How to Identify Chlorine Using Isotope Peaks in Mass Spectra

Your mass spectrum shows two molecular-ion peaks at \(m/z=78\) and \(m/z=80\). The second peak is about one-third the height of the first.

Are these two different fragments, or could they actually be the same compound?

The pattern is a strong clue that the compound contains chlorine. Naturally occurring chlorine has two main stable isotopes, chlorine-35 and chlorine-37. Because they have different masses but occur in a predictable ratio, they produce a recognisable pattern in a mass spectrum.

The quickest way to identify chlorine is to look for molecular-ion peaks separated by 2 mass units with an intensity ratio of approximately 3:1. If the molecule contains two chlorine atoms, look for three peaks with an approximate ratio of 9:6:1.

You can then examine the fragment ions to determine whether chlorine remains in the smaller pieces of the molecule.

The important distinction is that an isotope pattern identifies chlorine in a particular ion. It does not, by itself, establish the complete molecular structure.

01Why chlorine produces a distinctive isotope pattern

Chlorine has two naturally occurring stable isotopes:

• Chlorine-35, written as \({}^{35}\mathrm{Cl}\).
• Chlorine-37, written as \({}^{37}\mathrm{Cl}\).

They both contain 17 protons, but chlorine-37 has two additional neutrons. This makes chlorine-37 approximately 2 mass units heavier.

Their natural abundances are approximately:

These are representative isotopic abundances reported by the Commission on Isotopic Abundances and Atomic Weights (CIAAW). Natural samples can show small variations.

You can find the underlying isotope data in the CIAAW chlorine reference.

For a quick HSC calculation, round these abundances to:

• Chlorine-35: 75%
• Chlorine-37: 25%

This gives an approximate abundance ratio of:

\[ 75:25=3:1 \]

The two isotopes are chemically the same element, so they can occupy the same position in the same molecular structure. However, they produce ions with different masses.

If you need to revisit why atoms of the same element can have different masses, see Stable and Unstable Isotopes Explained for HSC Chemistry.

What happens when a molecule contains one chlorine atom?

Imagine a compound containing exactly one chlorine atom.

Most molecules will contain chlorine-35 because it is more abundant. A smaller proportion will contain chlorine-37.

When these molecules are ionised in a mass spectrometer, they produce two versions of the same molecular ion.

• The ion containing chlorine-35 produces the molecular-ion peak, \(M\).
• The ion containing chlorine-37 produces a second peak, \(M+2\).

The second peak is 2 mass units higher because chlorine-37 is 2 mass units heavier than chlorine-35.

The relative intensities should be approximately 3:1.

For example, if the molecular-ion peak has a relative intensity of 100, the chlorine-37 peak should have an intensity of approximately 32.

We can calculate this more accurately using the natural abundances:

\[ \frac{I_{M+2}}{I_M} = \frac{24.24}{75.76} \]

\[ =0.320 \]

Therefore:

\[ M:M+2 \approx 100:32 \]

This is the characteristic isotope pattern for one chlorine atom.

Remember that \(M\) refers to the molecular ion containing the lighter chlorine isotope. It does not mean that every molecule in the sample has exactly that mass.

02How the mass spectrometer produces these peaks

Before interpreting the pattern, it helps to understand where the measurements come from.

In a simplified electron-impact mass spectrometer, the process follows this sequence:

1. The sample is vaporised and enters the ionisation chamber.
2. A beam of electrons removes an electron from the molecules, producing positive molecular ions.
3. The positive ions are accelerated by electric fields.
4. An analyser separates the ions according to their mass-to-charge ratios.
5. A detector records the abundance of ions reaching it.

In a magnetic-sector instrument, for example, the ions travel through a magnetic field and follow curved paths. Their trajectories depend on their mass-to-charge ratios.

The quantity measured is the mass-to-charge ratio:

\[ m/z=\frac{\text{mass of ion}}{\text{charge number}} \]

Here:

• \(m\) is the mass of the ion.
• \(z\) is the magnitude of its charge.

For the singly charged ions used in the examples in this article, \(z=1\). Therefore, the numerical \(m/z\) value approximately matches the nominal mass of the ion.

This assumption matters. If an ion has a charge of \(+2\), replacing chlorine-35 with chlorine-37 changes its \(m/z\) by approximately 1 rather than 2.

The mass spectrum displays:

• Horizontal axis: \(m/z\).
• Vertical axis: relative abundance.

