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How do I compute the +2 charge state m/z for a peptide of 4813.5 Da?

Asked 29 Mar 2026Modified 2 months agoViewed 9.9k times
19

Numbers first: +2 · 4813.5 Da.

This should be a straightforward calculation and I keep getting two different answers.

The numbers are arbitrary; the method is what I am after.

Can someone walk through the arithmetic step by step?

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NP
askednet_peptide12k1529 Mar 2026
4Voting to keep this open — it is more specific than it first looks. – dead_volume 6 months ago
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3 Answers

Accepted answer first, then by votes
84

Accepted answer

m/z = 2407.76 at 2+. Electrospray charges a peptide by adding protons, so the observed ion is the neutral mass plus 2 protons, all divided by the charge: (4813.5 + 2 × 1.00728) ÷ 2 = 4815.515 ÷ 2 = 2407.76. 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 1605.51, 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.

Start from what electrospray ionisation does: it ionises the peptide without fragmenting it, creating singly or multiply charged species that the mass analyser then separates by their mass-to-charge ratio.

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

What each test answers

TestAnswersDoes NOT answer
RP-HPLC, area %What fraction of detected material is the targetHow much target is present
Quantified contentMilligrams of peptide per vialWhat the impurities are
ESI-MS identityWhether the molecular weight matchesPurity, or isomeric substitution
Peptide mappingSequence, localised to a fragmentQuantity
Karl FischerWater content of the solidSolvent content
LAL endotoxinPyrogen load in EU/mgSterility
Sterility testGrowth in defined media over 14 daysEndotoxin, or bioburden count

In practice, 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.

Electrospray ionisation soft-ionisation behaviour is well-characterised and standards exist for m/z calibration and mass accuracy assessment.

Worth noting that source contamination is common and silent, so a result that looks too good to be true often is.

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

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TG
answered · acceptedtandem_gradient61k24827 Apr 2026
3Adding a vote because this deserves more of them. – g_paskevicius 9 months ago
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34

On the detail: tandem mass spectrometry fragments the ions and measures the fragment masses, which provides sequence information and is the best tool for confirming identity.

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

Specifically, the m/z accuracy achievable depends on the mass analyser type — quadrupole gives low accuracy, time-of-flight gives moderate accuracy, and Orbitrap gives high accuracy.

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 limitation is that mass spectrometry tells you the mass and almost nothing else, so it needs to be paired with chromatography or other identity information.

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

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TG
answeredtandem_gradient61k24816 Apr 2026
6The distinction between purity and content cannot be repeated often enough here. – micron22 3 days ago
7Two of us submitted the same lot to different laboratories and got results a tenth apart. – Dr_Rosalind_Achebe 2 months ago
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26

Concretely, scrambled disulfides have the same mass as correctly formed ones, so mass spectrometry alone cannot detect a scrambling failure.

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.

A mass shift of plus one usually means deamidation at asparagine or glutamine, which creates a secondary amine instead of an amide and changes the mass by exactly one.

False positives from contamination are common in mass spectrometry work, and running a blank between every sample and a solvent background are standard practice.

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

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

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AM
answeredaine_mulcahy28k2719 May 2026

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