Finding Molecular Formulae from Percentage Composition and Mass Spectra

Two compounds can have exactly the same percentage composition but different molecular formulae. For example, both \(\ce{C2H4O2}\) and \(\ce{C6H12O6}\) contain approximately 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass.

So how do you decide which formula belongs to an unknown compound?

You need two different types of evidence. Percentage composition tells you the simplest ratio of atoms, while molecular mass tells you how many times that ratio repeats.

The method is:

1. Use percentage composition to calculate the empirical formula.
2. Use the molecular-ion peak in a mass spectrum to determine the molecular mass.
3. Divide the molecular mass by the empirical formula mass.
4. Multiply every empirical formula subscript by the resulting whole number.

Mass fragments and isotope patterns provide additional evidence. They help you identify the correct molecular-ion peak and check whether your proposed formula makes chemical sense.

01Why molecular mass is needed

Percentage composition tells you how much of a compound’s mass comes from each element. It does not directly tell you how many atoms of each element are present in one molecule.

For example, consider these two compounds:

Both compounds have the same empirical formula, \(\ce{CH2O}\).

However, Compound A contains two empirical formula units per molecule, while Compound B contains six.

This is why calculating an empirical formula is only the first part of the problem.

You need an independent measurement of molecular mass to determine the actual molecular formula.

02How mass spectrometry measures molecular mass

Mass spectrometry is an analytical technique that measures the mass-to-charge ratio of ions.

It is particularly useful because it can provide information about:

• The molecular mass of an unknown compound.
• The masses of smaller fragments produced when molecules break apart.
• The presence of elements with distinctive isotopic patterns, particularly chlorine and bromine.

How the apparatus works

Consider a simplified electron-ionisation mass spectrometer.

The sample enters through an inlet and is vaporised. The resulting gas passes into an ionisation chamber, where a beam of high-energy electrons strikes the molecules.

An electron can be removed from a molecule, producing a positively charged particle called a molecular ion.

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

Here:

• \(\mathrm{M}\) represents the original neutral molecule.
• \(e^-\) represents an electron.
• \(\mathrm{M}^{+\bullet}\) is the molecular ion, which has a positive charge and an unpaired electron.

The molecular ion has almost the same mass as the original molecule because the mass of the removed electron is negligible for these calculations.

After ionisation, the positively charged ions are accelerated using electric fields.

They then enter a mass analyser. In a magnetic-sector instrument, a magnetic field bends their paths according to their mass-to-charge ratios.

For ions with the same charge, lighter ions follow a tighter curved path than heavier ions.

A detector records the ions that reach it. The instrument uses these signals to produce a mass spectrum.

What does the mass spectrum show?

A mass spectrum is a graph with:

• Horizontal axis: mass-to-charge ratio, written as \(m/z\).
• Vertical axis: relative abundance of the detected ions.

The charge number is represented by \(z\). In the standard HSC-style examples below, the detected ions have a single positive charge, so \(z=+1\).

Therefore, their \(m/z\) values are numerically approximately equal to their masses in atomic mass units (u), also called daltons (Da).

For example, an ion with a mass of approximately 60 Da and a single positive charge appears at:

\[
\frac{m}{z}=\frac{60}{1}=60
\]

The tallest peak is called the base peak. Its relative abundance is defined as 100%.

However, the base peak is not necessarily the molecular ion.

This is one of the most important distinctions when interpreting mass spectra.

Molecular ions versus fragment ions

There are two important types of peaks to recognise.

Molecular-ion peak (M):

This represents the intact molecule after it has lost an electron. In electron ionisation, it is commonly written as \(\mathrm{M}^{+\bullet}\).

For a singly charged molecular ion, its \(m/z\) value gives the approximate molecular mass.

Fragment-ion peaks:

Electron ionisation can provide enough energy for bonds within the molecular ion to break.

This produces smaller positively charged fragments, which are also detected.

For example, a molecular ion at \(m/z=60\) might produce a fragment at \(m/z=43\).

The difference is:

\[
60-43=17
\]

This tells us that the detected fragment is 17 mass units lighter than the molecular ion.

