Why Relative Abundance Is Not the Same as Concentration

Suppose the molecular-ion peak in a mass spectrum has a relative abundance of 20%, while a fragment peak has a relative abundance of 100%. Does that mean 20% of the original molecules remained intact and 80% fragmented?

No. It also doesn’t mean there is five times as much fragment as molecular ion in the original sample.

Relative abundance measures the strength of a detected ion signal compared with the strongest signal in the same mass spectrum. It does not directly measure the concentration of the original substance.

This distinction matters because a mass spectrometer does not simply count molecules in a sample. It first converts molecules into ions, separates those ions, and measures the ions that reach its detector. Each stage influences the final signal.

01Start with what the instrument measures

Imagine analysing the same compound twice. In the second experiment, five times as much of the compound enters the instrument.

You might predict that every peak becomes five times taller. That prediction is reasonable if the instrument responds linearly to the additional material.

However, mass spectra are commonly displayed using relative abundance, where the tallest peak is assigned a value of 100%. If all the ion signals increase by the same factor, the relative peak heights remain unchanged.

Consider these illustrative measurements:

Here, nA means nanoamperes, a unit of electrical current.

Assuming identical injection volumes, sample matrices, and instrument conditions, Experiment B contains five times the analyte concentration.

Yet both normalised mass spectra have exactly the same relative abundances.

This is the first important limitation: normalising a spectrum removes information about the overall size of the original signal.

To understand the other limitations, we need to follow what happens inside the instrument.

Following a molecule through a mass spectrometer

Consider a simplified electron-impact mass spectrometer with a magnetic-sector analyser.

The sample enters through an inlet and is vaporised. The gas-phase molecules then enter an ionisation chamber.

An electron beam passes through this chamber. When a high-energy electron collides with a molecule, it can remove one of the molecule’s electrons, producing a positive molecular ion.

We can represent this process as:

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

Here:

• \(\mathrm{M}\) represents the original molecule.
• \(e^-\) represents an electron.
• \(\mathrm{M}^{+\bullet}\) is the molecular ion, a positively charged radical.

The molecular ion may then break apart into smaller ions and neutral particles.

\[ \mathrm{M}^{+\bullet}\rightarrow \mathrm{A}^{+}+\mathrm{B}^{\bullet} \]

The positive fragment ion, \(\mathrm{A}^{+}\), can be detected. The neutral fragment, \(\mathrm{B}^{\bullet}\), cannot be detected by the mass analyser as an ion.

The remaining positive ions are accelerated by an electric field and enter the mass analyser. In the simplified magnetic-sector model, a magnetic field bends their paths according to their mass-to-charge ratios.

Finally, the ions reach a detector, which produces an electrical signal.

What should this apparatus show?

If two groups of ions have different mass-to-charge ratios, they should follow different paths through the analyser and reach different detector positions.

That is what the instrument is designed to achieve. The resulting spectrum identifies the mass-to-charge ratios of detected ions and records the strength of their signals.

But notice what the detector actually measures: ions that have successfully formed, passed through the analyser, and reached the detector.

It does not directly measure every original molecule in the sample.

This is the experimental reason why peak intensity and original concentration are not interchangeable.

02Understanding relative abundance in a mass spectrum

Before interpreting peak heights, make sure you understand the two axes.

• The horizontal axis represents the mass-to-charge ratio, written as \(m/z\).
• The vertical axis represents relative abundance, usually expressed as a percentage.

The mass-to-charge ratio is the mass of an ion divided by its charge number.

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

For the singly charged positive ions used in our examples, \(z=1\). Therefore, the numerical \(m/z\) value is approximately equal to the ion’s mass in atomic mass units (u).

The tallest peak is called the base peak. It is assigned a relative abundance of 100%.

Every other peak is expressed relative to this peak.

\[ \text{Relative abundance (\%)} = \frac{\text{Signal of selected peak}} {\text{Signal of base peak}} \times100 \]

Importantly, the relative abundances of all the peaks do not need to add up to 100%.

