A vial is sold by mass and an assay is specified by concentration of molecules. Those are different quantities, and the bridge between them is molecular weight. This is the calculation that decides whether a 5 mg/mL stock is a convenient source of a 10 µM working solution or a source of a systematic error repeated across every plate, and it is the one most often done from memory.
Quick answer
How do you convert mg/mL to molarity for a peptide? Divide the mass concentration in mg/mL by the molecular weight in g/mol to get molarity in mol/L. Multiply by 1,000 for mM and by 1,000,000 for µM. A 5 mg/mL solution of a peptide with a molecular weight of 1,419.53 g/mol is 5 ÷ 1,419.53 = 0.00352 M, or 3.52 mM. Take the molecular weight from the lot certificate, and correct the mass first for net peptide content if the true molar concentration matters.
Why mass concentration is not enough
Mass tells you how much material is present. Molarity tells you how many molecules. For anything involving a receptor, a binding site or a stoichiometric interaction, molecules are the unit that matters, because a receptor does not count milligrams.
The practical consequence is that two solutions at identical mass concentration can differ in molar concentration by an order of magnitude. A 1 mg/mL solution of a small tripeptide contains roughly ten times as many molecules as a 1 mg/mL solution of a large multi-agonist peptide, purely because the individual molecules weigh ten times less.
| Molecular weight (g/mol) | Molarity of a 1 mg/mL solution | In µM |
|---|---|---|
| 400 | 2.50 mM | 2,500 µM |
| 500 | 2.00 mM | 2,000 µM |
| 1,000 | 1.00 mM | 1,000 µM |
| 1,400 | 0.714 mM | 714 µM |
| 2,000 | 0.500 mM | 500 µM |
| 3,000 | 0.333 mM | 333 µM |
| 4,000 | 0.250 mM | 250 µM |
| 5,000 | 0.200 mM | 200 µM |
This is also the reason cost per milligram cannot be compared between different compounds, a point developed in cost per milligram. A milligram is not a fixed number of molecules.
The formula, and its rearrangements
Molarity is moles per litre. Because a concentration in mg/mL is numerically identical to one in g/L, the conversion is a single division.
| To find | Formula | Units |
|---|---|---|
| Molarity | M = C ÷ MW | C in mg/mL, MW in g/mol, result in mol/L |
| Millimolar | mM = (C ÷ MW) × 1,000 | |
| Micromolar | µM = (C ÷ MW) × 1,000,000 | |
| Mass concentration needed | C = M × MW | M in mol/L, result in mg/mL |
| Volume of stock for a dilution | V₁ = (C₂ × V₂) ÷ C₁ | Any consistent concentration units |
The arithmetic of getting to a mass concentration in the first place, from a vial of powder and a measured volume of diluent, is covered in peptide reconstitution and concentration.
Where the molecular weight comes from
Take it from the lot certificate, not from a general reference and not from memory. The certificate reports the molecular weight alongside the sequence and the mass spectrometry result, which is the same figure the analysis was judged against. Two details on that line are worth reading carefully.
- Average or monoisotopic. Average molecular weight uses the natural isotopic mix and is the figure to use for solution preparation. Monoisotopic mass uses the lightest isotope of each element and appears in mass spectrometry. For a peptide of a few thousand daltons the two differ by a meaningful amount, and using the wrong one introduces a small systematic offset.
- Free peptide or salt. The molecular weight quoted is normally that of the free peptide, while the powder in the vial is usually a salt. These are not the same mass, and the difference is handled by net peptide content rather than by adjusting the molecular weight.
The vocabulary on that part of the certificate is defined in the COA glossary, and reading the document as a whole is covered in how to read a certificate of analysis.
The correction almost every calculation skips
A vial labelled 10 mg contains 10 mg of lyophilized material, of which only part is peptide. The remainder is counter-ion from purification, residual water and residual solvent. Net peptide content is the share that is peptide, commonly 70 to 90 percent for a trifluoroacetate salt, and it is reported on the certificate.
Molarity calculated from the label mass therefore overstates the true concentration by exactly the amount that is not peptide. The full chain is short and worth doing in order.
- Label mass × net peptide content = true peptide mass.
- True peptide mass ÷ volume = true mass concentration in mg/mL.
- True mass concentration ÷ molecular weight = molarity.
Why the counter-ion is there at all, and why it differs between preparations, is explained in TFA vs acetate. The distinction between content and purity, which are routinely confused, is in net peptide content vs purity.
A note on precision. For comparative work where every condition uses the same lot, this correction cancels out and can reasonably be omitted, provided the omission is recorded. For work where an absolute molar value is reported, or where results are compared against published figures, it does not cancel and should be applied.
