Accepted answer
Nothing is wrong with either report. Do the arithmetic in the direction the instrument works — neutral mass to m/z, one charge state at a time — and both spectra fall out exactly.
The only equation you need
For a peptide that picks up z protons in the electrospray source:
m/z = (M + z x 1.00728) / z
M is the neutral mass, and 1.00728 Da is the mass of a proton (hydrogen atom 1.00794 minus an electron 0.00055; if you use 1.008 you will be off by about 0.001 per charge, which nobody cares about at unit resolution).
Semaglutide, average mass 4113.64 Da
Molecular formula C187H291N45O59. Take the average mass and grind through z = 2 to 6:
- z = 2: (4113.64 + 2.015) / 2 = 4115.655 / 2 = 2057.83
- z = 3: (4113.64 + 3.022) / 3 = 4116.662 / 3 = 1372.22
- z = 4: (4113.64 + 4.029) / 4 = 4117.669 / 4 = 1029.42
- z = 5: (4113.64 + 5.036) / 5 = 4118.676 / 5 = 823.74
- z = 6: (4113.64 + 6.044) / 6 = 4119.684 / 6 = 686.61
Your three observed peaks were 1372.2, 2057.8 and 1029.4. That is [M+3H]3+, [M+2H]2+ and [M+4H]4+, matching to within 0.03 Da. Identity confirmed.
Tirzepatide, average mass 4813.53 Da
Formula C225H348N48O68.
- z = 3: (4813.53 + 3.022) / 3 = 4816.552 / 3 = 1605.52
- z = 4: (4813.53 + 4.029) / 4 = 4817.559 / 4 = 1204.39
- z = 5: (4813.53 + 5.036) / 5 = 4818.566 / 5 = 963.71
Observed 1605.5, 1204.4, 963.7. Same story.
Going backwards from a peak you have not assigned
If you know the charge, rearrange: M = z x (m/z) - z x 1.00728. For the tirzepatide 4+ peak: M = 4 x 1204.39 - 4 x 1.00728 = 4817.56 - 4.03 = 4813.53.
If you do not know the charge, you can get it from any two adjacent charge states. Call the lower-charge (higher m/z) peak m1 and the next one m2:
z1 = (m2 - 1.00728) / (m1 - m2)
For tirzepatide with m1 = 1605.52 and m2 = 1204.39: (1204.39 - 1.007) / (1605.52 - 1204.39) = 1203.383 / 401.13 = 3.00. So m1 is the 3+ ion. That trick is worth memorising because it lets you assign a spectrum with no software at all.
On a high-resolution instrument there is a second route: the spacing between isotope peaks within one charge envelope is 1/z. Spacing of 0.333 Da means 3+, 0.250 Da means 4+. If a report claims 4+ and the isotopes are 0.5 Da apart, the assignment is wrong.
Your 4113.6 versus 4111.1
Average mass sums the isotope-weighted atomic masses; monoisotopic mass uses only the lightest isotope of each element. For semaglutide:
| Quantity | Semaglutide | Tirzepatide |
| Average mass (Da) | 4113.64 | 4813.53 |
| Monoisotopic mass (Da) | 4111.12 | 4810.52 |
| Difference | 2.52 | 3.01 |
At roughly 4000 Da the gap is about 2.5 to 3 Da and it grows with molecular size, because a bigger molecule contains more carbons and therefore more chances to carry a carbon-13. Report 1 quoted average, report 2 quoted monoisotopic. Both are correct and neither told you which, which is a documentation failure rather than an analytical one.
Practical rule: low-resolution instruments (single quadrupole, ion trap) cannot resolve the isotope envelope on a 4 kDa peptide, so the centroid they report is effectively the average mass. High-resolution instruments (TOF, Orbitrap) resolve it and quote monoisotopic. If a COA gives you a mass to two decimal places, it is a high-resolution number and it should be monoisotopic; if it gives you a whole number, ask which convention was used.
edited 10 May 2025 by Dr_Yusuf_Adeyemi — clarified the distinction between purity and content
The z1 = (m2 - 1.00728)/(m1 - m2) formula has saved me twice on reports with no charge annotation. – sian_llewellyn 4 months ago 8Adding to the last paragraph: sodium adducts sit +21.98 above the protonated ion per charge, which trips people up on the 2+ envelope. – triple_agonist_q 2 months ago add a comment