Pharmaceutical Analysis

Percent Purity (Assay) Calculator

Works out the % purity of a drug sample from a titration, and checks it against assay limits.

Purity

Enter the sample weight, titrant volume, normality and equivalent factor.

Titration data

Sample weight (W)

Accurately weighed sample, in the same mass unit as E.

Titrant volume (V)

Burette reading at the end point (blank-corrected).

Titrant normality (N)

Standardised strength of the titrant, e.g. 0.1 N NaOH.

Equivalent factor (E)

g/mEq

Weight of analyte equivalent to 1 mEq of titrant — MW ÷ n ÷ 1000.

Dilution factor (D)

1 if you titrated the whole sample; e.g. 5 if you titrated a 1-in-5 aliquot. Blank or 0 counts as 1.

Try an example

Pharmacopoeial limits

USP90 – 110%

BP95 – 105%

IP98 – 102%

This calculator grades against 98–102% (compliant), 95–98% (borderline) and anything else (fails). Always check the individual monograph.

% Purity = (V × N × E × D × 100) ÷ W

V = titrant volume (mL), N = titrant normality (mEq/mL), E = equivalent factor — the milliequivalent weight of the analyte (for molecular weight M reacting with n equivalents, E = M ÷ n), D = dilution factor, W = sample weight.

V × N gives the milliequivalents of titrant used; multiplying by E converts that to the mass of analyte it reacted with. Dividing by the weight taken and multiplying by 100 gives the percentage.

Sources: USP General Chapter <541>, European Pharmacopoeia 2.5.

Worked example: Iodometric assay of ascorbic acid

0.1500 g of ascorbic acid (C₆H₈O₆, M = 176.12 g/mol) is dissolved and titrated directly with 0.1 N iodine (0.05 M I₂). The blank-corrected end point is 16.85 mL. Each molecule loses 2 electrons when iodine oxidises it to dehydroascorbic acid. Find the % purity.

Sample weight (W): 0.1500 g
Titrant volume (V): 16.85 mL
Titrant normality (N): 0.1 N
Equivalent factor (E): 0.08806 g/mEq
Dilution factor (D): 1

Equivalent factor
E = M ÷ n ÷ 1000 = 176.12 ÷ 2 ÷ 1000 = 0.08806 g/mEq
Milliequivalents of titrant
V × N = 16.85 mL × 0.1 mEq/mL = 1.6850 mEq
Analyte found
V × N × E × D = 1.6850 × 0.08806 × 1 = 0.1484 g
Percent purity
% purity = (16.85 × 0.1 × 0.08806 × 1 × 100) ÷ 0.1500 = 98.92 %

% purity = 98.92 % — inside the 98–102 % band this calculator grades against.

The 0.1500 g taken contained 0.1484 g of material that reacted with iodine as ascorbic acid. That is what an assay measures: titratable analyte, not identity. Any impurity that also reduces iodine is counted as ascorbic acid, so a 'compliant' figure does not show the sample is free of impurities — that needs the monograph's identification and related-substance tests. The 98–102 % band is this page's teaching range; the individual monograph's limits decide compliance.

Common mistakes

Forgetting n in the equivalent factor. For ascorbic acid with iodine n = 2, so E = 0.08806 g/mEq; using M ÷ 1000 = 0.17612 doubles the answer (197.84 % here).

Mixing mass units: W in mg with E in g/mEq (or the reverse) makes the result 1000 times too small or too large. Keep both in g, or both in mg.

Confusing molarity with normality. 0.05 M I₂ is 0.1 N, because each I₂ takes 2 electrons; entering 0.05 as N halves the result.

Using the raw burette reading. Subtract the blank first, and for a back-titration enter the volume of titrant that actually reacted with the sample.

Leaving D at 1 after titrating an aliquot. If the sample was made up to 100 mL and 20 mL was titrated, D = 5.

Using the nominal normality of the titrant instead of its standardised value (e.g. 0.1 N on the label, 0.0987 N by standardisation).

Assumptions and limitations

Assumes one known, stoichiometric reaction between titrant and analyte; side reactions or interfering substances change the result.

Assumes a sharp, correctly detected end point. Indicator error and over-titration add directly to V.

The result is reported on the weight taken, as is; drying-loss or anhydrous-basis corrections required by a monograph are not applied.

The compliance bands (98–102 %, 95–98 %) are general teaching ranges, not the limits of any particular monograph.

Beer-Lambert Law CalculatorThe spectrophotometric route to an assay, when the analyte is measured by absorbance instead of titration.Accuracy & % Recovery CalculatorCheck the assay method itself: recover a known amount and report mean recovery and %RSD.

References

Beckett AH, Stenlake JB. Practical Pharmaceutical Chemistry, Part 1. 4th ed. London: Athlone Press; 1988.
Harris DC, Lucy CA. Quantitative Chemical Analysis. 10th ed. New York: W. H. Freeman (Macmillan Learning); 2019.

Common questions

The same mass unit. If E is written in g/mEq (e.g. 0.1802 for aspirin), enter W in grams; if E is in mg/mEq (180.2), enter W in milligrams. Mixing them makes the result 1000 times too big or too small.

Most monographs state it directly, e.g. "each mL of 0.1 M NaOH is equivalent to 18.02 mg of C₉H₈O₄". Otherwise divide the molecular weight by the number of equivalents that react per molecule.

When you dissolve the sample, make it up to volume, and titrate only part of it. If you dissolved in 100 mL and titrated a 20 mL aliquot, D = 5.

Usually a unit mismatch between W and E, a normality that was not standardised, or an overshot end point. Results slightly over 100% can be real within assay error, which is why limits such as 98–102% extend above 100.

Prepared by the PharmaWallah student team. Verify results with your textbook or instructor. Report an error · Editorial policy