Worked example one: a 10 mg vial
A 10 mg vial of BPC-157, molecular weight 1,419.53 g/mol, reconstituted with 2 mL of bacteriostatic water.
- Nominal mass concentration: 10 mg ÷ 2 mL = 5 mg/mL.
- Nominal molarity: 5 ÷ 1,419.53 = 0.00352 mol/L = 3.52 mM.
- With a reported net peptide content of 85 percent: 10 × 0.85 = 8.5 mg, so 4.25 mg/mL.
- Corrected molarity: 4.25 ÷ 1,419.53 = 0.00299 mol/L = 2.99 mM.
The nominal and corrected figures differ by 15 percent, which is the net peptide content doing exactly what it says it does.
Worked example two: preparing a target concentration
A 50 mg vial of GHK-Cu, molecular weight 403.93 g/mol, reconstituted with 5 mL.
- Mass concentration: 50 mg ÷ 5 mL = 10 mg/mL.
- Stock molarity: 10 ÷ 403.93 = 0.02476 mol/L = 24.76 mM, or 24,760 µM.
- To prepare 10 mL of a 100 µM working solution: V₁ = (100 µM × 10 mL) ÷ 24,760 µM = 0.0404 mL, which is 40.4 µL of stock made up to 10 mL.
Note how much more molar a small peptide is at the same mass concentration. At 403.93 g/mol, GHK-Cu at 10 mg/mL is seven times more concentrated in molecules than BPC-157 would be at the same 10 mg/mL.
Two errors worth naming
- Mixing prefixes. A stock in mM and a target in µM differ by a factor of a thousand. Writing both in the same unit before dividing removes the most common source of a thousand-fold error.
- Assuming a molecular weight. Analogues and modified sequences differ from the parent molecule, sometimes by only a few daltons and sometimes considerably. An amidated C-terminus, an acetylated N-terminus or a single substituted residue all change the figure. The certificate is authoritative; a search result is not.
If the sequence notation on the certificate is unfamiliar, how to read a peptide sequence covers the one- and three-letter codes and how modifications are written.
What to record
The molecular weight used and whether it was average or monoisotopic, its source, the net peptide content applied and whether the correction was made, the resulting molar concentration with its unit, and the lot number that ties all of it to a certificate. Revoryn Labs publishes available batch documentation, including sequence and analytical results, in its public COA ledger, matched to the lot printed on the vial.
Revoryn Labs does not publish usage instructions, protocols or administration guidance of any kind for any product. The calculations here describe laboratory concentration, and nothing more.
Key takeaways
- Molarity is mass concentration divided by molecular weight: mg/mL ÷ g/mol = mol/L.
- Multiply by 1,000 for mM and by 1,000,000 for µM.
- At equal mass concentration, a small peptide is far more molar than a large one.
- Use the average molecular weight from the lot certificate, not the monoisotopic mass.
- Correct the mass for net peptide content before converting, whenever an absolute molar value matters.
- Convert both figures to the same unit before dividing; mixed prefixes cause thousand-fold errors.
Frequently asked questions
How do I convert mg/mL to µM for a peptide?
Divide the mass concentration in mg/mL by the molecular weight in g/mol, then multiply by 1,000,000. At 2 mg/mL with a molecular weight of 1,000 g/mol, the result is 2,000 µM.
What molecular weight should I use?
The average molecular weight of the free peptide, taken from the lot certificate. Monoisotopic mass belongs to the mass spectrometry result and is not the figure for preparing solutions.
Does net peptide content change the molarity?
Yes. Molarity calculated from the label mass assumes the powder is entirely peptide. Applying net peptide content first typically lowers the result by 10 to 30 percent.
Why do two peptides at the same mg/mL have different molarity?
Because molarity counts molecules and mass does not. A smaller molecule means more molecules in the same mass, so a low molecular weight peptide is more molar at identical mass concentration.
Is mg/mL the same as g/L?
Yes, numerically. Both the mass and the volume scale by a thousand, so the ratio is unchanged, which is what makes the conversion a single division by molecular weight.
Sources
- D’Hondt M, Bracke N, Taevernier L, et al. Related impurities in peptide medicines. J Pharm Biomed Anal. 2014;101:2-30. doi:10.1016/j.jpba.2014.06.012
- Merck / Sigma-Aldrich. Synthetic Peptide Handling & Storage Protocol. sigmaaldrich.com
- Bachem. Quality Control of Amino Acids & Peptides: A Guide. bachem.com
For laboratory and in-vitro research use only. Not for human or veterinary use.
For laboratory and in-vitro research use only. Not for human or veterinary use. See our research-use disclaimer.