The base peak is simply the strongest detected ion signal, not necessarily the molecular ion.

Three types of peaks you need to recognise

A mass spectrum can contain several different types of ions. Their peak heights have different chemical explanations.

The molecular-ion peak is particularly useful because, when correctly identified, it gives information about the mass of the intact molecule.

For a singly charged molecular ion, its \(m/z\) value is approximately equal to the molecular mass in u.

However, its relative abundance does not tell you what percentage of the original sample remains unfragmented.

03Why fragment peaks can be much stronger than molecular-ion peaks

Imagine two possible outcomes when a molecule is ionised.

In the first, most molecular ions remain intact long enough to reach the detector.

In the second, many molecular ions break apart, producing stable fragment ions.

You might expect the first molecule to have a stronger molecular-ion peak and the second to have stronger fragment peaks. That is a reasonable prediction.

However, the intensity of a fragment also depends on how readily it forms and how stable it is.

Some fragmentation pathways produce particularly stable positive ions. These ions can produce strong signals because they are formed relatively readily and survive long enough to reach the detector.

Other fragments may be unstable and break apart again.

The detector therefore records the combined result of ion formation, fragmentation, ion transmission, and detection.

A strong fragment peak is evidence that the corresponding ion is readily produced and detected under those conditions. It is not a direct measurement of the concentration of the original compound.

There is another important limitation: a fragment’s \(m/z\) value does not always identify its structure uniquely.

For example, two possible ions are:

\[ \mathrm{CH_2OH}^{+} \]

and

\[ \mathrm{CH_3O}^{+} \]

Both have a nominal mass of 31, so both can produce a peak at \(m/z=31\).

A peak at \(m/z=31\) alone cannot distinguish these two structures.

Fragment assignments should be checked against the molecular formula, possible fragmentation pathways, and other peaks in the spectrum.

For more practice with this constraint, see How to Check Whether a Fragment Fits a Molecular Formula.

04Isotope patterns: When peak ratios reveal something different

There is another use of relative abundance that you need to keep separate from the usual mass-spectrum definition.

Elements can exist as different isotopes. Isotopes have the same number of protons but different numbers of neutrons.

For example, naturally occurring chlorine contains two main isotopes:

• Chlorine-35.
• Chlorine-37.

Their natural abundances are approximately 3:1.

If a molecule contains one chlorine atom, its molecular-ion region can show two peaks separated by two mass units.

The first peak, \(M\), corresponds to the molecular ion containing chlorine-35.

The second peak, \(M+2\), corresponds to the molecular ion containing chlorine-37.

Their approximate intensity ratio is:

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

This ratio is related to the natural isotope proportions of chlorine.

Notice the difference.

The ratio is telling us about the isotopic composition of the chlorine atoms in the molecule. It is not telling us the concentration of the compound in the original sample.

Other elements produce different characteristic isotope patterns.

These patterns are useful evidence when identifying elements in an unknown compound.

However, an isotope pattern is not automatically proof of a complete molecular structure. Other isotopes, overlapping peaks, and alternative ion assignments must still be considered.

It is also possible for an isotope peak from one ion to overlap with a fragment or molecular-ion peak from another species.

For a more detailed comparison, see How to Distinguish Isotopic Peaks from Fragment Peaks.

05Can mass spectrometry measure concentration?

Yes. Mass spectrometry can be used for quantitative analysis.

The important distinction is that concentration must be determined using an appropriate calibration method. It cannot simply be read from the relative abundance column of an ordinary mass spectrum.

Consider two different compounds at the same concentration.

One compound may ionise efficiently and produce a strong molecular-ion signal. The other may ionise less efficiently or fragment more readily, producing a much weaker molecular-ion signal.

Their concentrations could be identical despite their different peak heights.

This difference is described through their response factors: how strongly the instrument responds to a given amount of each substance under specified conditions.

How calibration solves the problem

A common quantitative approach is to prepare standards containing known concentrations of the substance being measured.

The instrument measures the signal produced by each standard.

These measurements are used to construct a calibration curve.

