How to Identify Isotopes, Isobars and Ions from a Table

An early experiment on neon produced two signals: a strong one corresponding to a mass near 20 atomic mass units and a weaker one near 22. Did this mean neon contained two different isotopes, or could the same kind of atom produce both signals by losing different numbers of electrons?

To identify isotopes, isobars, and ions from a table, compare three things:

• Protons: The same number of protons but different numbers of neutrons means two particles are isotopes of the same element.
• Mass number: The same mass number but different numbers of protons means two particles are isobars.
• Electrons: A different number of electrons and protons means the particle is an ion.

The important part is that these labels answer different questions. Isotopes and isobars describe relationships between nuclei. Ion describes the electrical charge of a particle.

A particle can be an ion and still be one of a pair of isotopes or isobars. That’s where many table questions become tricky.

01Read the notation before comparing particles

Suppose a question gives you this particle:

\[
{}^{40}_{18}\mathrm{Ar}^{+}
\]

The three numerical positions tell you different things.

The general notation is:

\[
{}^{A}_{Z}\mathrm{X}^{q}
\]

Here:

• \(A\) is the mass number: the total number of protons and neutrons.
• \(Z\) is the atomic number: the number of protons.
• \(\mathrm{X}\) is the chemical symbol.
• \(q\) is the signed ionic charge, expressed as a multiple of the elementary charge.

Mass number and particle counts are whole numbers without units.

For any monatomic atom or ion, use:

\[
\begin{aligned}
p &= Z\\
n &= A-Z\\
q &= p-e
\end{aligned}
\]

Here \(p\), \(n\), and \(e\) represent the numbers of protons, neutrons, and electrons respectively.

For \({}^{40}_{18}\mathrm{Ar}^{+}\):

• Protons: \(18\)
• Neutrons: \(40-18=22\)
• Electrons: \(18-1=17\)

The positive charge means one electron has been removed. It does not mean the nucleus gained a proton.

Why the sign matters

Imagine two particles with 12 protons. One has 10 electrons, and the other has 14.

Predict their charges before calculating.

For the first particle:

\[
q=12-10=+2
\]

For the second:

\[
q=12-14=-2
\]

The first is a \(2+\) cation because it has lost two electrons. The second is a \(2-\) anion because it has two extra electrons.

The sign follows directly from the difference between positive and negative charges. In real chemistry, whether a particular ion is stable is a separate question.

One useful warning: if a table gives only protons and neutrons, you can identify the nuclide, but you cannot determine its ionic charge without an electron count or charge information.

02The decision method: compare protons first, then mass number

Consider two particles, X and Y.

Notice that the ion test is separate. You can apply it to every row, regardless of the relationships between rows.

Test the method on a mixed table

Here are four illustrative particle records.

First calculate each mass number:

\[
\begin{aligned}
A_P &=18+22=40\\
A_Q &=18+18=36\\
A_R &=20+20=40\\
A_S &=18+22=40
\end{aligned}
\]

Now compare the rows.

P and Q are isotopes. Both contain 18 protons, so both are argon. Their neutron numbers differ, producing argon-40 and argon-36.

P and R are isobars. Both have mass number 40, but P contains 18 protons while R contains 20. They are argon-40 and calcium-40.

P and S are the same nuclide in different charge states. They have identical nuclei. P is \({}^{40}_{18}\mathrm{Ar}^{+}\), while S is neutral argon-40. They are not different isotopes.

Q and R are neither isotopes nor isobars. Both have 18 electrons, but their proton numbers and mass numbers differ.

R and S provide another useful comparison: they are isobars even though both have 18 electrons.

This is why you should never classify isotopes or isobars using electron number alone.

For a closer explanation of why isotopes remain the same element, see Why Isotopes Share a Position in the Periodic Table.

03How early mass spectrographs exposed the isotope problem

The distinction between isotopes and ions became important when scientists began measuring individual atomic masses.

Ordinary chemical measurements gave neon an average atomic mass near \(20.2\,\mathrm{u}\), where \(\mathrm{u}\) is the unified atomic mass unit.

But an individual nucleus cannot contain 20.2 protons and neutrons.

Was this simply a limitation of chemical measurements, or did neon contain atoms with genuinely different masses?

Thomson’s positive-ray experiment

J. J. Thomson investigated positive ions using an apparatus that could separate particles according to their mass-to-charge ratios.

The path through the apparatus was:

1. A gas was introduced into a discharge tube at low pressure.
2. An electric discharge produced positive ions.
3. The ions accelerated towards a cathode and passed through a narrow opening, forming a beam.
4. Electric and magnetic fields deflected the moving ions in different directions.
5. The ions struck a photographic plate, leaving curved traces.

