How to Find Molecular Mass from a Mass Spectrum

An unknown organic compound produces five peaks in an electron-ionisation mass spectrum: \(m/z=15\), 29, 43, 57, and 72. The tallest peak is at 43, while the peak at 72 is barely visible.

Which peak tells you the molecular mass?

It is usually the peak at 72, not the tallest one. The tallest peak is called the base peak, and it often represents a fragment of the original molecule rather than the intact molecule.

To find molecular mass from a mass spectrum, you need to identify the molecular ion peak, distinguish it from fragment peaks, and check whether isotope patterns have shifted the apparent position of the parent ion.

For a standard electron-ionisation (EI) mass spectrum containing singly charged ions, a molecular ion peak at \(m/z=72\) indicates a nominal molecular mass of approximately \(72\ \mathrm{u}\), giving a relative molecular mass of \(M_r=72\).

Let’s work through how to recognise that peak, what the other peaks tell you, and what to do when the molecular ion is difficult to identify.

01What does a mass spectrometer actually measure?

A mass spectrometer measures the mass-to-charge ratio of ions. It does not directly weigh neutral molecules.

This distinction explains why a molecule must first be ionised before its mass can be measured.

In a typical electron-ionisation mass spectrometer, the process follows four main stages.

Step 1: Vaporisation

A small sample of the compound enters the instrument and is converted into a gas.

The sample travels into a chamber maintained under a high vacuum. This reduces collisions between particles as they move through the instrument.

Step 2: Ionisation

An electron beam passes through the chamber and collides with the gaseous molecules.

An electron can knock an electron out of a molecule, producing a positively charged ion:

\[
\mathrm{M}+e^-\rightarrow\mathrm{M}^{+\cdot}+2e^-
\]

Here:

• \(\mathrm{M}\) represents the original neutral molecule.
• \(e^-\) represents an electron.
• \(\mathrm{M}^{+\cdot}\) represents the positively charged molecular ion.

The molecular ion has lost one electron but has not yet broken apart.

Step 3: Separation

The positive ions are accelerated by an electric field and enter a mass analyser.

In a common magnetic-sector instrument, a magnetic field bends the paths of the moving ions. The amount of bending depends on their mass-to-charge ratios.

Ions with different \(m/z\) values follow different paths, allowing them to be separated.

Step 4: Detection

The separated ions reach a detector, which records their signals.

The instrument uses these signals to produce a mass spectrum.

The important observation is that the instrument detects ions, not intact neutral molecules.

Some molecules survive ionisation without breaking apart. Others fragment into smaller positively charged ions.

That gives us two important types of peaks.

02Molecular ion peaks versus fragment peaks

Imagine that a molecule has a nominal mass of 72.

If it loses one electron but remains intact, its molecular ion has approximately the same mass as the original molecule.

We call this the molecular ion peak, represented by \(M^{+\cdot}\).

However, the ionisation process can also provide enough energy to break covalent bonds within the molecule.

This produces smaller positively charged particles called fragment ions.

Each fragment ion can produce its own peak in the mass spectrum.

The distinction is straightforward:

The base peak is not necessarily the molecular ion

This is one of the most common mistakes when interpreting mass spectra.

The base peak is simply the tallest peak in the spectrum. It is assigned a relative abundance of 100%.

Other peaks are measured relative to it.

A base peak at \(m/z=43\) means that the ion producing that signal was detected more strongly than the other ions.

It does not mean the original molecule has a molecular mass of 43.

Consider the illustrative spectrum introduced earlier.

The peak at 43 is the base peak because it has the greatest relative abundance.

However, the peak at 72 is the candidate molecular ion peak because it is the highest-mass ion in this simplified spectrum and is consistent with the intact molecule.

Therefore:

\[
M_r=72
\]

The molecular mass is approximately \(72\ \mathrm{u}\).

Notice that the molecular ion peak can be much smaller than the base peak. That is completely normal.

What can a fragment peak tell you?