The difference may correspond to the loss of a small neutral species or radical, depending on the fragmentation pathway.

The fragment is not another intact molecule. It is a smaller ion produced from the original molecular ion.

A useful prediction is that fragment ions from a singly charged molecular ion will generally appear at lower \(m/z\) values than the parent ion.

This is why you should look towards the high-mass end of the spectrum for the molecular-ion cluster, rather than automatically selecting the tallest peak.

There is one limitation: some compounds fragment so readily that their molecular-ion peak is extremely weak or absent. If the intact molecular ion cannot be identified, you cannot simply use the base peak as a substitute for its molecular mass.

What happens if the molecular ion is weak?

The energetic electron-ionisation process is useful because it produces fragments, but it can also make the original molecular ion difficult to detect.

Chemists can use softer ionisation techniques to reduce fragmentation and preserve more intact molecular ions.

An example is protonation, which produces an ion written as \([\mathrm{M+H}]^+\).

Its \(m/z\) value is approximately one unit greater than the neutral molecule’s molecular mass.

This is an extension to the standard electron-ionisation calculations in this article. Do not confuse an explicitly labelled \([\mathrm{M+H}]^+\) peak with an electron-ionisation molecular-ion peak.

03Calculate the empirical formula from percentage composition

Before using molecular mass, you need the empirical formula.

The empirical formula gives the simplest whole-number ratio of atoms in a compound.

The calculation is the same whether the percentages come from elemental analysis or are supplied directly in an examination question.

If you need to revise the percentage-to-mole calculation itself, see tutorgum’s guides to Percentage Composition and Empirical Formulae and Percentage Composition by Mass.

Step 1: Assume a 100 g sample

If a compound contains:

• 40.0% carbon
• 6.7% hydrogen
• 53.3% oxygen

Assume you have a 100 g sample.

The masses become:

• Carbon = 40.0 g
• Hydrogen = 6.7 g
• Oxygen = 53.3 g

This is a convenient assumption because each percentage becomes the corresponding mass in grams.

Step 2: Convert each mass into moles

Use:

\[
n=\frac{m}{M}
\]

where:

• \(n\) is the amount in moles (mol).
• \(m\) is the mass in grams (g).
• \(M\) is the molar mass in grams per mole (\(\mathrm{g\,mol^{-1}}\)).

Divide each element’s mass by its molar mass.

Step 3: Divide by the smallest mole amount

This gives the simplest mole ratio.

If the result is close to a whole number, round appropriately.

If you obtain a ratio such as \(1:1.5:1\), multiply every value by 2 to obtain \(2:3:2\).

Do not round values prematurely. A ratio of \(1:1.49\) is not the same as \(1:1\).

Step 4: Write the empirical formula

Use the resulting whole-number mole ratio as the subscripts.

For example:

\[
\mathrm{C:H:O}=1:2:1
\]

gives:

\[
\ce{CH2O}
\]

You now have the empirical formula. The molecular mass will tell you how many empirical formula units are present in one molecule.

04Combine empirical formula mass and molecular mass

This is the key calculation.

First, calculate the empirical formula mass by adding the atomic masses of all the atoms in the empirical formula.

For \(\ce{CH2O}\):

\[
\begin{aligned}
M_{\mathrm{emp}}
&=12.01+2(1.008)+16.00\\
&=30.026\ \mathrm{g\,mol^{-1}}
\end{aligned}
\]

Now compare this with the molecular mass of the unknown compound.

Use:

\[
k=\frac{M_{\mathrm{mol}}}{M_{\mathrm{emp}}}
\]

where:

• \(k\) is the number of empirical formula units per molecule.
• \(M_{\mathrm{mol}}\) is the molar mass of the compound.
• \(M_{\mathrm{emp}}\) is the empirical formula mass.

Both masses must use the same units.

The value of \(k\) should be a positive whole number.

