Accepted answer
95.2 per cent is a statement about area, and the other 4.8 per cent is everything the detector saw and did not assign to your peak. Read it as 95.2 of every 100 units of peak area at whatever wavelength was used, not as 95.2 per cent of the mass in the vial. With the sampling plan attached you can at least see how the figure was produced, which is the difference between a measurement and a claim. What it still does not tell you is content: how many milligrams are actually there.
The single most important distinction is between what purity measures — the fraction of detected material that is your target — and what you actually want to know — how much of the material in the vial is your target.
Integration of the shoulder — whether you use perpendicular drop or tangent skim — determines what area gets assigned to the main peak versus the impurity table.
Reconciling gross mass to label claim
| Component | Typical share | Counted in purity? | Counted in content? |
|---|
| Target peptide | 88–94 % | Yes, as main peak | Yes |
| Related impurities | 1–3 % | Yes, as other peaks | No |
| Counter-ion (TFA or acetate) | 2–8 % | No | No |
| Residual water | 2–6 % | No | No |
| Bulking agent, if present | 0–40 % | No | No |
Column pore size affects mass transfer — a 100 Angstrom packing on a 5 kDa peptide restricts diffusion, broadening the peak and potentially hiding small impurities in the shoulders.
The ICH Q3A impurity thresholds and the relevant pharmacopoeial chapters all specify method validation requirements that almost no research-grade certificate claims to meet.
Compare purity within a single laboratory on the same method, never across laboratories.
edited 15 Oct 2025 by RP_C18 — expanded the table to cover the lower concentration
7Adding for future readers: the certificate should carry the lot number, not just a batch code. – lyoph_cake 5 months ago 8Adding a vote because this deserves more of them. – pip_okonjo 7 months ago add a comment