Fragment peaks are not just unwanted signals. They provide information about how the molecule is put together.

For example, suppose the molecular ion has \(m/z=72\), while a fragment ion has \(m/z=43\).

The difference is:

\[
72-43=29
\]

This suggests that a neutral fragment with a nominal mass of 29 has been lost.

One possible explanation is the loss of an ethyl radical, \(\ce{C2H5^{\cdot}}\), leaving a positively charged fragment.

However, do not jump straight from a fragment mass to a unique structural assignment.

For example, two possible ions with nominal mass 43 are:

• \(\ce{C3H7+}\)
• \(\ce{C2H3O+}\)

Both produce a peak at \(m/z=43\).

The mass spectrum alone cannot distinguish these possibilities at nominal mass resolution.

Fragmentation patterns support structural identification, but they do not automatically prove a particular structure.

If you need to refresh how covalent bonds hold atoms together within individual molecules, see Covalent Molecular Structures and Intermolecular Forces.

03How to calculate molecular mass from the molecular ion peak

The horizontal axis of a mass spectrum is labelled \(m/z\).

Here:

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

The molecular ion produced by ordinary electron ionisation usually has a charge of \(+1\).

Therefore, for a singly charged ion:

\[
\frac{m}{z}=\frac{m}{1}=m
\]

The numerical \(m/z\) value therefore corresponds approximately to the molecular mass in unified atomic mass units (\(\mathrm{u}\)).

The mass of the electron removed during ionisation is negligible for the nominal masses used in HSC calculations.

For example:

Remember the distinction between these quantities:

• Molecular mass is expressed in \(\mathrm{u}\).
• Relative molecular mass, \(M_r\), has no unit.
• Molar mass is expressed in \(\mathrm{g\,mol^{-1}}\).

For example, a molecule with a nominal molecular mass of \(72\ \mathrm{u}\) has \(M_r=72\) and a nominal molar mass of \(72\ \mathrm{g\,mol^{-1}}\).

The relationship above assumes a singly charged molecular ion. If an ion has a charge of \(+2\), its \(m/z\) value is approximately half its mass in \(\mathrm{u}\).

For standard HSC-style EI questions, the molecular ions are generally treated as singly charged.

04How isotope patterns help you identify the molecular ion

Sometimes finding the molecular ion is slightly more complicated.

A molecule can contain atoms that naturally exist as different isotopes.

Isotopes have the same number of protons but different numbers of neutrons. This means they have different masses.

When molecules containing these isotopes are ionised, they can produce several molecular ion peaks rather than just one.

Two particularly useful elements for mass spectrum interpretation are chlorine and bromine.

Chlorine isotope patterns

Chlorine has two naturally occurring stable isotopes that are particularly important in mass spectrometry:

• Chlorine-35 (\(^{35}\mathrm{Cl}\))
• Chlorine-37 (\(^{37}\mathrm{Cl}\))

Their approximate natural abundances are:

This gives an approximate abundance ratio of:

\[
^{35}\mathrm{Cl}:^{37}\mathrm{Cl}=3:1
\]

The isotopes differ in mass by two units.

Therefore, a molecule containing one chlorine atom produces two molecular ion peaks separated by two \(m/z\) units.

Their approximate relative intensities are:

\[
M:M+2=3:1
\]

For example, a molecule with a molecular ion peak at \(m/z=78\) and one chlorine atom would be expected to produce another peak at \(m/z=80\).

The peak at 80 is not a second molecular mass.

It represents the same molecule containing the heavier chlorine isotope.

Bromine isotope patterns

Bromine also has two important stable isotopes:

• Bromine-79 (\(^{79}\mathrm{Br}\))
• Bromine-81 (\(^{81}\mathrm{Br}\))

Their natural abundances are approximately equal.

This produces an approximate ratio of:

\[
^{79}\mathrm{Br}:^{81}\mathrm{Br}=1:1
\]

A molecule containing one bromine atom therefore produces two molecular ion peaks separated by two \(m/z\) units, with approximately equal intensities.