Multiply every empirical formula subscript by \(k\):

\[
\text{Molecular formula}=
(\text{Empirical formula})_k
\]

For example, if \(k=2\):

\[
(\ce{CH2O})_2=\ce{C2H4O2}
\]

If \(k=6\):

\[
(\ce{CH2O})_6=\ce{C6H12O6}
\]

An important detail about molecular mass and isotope peaks

The periodic table gives average atomic masses because elements occur naturally as mixtures of isotopes.

A mass spectrum, however, can show separate peaks for individual isotopes.

For example, chlorine has two important naturally occurring isotopes:

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

A compound containing chlorine can therefore have a molecular-ion peak at one mass and another peak two mass units higher.

This means you must be careful when comparing the molecular-ion peak with a formula mass calculated using average atomic masses.

For a chlorine-containing compound, the lower molecular-ion peak represents the isotopologue containing chlorine-35. Its higher-mass partner represents the corresponding chlorine-37 isotopologue.

Use average atomic masses when calculating the average molar mass from the periodic table. Use isotope-specific masses when interpreting individual isotope peaks.

At ordinary HSC calculation precision, these values should be consistent with the same molecular formula.

05Use isotope patterns to check your formula

Isotopes are atoms of the same element with different numbers of neutrons.

They have the same chemical properties in most reactions, but different masses.

This difference creates characteristic patterns in mass spectra.

Chlorine: approximately 3:1

Natural chlorine contains approximately:

• 75% chlorine-35.
• 25% chlorine-37.

Therefore, a molecular ion containing one chlorine atom produces two main isotope peaks:

\[
M:M+2\approx3:1
\]

For example, if the molecular-ion peak is at \(m/z=78\), a second peak should appear at \(m/z=80\), with approximately one-third of its intensity.

Bromine: approximately 1:1

Natural bromine contains approximately:

• 51% bromine-79.
• 49% bromine-81.

These isotopes occur in almost equal proportions.

A compound containing one bromine atom therefore produces two main molecular-ion peaks:

\[
M:M+2\approx1:1
\]

If the molecular-ion peak is at \(m/z=122\), expect a second peak at \(m/z=124\) with approximately the same intensity.

What if there are two chlorine atoms?

This is where isotope patterns become particularly useful.

If a molecule contains two chlorine atoms, three combinations are possible:

1. Both chlorine atoms are chlorine-35.
2. One is chlorine-35 and the other is chlorine-37.
3. Both chlorine atoms are chlorine-37.

The resulting molecular-ion peaks appear at:

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

Their approximate relative intensities are:

\[
9:6:1
\]

This comes from the probabilities of the different isotope combinations.

Using the approximate 3:1 abundance ratio:

• Two chlorine-35 atoms: \(3\times3=9\).
• One of each isotope: \(3\times1+1\times3=6\).
• Two chlorine-37 atoms: \(1\times1=1\).

The approximate isotope ratios for chlorine and bromine are also documented in NIST’s reference on chlorine-bromine isotope intensities.

The important limitation of isotope patterns

An isotope pattern tells you about the isotopes in the particular ion being detected.

It does not automatically tell you how many halogen atoms are present in the entire original molecule.

For example, a molecule containing two chlorine atoms can break apart and produce a fragment that retains only one chlorine atom.

That fragment will show a 3:1 isotope pattern, even though the original molecule contained two chlorine atoms.

This distinction becomes particularly important when interpreting more complicated spectra.

06Use fragment masses as supporting evidence

Fragment peaks help you test whether your proposed molecular formula is reasonable.

The mass difference between the molecular ion and a fragment can suggest which part of the molecule has been lost.

For singly charged ions:

\[
\Delta m=m_{\mathrm{M}}-m_{\mathrm{fragment}}
\]

For example, consider:

• Molecular ion at \(m/z=60\).
• Fragment ion at \(m/z=43\).

The difference is:

\[
\Delta m=60-43=17
\]

A loss of 17 mass units is consistent with the loss of an \(\mathrm{OH}^{\bullet}\) radical in an appropriate fragmentation pathway.

The resulting fragment could be \(\ce{C2H3O+}\).

However, a peak at \(m/z=43\) does not uniquely identify this ion.

Other ions, such as \(\ce{C3H7+}\), also have a nominal mass of 43.

