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Peptide Stock Solutions: Molarity and Dilution Math With Worked Examples
Most bench errors with peptides are not chemical. They are arithmetic errors, or hidden assumptions about how much of the powder is actually peptide. This guide walks through the basic calculations with numbers you can check yourself.
Key takeaways
- Moles equal mass divided by molecular weight.
- Molarity equals moles divided by volume.
- Correct for net peptide content before calculating.
- Use C1V1 equals C2V2 for dilutions.
Step 1: mass to moles
Take BPC-157, whose molecular weight is about 1,419.5 g/mol. A 10 mg vial, treating the whole mass as peptide, contains:
- 10 mg = 0.010 g
- 0.010 g divided by 1,419.5 g/mol = 0.00000704 mol = 7.04 micromoles
Step 2: molarity of the stock
Dissolve it in 2.0 mL: 7.04 micromoles divided by 2.0 mL equals 3.52 micromoles per mL, which is 3.52 millimolar (mM). Another way to see it: 5 mg per mL divided by 1,419.5 gives 3.52 mM.
Step 3: the net content correction
The powder is not 100 percent peptide, because of water and counterions (see net peptide content versus purity). If the report states a net peptide content of 80 percent, the actual peptide mass is 8 mg, or 5.63 micromoles, and in 2.0 mL the stock is 2.82 mM. Ignoring the correction would put every downstream concentration about 25 percent too high. Net content is measured by methods such as amino acid analysis.
Step 4: dilute with C1V1 equals C2V2
You want 10 micromolar (10 uM) working solution from the 2.82 mM stock. The dilution factor is 2,820 uM divided by 10 uM, or 282. For 1 mL of working solution you need 1,000 uL divided by 282, or about 3.5 uL of stock plus 996.5 uL of diluent. Because pipetting 3.5 uL is imprecise, make it in two steps: 1 to 20 first (50 uL stock plus 950 uL diluent giving 141 uM), then 1 to 14.1 (about 71 uL plus 929 uL giving 10 uM).
Step 5: from mg per mL to micromolar
A shortcut: concentration in uM equals mg per mL multiplied by 1,000,000, divided by the molecular weight in g/mol, all divided by 1,000. More simply, 1 mg/mL of a 1,419.5 g/mol peptide is 0.704 mM or 704 uM.
Serial dilution
A tenfold series takes 100 uL of the previous tube into 900 uL of diluent each time. Five steps from 1 mM reaches 10 nM. Errors compound, so a poor first pipetting step affects every tube, and adsorption at the low end (see peptide adsorption) makes the lowest concentrations less accurate than the arithmetic says.
Checklist to avoid common mistakes
- Use the molecular weight of the exact form: salt form, hydrate and any metal adds mass.
- Correct for net peptide content and note it in the notebook.
- Use calibrated pipettes and pipette volumes above about 2 uL where possible.
- Do not mix up mg and mcg, or mM and uM.
- Record the lot, the diluent and the date.
The bottom line
The chemistry may be complicated, but the arithmetic is simple. Doing it carefully and writing the numbers down is one of the cheapest ways to improve reproducibility.
Related reading
- Net Peptide Content vs. Purity: Why 99% Does Not Mean 99% Peptide
- Why Peptide Seems to Vanish: Surface Adsorption on Glass and Plastic
- Amino Acid Analysis: How Peptide Content Is Measured Without UV
- Browse the full Research Library
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For laboratory research use only. Not for human or veterinary use, and not intended to diagnose, treat, cure or prevent any condition. This article is educational and is not medical advice.