carbon and hydrogen but no chlorine. It therefore does not produce the characteristic chlorine isotope pair at 57 and 59.

The important observation is that the isotope pattern belongs to the fragment ion itself. A chlorine-containing parent molecule does not mean every fragment must contain chlorine.

Question 4

An unknown compound contains only carbon, hydrogen, and oxygen.

Its EI mass spectrum provides the following information:

• Molecular ion at \(m/z=72\).
• M+1 isotope peak at \(m/z=73\), with approximately 4.4% of the molecular ion’s relative intensity.
• A prominent fragment at \(m/z=43\).

Assume the M+1 intensity is dominated by naturally occurring carbon-13, contributing approximately 1.1% per carbon atom.

a) Estimate the number of carbon atoms in the original molecule.

b) Determine its molecular formula.

c) Give two possible fragment-ion formulas at \(m/z=43\) that satisfy the parent molecule’s atom inventory.

d) Explain why the mass spectrum data provided do not uniquely identify the fragment’s elemental composition.

Solution 4

The molecule most likely contains four carbon atoms and has the formula \(\ce{C4H8O}\). However, the fragment at \(m/z=43\) cannot be assigned a unique formula from nominal mass alone.

Step 1: Estimate the number of carbon atoms.

Each carbon atom contributes approximately 1.1% to the M+1 peak through carbon-13.

Therefore:

\[ \text{Number of C atoms} \approx \frac{4.4}{1.1} \]

\[ \boxed{4\text{ carbon atoms}} \]

Step 2: Determine the molecular formula.

The nominal molecular mass is 72 u.

Four carbon atoms contribute:

\[ 4(12)=48\ \mathrm{u} \]

The remaining mass is:

\[ 72-48=24\ \mathrm{u} \]

Since the molecule contains oxygen, one oxygen atom contributes 16 u.

The remaining mass is:

\[ 24-16=8\ \mathrm{u} \]

This corresponds to eight hydrogen atoms.

Therefore:

\[ \boxed{\ce{C4H8O}} \]

Step 3: Find two possible fragment formulas.

The first possibility is:

\[ \ce{C3H7+} \]

Its mass is:

\[ 3(12)+7=43\ \mathrm{u} \]

The second possibility is:

\[ \ce{C2H3O+} \]

Its mass is:

\[ 2(12)+3+16=43\ \mathrm{u} \]

Both have the correct nominal mass.

Step 4: Check the parent atom inventory.

If the fragment is \(\ce{C3H7+}\), the remaining neutral composition is \(\ce{CHO^.}\), with a nominal mass of 29 u.

If the fragment is \(\ce{C2H3O+}\), the remaining neutral composition is \(\ce{C2H5^.}\), also with a nominal mass of 29 u.

Both are atom-balanced possibilities.

Consequently, the molecular ion, M+1 peak, and nominal fragment mass do not distinguish these two assignments.

You would need further evidence, such as additional fragment peaks, a more specific fragmentation pattern, or complementary structural information.

The M+1 calculation also remains an estimate rather than an exact carbon count, particularly when other isotope contributions or overlapping peaks are significant.

Question 5

Challenge: charge state and isotope patterns

A compound has the molecular formula:

\[ \ce{C4H8Cl2} \]

Its nominal molecular mass is 126 u.

A mass spectrometer is capable of detecting both singly and doubly charged positive ions.

The spectrum contains an isotope cluster at:

No corresponding cluster is observed at 126, 128, or 130.

A student confidently concludes:

The peak at 63 must be a fragment ion with a mass of 63 u. The molecule has therefore lost 63 u during fragmentation.

Evaluate this conclusion.

Determine the most defensible assignment of the observed ion, using both its isotope pattern and mass-to-charge ratio. Explain why the apparent mass difference is misleading.

Solution 5

The most defensible assignment is a doubly charged molecular ion, \(\mathrm{C_{4}H_{8}Cl_{2}}^{2+}\), rather than a singly charged fragment with a mass of 63 u.

The key is that the isotope spacing and isotope ratios must be interpreted together.

Step 1: Work out the expected molecular isotope pattern.

The molecule contains two chlorine atoms.

Using chlorine-35 for both atoms gives:

\[ m=4(12)+8+2(35) \]

\[ =126\ \mathrm{u} \]

Replacing one chlorine-35 atom with chlorine-37 increases the mass by two units:

\[ 128\ \mathrm{u} \]

Replacing both chlorine-35 atoms gives:

\[ 130\ \mathrm{u} \]

The expected isotope pattern for two chlorine atoms is approximately:

\[ 9:6:1 \]

Therefore, the molecular ion isotopologues have nominal masses of 126, 128, and 130 u, with those relative proportions.

Step 2: Consider what happens if the molecular ion has a charge of +2.

The mass-to-charge equation is:

\[ m/z=\frac{m_{\text{ion}}}{|z|} \]

For the first isotopologue:

\[ m/z=\frac{126}{2}=63 \]

For the second:

\[ m/z=\frac{128}{2}=64 \]

For the third:

\[ m/z=\frac{130}{2}=65 \]

These are exactly the three observed peaks.

Their relative intensities also match the expected 9:6:1 pattern for two chlorine atoms.

Step 3: Check the isotope spacing.

This is the decisive evidence.

Replacing chlorine-35 with chlorine-37 increases the ion’s actual mass by 2 u.

However, because the ion has a charge of +2, its measured \(m/z\) increases by only:

\[ \Delta(m/z)=\frac{2}{2}=1 \]

That explains the equal spacing between 63, 64, and 65.

For a singly charged ion, the same chlorine isotope substitutions would produce peaks separated by two \(m/z\) units.

Step 4: Evaluate the student’s conclusion.

The student has assumed that \(m/z=63\) means an ion mass of 63 u.

That assumption is incorrect because the charge state has not been considered.

The observed cluster is consistent with an ion of mass 126 u and charge +2. It contains both chlorine atoms and the complete carbon-hydrogen framework of the original molecule.

There is therefore no evidence of a 63 u neutral loss.

The most defensible assignment is:

\[ \boxed{\mathrm{C_{4}H_{8}Cl_{2}}^{2+}} \]

This question extends the usual introductory EI model, which generally assumes singly charged ions. The important lesson is that isotope spacing can provide evidence about charge state as well as elemental composition.

Once you can assign individual fragment formulas, the next step is to connect them into a complete fragmentation pattern. Several consistent fragment assignments provide much stronger structural evidence than one isolated peak.

You can then compare that evidence with other analytical techniques, particularly infrared spectroscopy and NMR, to narrow down the possible molecular structures.

For the underlying physics of mass-to-charge measurements, see How a Mass Spectrometer Separates Ions by Mass-to-Charge Ratio and Thomson’s Charge-to-Mass Experiment: HSC Physics Explained.