How to Read Isotope Peaks: Mass, Charge and Abundance

Neon has an average atomic mass of approximately 20.20, yet its mass spectrum contains a strong peak near 20 and a weaker peak near 22. Why doesn’t the sample produce just one peak at 20.20?

The answer is that the sample contains different isotopes of neon. Each isotope produces its own signal, and the signals reveal two different pieces of information.

Read the horizontal position of a peak to determine its mass-to-charge ratio (\(m/z\)). Read its vertical height to determine its relative intensity, which indicates relative abundance when comparing equivalent isotope ions.

A taller peak means that isotope is more abundant. It does not mean that the isotope is heavier.

Let’s work through how to interpret these peaks, including the effect of ionic charge, and how to avoid confusing mass number, actual isotope mass, and relative abundance.

01How mass spectrometry revealed different isotopes

The problem of identifying isotopes became much easier once scientists could measure the masses of individual atoms rather than relying on the average mass of a large sample.

An important step came from the work of Francis Aston, who developed the mass spectrograph and used it to investigate isotopes of elements such as neon.

The experiment: separating atoms by mass

Imagine a sample of neon gas entering a low-pressure discharge tube.

An electrical discharge removes electrons from some neon atoms, producing positively charged ions. These ions then travel through the apparatus as a narrow beam.

The path through Aston’s mass spectrograph can be understood in this order:

1. Ion source: Neon gas is introduced into a low-pressure tube, where an electrical discharge produces positive ions.
2. Narrow slits: The ions pass through two narrow, parallel openings to form a controlled beam.
3. Electric field: The beam passes between electrically charged plates, which deflect the ions.
4. Magnetic field: A magnetic field bends the moving ions in the opposite direction.
5. Photographic plate: The separated ions reach a recording surface, producing lines at different positions.

Before looking at the results, make a prediction.

If every neon atom had exactly the same mass and charge, how many distinct positions would you expect the ions to reach?

You would expect one main position. However, if neon contained isotopes with different masses, their ions would follow slightly different paths and produce separate signals.

That is what the experiment revealed.

The two prominent lines corresponded to neon isotopes with approximate mass numbers 20 and 22.

The lighter isotope produced the stronger signal. This indicated that neon-20 was more abundant than neon-22.

Why the original approach needed improvement

Aston initially investigated whether the different neon components could be separated using fractional distillation over charcoal cooled with liquid air.

Even after thousands of operations, this approach did not produce convincing separation. The isotopes had almost identical chemical properties, and the difference in their physical properties was very small.

An earlier positive-ray method developed by J. J. Thomson could produce separate traces, but its mass measurements were only accurate to about 1%.

Aston improved the design using narrow parallel slits and electromagnetic focusing. This produced sharper, better-separated signals and allowed much more accurate mass measurements.

The improved instrument provided convincing evidence that neon contained atoms with different masses.

The historical details are described in Aston’s 1922 Nobel lecture on mass spectra and isotopes.

A useful scale estimate: what does an average mass of 20.20 tell us?

Suppose we simplify the historical neon sample to two isotopes:

• Neon-20, with a mass number of 20.
• Neon-22, with a mass number of 22.

Let \(x\) represent the fraction of neon-22.

The average mass is approximately:

\[
20.20=20(1-x)+22x
\]

Expanding:

\[
20.20=20+2x
\]

\[
x=0.10
\]

Therefore, approximately:

• 90% of the atoms are neon-20.
• 10% of the atoms are neon-22.

This is a simplified estimate using rounded isotope masses and a two-isotope model.

Notice the important distinction: the average atomic mass is 20.20, but there is no isotope with a mass number of 20.20.

The average describes the mixture. The individual peaks describe the isotopes within it.

02What the two axes of an isotope spectrum actually mean

A mass spectrum contains two axes. You must interpret them separately.

1. Horizontal position: mass-to-charge ratio

The horizontal position tells you the mass-to-charge ratio, written as:

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

Here:

• \(m\) is the mass of the ion.
• \(z\) is the magnitude of its charge, expressed as a number of elementary charges.

For example:

• A singly charged ion has \(z=1\).
• A doubly charged ion has \(z=2\).
• A triply charged ion has \(z=3\).

For a singly charged isotope ion, the numerical \(m/z\) value is approximately equal to its isotope mass in atomic mass units.

This is why a singly charged magnesium-24 ion produces a peak near \(m/z=24\).

However, a doubly charged magnesium-24 ion produces a peak near \(m/z=12\).

