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
| Component | Typical share | Counted in purity? | Counted in content? |
|---|
| Target peptide | 88–94 % | Yes, as main peak | Yes |
| Related impurities | 1–3 % | Yes, as other peaks | No |
| Counter-ion (TFA or acetate) | 2–8 % | No | No |
| Residual water | 2–6 % | No | No |
| Bulking agent, if present | 0–40 % | No | No |
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