Compare the two patterns:

This is a useful distinction in HSC Chemistry questions.

What happens when there are two halogen atoms?

The isotope pattern becomes more interesting when a molecule contains two chlorine or two bromine atoms.

For two chlorine atoms, three combinations are possible:

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

These produce three molecular ion peaks:

\[
M,\quad M+2,\quad M+4
\]

Their approximate intensity ratio is:

\[
9:6:1
\]

For two bromine atoms, the corresponding pattern is approximately:

\[
1:2:1
\]

The important point is that you should examine the entire isotope cluster, not just two neighbouring peaks.

The M+1 peak and estimating the number of carbon atoms

There is another isotope pattern worth recognising.

Carbon has two naturally occurring isotopes:

• Carbon-12, which makes up approximately 98.9% of natural carbon.
• Carbon-13, which makes up approximately 1.1%.

If a molecule contains carbon atoms, a small proportion of its molecular ions will contain one \(^{13}\mathrm{C}\) atom instead of \(^{12}\mathrm{C}\).

This creates a small peak at \(M+1\).

For relatively small organic molecules, the approximate relationship is:

\[
\frac{I_{M+1}}{I_M}\times100\%
\approx1.1n_C\%
\]

where:

• \(I_{M+1}\) is the intensity of the \(M+1\) peak.
• \(I_M\) is the intensity of the molecular ion peak.
• \(n_C\) is the number of carbon atoms.

For example, if the \(M+1\) peak has approximately 3.3% of the intensity of the molecular ion peak:

\[
n_C\approx\frac{3.3}{1.1}=3
\]

The molecule therefore contains approximately three carbon atoms.

This is an estimate rather than an exact rule. Other naturally occurring isotopes can also contribute to the \(M+1\) signal, so the approximation works best for relatively simple organic compounds.

05A reliable method for finding molecular mass

You can now combine the information into a consistent method.

When given a mass spectrum in an HSC Chemistry question, follow these steps.

Step 1: Check what type of mass spectrum you have.

For the standard method in this article, assume electron ionisation and singly charged positive ions.

If a different ionisation method is specified, check whether the detected ion is actually the molecular ion or an ion such as \([M+H]^+\).

Step 2: Look towards the higher \(m/z\) values.

The molecular ion is usually found towards the high-mass end of an EI spectrum.

Do not automatically select the tallest peak.

Step 3: Check for isotope patterns.

Look for:

• An \(M+1\) peak, which can help estimate the number of carbon atoms.
• An \(M+2\) peak with approximately a 3:1 ratio, suggesting one chlorine atom.
• An \(M+2\) peak with approximately a 1:1 ratio, suggesting one bromine atom.
• Multiple peaks at \(M\), \(M+2\), and \(M+4\), which may indicate multiple halogen atoms.

Identify the molecular ion using the isotope cluster rather than assuming the furthest-right peak is the parent molecule.

Step 4: Calculate the molecular mass.

For a singly charged molecular ion:

\[
\text{Molecular mass}\approx m/z
\]

Use the lower-mass member of the molecular ion isotope cluster as the nominal molecular mass.

Step 5: Use fragment peaks as supporting evidence.

Compare fragment masses with the molecular ion mass.

A difference can suggest the mass of a neutral species lost during fragmentation.

Use this information to support possible structural assignments, not to claim that one peak proves an entire molecular structure.

Step 6: Check whether the molecular ion is actually present.

Sometimes the molecular ion fragments too readily to produce a detectable peak.

If there is no convincing molecular ion, you cannot simply assume that the highest observed fragment is the molecular mass.

You need additional evidence or another analytical method.

06What mass spectrometry cannot tell you by itself

A mass spectrum is extremely useful, but it has limitations.

There are three situations where you need to be particularly careful.

The molecular ion peak might be missing

Electron ionisation can provide enough energy to fragment a molecule extensively.