The height of a peak tells you how abundant that particular ion is compared with the other detected ions.

For a broader explanation of molecular ions, fragmentation, and interpreting mass spectra, see Mass Spectrometry of Organic Compounds: HSC Chemistry Guide.

03What if the molecule contains two chlorine atoms?

This is where the isotope pattern becomes even more useful.

A molecule containing two chlorine atoms can have three different combinations of chlorine isotopes.

Notice that the middle combination can occur in two ways:

• The first chlorine is chlorine-35 and the second is chlorine-37.
• The first chlorine is chlorine-37 and the second is chlorine-35.

Both combinations produce the same nominal mass.

This is why the middle peak is relatively large.

Calculating the expected intensity ratio

Using the simplified natural abundances:

\[ P({}^{35}\mathrm{Cl})=0.75 \]

\[ P({}^{37}\mathrm{Cl})=0.25 \]

The probability of getting two chlorine-35 atoms is:

\[ 0.75\times0.75=0.5625 \]

The probability of getting one isotope of each type is:

\[ 2(0.75)(0.25)=0.375 \]

The probability of getting two chlorine-37 atoms is:

\[ 0.25\times0.25=0.0625 \]

Therefore:

\[ 0.5625:0.375:0.0625 \]

Dividing each value by \(0.0625\):

\[ \boxed{9:6:1} \]

So a molecule containing two chlorine atoms should produce three molecular-ion peaks with approximate relative intensities of:

\[ \boxed{M:M+2:M+4=9:6:1} \]

Using the more precise representative abundances gives approximately \(9.8:6.3:1\). The familiar 9:6:1 ratio is a convenient HSC approximation.

Quick comparison: one chlorine versus two chlorine atoms

The decision is straightforward:

• Two peaks separated by 2 mass units, with a ratio close to 3:1, suggest one chlorine atom.
• Three peaks separated by 2 mass units, with a ratio close to 9:6:1, suggest two chlorine atoms.

These patterns describe the chlorine atoms in the particular ion being measured.

04How to use fragment peaks to identify chlorine

So far, we’ve looked at molecular ions. But what happens when the molecule breaks apart?

During electron-impact ionisation, some molecular ions receive enough energy to fragment into smaller ions.

These fragments provide another opportunity to identify chlorine.

The key is to check whether the fragment still contains a chlorine atom.

A fragment that retains chlorine

Suppose a molecular ion breaks apart and produces a fragment containing one chlorine atom.

That fragment can also exist in two isotopic forms:

• One containing chlorine-35.
• One containing chlorine-37.

The resulting fragment peaks will be separated by approximately 2 mass units, with an intensity ratio close to 3:1.

For example:

This suggests that the fragment contains one chlorine atom.

One possible assignment is:

\[ \ce{CH2^{35}Cl+} \]

Its nominal mass is:

\[ 12+2+35=49 \]

Replacing chlorine-35 with chlorine-37 gives:

\[ 12+2+37=51 \]

The two peaks therefore represent isotopic versions of the same fragment ion.

A fragment that loses chlorine

Now consider a different fragmentation pathway.

Suppose the molecular ion breaks apart and the resulting positive fragment does not contain chlorine.

That fragment will not show the characteristic chlorine-35/chlorine-37 pair.

For example, a fragment at \(m/z=43\) could be:

\[ \ce{C3H7+} \]

Its nominal mass is:

\[ (3\times12)+(7\times1)=43 \]

There is no chlorine atom in this fragment.

Therefore, there is no chlorine-derived isotope partner at \(m/z=45\).

This gives us an important distinction:

A fragment’s isotope pattern tells you about the composition of that fragment, not necessarily the total number of chlorine atoms in the original molecule.

05A practical method for interpreting chlorine isotope peaks

When you receive an unfamiliar mass spectrum, use the following method.

Step 1: Locate the molecular-ion peak.

Look for the molecular ion, \(M^{+\bullet}\), if it is present.

Do not automatically choose the tallest peak. The tallest peak is the base peak, which may be a much smaller fragment.

Step 2: Look for peaks separated by 2 mass units.

Check whether a peak has another peak approximately 2 \(m/z\) units higher.

For singly charged ions, this is the expected spacing for chlorine-35 and chlorine-37.

Step 3: Compare the relative intensities.

Use the ratio to estimate the number of chlorine atoms.

• Approximately 3:1 suggests one chlorine.
• Approximately 9:6:1 suggests two chlorines.