An internal standard can also be added in a known amount to both the standards and unknown samples. The ratio of the analyte signal to the internal-standard signal helps account for variations in sample introduction and instrument response.

The unknown concentration can then be estimated by comparing its measured signal ratio with the calibration curve.

For this method to be reliable, the standards and unknown should be analysed under appropriately matched conditions.

Factors such as sample preparation, matrix effects, ionisation efficiency, and detector response must be controlled or accounted for.

The key distinction is:

• Relative abundance describes a peak compared with the base peak in the same spectrum.
• Absolute signal describes the actual measured ion response.
• Concentration describes the amount of substance per unit volume, such as mol L\(^{-1}\).
• A calibrated quantitative method connects the measured signal to concentration.

Without calibration or another validated quantitative method, relative abundance alone is insufficient to determine concentration.

06Questions and solutions

The following questions build from basic relative-abundance calculations to situations involving isotope overlap, fragmentation, and quantitative analysis.

Question 1

An illustrative mass spectrum has a base peak at \(m/z=43\), with a detector current of 5.0 nA.

The molecular-ion peak occurs at \(m/z=58\), with a detector current of 1.0 nA.

Calculate the relative abundance of the molecular-ion peak and explain what the result means.

Solution 1

The molecular-ion peak has a relative abundance of 20%. This means its detected signal is one-fifth as strong as the base peak, not that 20% of the original molecules remain intact.

Step 1: Identify the base peak.

The base peak has a current of 5.0 nA.

Step 2: Apply the relative abundance equation.

\[ \text{Relative abundance} = \frac{\text{Molecular-ion current}} {\text{Base-peak current}}\times100 \]

Substitute the measurements:

\[ =\frac{1.0}{5.0}\times100 \]

\[ =20\% \]

Step 3: Explain the result.

The molecular-ion signal is 20% of the base-peak signal.

It would be incorrect to conclude that 80% of the molecules fragmented and 20% remained intact. Relative abundance is a comparison of detected ion signals, not a direct measurement of the original molecules.

Question 2

An unknown compound produces the following illustrative molecular-ion isotope peaks:

Assume the peaks are correctly assigned, the ions are singly charged, and there are no significant overlapping contributions.

(a) Calculate the ratio of the molecular-ion peaks.

(b) Identify the element whose presence is supported by this pattern.

(c) Explain why this result does not establish the compound’s complete molecular structure or its concentration.

Solution 2

The isotope ratio is 3.15:1, which strongly supports the presence of one chlorine atom in the molecular ion. It does not uniquely identify the compound or determine its concentration.

Step 1: Calculate the molecular-ion isotope ratio.

\[ M:M+2=6.3:2.0 \]

Divide both values by 2.0:

\[ M:M+2=3.15:1 \]

Step 2: Compare the result with the expected isotope pattern.

One chlorine atom produces an approximate molecular-ion isotope ratio of:

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

The experimental ratio of 3.15:1 is consistent with this pattern.

Step 3: Explain what can be concluded.

The isotope pattern provides evidence for one chlorine atom in the molecular ion.

However, many different compounds can contain one chlorine atom. The isotope pattern does not reveal the complete arrangement of atoms or uniquely identify the molecular structure.

It also does not measure concentration. The ratio reflects the relative abundance of chlorine isotopes, rather than the amount of compound in the original sample.

Question 3

The same compound is analysed in two separate experiments.

Both experiments use the same injection volume, sample matrix, instrument settings, and linear-response range.

The measured detector currents are:

Assume the absolute ion current is proportional to the analyte concentration under these conditions.

(a) Calculate the relative abundance of the molecular-ion peak in each experiment.

(b) Determine the concentration ratio between the two experiments.

(c) A student claims that the two samples must have equal concentrations because their molecular-ion peaks have the same relative abundance. Explain the error.

Solution 3

Both experiments have a molecular-ion relative abundance of 20%, but Experiment B has four times the concentration of Experiment A. Identical relative abundances do not establish equal concentrations.