If the molecular ion breaks apart before reaching the detector, there may be no clearly identifiable \(M\) peak.

In that situation, the highest observed \(m/z\) value is not necessarily the molecular mass.

Modern mass spectrometry can use softer ionisation techniques to reduce fragmentation.

For example, electrospray ionisation can produce a protonated molecule, \([M+H]^+\), rather than the radical cation produced by ordinary EI.

This can help preserve information about the original molecular mass.

However, you must account for the additional proton when interpreting the measured \(m/z\).

This is an extension beyond the standard EI interpretation used in the worked examples.

Different molecules can have the same molecular mass

A molecular mass of 72 does not automatically mean that the compound has a particular molecular formula.

For example, the following formulas all have a nominal molecular mass of 72:

\[
\ce{C4H8O}
\]

\[
\ce{C3H4O2}
\]

\[
\ce{C3H8N2}
\]

The molecular ion peak establishes the mass, but it does not independently identify which formula is correct.

Other information, such as elemental composition, isotope patterns, infrared spectroscopy, and NMR spectroscopy, can help distinguish the possibilities.

Fragment peaks do not uniquely identify structures

A peak at \(m/z=43\) does not prove that the molecule contains a particular functional group.

Different ions can have the same nominal mass.

You need to consider several peaks together, their possible fragmentation pathways, and other analytical evidence.

The strongest interpretation is one that explains the molecular ion, isotope pattern, and important fragment peaks consistently.

07Questions and solutions

The following questions progress from straightforward molecular-ion identification to more demanding interpretation of isotope patterns and fragmentation.

All numerical spectra are illustrative.

Question 1

An unknown compound produces the following electron-ionisation mass spectrum.

Assume the molecular ion is present and all ions have a charge of \(+1\).

(a) Identify the molecular ion peak.

(b) Calculate the molecular mass and relative molecular mass.

(c) Identify the base peak and explain why it does not represent the molecular mass.

Solution 1

The molecular ion peak is at \(m/z=46\), giving a molecular mass of approximately \(46\ \mathrm{u}\) and a relative molecular mass of 46. The base peak is at \(m/z=31\).

(a) The molecular ion is the highest-mass ion in this simplified spectrum:

\[
M^{+\cdot}=m/z=46
\]

(b) Since the molecular ion has a charge of \(+1\):

\[
\text{Molecular mass}=46\ \mathrm{u}
\]

Therefore:

\[
M_r=46
\]

(c) The base peak is at \(m/z=31\), with a relative abundance of 100%.

It represents the most strongly detected ion in the spectrum.

It does not represent the original molecule because it may be a fragment formed during ionisation.

The important distinction is between the mass of the intact molecule and the abundance of an individual fragment ion.

Question 2

An unknown organic compound has the following illustrative mass spectrum.

Assume the compound contains only carbon, hydrogen, and chlorine. The molecular ion is present, and all ions have a charge of \(+1\).

(a) Identify the molecular ion peak.

(b) Determine whether the compound contains chlorine.

(c) Estimate the number of carbon atoms.

(d) Determine the molecular formula.

Solution 2

The molecular ion peak is at \(m/z=64\). The isotope pattern indicates one chlorine atom, and the estimated molecular formula is \(\ce{C2H5Cl}\).

(a) The molecular ion cluster consists of peaks at 64, 65, and 66.

The peak at 64 represents \(M\), while the peak at 66 represents \(M+2\).

Therefore:

\[
\text{Molecular mass}\approx64\ \mathrm{u}
\]

(b) Compare the molecular ion and \(M+2\) intensities:

\[
\frac{I_M}{I_{M+2}}
=
\frac{100}{32}
\]

\[
=3.125
\]

This is approximately a 3:1 ratio.

It indicates the presence of one chlorine atom.

