Accepted answer
On a 60 mg vial every percentage point of water and counter-ion is 0.6 mg, so the answer is a multiplication once you have the net peptide content off the certificate. Settle first whether the 60 mg on the label is gross fill or net peptide, because the two differ by exactly the quantity being asked about. If the certificate quotes net peptide content of 85 per cent, a 60 mg gross fill holds 51 mg of peptide; at 78 per cent it holds 46.8 mg. That 4.2 mg gap is 7 per cent of the label, larger than any purity difference anybody argues about, and it is invisible to a purity figure because water and acetate are not impurity peaks. Amino acid analysis or a nitrogen determination gives you the number; an HPLC area per cent never will.
The honest answer is that you cannot know for certain what you have without a content assay, and the purity number alone is not enough.
If the standard and sample have different absorption coefficients at the detection wavelength, the response factors differ and the inference fails.
Worth being precise here: for peptides at 214 nanometres the response is roughly proportional to the number of peptide bonds, so truncation impurities have lower response factors and overestimate content.
Where content data have been published from testing services on common peptides, the spread between services on identical material is typically a few per cent.
Ask for both the purity and the content, and do not accept purity alone.
edited 14 Mar 2026 by kwn_analytical — tightened the wording; no substantive change