The proposed fragment must be consistent with the molecular formula and the possible fragmentation pathway.

This gives you a useful hierarchy of evidence:

Do not claim that one fragment peak proves a particular structure. Several structural isomers can produce the same fragment ion.

07A reliable decision process

When you are given percentage composition and a mass spectrum in an HSC question, work through the evidence in this order.

If the calculated value of \(k\) is nowhere near a whole number, stop and investigate.

You may have:

• Selected a fragment peak instead of the molecular ion.
• Misread an isotope peak as a second compound.
• Made an error in the empirical formula.
• Compared an isotope-specific mass with an average atomic mass.
• Used data that do not belong to the same pure compound.

The molecular formula should agree with all the available evidence, not just one calculation.

08Worked examples

The following numerical data and spectra are illustrative. They are original teaching examples, not records of actual laboratory measurements.

09Questions and solutions

The following questions increase in difficulty. The first tests the basic calculation, while the later questions require you to combine several types of mass-spectrometry evidence.

Question 1

An unknown organic compound contains 62.0% carbon, 10.4% hydrogen, and 27.6% oxygen by mass.

Its electron-ionisation mass spectrum shows a molecular-ion peak at \(m/z=58\).

Determine its empirical formula and molecular formula.

Use:

• C = \(12.01\ \mathrm{g\,mol^{-1}}\)
• H = \(1.008\ \mathrm{g\,mol^{-1}}\)
• O = \(16.00\ \mathrm{g\,mol^{-1}}\)

Solution 1

The empirical and molecular formulae are both \(\ce{C3H6O}\).

First, assume a 100 g sample:

\[
m(\ce{C})=62.0\ \mathrm{g}
\]

\[
m(\ce{H})=10.4\ \mathrm{g}
\]

\[
m(\ce{O})=27.6\ \mathrm{g}
\]

Convert each mass into moles:

\[
\begin{aligned}
n(\ce{C})&=\frac{62.0}{12.01}=5.162\\
n(\ce{H})&=\frac{10.4}{1.008}=10.317\\
n(\ce{O})&=\frac{27.6}{16.00}=1.725
\end{aligned}
\]

Divide each amount by 1.725 mol:

\[
\ce{C:H:O}=2.99:5.98:1
\]

This simplifies to:

\[
3:6:1
\]

Therefore:

\[
\text{Empirical formula}=\ce{C3H6O}
\]

Calculate its formula mass:

\[
\begin{aligned}
M_{\mathrm{emp}}
&=3(12.01)+6(1.008)+16.00\\
&=58.078\ \mathrm{g\,mol^{-1}}
\end{aligned}
\]

The molecular-ion peak at \(m/z=58\) indicates a molecular mass of approximately 58 Da.

Therefore:

\[
k=\frac{58.08}{58.078}\approx1
\]

The molecular formula is:

\[
\boxed{\ce{C3H6O}}
\]

The molecular formula is the same as the empirical formula because there is only one empirical formula unit per molecule.

Question 2

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

Its elemental composition is:

• 26.3% carbon
• 3.7% hydrogen
• 11.7% oxygen
• 58.3% bromine

The mass spectrum shows two molecular-ion peaks at \(m/z=136\) and \(m/z=138\), with approximately equal intensities.

A fragment peak also appears at \(m/z=57\).

Determine the molecular formula. Explain the isotope pattern and what the fragment peak suggests.

Use:

• C = \(12.01\)
• H = \(1.008\)
• O = \(16.00\)
• Br = \(79.90\)

Solution 2

The molecular formula is \(\ce{C3H5BrO}\).

The approximately equal molecular-ion peaks indicate the presence of one bromine atom.

First, assume a 100 g sample.