(c) The \(M+1\) peak provides an estimate of the number of carbon atoms.

\[
\frac{I_{M+1}}{I_M}\times100\%
=
\frac{2.2}{100}\times100\%
\]

\[
=2.2\%
\]

Therefore:

\[
n_C\approx\frac{2.2}{1.1}=2
\]

The compound contains approximately two carbon atoms.

(d) We know the compound contains two carbon atoms and one chlorine atom.

Their combined nominal mass is:

\[
2(12)+35=59
\]

The remaining mass is:

\[
64-59=5
\]

Therefore, the molecular formula is:

\[
\boxed{\ce{C2H5Cl}}
\]

Check:

\[
M_r=2(12)+5(1)+35=64
\]

The formula matches the molecular ion mass and isotope pattern.

The peak at 66 is particularly important. Ignoring it would remove the strongest evidence for chlorine.

Question 3

An unknown organic compound produces the following illustrative EI mass spectrum.

Assume the compound contains only carbon, hydrogen, and bromine. The molecular ion is present, and all ions have a charge of \(+1\).

(a) Identify the molecular ion peak and calculate the molecular mass.

(b) Explain what the peaks at 136 and 138 suggest.

(c) Estimate the number of carbon atoms.

(d) Determine the molecular formula.

(e) Explain why the fragment at \(m/z=57\) is useful evidence, without claiming that it proves the complete structure.

Solution 3

The molecular ion is at \(m/z=136\), giving a molecular mass of approximately \(136\ \mathrm{u}\). The isotope pattern indicates one bromine atom, and the molecular formula is \(\ce{C4H9Br}\).

(a) The molecular ion cluster appears at 136, 137, and 138.

The peak at 136 represents \(M\), while the peak at 138 represents \(M+2\).

Therefore:

\[
\text{Molecular mass}\approx136\ \mathrm{u}
\]

(b) Compare the intensities of the two main molecular ion peaks:

\[
\frac{I_M}{I_{M+2}}
=
\frac{34}{33}
\]

\[
\approx1.03
\]

This is approximately a 1:1 ratio.

That is characteristic of a molecule containing one bromine atom.

The separate peaks at 79 and 81 also have approximately equal intensities, which is consistent with bromine-containing fragment ions.

(c) Estimate the number of carbon atoms using the \(M+1\) peak:

\[
\frac{I_{M+1}}{I_M}\times100\%
=
\frac{1.5}{34}\times100\%
\]

\[
\approx4.41\%
\]

Therefore:

\[
n_C\approx\frac{4.41}{1.1}=4
\]

The molecule contains approximately four carbon atoms.

(d) We know the compound contains four carbon atoms and one bromine atom.

Their combined nominal mass is:

\[
4(12)+79=127
\]

The remaining mass is:

\[
136-127=9
\]

Therefore, the molecular formula is:

\[
\boxed{\ce{C4H9Br}}
\]

Check:

\[
M_r=4(12)+9(1)+79=136
\]

(e) The fragment at \(m/z=57\) is consistent with a positively charged \(\ce{C4H9+}\) ion.

One possible fragmentation pathway involves the loss of a neutral bromine-containing fragment from the original molecular ion.

The difference in mass is:

\[
136-57=79
\]

This is consistent with the loss of a bromine-79 atom as a neutral radical.

The fragment therefore provides supporting evidence for a brominated organic compound.

However, it does not identify the precise arrangement of atoms in the original molecule. Different structural isomers can produce the same fragment mass.

Question 4

An unknown organic compound produces the following EI mass spectrum.

No convincing molecular ion peak is detected.

A student argues:

“The highest \(m/z\) value is 71, so the molecular mass must be \(71\ \mathrm{u}\).”

(a) Evaluate the student’s conclusion.

(b) Explain why the molecular ion might be missing.

(c) Describe one experimental approach that could help determine the molecular mass.

(d) Explain what information the fragment peaks can still provide.

Solution 4

The student’s conclusion is not justified. The molecular mass cannot be determined confidently from this EI spectrum alone because the molecular ion has not been identified.