The isotope has not become lighter. Its charge has changed.

In general, for ions whose masses are close to their integer mass numbers:

\[
\frac{m}{z}\approx\frac{A}{z}
\]

where \(A\) is the mass number.

2. Vertical height: relative abundance

The vertical axis tells you how strong the signal is for each ion.

When comparing isotopes of the same element with the same charge, a stronger signal generally indicates a greater abundance of that isotope.

For example, consider three peaks with relative intensities of 100, 20, and 10.

The first isotope produces the strongest signal. It is the most abundant of the three, assuming comparable detection responses.

However, there is an important distinction between relative intensity and percentage abundance.

A mass spectrum often sets its tallest peak to 100. This does not mean that the tallest peak represents 100% of all the ions.

To calculate percentage abundance, you must first add the intensities.

For the three peaks:

\[
I_{\text{total}}=100+20+10=130
\]

The percentage abundance of the first isotope is:

\[
\frac{100}{130}\times100=76.9\%
\]

The other two percentages are:

\[
\frac{20}{130}\times100=15.4\%
\]

\[
\frac{10}{130}\times100=7.7\%
\]

The percentages now add to 100%.

Remember: If the vertical axis already gives percentage abundance, you do not need to perform this conversion. The distinction matters when the axis is labelled relative intensity.

03How to read isotope notation and ionic charge together

A common source of mistakes is reading isotope notation correctly but forgetting to account for the charge of the ion.

Consider the following magnesium ion:

\[
{}^{25}_{12}\mathrm{Mg}^{2+}
\]

Each part gives you different information.

The mass number is the total number of protons and neutrons:

\[
A=Z+N
\]

Rearranging:

\[
N=A-Z=25-12=13
\]

The \(2+\) charge tells us that the magnesium ion has lost two electrons.

It does not mean that the mass number has decreased by two.

This gives us an important distinction:

• Changing the isotope changes the number of neutrons and therefore the mass.
• Changing the charge changes the number of electrons and therefore the mass-to-charge ratio.

The number of protons remains unchanged.

If isotope notation itself is still confusing, revise Stable and Unstable Isotopes Explained for HSC Chemistry.

04Mass number is not exactly the same as isotope mass

You will often see mass spectrum peaks at approximately 24, 25, or 26 for magnesium isotopes.

These values are useful approximations, but there is a subtle distinction.

The mass number is the total number of protons and neutrons. It is always a whole number.

The actual isotopic mass is a measured mass expressed in atomic mass units (u). It is not necessarily a whole number.

For example, an isotope with mass number 35 has an actual isotopic mass slightly different from 35 u.

This difference arises because the masses of the particles inside a nucleus do not simply add together to give the exact mass of the nucleus.

For typical HSC calculations, use the isotope masses supplied in the question. If only integer mass numbers are provided, use those as approximations.

Do not automatically treat a measured decimal mass as a mass number.

06Recognising isotope patterns in chlorine and bromine

So far, we have considered individual atoms being ionised.

However, mass spectrometry can also produce molecular ions. These contain several atoms, and the isotopes of those atoms create characteristic groups of peaks.

Two particularly useful examples involve chlorine and bromine.

One chlorine atom: the 3:1 pattern

Natural chlorine contains two major isotopes:

• Chlorine-35.
• Chlorine-37.

Their approximate relative abundances produce a ratio of 3:1.

Suppose a molecular ion contains one chlorine atom.

If it contains chlorine-35, its molecular ion appears at mass \(M\).

If it contains chlorine-37 instead, the molecular ion appears two mass units higher, at \(M+2\).

Therefore, the expected pattern is:

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

These peaks do not indicate two different compounds.

They represent the same molecular structure containing different isotopes of chlorine.

One bromine atom: the 1:1 pattern

Bromine also has two major isotopes:

• Bromine-79.
• Bromine-81.

Their natural abundances are approximately equal.

A molecule containing one bromine atom therefore produces two molecular-ion peaks separated by two mass units, with approximately equal intensities.

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

This gives you a useful diagnostic comparison:

Do not confuse the \(+2\) in \(M+2\) with a \(2+\) ionic charge.

They describe different things:

• \(M+2\) means the peak is two mass-to-charge units above the molecular ion’s lighter isotope peak.
• \(2+\) means the ion carries two positive charges, which halves its \(m/z\) value compared with an equivalent singly charged ion.

What happens when a molecule contains two chlorine atoms?

This is where isotope patterns become more interesting.

