Potency Calculation (Worked Example)
A step-by-step numerical walkthrough of an API potency calculation on the dispensing floor — from CoA parsing through basis matching, moisture correction, salt-to-base conversion, and the actual mass the operator weighs.
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This is the calculation regulators reconstruct first when the batch record is questioned.
On this page · 9 sections
How does Potency Calculation (Worked Example) apply to your shop floor?
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01The scenario
The recipe (MMR) calls for 50.000 kg of metoprolol on a free-base, anhydrous basis. The lot picked from inventory is metoprolol tartrate. The supplier CoA reports assay 98.6% on anhydrous basis and LOD 0.4% as-is. Salt stoichiometry: 2:1 (two metoprolol molecules per tartrate). MW(base) = 267.36; MW(salt) = 684.81.
| Input | Value | Source |
|---|---|---|
| Recipe target | 50.000 kg (free base, anhydrous) | MMR |
| Assay | 98.6% (anhydrous) | CoA |
| LOD | 0.4% (as-is) | CoA |
| Salt form | Tartrate, 2:1 | Item master |
| MW(base) / MW(salt) | 267.36 / 684.81 | Item master |
02The four-factor stack (order matters)
- Salt-to-base conversion — recipe says 'free base', material is a salt.
- Basis alignment — recipe is anhydrous; assay is anhydrous. No conversion needed here.
- Potency correction — apply the actual assay factor.
- Moisture correction — bring anhydrous target back to as-is (what the operator weighs).
03Step 1 — Salt-to-base conversion
SBF = MW(salt) / MW(base) / stoichiometry = 684.81 / 267.36 / 2 = 1.28089.
Mass after SBF = 50.000 × 2 × 267.36 / 684.81 = 39.036 kg of tartrate would deliver the base mass at 100% purity — but that direction is wrong for a 2:1 salt. Restated correctly: the tartrate salt contains 2 × 267.36 / 684.81 = 78.07% metoprolol base by mass. To deliver 50.000 kg of base, dispense 50.000 / 0.78071 = 64.043 kg of pure tartrate.
04Step 2 — Basis alignment
Recipe target: anhydrous. Assay: anhydrous. No basis conversion. The 64.043 kg from Step 1 is the anhydrous-basis mass of pure tartrate needed.
05Step 3 — Potency correction
The lot assays at 98.6%, not 100%. Apply PF = 100 / 98.6 = 1.01420.
Mass after PF = 64.043 × (100 / 98.6) = 64.953 kg (anhydrous basis, as-received tartrate).
06Step 4 — Moisture correction
The 64.953 kg is the anhydrous mass. But the operator is weighing wet material with 0.4% LOD. Divide by (1 − LOD) to get the as-is mass.
As-charged mass = 64.953 / (1 − 0.004) = 65.214 kg.
07Full audit trail entry
| Field | Value |
|---|---|
| Nominal target (MMR) | 50.000 kg metoprolol (free base, anhydrous) |
| Lot selected | MT-2026-0417 (metoprolol tartrate) |
| SBF applied | 1 / 0.78071 = 1.28089 |
| PF applied | 100 / 98.6 = 1.01420 |
| Moisture correction | 1 / (1 − 0.004) = 1.00402 |
| Combined factor | 1.30440 |
| Adjusted charge (as-is) | 65.214 kg |
| Weighed (actual) | 65.209 kg |
| Delta | −0.005 kg (−0.008%) |
| Tolerance band | ±0.10% |
| Status | Accepted; e-signed operator + verifier |
08Where this goes wrong in the real world
- Operator forgets SBF and dispenses 50.000 kg of tartrate → batch is 21.9% sub-potent.
- Spreadsheet applies moisture before assay → answer is off by 0.4% (small here, catastrophic with wet materials).
- CoA parsed as 'as-is' assay when it was reported anhydrous → double moisture correction; 0.4% over-charge.
- 2:1 stoichiometry treated as 1:1 → 2× under-charge on the tartrate.
- Free-text CoA field 'purity' interpreted as assay → conflates purity (impurity total) with active content.
09How V5 does this
Frequently asked questions
Q.What if the CoA reports assay to 3 decimals but the scale reads to 2?+
Carry full precision through the calculation; round only the final as-charged mass to the scale's readable precision. Truncating intermediates cascades error across the four factors.
Q.What tolerance band applies?+
Recipe-defined. Common bands are ±0.1% for high-potency APIs, ±0.5% for standard APIs, ±1.0% for excipients. V5 checks the actual weighed mass against the as-charged target ± band.
Q.Does the same math apply to biologics?+
The stack changes — bioactivity units (IU) replace mass assay, but the principle is identical: normalise for reference, correct for lot, adjust to as-is charge.
Q.What if two of the four factors are '100%' (no correction)?+
The math still runs; the neutral factors resolve to 1.0. Never hard-code 'no correction needed' into the recipe — that hides the case where a future lot actually needs it.
Primary sources
Further reading
Want to see how Potency Calculation (Worked Example) could fit into your own records and workflows? Explore the related V5 pages or talk to our team about what applies to your operation.