One molecule can make many peaks

Lesson 2 · 30 minutes
Prerequisite: A spectrum shows ions, not molecules.
Mathematics: Arithmetic; optional binomial model and derivation.
Audience: Beginner, with practitioner and expert extensions.

Several peaks can be related to one analyte. Your task is to propose a relationship, calculate what it predicts, and look for evidence that could disprove it.

By the end, you will be able to:

• distinguish an isotope substitution, a charge change, an adduct, and a fragment;
• predict their effects on m/z under explicit assumptions;
• interpret a carbon-only isotope model without confusing it with a full molecular prediction;
• state why a matching mass difference is evidence for a proposal rather than proof of identity.

Predict

A teaching sample contains a small molecule S with formula C₆H₁₂O₆ and neutral monoisotopic mass 180.06338810418 Da. The formula does not specify which structural isomer it is. The mass is calculated from neutral isotope masses, not measured here. NIST carbon, hydrogen, oxygen

Suppose you see features near m/z 163.0601, 181.0707, 182.0740, and 203.0526. Must the sample contain four compounds?

• A. Yes. Every peak represents a different original compound.
• B. No. Several ion relationships could connect these features.
• C. No. Therefore all four must come from S.

Explain

Ask what changed

A sodium adduct contains the analyte and Na⁺; it is written [M+Na]⁺. A fragment ion is a charged dissociation product. These labels describe different chemical relationships, so they require different calculations. IUPAC adduct ion, IUPAC fragment ion

One analyte is a population, not one indestructible particle

The lesson title refers to one molecular species across many molecules in the sample. Different members can contain different isotopes, acquire different charges or adducts, and follow different fragmentation routes. It does not mean that one individual ion generates an entire isotope envelope.

A neutral fragment does not appear as an ordinary mass-spectrum peak merely because a bond broke; a detected product must carry charge. Nor must all sample ions come from the intended analyte. Solvent, contaminants, matrix components, and ions from coeluting substances can contribute. You are testing relationships within a population of signals.

Keep two uses of “M” distinct: in [M+H]⁺, M denotes a neutral molecule; in informal “M+1” isotope-envelope language, the label often denotes a nominal isotope increment. A label such as “M+1” is not enough to specify an ion formula or its charge. Write the actual proposed relationship when calculating.

Isotope spacing tests charge

For otherwise identical ions differing by one ¹²C → ¹³C substitution, let Δx be the numerical m/z shift and n the charge magnitude:

Δx = 1.00335483507 / n

That gives approximately 1.003355, 0.501677, or 0.334452 for n = 1, 2, or 3. This is a calculation for one specified substitution. Other elements contribute their own isotope shifts; a real envelope can combine several substitutions and unresolved fine structure. NIST isotope tables

Monoisotopic does not mean tallest

Monoisotopic mass is calculated using the most abundant isotope of each element. It is a composition convention, not an instruction to choose the tallest measured peak. IUPAC monoisotopic mass

In the model below, “zero ¹³C” means every carbon atom is ¹²C. As carbon count increases, the chance of at least one ¹³C atom increases. The zero-¹³C member can therefore cease to be the most probable member.

A second isotope ruler: chlorine

The carbon increment is useful, but it is not a universal distance between all isotope peaks. NIST gives approximately 1.99704992 Da for replacement of ³⁵Cl by ³⁷Cl. For a singly charged ion containing one chlorine, and ignoring other elements' isotope contributions, the two corresponding members have an approximate abundance ratio of 75.76:24.24, or 100:32. NIST chlorine isotope masses and representative abundances

At charge magnitude 2, that chlorine substitution gives about 0.998525 in m/z. A gap near 1 can therefore have a different explanation from a singly charged one-carbon substitution. Examine the complete envelope and its chemistry rather than using one rounded distance as a universal charge detector. The chlorine ratio here is a one-chlorine model, not the full isotope pattern of any unspecified compound.

Manipulate

Make a prediction, then change one input in this model. The equations and limitations travel with the plot.

Change charge while holding the analyte fixed

Return to the hypothetical 1000.0000 Da peptide P from lesson 1. Predict both the peak position and the one-¹³C spacing when you change from [P+H]⁺ to [P+2H]²⁺.

Change carbon count in a teaching model

Use an illustrative ¹³C probability p = 0.0107 and ¹²C probability 1−p = 0.9893, based on the representative NIST carbon composition. Natural isotope composition can vary between materials. NIST carbon data

Model assumptions: Each of N carbon atoms independently has the same probability p of being ¹³C. This carbon-only binomial model omits all other isotope contributions, including hydrogen, nitrogen, oxygen, sulfur, chlorine, and bromine. It also does not model ionization, detector response, or peak width. The formula C₆H₁₂O₆ contains oxygen and hydrogen; their isotope contributions are deliberately omitted here.

Prediction: If N changes from 6 to 100, must the zero-¹³C peak remain tallest?

The table shows probabilities, not measured peak heights. Only the first three members are shown. Their unrounded retained probabilities sum to 0.999976 for N = 6 and 0.907370 for N = 100; the remainder lies in members with three or more ¹³C atoms. Do not silently rescale a truncated distribution to imply it is complete.

Optional derivation

For k carbon-13 atoms among N carbons:

P(k) = choose(N, k) × p^k × (1−p)^(N−k)

The binomial coefficient counts the ways to place k carbon-13 atoms. Each placement has the probability given by the two powers. Summing k = 0 through N gives 1.