(a) The highest observed peak position is \(m/z=71\).

However, this only establishes that an ion with a mass-to-charge value of 71 was detected.

It does not establish that the ion represents the intact molecule.

The ion could be a fragment produced from a larger molecular ion.

Therefore, reporting \(M_r=71\) would be an unsupported conclusion.

(b) The molecular ion may be missing because electron ionisation provides enough energy for extensive fragmentation.

The original molecular ion may break apart before it reaches the detector.

Some compounds produce particularly weak molecular ion peaks, making them difficult to identify.

(c) One possible approach is to use a softer ionisation technique, such as electrospray ionisation, if the compound is suitable.

This can produce a protonated molecular ion rather than causing extensive fragmentation.

For example, if a singly charged ion is explicitly identified as \([M+H]^+\) at \(m/z=89\), the neutral molecule has a nominal molecular mass of approximately:

\[
89-1=88\ \mathrm{u}
\]

The proton contributes approximately one mass unit.

This provides an independent estimate of the original molecular mass.

(d) The fragment peaks still provide useful structural information.

For example, peaks at 29, 43, and 57 may be consistent with different carbon-containing fragments.

Comparing their masses and possible neutral losses can help identify features of the original molecule.

However, the fragment peaks cannot independently establish the complete molecular formula or structure.

The key lesson is that an absent molecular ion is a limitation of the measurement, not proof that the molecule has the same mass as its largest observed fragment.

Question 5

A particularly challenging unknown compound is analysed using electron ionisation.

Independent elemental analysis establishes that the compound contains only carbon, hydrogen, and chlorine.

The mass spectrum produces the following illustrative data:

Assume all ions have a charge of \(+1\).

Two students propose different interpretations.

• Student A argues that the molecular ion is at \(m/z=100\), with the peak at 102 representing the \(M+2\) chlorine isotope.
• Student B argues that the molecular ion is at \(m/z=98\), with peaks at 100 and 102 forming a multiple-chlorine isotope pattern.

Determine which interpretation is correct.

Your answer must:

(a) Identify the molecular ion and calculate the molecular mass.

(b) Determine how many chlorine atoms are present using the complete isotope pattern.

(c) Estimate the number of carbon atoms using the \(M+1\) peak.

(d) Determine the molecular formula.

(e) Explain what the fragment peaks at 63 and 65 reveal about the fragmentation process.

(f) State whether the mass spectrum uniquely identifies the structural isomer.

Solution 5

Student B is correct. The molecular ion is at \(m/z=98\), the compound contains two chlorine atoms, and its molecular formula is \(\ce{C2H4Cl2}\). The fragment pattern supports the loss of one chlorine-containing fragment, but the spectrum cannot distinguish the structural isomers.

(a) Identify the molecular ion.

The molecular ion cluster contains peaks at:

\[
98,\quad99,\quad100,\quad101,\quad102
\]

The important observation is that the peaks at 98, 100, and 102 have relative intensities of:

\[
100:66:11
\]

The molecular ion is at \(m/z=98\).

Therefore:

\[
\text{Molecular mass}\approx98\ \mathrm{u}
\]

and:

\[
M_r=98
\]

The peaks at 100 and 102 are heavier isotope combinations of the same molecular ion.

(b) Determine the number of chlorine atoms.

For two chlorine atoms, the expected isotope pattern is approximately:

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

The observed pattern is:

\[
100:66:11
\]

Dividing by approximately 11 gives:

\[
9.1:6:1
\]

This is an excellent match.

Therefore, the molecule contains two chlorine atoms.

The interpretation follows from the entire three-peak cluster, not simply from finding two peaks separated by two mass units.

Student A’s interpretation is tempting because a single chlorine atom produces an \(M\) and \(M+2\) pair.

However, if 100 were the molecular ion and 102 its \(M+2\) peak, their expected ratio for one chlorine atom would be approximately 3:1.