What should happen if all neon ions have the same mass and the same charge?

They should produce one trace corresponding to that mass-to-charge ratio, although different ion speeds can spread the impacts along a curve.

In 1912, measurements involving neon revealed two traces. One corresponded approximately to mass 20 and a weaker one to mass 22, assuming singly charged atomic ions.

That observation challenged the idea that every atom of the same element must have exactly the same mass.

But there was a problem with the interpretation.

The detector measured mass-to-charge ratio, not isotope identity

A mass spectrograph separates ions according to their mass-to-charge ratio, written \(m/z\).

For positive ions, the commonly reported numerical value is approximately:

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

Here:

• \(m\) is the ion’s mass, expressed numerically in atomic mass units.
• \(z\) is the positive integer charge number, such as 1 or 2.
• \(A\) is the mass number, used as an approximation to the mass in \(\mathrm{u}\).

The approximation works for introductory isotope problems, but mass number is not an exact measurement of atomic mass.

For example:

Four different signals can therefore come from just two isotopes.

A signal near \(m/z=22\) might also come from a different species, such as a doubly charged molecule with mass near \(44\,\mathrm{u}\).

A separate signal does not automatically prove the existence of another isotope.

Aston’s revised apparatus

Francis William Aston worked on the neon problem with Thomson and later developed a more accurate mass spectrograph in 1919.

Earlier attempts to separate neon into lighter and heavier fractions had produced limited evidence. More accurate individual mass measurements were needed.

Aston’s design replaced the very narrow cylindrical passage with parallel slits and arranged successive electric and magnetic fields to improve the focusing of ions.

The ions moved from the source through the slits, between electric deflection plates, through a further aperture, between magnetic poles, and finally onto a photographic plate.

The two fields bent ions in opposite directions. By carefully arranging the apparatus, ions with the same mass-to-charge ratio but slightly different speeds could be brought towards the same position on the plate.

This produced sharper mass measurements than Thomson’s earlier arrangement.

Aston’s measurements distinguished neon components near mass 20 and 22. They also produced signals at 10 and 11 from doubly charged ions.

The evidence supported a mixture of neon isotopes rather than four different atomic masses. The conclusion depended on interpreting the charge states and identifying the chemical species, not just counting lines.

Aston described the refinement and these observations in his 1922 Nobel lecture, Mass Spectra and Isotopes.

A small calculation: why neon’s average could be 20.2

Suppose, as a simplified illustration, a neon sample contains:

• 90% neon-20
• 10% neon-22

If we approximate the isotope masses by their mass numbers in \(\mathrm{u}\), the mean atomic mass is:

\[
\begin{aligned}
\bar m&\approx0.90(20\,\mathrm{u})\\
&\quad+0.10(22\,\mathrm{u})\\
&=20.2\,\mathrm{u}
\end{aligned}
\]

Here \(\bar m\) is the mean mass of an atom in the sample.

That value is an average across many atoms. It does not describe a single neon atom with mass number 20.2.

The 90:10 mixture is a rounded teaching illustration, not a precise modern measurement of natural neon abundance.

This is the experimental reason isotope tables must distinguish individual particles from mixture averages.

06What the numbers cannot tell you

A particle table can answer some questions exactly and leave others unresolved.

One final distinction is worth remembering.

Two isobars share a mass number, but they do not necessarily have exactly equal measured masses. Mass number counts nucleons, while measured atomic mass depends on the actual mass of the particle.

In a basic table, identifying isobars is a whole-number comparison. In high-resolution mass spectrometry, small differences in actual masses can provide additional evidence.

07Questions and solutions

The questions start with direct particle-count comparisons and move towards interpreting experimental evidence when several species contribute to the same measured signal.

Question 1

Three particles have the following compositions.

a) Write the complete nuclide symbol for each particle, including charge where applicable.

b) Identify the isotope pair and the isobar pair.

c) Identify which particle is an ion and explain its charge.

Solution 1

A is neutral carbon-12, B is a carbon-14 anion, and C is neutral nitrogen-14. A and B are isotopes, B and C are isobars, and B is the only ion.

For A:

\[
\begin{aligned}
A_A&=6+6=12\\
q_A&=6-6=0
\end{aligned}
\]

Therefore:

\[
{}^{12}_{6}\mathrm{C}
\]

For B:

\[
\begin{aligned}
A_B&=6+8=14\\
q_B&=6-7=-1
\end{aligned}
\]

Therefore:

\[
{}^{14}_{6}\mathrm{C}^{-}
\]

For C:

\[
\begin{aligned}
A_C&=7+7=14\\
q_C&=7-7=0
\end{aligned}
\]

Therefore:

\[
{}^{14}_{7}\mathrm{N}
\]

A and B have the same proton number, 6, but different neutron counts. They are isotopes of carbon.