Step 1: Calculate the relative abundance in Experiment A.

\[ \frac{0.60}{3.0}\times100=20\% \]

Step 2: Calculate the relative abundance in Experiment B.

\[ \frac{2.40}{12.0}\times100=20\% \]

Both experiments therefore produce the same relative-abundance value.

Step 3: Compare the absolute detector currents.

The base-peak currents are:

\[ \frac{12.0}{3.0}=4 \]

The molecular-ion currents also give:

\[ \frac{2.40}{0.60}=4 \]

Both measured signals have increased by a factor of four.

Step 4: Determine the concentration ratio.

Because the injection volumes and experimental conditions are identical, and the question states that the response is linear:

\[ \frac{c_B}{c_A}=4 \]

Therefore:

\[ \boxed{c_A:c_B=1:4} \]

Experiment B has four times the concentration of Experiment A.

Step 5: Explain the student’s error.

The student has confused normalised signal with absolute signal.

Relative abundance compares each peak with the base peak. If both signals increase by the same factor, their ratio remains unchanged.

The absolute currents provide additional information here because the experimental conditions and linear response are specified.

Without those assumptions, the relative-abundance values alone would not establish the concentration ratio.

Question 4

Two structural isomers have the molecular formula \(\mathrm{C_3H_8O}\).

Their illustrative mass spectra contain the following relative abundances:

A student makes the following claims:

• Isomer X must be propan-1-ol because its base peak occurs at m/z 31.
• Isomer Y must be propan-2-ol because its base peak occurs at m/z 45.

Evaluate these claims. Explain what the spectra support and why the individual peaks do not conclusively establish either structure.

Solution 4

The spectra support different fragmentation patterns for the two isomers, but the proposed structures cannot be established conclusively from the base peaks alone.

Step 1: Interpret the molecular-ion peaks.

Both compounds have a molecular formula of \(\mathrm{C_3H_8O}\).

Their molecular mass is:

\[ M_r=(3\times12)+(8\times1)+16 \]

\[ =60 \]

Therefore, a molecular-ion peak at \(m/z=60\) is consistent with both compounds, assuming singly charged molecular ions.

This does not distinguish the isomers because they have the same molecular formula and molecular mass.

Step 2: Interpret the base peak of Isomer X.

Isomer X has its strongest signal at \(m/z=31\).

This suggests that an ion with a nominal mass of 31 is produced relatively readily and survives sufficiently well to reach the detector.

Possible ionic formulas with this nominal mass include:

\[ \mathrm{CH_2OH}^{+} \]

and

\[ \mathrm{CH_3O}^{+} \]

The peak alone does not establish which ion is present or which original structure produced it.

Step 3: Interpret the base peak of Isomer Y.

Isomer Y has its strongest signal at \(m/z=45\).

This indicates that an ion with a mass-to-charge ratio of 45 produces the strongest detected signal in this spectrum.

However, the mass-to-charge ratio does not uniquely establish the ion’s structure or its formation pathway.

Step 4: Evaluate the proposed identifications.

The different relative abundances are useful evidence that the compounds undergo different fragmentation processes.

However, a base peak does not uniquely identify a structural isomer.

Other fragment assignments, possible rearrangements, and the full pattern of peaks must be considered.

The molecular-ion mass, isotope information, plausible fragmentation pathways, and additional chemical evidence should be used together.

The correct conclusion is that the spectra support different fragmentation behaviour, but the individual base peaks do not prove the proposed structures.

Question 5

An unknown mixture contains two compounds, A and B.

Both compounds produce singly charged molecular ions.

The following information is available:

• Compound A has a molecular ion at \(m/z=100\).
• Compound A contains exactly one chlorine atom.
• Its molecular-ion isotope ratio is approximately \(M:M+2=3:1\).
• Compound B has a molecular ion at \(m/z=102\).
• Compound B contains no chlorine.
• The observed peak at \(m/z=102\) may contain contributions from both compounds.

The mass spectrum has a base peak at \(m/z=43\), with an absolute detector current of 20.0 nA.