The observed ratio is:

\[
\frac{66}{11}=6:1
\]

Furthermore, that interpretation fails to explain the strong peak at 98.

The complete isotope pattern supports two chlorine atoms with \(M=98\).

(c) Estimate the number of carbon atoms.

The \(M+1\) peak is at \(m/z=99\).

Its intensity is 2.2%, compared with 100% for the molecular ion peak.

Therefore:

\[
\frac{I_{M+1}}{I_M}\times100\%=2.2\%
\]

Estimate the number of carbon atoms:

\[
n_C\approx\frac{2.2}{1.1}
\]

\[
n_C\approx2
\]

The molecule contains approximately two carbon atoms.

(d) Determine the molecular formula.

We have established that the compound contains:

• Two carbon atoms.
• Two chlorine atoms.
• Hydrogen atoms.

Calculate the mass contributed by carbon and chlorine:

\[
2(12)+2(35)=94
\]

The remaining mass is:

\[
98-94=4
\]

Therefore, the compound contains four hydrogen atoms.

The molecular formula is:

\[
\boxed{\ce{C2H4Cl2}}
\]

Check:

\[
\begin{aligned}
M_r&=2(12)+4(1)+2(35)\\
&=24+4+70\\
&=98
\end{aligned}
\]

This agrees with the molecular ion peak.

(e) Interpret the fragment peaks.

Consider the peaks at 63 and 65.

Their relative intensities are:

\[
42:14=3:1
\]

This indicates a fragment containing one chlorine atom.

A possible fragment is:

\[
\ce{C2H4Cl+}
\]

Its nominal mass is:

\[
2(12)+4(1)+35=63
\]

The corresponding heavier chlorine isotope produces the peak at 65.

The difference between the original molecular ion and this fragment is:

\[
98-63=35
\]

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

The peaks at 35 and 37 provide further evidence of chlorine-containing ions.

Together, these observations support a molecule containing two chlorine atoms, with fragmentation producing ions that retain one chlorine atom.

(f) Can we identify the structural isomer?

No. The molecular formula \(\ce{C2H4Cl2}\) is consistent with more than one structural arrangement.

For example:

• 1,1-dichloroethane.
• 1,2-dichloroethane.

Both have the same molecular formula and nominal molecular mass.

Their fragmentation patterns may differ, but the simplified data provided do not establish a unique structure.

Additional evidence, particularly NMR and infrared spectroscopy, would be needed to distinguish the possibilities.

The important reasoning is that the molecular ion isotope cluster establishes the molecular mass and provides evidence for the number of chlorine atoms. Fragment peaks then provide additional structural clues, but they do not automatically identify the complete molecule.

08What to remember when interpreting a mass spectrum

When a question asks you to find molecular mass from a mass spectrum, the first thing to identify is the molecular ion peak.

For standard singly charged EI spectra:

\[
\boxed{\text{Molecular mass}\approx m/z\text{ of }M^{+\cdot}}
\]

But identifying that peak requires more than choosing the tallest or furthest-right signal.

Remember these key distinctions:

Also remember that the molecular ion may be weak or absent. If you cannot identify it confidently, do not invent a molecular mass from the highest fragment.

09Where to go next

Finding molecular mass is an important part of identifying an unknown organic compound, but it is only one piece of the evidence.

Mass spectrometry helps establish molecular mass, isotope composition, and possible fragmentation pathways.

Infrared spectroscopy can then help identify functional groups, while proton and carbon-13 NMR provide information about the different chemical environments and connectivity of atoms.

This is how several analytical techniques work together to build a convincing molecular structure rather than relying on one promising-looking peak.

Syllabus note: This article follows the Chemistry Stage 6 Syllabus (2017), Module 8, which includes mass spectrometry in the analysis of organic substances. The 2017 syllabus remains in use until implementation of the replacement Chemistry 11-12 Syllabus (2025), beginning in 2028. Students in later cohorts should check their school’s current course program. See NESA’s syllabus information and 2017 syllabus errata.