Accepted answer
m/z = 1204.38 at 4+. Electrospray charges a peptide by adding protons, so the observed ion is the neutral mass plus 4 protons, all divided by the charge: (4813.5 + 4 × 1.00728) ÷ 4 = 4817.529 ÷ 4 = 1204.38. 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 963.71, 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.
It helps to be literal here: 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.
Electrospray ionisation creates multiple charge states of the same peptide — a 4 kDa peptide might appear at +2, +3 and +4 — and all of them must be accounted for in the spectrum.
In practice, a mass shift of plus sixteen usually means oxidation at methionine or tryptophan, which is common in peptides and often comes from sample handling rather than synthesis failure.
False positives from contamination are common in mass spectrometry work, and running a blank between every sample and a solvent background are standard practice.
Always run a blank between samples and check for carry-over.
5The distinction between purity and content cannot be repeated often enough here. – jana_horakova 6 months ago 4The system-suitability data is the part that tells you whether to believe the rest. – Dr_Bram_Verhoeven 4 months ago add a comment