Convert the percentages into moles:

\[
\begin{aligned}
n(\ce{C})&=\frac{26.3}{12.01}=2.190\\
n(\ce{H})&=\frac{3.7}{1.008}=3.671\\
n(\ce{O})&=\frac{11.7}{16.00}=0.7313\\
n(\ce{Br})&=\frac{58.3}{79.90}=0.7297
\end{aligned}
\]

Divide each amount by the smallest value, 0.7297 mol:

\[
\ce{C:H:O:Br}
=
3.00:5.03:1.00:1.00
\]

The empirical formula is:

\[
\ce{C3H5BrO}
\]

Calculate its formula mass:

\[
\begin{aligned}
M_{\mathrm{emp}}
&=3(12.01)+5(1.008)+79.90+16.00\\
&=136.97\ \mathrm{g\,mol^{-1}}
\end{aligned}
\]

The average molar mass agrees with the molecular mass indicated by the isotope-specific peaks at 136 and 138.

Therefore:

\[
k=\frac{136.97}{136.97}=1
\]

The molecular formula is:

\[
\boxed{\ce{C3H5BrO}}
\]

The isotope pattern provides an important independent check.

Bromine-79 and bromine-81 occur in approximately equal natural abundances. This produces molecular-ion peaks separated by two mass units with approximately equal intensities.

The fragment peak at \(m/z=57\) is also consistent with the proposed formula.

The difference between the molecular ion and fragment is:

\[
136-57=79
\]

This corresponds to the loss of a bromine-79 radical.

A possible fragment composition is:

\[
\ce{C3H5O+}
\]

However, this fragmentation evidence does not establish one unique molecular structure. It supports a possible pathway consistent with the molecular formula.

Question 3

An unknown compound contains:

• 54.5% carbon
• 9.1% hydrogen
• 36.4% oxygen

Its electron-ionisation mass spectrum has a very weak peak at \(m/z=88\), while the base peak occurs at \(m/z=43\).

Determine the molecular formula.

Explain why using the base peak as the molecular mass would give an incorrect result.

Use:

• C = \(12.01\)
• H = \(1.008\)
• O = \(16.00\)

Solution 3

The molecular formula is \(\ce{C4H8O2}\). The peak at \(m/z=88\) represents the weak molecular ion, not the much taller fragment peak at 43.

Assume a 100 g sample.

Convert the percentages into moles:

\[
\begin{aligned}
n(\ce{C})&=\frac{54.5}{12.01}=4.538\\
n(\ce{H})&=\frac{9.1}{1.008}=9.028\\
n(\ce{O})&=\frac{36.4}{16.00}=2.275
\end{aligned}
\]

Divide by the smallest amount:

\[
\ce{C:H:O}=1.995:3.968:1
\]

This gives the simplest ratio:

\[
2:4:1
\]

Therefore, the empirical formula is:

\[
\ce{C2H4O}
\]

Calculate its empirical formula mass:

\[
\begin{aligned}
M_{\mathrm{emp}}
&=2(12.01)+4(1.008)+16.00\\
&=44.052\ \mathrm{g\,mol^{-1}}
\end{aligned}
\]

The molecular-ion peak occurs at \(m/z=88\).

Assuming a singly charged ion:

\[
M_{\mathrm{mol}}\approx88\ \mathrm{g\,mol^{-1}}
\]

Calculate the multiplier:

\[
\begin{aligned}
k&=\frac{88.0}{44.052}\\
&\approx2
\end{aligned}
\]

Multiply every empirical formula subscript by 2:

\[
(\ce{C2H4O})_2
\]

Therefore:

\[
\boxed{\ce{C4H8O2}}
\]

The tempting mistake is to select the tallest peak at \(m/z=43\).

However, the base peak represents the most abundant detected ion, not the intact molecule. A molecule can fragment readily and produce a very intense fragment while leaving only a small number of intact molecular ions.

Also, a fragment at \(m/z=43\) could have different possible ion compositions. It cannot independently establish the molecular mass.

The weak molecular-ion peak at 88 is the relevant evidence for the molecular formula.

Question 4

Two pure compounds, A and B, contain only carbon, hydrogen, and oxygen.

Both have the following percentage composition:

• 40.0% carbon
• 6.7% hydrogen
• 53.3% oxygen

Their electron-ionisation mass spectra show:

Determine the molecular formula of each compound.

Explain why identical percentage compositions and identical base-peak positions do not establish that the two compounds are the same substance.