A molecule containing two chlorine atoms can have three different combinations:

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

Assuming the isotope proportions are approximately 75% chlorine-35 and 25% chlorine-37, the probabilities are:

The factor of 2 in the middle calculation matters.

There are two ways to arrange one chlorine-35 and one chlorine-37 atom, and both produce the same molecular mass.

The resulting peak ratio is:

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

This pattern is particularly useful because it shows that the number of isotope peaks depends not only on the number of isotopes available, but also on how many atoms of that element are present in the molecule.

07A reliable method for reading isotope peaks

When you encounter a mass spectrum in an HSC question, work through these decisions in order.

1. Read the horizontal axis. Identify the \(m/z\) values of the peaks.
2. Identify the ion charge. Determine whether the ions are \(1+\), \(2+\), or another charge state.
3. Determine the approximate ion masses. Account for the charge before assigning mass numbers.
4. Identify the isotopes. Use the element symbol and atomic number to write the isotope notation.
5. Compare peak heights. Identify the most abundant isotope or isotopic combination.
6. Check the vertical-axis label. Decide whether the data are percentage abundances or relative intensities.
7. Calculate the weighted average if required. Use fractional abundances rather than unconverted relative intensities.

A few common mistakes are worth checking before you move on.

08Questions and solutions

The following questions build from straightforward isotope identification to more demanding problems involving charge, relative intensity, and molecular isotope combinations.

Question 1

A sample of oxygen produces three singly charged atomic ions with the following percentage abundances:

Oxygen has atomic number 8.

a) Write the isotope notation for each ion.

b) Calculate the number of neutrons in each isotope.

c) Identify the most abundant isotope and explain how the spectrum tells you this.

Solution 1

The three isotopes are oxygen-16, oxygen-17, and oxygen-18, and oxygen-16 is the most abundant.

Because the ions are singly charged, their \(m/z\) values approximately equal their mass numbers.

a) Isotope notation

Using the atomic number of oxygen, \(Z=8\):

\[
{}^{16}_{8}\mathrm{O}^{+}
\]

\[
{}^{17}_{8}\mathrm{O}^{+}
\]

\[
{}^{18}_{8}\mathrm{O}^{+}
\]

b) Number of neutrons

Use:

\[
N=A-Z
\]

For oxygen-16:

\[
N=16-8=8
\]

For oxygen-17:

\[
N=17-8=9
\]

For oxygen-18:

\[
N=18-8=10
\]

c) Most abundant isotope

Oxygen-16 is the most abundant because its peak has the greatest relative abundance, at 96%.

The vertical axis represents percentage abundance, so the values already indicate the proportion of each isotope in the sample.

The peak at 16 is tallest because oxygen-16 makes up the greatest proportion of the sample, not because it has the greatest mass.

Question 2

A sample of copper produces two singly charged atomic ions.

The mass spectrum gives the following illustrative relative intensities:

Copper has atomic number 29.

a) Identify the two isotopes and write their isotope notation.

b) Calculate the percentage abundance of each isotope.

c) Estimate the relative atomic mass of the copper sample using the supplied mass numbers.

d) Explain why the relative intensities cannot be interpreted directly as percentage abundances.

Solution 2

The isotopes are copper-63 and copper-65, with approximate abundances of 69.4% and 30.6%, respectively. The estimated relative atomic mass is 63.61.

a) Identify the isotopes

The ions are singly charged, so their mass numbers are approximately equal to their \(m/z\) values.

The isotope notation is:

\[
{}^{63}_{29}\mathrm{Cu}^{+}
\]

\[
{}^{65}_{29}\mathrm{Cu}^{+}
\]

b) Calculate percentage abundances

First, calculate the total relative intensity:

\[
I_{\text{total}}=100+44=144
\]

For copper-63:

\[
\frac{100}{144}\times100=69.4\%
\]

For copper-65:

\[
\frac{44}{144}\times100=30.6\%
\]

c) Calculate relative atomic mass

Use the weighted average:

\[
A_r=\frac{63(100)+65(44)}{144}
\]

\[
A_r=\frac{6300+2860}{144}
\]

\[
\boxed{A_r\approx63.61}
\]

The result is closer to 63 than 65 because copper-63 is more abundant.

d) Explain the relative intensity

The values 100 and 44 are relative intensities, not percentages.

The tallest peak has been assigned a relative intensity of 100, but the combined intensity is 144.

The intensities must therefore be divided by their total before calculating percentage abundances.

This distinction is essential whenever a question provides relative intensities rather than percentages.

Question 3

A calcium sample produces three doubly charged atomic ions.