Step 4: Repeat the process for important fragment peaks.

Look for fragment-ion pairs that show the same chlorine isotope pattern.

This helps identify which parts of the molecule retain chlorine after fragmentation.

Step 5: Check whether the proposed interpretation makes chemical sense.

The proposed molecular formula must agree with the molecular-ion mass.

The fragment formulas must also be consistent with the atoms available in the original molecule.

Step 6: Avoid claiming a unique structure from isotope peaks alone.

The isotope pattern is evidence for the number of chlorine atoms in an ion. Different structural isomers can have the same molecular formula and similar isotope patterns.

You need the remaining fragment evidence, and potentially other structural techniques, to narrow down the structure.

06A common misconception: every pair separated by 2 must contain chlorine

A pair of peaks separated by 2 mass units is not automatically evidence of chlorine.

Consider a chlorine-free organic compound with molecular formula \(\ce{C4H10O}\).

Its molecular ion has a nominal mass of:

\[ (4\times12)+(10\times1)+16=74 \]

Suppose its spectrum contains the following peaks:

A student might look at the peaks at 43 and 45 and conclude that they indicate chlorine.

That conclusion is incorrect.

There are three problems with this interpretation:

1. The peaks at 43 and 45 are fragment peaks, not the molecular-ion pair.
2. The molecular ion is at 74, and there is no corresponding molecular-ion peak at 76.
3. Independent formula evidence identifies the compound as containing only carbon, hydrogen, and oxygen.

The similar 3:1 intensity ratio is coincidental. The two fragment peaks do not represent chlorine isotopologues.

This is why you should never identify chlorine using peak spacing alone.

You should consider both the position of the peaks and their relative intensities, then check whether the proposed ion composition is chemically possible.

Other isotopes can also contribute to peaks at higher masses. For example, sulfur-34 and combinations of heavier carbon isotopes can contribute to nominal \(M+2\) signals.

For ordinary HSC-style interpretation, the strong and repeated chlorine pattern is much more useful than an isolated peak.

07When the molecular ion is missing

Sometimes the molecular-ion peak is extremely weak or absent.

This creates a problem: you cannot directly examine the full molecular-ion isotope pattern.

However, you can still use chlorine-containing fragments.

Suppose a spectrum contains:

• A pair at \(m/z=49\) and \(m/z=51\), with a ratio close to 3:1.
• Another pair at \(m/z=63\) and \(m/z=65\), also with a ratio close to 3:1.

Both fragment families suggest ions containing one chlorine atom.

But can you conclude that the original molecule contains two chlorine atoms?

No.

The same chlorine atom can be retained in different fragment ions produced by different fragmentation events. You are not observing both fragments leaving the same individual molecule simultaneously.

Without a reliable molecular-ion pattern or other independent evidence, you cannot establish the total number of chlorine atoms in the original compound.

Improving the experiment: separating mixtures

There is another limitation worth recognising.

A mass spectrum records the ions reaching the detector. If the sample contains multiple compounds, their signals can overlap.

One way to address this for suitable volatile compounds is to connect a gas chromatograph to the mass spectrometer.

The sample first travels through a chromatography column. Different compounds emerge at different times before entering the mass spectrometer.

This helps separate their individual spectra and makes it easier to determine whether an isotope pattern belongs to one compound or results from overlapping signals from multiple compounds.

This is an important limitation of interpreting isotope patterns: a convincing pattern is strong evidence, but the sample’s purity and the identity of overlapping ions still matter.

08Questions and solutions

The following questions build from identifying one chlorine atom to interpreting more complicated molecular and fragment-ion evidence.

Question 1

A singly charged fragment ion produces two peaks at \(m/z=35\) and \(m/z=37\).

Their relative intensities are 100 and 32, respectively.

Explain what these peaks suggest about the fragment ion. Can you determine the total number of chlorine atoms in the original molecule?

Solution 1

The fragment ion contains one chlorine atom, but the total number of chlorine atoms in the original molecule cannot be determined from these peaks alone.

The two peaks are separated by 2 mass units:

\[ 37-35=2 \]

This matches the nominal mass difference between chlorine-35 and chlorine-37.

Now compare their intensities:

\[ 100:32\approx3:1 \]

This agrees with the natural abundance ratio of chlorine’s two main stable isotopes.

Therefore, the peaks are consistent with \(\ce{^{35}Cl+}\) and \(\ce{^{37}Cl+}\).