For N = 6, a display that sets the zero-¹³C member to 100 gives the one-¹³C member approximately 6.49, because 100 × P(1)/P(0) ≈ 6.49. Its probability is approximately 6.08%. These differ because one uses the largest peak as denominator and the other uses the complete probability distribution.

Use the model backwards, carefully

Dividing the one-¹³C probability by the zero-¹³C probability cancels most terms in the binomial equation:

P(1)/P(0) = N × p/(1−p)

For p = 0.0107, each carbon contributes about 0.010816 to this ratio. If a clean carbon-only model gives a one-to-zero ratio of 0.0649, then N ≈ 0.0649 × 0.9893 / 0.0107 ≈ 6. This is an estimate conditional on the model, not a universal carbon-count measurement from a real M+1 peak. Other isotopes, overlap, labeling, saturation, and processing can invalidate that shortcut.

Apply

Worked example annotating the four features

Use S at 180.06338810418 Da. Assume singly charged ions for this candidate set. Use proton mass h ≈ 1.0072764665789 Da, Na⁺ mass ≈ 22.9892207021 Da, and neutral H₂O mass ≈ 18.01056468403 Da. The Na⁺ value subtracts one electron mass from the neutral ²³Na atomic mass; binding-energy mass corrections are negligible at the four-decimal display precision used here. NIST CODATA, sodium, hydrogen, oxygen

The sodium and protonated candidates differ by approximately 21.9819 in m/z. The water-loss candidate differs from [S+H]⁺ by approximately 18.0106. For a neutral loss L from an ion whose charge magnitude n is retained, the numerical shift is mass of L in Da / n. Losing water from a doubly charged ion would therefore shift m/z by approximately 9.0053, not 18.0106.

These are bookkeeping predictions. They do not guarantee that S forms every listed ion, or establish a fragmentation pathway from a mass difference alone. Use unrounded values for calculations, then round the display.

What can you claim

The four positions are consistent with the proposed set of ions from S under the stated assumptions. They do not establish that all four share an origin or identify which C₆H₁₂O₆ isomer is present.

To test the proposal, compare the isotope envelopes, check blanks, and examine whether the features track together through a separation. If tandem MS is available, isolate the proposed precursor and test the proposed product relationship. Coelution helps, but two unrelated compounds can coelute.

Practitioner extension

Before counting compounds, group candidate relationships and preserve alternatives. Record whether each link is based only on a calculated difference or is also supported by isotope, chromatographic, or fragmentation evidence.

Optional extension: recover adjacent charge states

Suppose two monoisotopic features belong to the same neutral molecule, carry only added protons, and have adjacent charges n and n+1. Let x be the higher m/z and y the lower. Then x = M₀/n + h and y = M₀/(n+1) + h. Equating the recovered neutral masses gives:

n = (y − h)/(x − y)

For x = 501.0073 and y = 334.3406, this gives n ≈ 2, so the candidates are 2+ and 3+. Each calculation recovers approximately 1000.0000 Da. Charge-state series are characteristic evidence in electrospray work with large biomolecules; the equation here follows from the stated protonation model. Fenn and colleagues, 1989

The integer result is a test of a hypothesis, not proof that any two peaks are adjacent charge states. Missing charge states, adducts, wrong isotope members, or unrelated ions can defeat the premise. Where isotope structure is resolved, check its spacing as additional evidence. Do not apply this equation to adjacent isotope peaks.

Expert extension

Suppose a carbon-only model matches a measured envelope. List what it has not tested: other isotope contributions, overlapping species, isotope labeling, transmission differences, and processing. A good fit to an incomplete model does not show that its omitted effects are absent.

Check your understanding

1 A second feature is 1.0034 higher in m/z

Which statement is best?

2 A doubly charged ion loses neutral water and keeps its charge

What is the predicted decrease in m/z?

3 The N = 6 model gives P(1) about 0.060838

Does it predict the full measured M+1 signal for C₆H₁₂O₆?

4 The tallest isotope peak is not the zero-¹³C peak

Must the spectrum be wrong?

5 Transfer challenge: charge or chemical loss?

A proposed precursor is [P+2H]²⁺ at m/z 501.0073. A second feature is at 492.0020 with an isotope spacing near 0.5017. A third is at 334.3406 with spacing near 0.3345. Which candidate relationship fits each feature? State a test that would strengthen each assignment.

6 Transfer challenge: challenge the neat annotation

The 181.0707 and 203.0526 features coelute in a sample. The 203.0526 feature also appears in a solvent blank at similar response. Does the 21.9819 separation establish a sodium adduct from the sample analyte? What would you investigate next?

Primary references

Additional primary inputs: NIST chlorine data for the specified substitution model, and Fenn et al., Science (1989), DOI 10.1126/science.2675315, for electrospray charge-state series. The examples and algebra above are original teaching calculations.

• NIST isotope data: carbon, hydrogen, oxygen, and sodium. Inputs to the original mass and binomial calculations; representative abundances are not universal constants.
• NIST: CODATA 2022, proton and electron masses.
• IUPAC Recommendations 2013, reproduced entries: adduct ion, fragment ion, and monoisotopic mass.

Next step

For each candidate ion, ask how the experiment could produce it. The ion source determines which kinds of relationships are plausible starting points.

Continue to lesson 3 — Where ions come from
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