Accepted answer
m/z = 1578.11 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: (4731.3 + 3 × 1.00728) ÷ 3 = 4734.322 ÷ 3 = 1578.11. 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 1183.83, 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.
More usefully, scrambled disulfides have the same mass as correctly formed ones, so mass spectrometry alone cannot detect a scrambling failure.
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.
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.
A correct mass is necessary for identity but not sufficient — you also need the chromatography to confirm it.