The observed relative abundances are:

Separate calibration experiments, using the same injection volume and instrument conditions, establish these response factors:

Assume that:

• The response is linear over the concentration range considered.
• There is no significant ion suppression.
• No other ions contribute to the peaks at m/z 100 or 102.
• The chlorine isotope ratio is approximately 3:1.
• The calibration factors apply to the injected unknown sample.

Determine the concentrations of A and B.

A student argues that B must be more concentrated because the peak at m/z 102 has a greater relative abundance than the peak at m/z 100.

Explain why this reasoning is incorrect.

Solution 5

Compound A is more concentrated, with a concentration of 4.0 mmol L\(^{-1}\), compared with 2.0 mmol L\(^{-1}\) for compound B. The student’s conclusion is incorrect because the m/z 102 peak contains contributions from both compounds, and the two compounds have different response factors.

This question requires three separate steps: converting relative abundance into absolute current, separating the overlapping isotope signal, and applying the calibration factors.

Step 1: Calculate the absolute current at m/z 100.

The base-peak current is 20.0 nA.

The relative abundance at m/z 100 is 60%.

Therefore:

\[ I_{100}=\frac{60}{100}\times20.0 \]

\[ \boxed{I_{100}=12.0\text{ nA}} \]

This signal comes from the molecular ion of compound A.

Step 2: Calculate the total absolute current at m/z 102.

The relative abundance at m/z 102 is 70%.

Therefore:

\[ I_{102}=\frac{70}{100}\times20.0 \]

\[ \boxed{I_{102}=14.0\text{ nA}} \]

However, this entire signal does not belong to compound B.

Step 3: Account for the chlorine isotope peak from compound A.

Compound A contains one chlorine atom.

Its molecular-ion isotope ratio is approximately:

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

The molecular-ion signal of A at m/z 100 is 12.0 nA.

Therefore, its expected isotope contribution at m/z 102 is:

\[ I_{A,M+2}=\frac{12.0}{3} \]

\[ \boxed{I_{A,M+2}=4.0\text{ nA}} \]

This is part of the observed signal at m/z 102.

Step 4: Determine the actual molecular-ion signal from compound B.

The total signal at m/z 102 is 14.0 nA.

Of this, 4.0 nA comes from the chlorine isotope of compound A.

The remaining signal belongs to the molecular ion of B:

\[ I_B=14.0-4.0 \]

\[ \boxed{I_B=10.0\text{ nA}} \]

Step 5: Calculate the concentration of compound A.

The response factor for A is:

\[ 3.0\text{ nA per mmol L}^{-1} \]

Using the measured molecular-ion current:

\[ c_A=\frac{12.0}{3.0} \]

\[ \boxed{c_A=4.0\text{ mmol L}^{-1}} \]

Step 6: Calculate the concentration of compound B.

The response factor for B is:

\[ 5.0\text{ nA per mmol L}^{-1} \]

Using its corrected molecular-ion current:

\[ c_B=\frac{10.0}{5.0} \]

\[ \boxed{c_B=2.0\text{ mmol L}^{-1}} \]

Step 7: Compare the concentrations.

The concentration ratio is:

\[ c_A:c_B=4.0:2.0 \]

\[ \boxed{c_A:c_B=2:1} \]

Compound A is twice as concentrated as compound B.

The student’s initial prediction was tempting because the observed peak at m/z 102 is taller than the peak at m/z 100.

However, that comparison ignores two important constraints:

1. The m/z 102 peak includes an isotope contribution from compound A.
2. Compound B produces a different detector response per unit concentration from compound A.

Even after correcting for the overlapping isotope peak, the stronger molecular-ion response of B per unit concentration means its smaller signal corresponds to a lower concentration.

This is why relative abundance, peak intensity, isotope patterns, and concentration must be interpreted as separate pieces of information.

The next step in interpreting an unknown mass spectrum is to establish which peaks belong to the molecular ion, which arise from isotopes, and which represent fragments. Once those assignments are reasonable, you can compare the complete fragmentation patterns of structural isomers without treating any single peak as definitive proof.