The mass spectrum contains the following illustrative data:

Calcium has atomic number 20.

a) Determine the mass number of each isotope.

b) Write the isotope notation for each ion.

c) Calculate the percentage abundance of each isotope.

d) Calculate the approximate relative atomic mass of the sample.

e) State the number of electrons in each ion.

Solution 3

The isotopes are calcium-40, calcium-41, and calcium-42, with an approximate relative atomic mass of 40.57. Each ion contains 18 electrons.

a) Determine the mass numbers

All ions have a \(2+\) charge.

Therefore:

\[
A\approx2\left(\frac{m}{z}\right)
\]

For the first peak:

\[
A=2(20)=40
\]

For the second peak:

\[
A=2(20.5)=41
\]

For the third peak:

\[
A=2(21)=42
\]

b) Write the isotope notation

Using calcium’s atomic number of 20:

\[
{}^{40}_{20}\mathrm{Ca}^{2+}
\]

\[
{}^{41}_{20}\mathrm{Ca}^{2+}
\]

\[
{}^{42}_{20}\mathrm{Ca}^{2+}
\]

c) Calculate percentage abundances

First, calculate the total intensity:

\[
I_{\text{total}}=100+50+25=175
\]

For calcium-40:

\[
\frac{100}{175}\times100=57.1\%
\]

For calcium-41:

\[
\frac{50}{175}\times100=28.6\%
\]

For calcium-42:

\[
\frac{25}{175}\times100=14.3\%
\]

d) Calculate relative atomic mass

\[
A_r=\frac{40(100)+41(50)+42(25)}{175}
\]

\[
A_r=\frac{4000+2050+1050}{175}
\]

\[
\boxed{A_r\approx40.57}
\]

The result is greater than 40 because the heavier isotopes contribute to the weighted average.

e) Calculate the number of electrons

A neutral calcium atom contains 20 electrons.

Each ion has lost two electrons:

\[
e^-=20-2
\]

\[
\boxed{e^-=18}
\]

All three isotopes have the same number of electrons because they have the same atomic number and ionic charge.

Question 4

An unknown organic compound produces a singly charged molecular-ion cluster with two peaks:

The compound contains one halogen atom. The non-halogen portion of its molecular formula is \(\mathrm{C_3H_7}\), with a nominal mass of 43.

a) Identify the halogen.

b) Determine the molecular formula of the compound.

c) Explain why two molecular-ion peaks are produced.

d) Explain why this pattern would be different if the compound contained bromine instead.

Solution 4

The halogen is chlorine, and the molecular formula is \(\mathrm{C_3H_7Cl}\). The two peaks represent molecular ions containing different chlorine isotopes.

a) Identify the halogen

The intensity ratio is:

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

This is characteristic of a molecule containing one chlorine atom.

Chlorine has two major isotopes, chlorine-35 and chlorine-37, which occur in an approximate 3:1 abundance ratio.

b) Determine the molecular formula

The non-halogen portion has a nominal mass of 43:

\[
m(\mathrm{C_3H_7})=3(12)+7(1)
\]

\[
=43
\]

Adding chlorine-35 gives:

\[
43+35=78
\]

Adding chlorine-37 gives:

\[
43+37=80
\]

Therefore, the molecular formula is:

\[
\boxed{\mathrm{C_3H_7Cl}}
\]

c) Explain the two peaks

Some molecular ions contain chlorine-35, while others contain chlorine-37.

The chlorine-37 isotope has two more neutrons, making its mass approximately two units greater.

Both ions have the same molecular structure and a single positive charge, but their masses differ.

This produces the peaks at \(m/z=78\) and \(m/z=80\).

d) Compare with bromine

Bromine-79 and bromine-81 occur in approximately equal abundances.

A molecule containing one bromine atom would therefore produce an \(M\) and \(M+2\) pair with approximately equal intensities.

The 3:1 ratio observed here is consistent with chlorine rather than bromine.

Question 5

A sample of chlorine gas is introduced into a mass spectrometer.

Assume that:

• The detected molecular ions are all singly charged \(\mathrm{Cl_2}^{+}\) ions.
• No fragmentation occurs.
• Only chlorine-35 and chlorine-37 are considered.
• Chlorine atoms combine randomly, and the detector responds equally to each isotopic combination.