The important distinction is that these are fragment-ion measurements. They establish that the detected fragment contains chlorine, but they do not reveal whether the original molecule contained one, two, or more chlorine atoms.

Question 2

An organic compound produces the following molecular-ion isotope pattern:

A major fragment also produces two peaks:

A student argues that the original molecule must contain only one chlorine atom because the fragment peaks have an approximate 3:1 ratio.

Explain why this conclusion is incorrect. Determine the likely number of chlorine atoms in the molecular ion and in the fragment ion.

Solution 2

The molecular ion contains two chlorine atoms, while the fragment ion contains one chlorine atom.

The student’s mistake is treating the fragment’s isotope pattern as though it describes the entire original molecule.

First, examine the molecular-ion peaks.

Their relative intensities are:

\[ 100:64:10 \]

This is close to the expected two-chlorine pattern:

\[ 9:6:1 \]

The three peaks at 100, 102, and 104 therefore support the presence of two chlorine atoms in the molecular ion.

Next, examine the fragment.

Its relative intensity ratio is:

\[ 45:14 \]

Dividing both values by 14:

\[ 3.21:1 \]

This is close to the expected 3:1 ratio for a fragment containing one chlorine atom.

The fragment therefore retains one of the two original chlorine atoms.

The other chlorine atom is not present in this detected fragment. It must leave with the other products of fragmentation, although the isotope pattern alone does not establish the exact fragmentation mechanism.

The two patterns are consistent with each other:

• Molecular ion: two chlorine atoms.
• Fragment ion: one chlorine atom.

Question 3

A mass spectrum has no detectable molecular-ion peak.

However, two fragment-ion families are observed:

Assume the fragment assignments are:

\[ \ce{CH2Cl+} \]

for family A, and:

\[ \ce{C2H4Cl+} \]

for family B.

A student concludes that the original molecule must contain at least two chlorine atoms because two different chlorine-containing fragments are present.

Evaluate this conclusion. Could a compound containing only one chlorine atom produce both fragment families?

Solution 3

The conclusion is incorrect. A molecule containing only one chlorine atom can produce both fragment families through alternative fragmentation pathways.

First, examine family A:

\[ \ce{CH2Cl+} \]

Its nominal mass is:

\[ 12+2+35=49 \]

Replacing chlorine-35 with chlorine-37 produces a fragment at \(m/z=51\).

The intensity ratio is:

\[ 100:32\approx3:1 \]

This supports one chlorine atom in family A.

Now examine family B:

\[ \ce{C2H4Cl+} \]

Its nominal mass is:

\[ (2\times12)+(4\times1)+35=63 \]

Its isotope partner appears at \(m/z=65\).

The intensity ratio is:

\[ 70:22\approx3.18:1 \]

This also supports one chlorine atom in the fragment.

However, these fragments do not have to be produced by the same original molecule in the same fragmentation event.

Consider a possible parent molecule:

\[ \ce{C3H7Cl} \]

It contains only one chlorine atom.

Different molecules of this compound can break at different carbon-carbon bonds.

One pathway can produce:

\[ \ce{CH2Cl+} \]

while another can produce:

\[ \ce{C2H4Cl+} \]

In both cases, the same chlorine atom from the original molecular structure is retained in the detected fragment.

The other atoms leave as neutral fragments.

Therefore, the strongest justified conclusion is that the original molecule contains at least one chlorine atom. Its total chlorine count cannot be established because the molecular-ion peak is missing.

This is a useful example of why you must distinguish the composition of individual fragment ions from the composition of the original molecule.

Question 4

A compound has the molecular formula:

\[ \ce{C2H4Cl2} \]

A student calculates its molecular mass using the average atomic mass of chlorine:

\[ M=24+4+2(35.45) \]

\[ M=98.9 \]

They therefore expect a molecular-ion peak at approximately \(m/z=99\).

However, the measured spectrum contains three molecular-ion peaks:

Explain why the student expects the wrong peak position. Identify the isotope combination responsible for each observed peak.

Solution 4

The student is incorrectly using the average atomic mass of chlorine to predict the mass of an individual molecular ion.

A mass spectrometer detects individual ions containing particular isotope combinations. It does not detect a molecule containing an average fraction of chlorine-35 and chlorine-37.

The formula contains two chlorine atoms, so three isotope combinations are possible.

First peak: \(m/z=98\)

Both chlorine atoms are chlorine-35.

\[ 24+4+(2\times35)=98 \]

This is the most abundant isotopologue because chlorine-35 is the more common isotope.