B and C have the same mass number, 14, but different proton numbers. They are isobars.

B is negatively charged because it contains seven electrons but only six protons. The extra electron produces a net charge of \(-1\).

Notice that B and C have the same electron count. That does not make them the same element: B contains six protons, while C contains seven.

Question 2

Three unknown particles, X, Y, and Z, each contain 28 electrons.

Additional information is available:

• X is neutral and has mass number 62.
• Y is neutral, has the same proton number as X, and contains two more neutrons.
• Z is a \(2+\) ion, has the same mass number as Y, and contains 28 electrons.

a) Determine the numbers of protons, neutrons, and electrons in all three particles. Write their complete nuclide symbols.

b) Identify the isotope pair and isobar pair.

c) X is then converted into a \(2+\) ion without changing its nucleus. Determine its new electron count. Does this change any isotope or isobar relationship between X, Y, and Z?

Solution 2

X is nickel-62, Y is nickel-64, and Z is zinc-64 with a \(2+\) charge. X and Y are isotopes; Y and Z are isobars. Converting X into an ion changes its electrons, but none of its nuclear relationships.

Since X is neutral:

\[
p_X=e_X=28
\]

Atomic number 28 is nickel.

Its neutron number is:

\[
n_X=62-28=34
\]

Therefore:

\[
X={}^{62}_{28}\mathrm{Ni}
\]

Y has the same 28 protons and two additional neutrons:

\[
n_Y=34+2=36
\]

Its mass number is:

\[
A_Y=28+36=64
\]

Y is neutral, so it also has 28 electrons.

\[
Y={}^{64}_{28}\mathrm{Ni}
\]

For Z, the charge gives the crucial missing information.

\[
q=p-e
\]

Rearrange:

\[
p=q+e
\]

Substitute the supplied values:

\[
p_Z=2+28=30
\]

Atomic number 30 is zinc.

Z has mass number 64, so:

\[
n_Z=64-30=34
\]

Therefore:

\[
Z={}^{64}_{30}\mathrm{Zn}^{2+}
\]

The complete comparison is:

X and Y are isotopes because they share 28 protons but have different neutron numbers.

Y and Z are isobars because they share mass number 64 but have different proton numbers.

When X loses two electrons:

\[
e_X=28-2=26
\]

It becomes:

\[
{}^{62}_{28}\mathrm{Ni}^{2+}
\]

X no longer has the same electron count as Z, but it remains an isotope of Y and is still neither an isotope nor an isobar of Z.

The tempting mistake is to think that ionisation changes the particle’s overall identity in every sense. Ionisation changes its electrical state, not its nucleus. Isotope and isobar relationships therefore remain unchanged.

Question 3

An illustrative positive-ion experiment investigates a neon sample contaminated with carbon dioxide.

The original spectrum contains peaks at nominal \(m/z\) values of 20, 22, and 44.

The only possible contributors are singly charged monatomic neon ions and carbon dioxide molecular ions with charges \(+1\) or \(+2\).

Carbon dioxide has a nominal molecular mass of \(44\,\mathrm{u}\).

A selective trap is then introduced before the ion source. Independent checks confirm that it removes all carbon dioxide without altering the neon sample.

The second spectrum contains peaks at 20 and 22, but no peak at 44. The remaining ions are independently verified as singly charged, monatomic neon ions.

a) Give two possible contributors to the original \(m/z=22\) signal.

b) Explain what the second spectrum establishes about neon.

c) Can the original peak heights at 20 and 22 be used directly to determine neon’s isotopic abundance? Explain.

Solution 3

The original \(m/z=22\) signal could contain both neon-22 ions and doubly charged carbon dioxide ions. The purified spectrum supports the presence of neon-20 and neon-22, but the original uncorrected peak heights cannot reliably establish their relative abundances.

For part (a), singly charged neon-22 has:

\[
m/z\approx22/1=22
\]

Doubly charged carbon dioxide has:

\[
m/z\approx44/2=22
\]

These chemically different particles can therefore contribute to the same nominal peak.

Singly charged carbon dioxide also explains a possible peak at:

\[
m/z\approx44/1=44
\]

For part (b), the verified removal of carbon dioxide removes that competing explanation.

The remaining signals come from singly charged monatomic neon ions, so the nominal mass numbers are 20 and 22.

Neon has atomic number 10. The neutron numbers are therefore:

\[
\begin{aligned}
n_{20}&=20-10=10\\
n_{22}&=22-10=12
\end{aligned}
\]

Both particles have 10 protons but different neutron numbers. They are isotopes.

