How to calculate peptide molarity for a stock solution

Methods · 4 min read · Updated November 2, 2026

To find the molarity of a peptide stock solution, convert peptide mass to moles with the molecular weight, then divide by the volume in liters. Use net peptide mass, not powder mass: 5.00 mg of powder at 72% net peptide content is 3.60 mg, which for a 1,123.26 g/mol peptide is 3.20 µmol, or 1.00 mM in 3.20 mL.

Which molecular weight should you use?

Use the average molecular weight of the free peptide, in g/mol. It reflects the natural isotope mix in a weighed sample. The monoisotopic mass is for comparing with mass spectra and runs slightly low. The examples here use an invented decapeptide, GAYSLKTEVR (C49H82N14O16), with an average molecular weight of 1,123.26 g/mol and a monoisotopic mass of 1,122.60 Da.

Do not combine the salt's molar mass with a net peptide content correction, because that subtracts the counter-ion twice. Either multiply the powder mass by net peptide content and divide by the free peptide's molecular weight, or work from the salt's molar mass if you know its exact counter-ion and water content. The first route is simpler and is used below.

How do you convert mass to moles and molarity?

Two relations do the work. Moles equal mass divided by molecular weight, n = m ÷ MW. Molarity equals moles divided by volume in liters, c = n ÷ V. Keep the prefixes consistent: milligrams divided by g/mol gives millimoles, and micromoles per milliliter equals millimolar.

Example 1. Dissolve 2.00 mg of net peptide to a final volume of 1.00 mL. Moles: 2.00 mg ÷ 1,123.26 g/mol = 1.78 µmol. Concentration: 1.78 µmol ÷ 1.00 mL = 1.78 mM.

How do you correct for net peptide content?

Lyophilized peptide is not all peptide. Counter-ions such as TFA or acetate and bound water make up the rest, and net peptide content is typically 70 to 85% for SCALE Performance material. Check whether the fill quantity on the vial refers to powder or to net peptide; the correction applies to powder.

Example 2. You want a 1.00 mM stock from a vial holding 5.00 mg of powder at 72% net peptide content. Peptide mass: 5.00 mg × 0.72 = 3.60 mg. Moles: 3.60 mg ÷ 1,123.26 g/mol = 3.20 µmol. Volume: 3.20 µmol ÷ 1.00 mM = 3.20 mL. Dissolving the entire contents of the vial avoids weighing out small amounts of a hygroscopic powder.

Skip the correction and the arithmetic gives 5.00 mg ÷ 1,123.26 g/mol = 4.45 µmol, so you would make the solution up to 4.45 mL for a nominal 1.00 mM. The real concentration would be 3.20 ÷ 4.45 = 0.72 mM, 28% below target, and every dilution made from it inherits the error.

Purity is a much smaller correction. To count only the target sequence, multiply by HPLC purity as well: 3.60 mg × 0.992 = 3.57 mg at 99.2%.

How do you dilute a stock to a working concentration?

Use C1V1 = C2V2. Example 3. For 1.00 mL of a 10.0 µM working solution from the 1.00 mM stock, V1 = (10.0 µM × 1.00 mL) ÷ 1,000 µM = 0.0100 mL. Add 10.0 µL of stock to 990 µL of assay buffer and mix.

Volumes under a few microliters are hard to pipette accurately. For lower concentrations, dilute in steps of 1 part to 9 parts buffer: 1.00 mM to 100 µM, then 10.0 µM, then 1.00 µM.

How do you convert between µg/mL and µM?

Use the molecular weight, minding the units: µM = (µg/mL × 1,000) ÷ MW, and µg/mL = (µM × MW) ÷ 1,000. Example 4. For the example peptide, 10.0 µg/mL is (10.0 × 1,000) ÷ 1,123.26 = 8.90 µM, and 10.0 µM is (10.0 × 1,123.26) ÷ 1,000 = 11.2 µg/mL.

Be clear whether a µg/mL figure refers to powder or to peptide. A solution of 1.00 mg/mL of powder at 72% net peptide content is 0.72 mg/mL of peptide, or 0.641 mM.

How can you check the concentration of a stock solution?

If the sequence contains Trp or Tyr, measure absorbance at 280 nm and apply the Beer-Lambert law, c = A ÷ (ε × l). The molar absorption coefficients are about 5,500 M⁻¹ cm⁻¹ per Trp and 1,490 M⁻¹ cm⁻¹ per Tyr. The example peptide has one Tyr, so a true 1.00 mM solution gives A280 = 1.49 in a 1 cm cell. Dilute it twofold to read about 0.75, where most spectrophotometers are more accurate.

A reading near 0.54 at that dilution would mean 0.72 mM, the uncorrected result from Example 2. Coefficients calculated from Tyr alone can be off by 10% or more, so use A280 as a check, not a definitive value. For sequences without Trp or Tyr, amino acid analysis is the usual way to measure concentration.

Questions

How do you calculate the molarity of a peptide solution?

Divide the peptide mass by its molecular weight to get moles, then divide by the solution volume in liters. Use net peptide mass, the powder mass multiplied by net peptide content.

Should I use the average or monoisotopic mass for peptide solutions?

Use the average molecular weight. The monoisotopic mass is for mass spectrometry and is slightly lower, about 0.06% for a 1.1 kDa peptide, so using it adds a small, avoidable error.

How do you convert µg/mL to µM for a peptide?

Multiply the µg/mL value by 1,000 and divide by the molecular weight in g/mol. For a 1,123.26 g/mol peptide, 10.0 µg/mL is 8.90 µM.

Do you need to correct for purity as well as net peptide content?

Only if you need the amount of the target sequence specifically. At 99% or higher HPLC purity the purity correction is under 1%, while the net peptide content correction is typically 15 to 30%.