Second peak: \(m/z=100\)

One chlorine atom is chlorine-35 and the other is chlorine-37.

\[ 24+4+35+37=100 \]

There are two ways to form this combination, so this peak is relatively intense.

Third peak: \(m/z=102\)

Both chlorine atoms are chlorine-37.

\[ 24+4+(2\times37)=102 \]

This is the least abundant combination.

The observed intensity ratio is:

\[ 100:64:10 \]

This agrees closely with the expected two-chlorine pattern of approximately:

\[ 9:6:1 \]

The student’s calculated value of 98.9 is an average molecular mass across the different isotopic combinations. It is not the mass of a particular isotopologue.

The observed peaks are therefore exactly what we would expect from a compound containing two chlorine atoms.

Question 5

An unknown sample produces three molecular-ion peaks at:

Assume the sample has not been separated into its individual components before mass spectrometry.

A student concludes that the sample must contain one pure compound with two chlorine atoms because the peaks match the 9:6:1 isotope pattern.

However, consider an alternative explanation involving two different compounds, each containing only one chlorine atom.

The proposed contributions are:

Assume that:

• Both compounds produce singly charged molecular ions.
• Each compound has the usual approximate 3:1 chlorine isotope ratio.
• The listed intensities are their contributions to the combined detector signal.
• All contributions at the same \(m/z\) are added together.

Answer the following:

1. Show how the two one-chlorine compounds could produce the observed 9:6:1 pattern.
2. Can the three observed peaks alone establish that a single molecule contains two chlorine atoms?
3. Describe an experimental change that could help distinguish the two explanations.

Solution 5

The observed 9:6:1 pattern does not prove that a single compound contains two chlorine atoms. A mixture of two different one-chlorine compounds can produce exactly the same combined pattern.

The key is that the detector adds the signals from all ions with the same \(m/z\). It does not automatically identify which original compound produced each ion.

Step 1: Calculate the combined intensity at \(m/z=100\).

Only compound A contributes to this peak.

\[ I_{100}=9 \]

Step 2: Calculate the combined intensity at \(m/z=102\).

This is where the overlap occurs.

Compound A contributes its chlorine-37 molecular-ion peak with an intensity of 3.

Compound B contributes its chlorine-35 molecular-ion peak with an intensity of 3.

Therefore:

\[ I_{102}=3+3=6 \]

Step 3: Calculate the combined intensity at \(m/z=104\).

Only compound B contributes to this peak.

\[ I_{104}=1 \]

The combined spectrum is therefore:

\[ \boxed{9:6:1} \]

This is identical to the approximate molecular-ion isotope pattern of a single compound containing two chlorine atoms.

Step 4: Decide what the experiment actually establishes.

The spectrum establishes that the detected signal is consistent with a two-chlorine molecular-ion pattern.

However, it does not establish that all three peaks belong to different isotopologues of one molecular species.

The two-compound explanation is also consistent with the data.

This is the important limitation: isotope patterns can overlap when multiple compounds contribute to the same mass spectrum.

Step 5: Improve the experimental design.

For suitable volatile compounds, gas chromatography-mass spectrometry (GC-MS) could help resolve the ambiguity.

The gas chromatograph separates the compounds before they enter the mass spectrometer.

If compounds A and B emerge at different retention times, their individual mass spectra can be recorded separately.

Each compound should then show its own one-chlorine isotope pair, rather than the overlapping three-peak pattern.

Therefore, the student needs additional evidence about sample purity or individual components before claiming that a single molecule contains two chlorine atoms.

09What to do with chlorine isotope evidence next

Chlorine isotope peaks give you a powerful way to constrain the molecular formula of an unknown compound.

For HSC-style questions, remember the central patterns:

• One chlorine atom: \(M\) and \(M+2\), approximately 3:1.
• Two chlorine atoms: \(M\), \(M+2\), and \(M+4\), approximately 9:6:1.
• A chlorine-containing fragment: its own isotope pair, approximately 3:1.
• A fragment without chlorine: no characteristic chlorine-derived isotope pair.

Most importantly, always identify whether you’re interpreting a molecular ion or a fragment ion.

Once you’ve established the likely number of chlorine atoms, your next task is to check whether the proposed molecular formula fits the molecular-ion mass and whether the major fragment ions can be formed from that formula.

That combination of isotope evidence, molecular mass, and fragmentation is much stronger than identifying a compound from one convenient peak.