The critical evidence is not simply that two peaks remain. It is the combination of sample purification, verified particle identity, known ionic charge, and two distinct masses.

For part (c), the original peak at 22 may have included carbon dioxide contributions. Its entire height cannot be assigned to neon-22.

Peak intensity can also depend on ion production and detection efficiency. Without suitable calibration and correction, signal height is not necessarily equal to isotope abundance.

The tempting conclusion is to read every peak as a separate isotope. The experiment shows why competing explanations must be removed before making that identification.

Question 4

A magnesium sample contains only magnesium-24 and magnesium-25.

Two mass spectra are recorded using different ionisation settings. Both settings produce a mixture of singly and doubly charged monatomic ions.

The sample composition remains fixed, and each run detects exactly 100 ions. Assume every atom produces one detected ion and all ion types have equal detection efficiency.

The illustrative measurements are:

A student claims the sample became richer in magnesium-25 because the peak at \(m/z=25\) increased from 10 to 15 counts.

a) Determine the percentage abundance of each isotope in both runs.

b) Calculate the approximate mean atomic mass of the sample.

c) Explain why the student’s claim is incorrect, despite the measured change in peak heights.

Solution 4

Both runs contain 80% magnesium-24 and 20% magnesium-25. The approximate mean atomic mass is \(24.2\,\mathrm{u}\). The changing peak heights reflect different proportions of \(+1\) and \(+2\) ions, not a change in isotopic composition.

First identify the possible mass-to-charge values.

For magnesium-24:

\[
\begin{aligned}
m/z\;(+1)&\approx24\\
m/z\;(+2)&\approx12
\end{aligned}
\]

For magnesium-25:

\[
\begin{aligned}
m/z\;(+1)&\approx25\\
m/z\;(+2)&\approx12.5
\end{aligned}
\]

For Run A, magnesium-24 contributes:

\[
30+50=80\text{ ions}
\]

Magnesium-25 contributes:

\[
10+10=20\text{ ions}
\]

The total is 100 ions, giving 80% and 20%.

For Run B:

\[
\begin{aligned}
\text{Mg-24 count}&=60+20=80\\
\text{Mg-25 count}&=5+15=20
\end{aligned}
\]

The same abundances are obtained.

The approximate mean atomic mass is:

\[
\begin{aligned}
\bar m&\approx0.80(24\,\mathrm{u})\\
&\quad+0.20(25\,\mathrm{u})\\
&=24.2\,\mathrm{u}
\end{aligned}
\]

This is the approximate average mass of an atom in the sample, not the mass number of a new isotope.

The student’s error is to compare only the singly charged magnesium-25 peak.

In Run B, fewer magnesium-25 ions are recorded at \(m/z=12.5\), while more appear at \(m/z=25\). Their combined count stays at 20.

The two ionisation settings redistribute signals between charge states. They do not create or destroy the magnesium isotopes.

Without the charge-state information and detection assumptions, the individual peak heights would not be sufficient to infer those isotope abundances reliably.

Question 5

An idealised mass-spectrometry source contains a deliberately prepared mixture of exactly three monatomic species:

\[
{}^{36}_{18}\mathrm{Ar},
\qquad
{}^{40}_{18}\mathrm{Ar},
\qquad
{}^{40}_{20}\mathrm{Ca}
\]

Every atom becomes exactly one detected positive ion, carrying either a \(+1\) or \(+2\) charge. The detector reports nominal \(m/z\) values using \(m/z\approx A/z\).

An initial spectrum is recorded. Calcium is then selectively removed before the ion source, without changing the incoming argon population or its ionisation behaviour.

The ionisation settings remain unchanged, and the detector counts are corrected for efficiency.

a) Identify every possible ion contributing to each original peak. Explain why some peaks contain contributions from more than one species.

b) Determine the original atom counts for argon-36, argon-40, and calcium-40.

c) Calculate the approximate mean atomic mass of the argon component alone.

d) Classify the relationships between the three neutral starting species. A student claims that a peak at \(m/z=20\) proves an isotope with mass number 20 was present. Evaluate this claim.

e) A higher-resolution measurement separates the original nominal \(m/z=40\) peak into two very closely spaced peaks. Another student argues that this is impossible because argon-40 and calcium-40 are isobars and must have exactly the same mass. Is the student correct?

Solution 5

The original mixture contains 50 argon-36 atoms, 70 argon-40 atoms, and 80 calcium-40 atoms. The argon component has an approximate mean atomic mass of \(38.33\,\mathrm{u}\). Two of the starting species are isotopes, two are isobars, and the overlapping peaks arise because mass-to-charge ratio is not a unique identifier of a nuclide.

