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How do I compute the +3 charge state m/z for a peptide of 4207.2 Da?

Asked 25 Jan 2026Modified 4 months agoViewed 2.9k times
5

Setup, so nobody has to ask: +3 · 4207.2 Da.

The units are where I keep going wrong, so please be explicit about them.

I have sanity-checked the order of magnitude and it seems right, which is not the same as being right.

Where is my error, and what is the correct working?

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askedpierce_count24k3825 Jan 2026

3 Answers

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13

m/z = 1403.41 at 3+. Electrospray charges a peptide by adding protons, so the observed ion is the neutral mass plus 3 protons, all divided by the charge: (4207.2 + 3 × 1.00728) ÷ 3 = 4210.222 ÷ 3 = 1403.41. The proton term is the one people drop, and because it is z protons over z charges it shifts m/z by 1.007 at every charge state — small, and far larger than the mass accuracy of the instrument. The neighbouring charge state sits at 1052.81, and seeing the two of them where they belong is better identity evidence than either one alone. Use the average mass against an average-mass calculation and the monoisotopic mass against a monoisotopic one; mixing them costs you a couple of daltons on a peptide this size.

On the detail: a mass shift of exactly zero with a shifted retention time points to an isomer — a scrambled disulfide or a racemised residue — which mass spectrometry alone cannot identify.

For a large peptide with multiple peaks in the mass spectrum, comparing the observed isotope pattern to the calculated pattern is a quick check that the formula matches.

Reconciling gross mass to label claim

ComponentTypical shareCounted in purity?Counted in content?
Target peptide88–94 %Yes, as main peakYes
Related impurities1–3 %Yes, as other peaksNo
Counter-ion (TFA or acetate)2–8 %NoNo
Residual water2–6 %NoNo
Bulking agent, if present0–40 %NoNo

It helps to be literal here: a monoisotopic mass includes only the lightest isotope of each element, while the average mass weights by natural isotope abundance, and small peptides use monoisotopic mass.

I would not trust a mass result without a good baseline and a blank injection check.

A correct mass is necessary for identity but not sufficient — you also need the chromatography to confirm it.

edited 20 Mar 2026 by tandem_gradient — updated for the 2026 guidance change

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TG
answeredtandem_gradient61k24812 Mar 2026
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8

Worth being precise here: scrambled disulfides have the same mass as correctly formed ones, so mass spectrometry alone cannot detect a scrambling failure.

A mass shift of minus eighteen usually means dehydration or a succinimide intermediate, which is pH-dependent and can be reversible.

More usefully, high-resolution mass spectrometry can distinguish a Lys-containing peptide from an Arg-containing peptide of similar mass because of the isotope difference.

The caveat is that a correct mass does not mean the peak is correct — isomers and co-eluting species can have the same m/z.

Always run a blank between samples and check for carry-over.

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ID
answeredines_delacruz16k1623 Mar 2026
7

Stated carefully, a D-amino-acid substitution has the same molecular weight as the L-form, so mass spectrometry cannot distinguish them without fragmenting the peptide.

A mass shift of minus one hundred and twenty-eight usually means a missing Gln or Lys residue from a synthesis deletion sequence.

Mechanically, the baseline noise on a mass spectrum sets the limit of detection, and a weak signal close to the noise is not reliable evidence for the presence of a species.

Peptide mapping — enzymatic digestion followed by tandem mass spectrometry — can confirm the primary sequence and is the method of choice when identity is ambiguous.

The practical summary: use mass spectrometry for identity, not for purity.

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KL
answeredkelvin_lam7.6k153 Apr 2026
7Thank you — this is the answer I was looking for. – nkem_obiora 7 months ago
8Two of us submitted the same lot to different laboratories and got results a tenth apart. – gradient_slope 9 months ago
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