Solution 4

Compound A has molecular formula \(\ce{C2H4O2}\), while Compound B has molecular formula \(\ce{C6H12O6}\).

Both have the same empirical formula, but different numbers of empirical formula units per molecule.

First, calculate the empirical formula from the common percentage composition.

Assume a 100 g sample:

\[
\begin{aligned}
n(\ce{C})&=\frac{40.0}{12.01}=3.331\\
n(\ce{H})&=\frac{6.7}{1.008}=6.647\\
n(\ce{O})&=\frac{53.3}{16.00}=3.331
\end{aligned}
\]

Divide by the smallest amount:

\[
\ce{C:H:O}=1:2:1
\]

Therefore:

\[
\text{Empirical formula}=\ce{CH2O}
\]

Its empirical formula mass is:

\[
\begin{aligned}
M_{\mathrm{emp}}
&=12.01+2(1.008)+16.00\\
&=30.026\ \mathrm{g\,mol^{-1}}
\end{aligned}
\]

Compound A

Its molecular-ion peak is at \(m/z=60\).

\[
\begin{aligned}
k_A&=\frac{60.0}{30.026}\\
&\approx2
\end{aligned}
\]

Multiply the empirical formula subscripts by 2:

\[
\boxed{\ce{C2H4O2}}
\]

Compound B

Its molecular-ion peak is at \(m/z=180\).

\[
\begin{aligned}
k_B&=\frac{180.0}{30.026}\\
&\approx6
\end{aligned}
\]

Multiply the empirical formula subscripts by 6:

\[
\boxed{\ce{C6H12O6}}
\]

The two compounds have identical percentage compositions because their molecular formulae contain the same simplest atom ratio.

Their molecular masses differ because one contains twice the number of empirical formula units of the other.

The identical base peaks at \(m/z=43\) do not change this conclusion.

Both compounds can produce fragment ions with a nominal mass of 43, even though their original molecular masses are very different.

This demonstrates an important limitation of mass spectrometry: a shared fragment peak does not mean two compounds have the same molecular formula or molecular structure.

Question 5

A pure organic compound is known to contain only carbon, hydrogen, oxygen, and chlorine.

Elemental analysis gives:

• 33.6% carbon
• 5.6% hydrogen
• 11.2% oxygen
• 49.6% chlorine

Its electron-ionisation mass spectrum contains two groups of peaks:

Group 1: Molecular-ion isotope cluster

Group 2: Fragment-ion isotope cluster

A student makes the following argument:

“The first group proves that the molecule contains two chlorine atoms. The second group has another 3:1 isotope pattern, so it must represent an additional chlorine atom. Therefore, the original molecule contains three chlorine atoms.”

Another student suggests that the second group must belong to a separate, mono-chlorinated impurity.

Determine the molecular formula and evaluate both interpretations.

Assume the sample is pure, the ions are singly charged, and all the listed peaks are attributable to the same compound and its fragments.

Use:

• C = \(12.01\)
• H = \(1.008\)
• O = \(16.00\)
• Cl = \(35.45\)

Solution 5

The molecular formula is \(\ce{C4H8Cl2O}\). The second isotope pattern does not indicate a third chlorine atom. It represents a fragment of the original molecule that retains one chlorine atom.

The key is to recognise that isotope patterns describe the composition of the detected ion, not automatically the composition of the original molecule.

Step 1: Determine the empirical formula from percentage composition.

Assume a 100 g sample.

\[
\begin{aligned}
n(\ce{C})&=\frac{33.6}{12.01}=2.798\\
n(\ce{H})&=\frac{5.6}{1.008}=5.556\\
n(\ce{O})&=\frac{11.2}{16.00}=0.700\\
n(\ce{Cl})&=\frac{49.6}{35.45}=1.399
\end{aligned}
\]

Divide every value by 0.700 mol:

\[
\ce{C:H:O:Cl}
=
4.00:7.94:1.00:2.00
\]

This gives:

\[
\boxed{\ce{C4H8Cl2O}}
\]

Step 2: Calculate the empirical formula mass.

\[
\begin{aligned}
M_{\mathrm{emp}}
&=4(12.01)+8(1.008)\\
&\quad+2(35.45)+16.00\\
&=143.004\ \mathrm{g\,mol^{-1}}
\end{aligned}
\]

Step 3: Interpret the molecular-ion isotope cluster.