The hardest part is recognising that two different ambiguities occur at once: different charge states can shift a species to another nominal \(m/z\), while different elements with the same mass number can contribute to the same nominal peak.

Part (a): Identify the possible peak assignments.

For argon-36:

\[
\begin{aligned}
m/z\;(+1)&\approx36\\
m/z\;(+2)&\approx18
\end{aligned}
\]

For argon-40:

\[
\begin{aligned}
m/z\;(+1)&\approx40\\
m/z\;(+2)&\approx20
\end{aligned}
\]

For calcium-40:

\[
\begin{aligned}
m/z\;(+1)&\approx40\\
m/z\;(+2)&\approx20
\end{aligned}
\]

Therefore:

The peaks at 20 and 40 each contain two possible contributors.

A single original spectrum cannot separate their counts because the nominal peak positions overlap.

Part (b): Use the controlled removal to recover the atom counts.

After calcium removal, the remaining counts must come from argon.

Because the argon population and its ionisation behaviour were held unchanged, we can subtract the corresponding measurements.

At \(m/z=20\):

\[
50-20=30
\]

Those 30 counts originally came from calcium-40 ions with charge \(+2\).

At \(m/z=40\):

\[
100-50=50
\]

Those 50 counts came from singly charged calcium-40 ions.

So the original calcium-40 count was:

\[
30+50=80
\]

Now determine the argon isotope counts from the calcium-free spectrum.

Argon-36 contributes at 18 and 36:

\[
20+30=50
\]

Argon-40 contributes at 20 and 40:

\[
20+50=70
\]

The original atom counts are:

The counts sum to 200, matching the original spectrum total of \(20+50+30+100=200\).

That agreement is a useful consistency check. It does not independently prove the assumptions, but it confirms that the accounting is complete under the stated model.

Part (c): Calculate the argon mean atomic mass.

Only the argon atoms belong in this average.

There are:

\[
50+70=120
\]

argon atoms.

Using nominal isotope masses:

\[
\begin{aligned}
\bar m_{\mathrm{Ar}}
&\approx\frac{50(36)+70(40)}{120}\\
&=\frac{4600}{120}\,\mathrm{u}\\
&\approx38.33\,\mathrm{u}
\end{aligned}
\]

The approximate mean atomic mass of this deliberately prepared argon component is \(38.33\,\mathrm{u}\).

Calcium-40 is excluded even though it contributes to some of the same nominal mass-spectrum peaks. It is a different element.

Including calcium in the argon average would be a classification error, not merely an arithmetic one.

Part (d): Classify the relationships and test the student’s claim.

Argon-36 and argon-40 are isotopes because both contain 18 protons but have different mass numbers and neutron counts.

Argon-40 and calcium-40 are isobars because both have mass number 40 but different proton numbers.

The \(m/z=20\) peak does not prove that a mass-number-20 isotope is present.

Both argon-40 and calcium-40 can produce ions at that nominal position when doubly charged:

\[
40/2=20
\]

No mass-number-20 isotope is needed to explain the observations.

There is another important point: none of the detected ions is the same electrical state as its neutral starting atom. Ionisation changes the electron count without changing the nuclear isotope or isobar classification.

Part (e): Explain why a high-resolution instrument can distinguish the isobars.

The second student’s claim is incorrect.

Equal mass numbers mean equal total numbers of protons and neutrons. They do not require exactly equal measured atomic masses.

Argon-40 and calcium-40 have different proton-neutron combinations, and their actual atomic masses differ slightly. An instrument with sufficient mass resolution can distinguish their singly charged ions even though both have nominal \(m/z=40\).

This is an extension of the basic HSC table method: actual isotope masses are not exactly their whole-number mass numbers.

The deeper experimental lesson is that neither a matching mass number nor a matching nominal mass-to-charge ratio establishes that two particles are identical.

You need to know which quantity was measured, what charge states were possible, and what independent evidence identifies the element.

08The next decision: what does the table actually establish?

When you see an unfamiliar particle table, use proton number to establish elemental identity, mass number to compare nuclei, and electron number to establish charge.

That method tells you which particles are isotopes, which are isobars, and which are ions. But it doesn’t automatically tell you what caused a measured signal, or which ions are present in an unknown chemical sample.

Mass spectrometry needs charge-state and species identification. An unknown aqueous solution needs chemical observations. For the second problem, the next useful method is How to Identify Two Unknown Ions Using Precipitation Tests.

The same habit applies to both: decide what the evidence actually measures before deciding what the substance must be.