The molecular-ion spectrum contains three peaks:

A separate bulk measurement gives a relative atomic mass of approximately 35.50.

a) Identify the isotope combination responsible for each peak.

b) Deduce the relative abundances of chlorine-35 and chlorine-37 from the molecular-ion peak pattern.

c) A student argues that the middle peak represents 37.5% chlorine-37 because its relative intensity is 6 out of a total of 16. Explain precisely why this interpretation is incorrect.

d) Calculate the relative atomic mass using your inferred isotope abundances. Check whether it agrees with the independent bulk measurement.

e) Suppose the instrument now detects doubly charged chlorine molecular ions, \(\mathrm{Cl_2}^{2+}\), instead. Predict the three new \(m/z\) values and their relative intensities.

Solution 5

The chlorine isotopes have an approximate abundance ratio of 3:1, corresponding to 75% chlorine-35 and 25% chlorine-37. The calculated relative atomic mass is 35.50, consistent with the independent measurement.

The key is to recognise that the middle molecular-ion peak represents two possible arrangements of the mixed-isotope molecule.

a) Identify the isotope combinations

Each chlorine atom can be either chlorine-35 or chlorine-37.

The three possible molecular ions are:

All ions are singly charged, so the nominal molecular masses correspond directly to the \(m/z\) values.

b) Deduce the isotope abundances

Let:

• \(p\) be the fractional abundance of chlorine-35.
• \(q\) be the fractional abundance of chlorine-37.

Since there are only two isotopes:

\[
p+q=1
\]

The probability of forming a molecule containing two chlorine-35 atoms is:

\[
p^2
\]

The probability of forming a molecule containing two chlorine-37 atoms is:

\[
q^2
\]

The probability of forming a mixed-isotope molecule is:

\[
2pq
\]

The factor of 2 accounts for the two possible arrangements:

• Chlorine-35 followed by chlorine-37.
• Chlorine-37 followed by chlorine-35.

The observed ratio is:

\[
p^2:2pq:q^2=9:6:1
\]

Compare the first and third terms:

\[
p^2:q^2=9:1
\]

Taking square roots:

\[
p:q=3:1
\]

Since the total fractional abundance is 1:

\[
p=\frac{3}{4}=0.75
\]

\[
q=\frac{1}{4}=0.25
\]

Therefore:

• Chlorine-35: 75%.
• Chlorine-37: 25%.

c) Explain the student’s error

The student’s calculation:

\[
\frac{6}{16}\times100=37.5\%
\]

is mathematically correct for the proportion of molecular ions represented by the middle peak.

However, it does not represent the percentage abundance of chlorine-37 atoms.

The middle peak contains molecules with one chlorine-35 atom and one chlorine-37 atom.

There are two possible arrangements that produce this same molecular mass.

Consequently, the middle peak represents a combination of two isotope arrangements, rather than the abundance of one isotope on its own.

The percentage of chlorine-37 atoms must instead be determined from the isotope probabilities:

\[
q=0.25
\]

This is an important distinction between the abundance of individual isotopes and the abundance of molecular combinations.

d) Verify the relative atomic mass

Use the weighted-average formula:

\[
A_r=35(0.75)+37(0.25)
\]

\[
A_r=26.25+9.25
\]

\[
\boxed{A_r=35.50}
\]

This agrees with the independent bulk measurement.

The agreement supports the interpretation of the molecular-ion pattern.

e) Predict the doubly charged molecular-ion spectrum

The molecular ions now carry a \(2+\) charge.

Their masses have not changed, but their mass-to-charge ratios are halved.

For the first combination:

\[
\frac{70}{2}=35
\]

For the mixed combination:

\[
\frac{72}{2}=36
\]

For the final combination:

\[
\frac{74}{2}=37
\]

The new spectrum is:

The relative intensities remain 9:6:1 because changing the charge state does not change the isotope combinations or their probabilities.

Only the horizontal positions change.

This is the final distinction to remember: isotope composition determines which masses are present and how frequently they occur, while ionic charge determines where those ions appear on the \(m/z\) axis.

09What to learn next

Reading isotope peaks correctly gives you two useful skills.

First, you can identify individual isotopes by combining their mass-to-charge ratios with their charge states and atomic numbers.

Second, you can use relative peak intensities to determine isotope abundances and calculate relative atomic mass.

The next step is applying these ideas to more complex mass spectra, where molecular-ion peaks, isotope patterns, and fragmentation peaks must all be interpreted together.

Keep the two questions separate whenever you analyse a spectrum:

• Where is the peak? This tells you about the ion’s mass and charge.
• How tall is the peak? This tells you about the relative strength of its signal.

Once you can answer both independently, isotope mixtures become much easier to interpret.