The first group contains peaks at:

\[
142,\quad144,\quad146
\]

Their relative intensities are:

\[
9:6:1
\]

This is the characteristic pattern for two chlorine atoms.

The three peaks correspond to:

• \(m/z=142\): two chlorine-35 atoms.
• \(m/z=144\): one chlorine-35 and one chlorine-37 atom.
• \(m/z=146\): two chlorine-37 atoms.

The lower molecular-ion peak at 142 is therefore consistent with the molecular formula containing two chlorine-35 isotopes.

The average molar mass, calculated using the periodic table, is approximately \(143.0\ \mathrm{g\,mol^{-1}}\).

This agrees with the empirical formula mass.

Therefore:

\[
k=\frac{143.0}{143.004}\approx1
\]

The molecular formula is:

\[
\boxed{\ce{C4H8Cl2O}}
\]

Step 4: Explain the second isotope pattern.

The second group contains:

\[
107,\quad109
\]

with relative intensities:

\[
3:1
\]

This is consistent with a detected ion containing one chlorine atom.

But the original molecule contains two chlorine atoms.

How can both observations be correct?

The molecular ion can fragment by losing one chlorine-35 radical.

The mass difference is:

\[
142-107=35
\]

The fragment at \(m/z=107\) can therefore be represented by:

\[
\ce{C4H8ClO+}
\]

This fragment retains one chlorine atom.

If that remaining chlorine is chlorine-37 instead of chlorine-35, the fragment appears two mass units higher:

\[
107+2=109
\]

The two fragment peaks therefore produce the expected 3:1 ratio.

Step 5: Evaluate the two students’ interpretations.

The first student is correct that the molecular-ion cluster indicates two chlorine atoms.

However, the student incorrectly assumes that the second isotope pattern represents another chlorine atom in the original molecule.

The 3:1 ratio at 107 and 109 belongs to a fragment containing one chlorine atom. It does not add another chlorine atom to the molecular formula.

The second student’s impurity explanation is also unnecessary and conflicts with the stated assumption that the sample is pure and the peaks belong to the same compound and its fragments.

The complete interpretation is:

• The original molecular ion contains two chlorine atoms.
• Its isotope pattern is approximately 9:6:1.
• Fragmentation removes one chlorine atom.
• The remaining fragment contains one chlorine atom.
• Its isotope pattern is approximately 3:1.

This is the crucial distinction: the number of chlorine atoms in a fragment does not have to equal the number of chlorine atoms in the original molecule.

09What to do with the formula once you have it

Combining percentage composition and molecular mass gives you the molecular formula, but it does not necessarily give you the complete molecular structure.

Different structural isomers can have:

• The same molecular formula.
• The same molecular mass.
• Similar fragment-ion peaks.

Fragmentation patterns provide useful structural clues, but they are evidence rather than definitive proof.

The next step is to combine the molecular formula with other analytical techniques, particularly infrared spectroscopy and proton or carbon-13 NMR spectroscopy.

These techniques provide additional information about functional groups and the arrangement of atoms within the molecule.

For an HSC Chemistry response, keep the reasoning in the correct order:

1. Use percentage composition to establish the empirical formula.
2. Identify the molecular ion rather than the base peak.
3. Use molecular mass to calculate the molecular formula.
4. Check chlorine and bromine isotope patterns.
5. Use fragment masses to support possible structural arrangements without claiming that one peak proves a unique structure.

Syllabus note (2026): Mass spectrometry is included in Module 8, Analysis of Organic Substances, of the 2017 Chemistry Stage 6 syllabus. NESA schedules implementation of the Chemistry 11-12 Syllabus (2025) from 2028. This article therefore treats mass spectrometry as 2017-syllabus content for current 2026 HSC students, rather than presenting it as a requirement of the 2025 syllabus. See the NESA 2017 syllabus version log and NESA’s 2025 syllabus implementation information.