Peptide Medix product catalog

ET
Editorial Team
August 16, 2026 5 min read

Molecular weight, moles and molarity are connected by three relationships — moles = mass ÷ MW, molarity = moles ÷ litres, and C₁V₁ = C₂V₂ for dilutions — and because molecular weight in daltons is numerically identical to grams per mole, every conversion is a single division. Concentration in mg/mL is what a vial label and a syringe give you; molarity is what receptor binding, enzyme kinetics and cell-culture protocols are written in. Converting between them is the arithmetic that stops two peptides at "the same concentration" from being at wildly different numbers of molecules. This guide works through every conversion with real catalogued molecular weights, covers the net-peptide-content correction almost everyone skips, and ends with a µmol-per-mg reference table. Research use only.

Molecular weight, moles and molarity: the three equations

  1. Moles from mass. mol = mass (g) ÷ MW (g/mol). Working in the units a peptide lab actually uses: µmol = mg ÷ MW × 1,000, or equivalently µmol per mg = 1,000 ÷ MW.
  2. Molarity from concentration. A solution at x mg/mL is x g/L, so molarity (mol/L) = (mg/mL) ÷ MW. Multiply by 1,000 for mM, by 1,000,000 for µM.
  3. Dilution. C₁V₁ = C₂V₂, in any consistent pair of units. Solve for the stock volume needed: V₁ = (C₂ × V₂) ÷ C₁.

Nothing else is required. The molarity calculator runs all three, but the arithmetic is worth being able to do on paper, because that is how you catch a calculator entered in the wrong units.

Worked example 1: vial to molarity

Take a 5 mg vial of BPC-157, molecular weight 1,419.55 Da.

  1. Reconstitute. 5 mg in 2 mL of diluent → 5 ÷ 2 = 2.5 mg/mL.
  2. Convert. 2.5 mg/mL = 2.5 g/L. Divide by 1,419.55 g/mol → 0.001761 mol/L = 1.76 mM.
  3. Express in µM if the protocol is written that way: 1,761 µM.
  4. Sanity check. µmol per mg for BPC-157 is 1,000 ÷ 1,419.55 = 0.704. The 5 mg vial therefore contains 3.52 µmol, and 3.52 µmol in 0.002 L is 1,761 µM. Same answer by a different route.

Worked example 2: molarity to mass

A cell-culture protocol calls for 10 mL of a 100 µM working solution.

  1. Moles required. 100 µM = 1 × 10⁻⁴ mol/L. In 0.010 L: 1 × 10⁻⁶ mol = 1 µmol.
  2. Mass required. 1 µmol × 1,419.55 g/mol = 1,419.55 µg = 1.42 mg.
  3. Or prepare it by dilution. From the 1.76 mM stock above: V₁ = (100 µM × 10 mL) ÷ 1,761 µM = 0.568 mL of stock, made up to 10 mL.

The second route is almost always preferable — weighing 1.42 mg accurately is harder than pipetting 0.568 mL, and it wastes less material.

Worked example 3: why equal mass is not equal moles

This is the failure mode the conversion exists to prevent.

  • GHK-Cu, 403.93 Da: 1 mg = 1,000 ÷ 403.93 = 2.476 µmol.
  • Semaglutide, 4,113.58 Da: 1 mg = 1,000 ÷ 4,113.58 = 0.243 µmol.

The same milligram of GHK-Cu supplies 10.2 times as many molecules as a milligram of semaglutide. Any comparison of two peptides framed in mg/mL is, for receptor-occupancy purposes, comparing different numbers of things. A blend at a 1:1 mass ratio has the same problem internally, which is why blends versus single vials matters for stoichiometric work.

Worked example 4: the net peptide content correction

Synthetic peptides are isolated as salts, so a vial labelled 5 mg contains 5 mg of powder — peptide plus counter-ion plus residual water and solvent. Net peptide content typically runs 70–90%, and it should be stated on the certificate.

  1. Uncorrected. 5 mg in 2 mL → 2.5 mg/mL → 1.761 mM, as above.
  2. Corrected at 85% net peptide content. Actual peptide = 5 × 0.85 = 4.25 mg. In 2 mL → 2.125 mg/mL. Divide by 1,419.55 → 1.497 mM.
  3. The error. (1.761 − 1.497) ÷ 1.761 = 15% high, propagated into every downstream dilution and every reported concentration.
  4. The fix. Either correct the calculation, or reconstitute with the corrected volume: to hit exactly 2.5 mg/mL of peptide from a vial holding 4.25 mg, add 4.25 ÷ 2.5 = 1.70 mL instead of 2 mL.

For qualitative work the correction rarely changes a conclusion. For anything reporting an IC₅₀, an EC₅₀ or a binding constant, ignoring it is a 15–30% systematic error. How to find the figure is covered in third-party testing explained.

Reference table: µmol per milligram

PeptideMW (Da)µmol per mg1 mg/mL equals
Epitalon390.352.5622.56 mM
GHK-Cu403.932.4762.48 mM
Ipamorelin711.851.4051.40 mM
Semax813.931.2291.23 mM
TB-500889.021.1251.12 mM
BPC-1571,419.550.7040.70 mM
MOTS-c2,174.550.4600.46 mM
Semaglutide4,113.580.2430.24 mM
Tirzepatide4,813.450.2080.21 mM
IGF-1 LR39,117.600.1100.11 mM

Where a molecular weight is not on the label, it can be computed from the sequence — sum the average residue masses and add 18.02 Da, as set out in how to read a peptide sequence.

Unit traps worth memorising

  • 1 mg/mL = 1 g/L. Getting this backwards by a factor of 1,000 is the most common single error.
  • 1 mg = 1,000 mcg. Vial labels use mg; working concentrations are often quoted in mcg/mL. The unit converter exists for this.
  • mM, µM and nM differ by 1,000 each. Cell-culture work usually lives in µM or nM; a reconstituted vial usually lives in mM.
  • Daltons and g/mol are the same number. A 1,419.55 Da peptide is 1,419.55 g/mol.
  • Volume added is not final volume. Adding 2 mL of diluent to a lyophilized cake gives very slightly more than 2 mL of solution; for peptide masses of a few milligrams the displacement is negligible, but it stops being negligible above roughly 50 mg per vial.
  • Copper and other complexes. GHK-Cu's 403.93 Da includes the copper ion. Calculating from the free tripeptide sequence alone gives 340.4 Da and a 19% error.

For the mg/mL and per-unit side of the same problem — choosing a diluent volume so that routine draws land on readable syringe marks — see reconstitution math explained and the reconstitution calculator.

Frequently Asked Questions

How do I convert mg/mL to molarity?
A solution at x mg/mL is x g/L, so molarity in mol/L is simply (mg/mL) divided by molecular weight in g/mol. BPC-157 at 2.5 mg/mL with a molecular weight of 1,419.55 gives 2.5 ÷ 1,419.55 = 0.001761 mol/L, or 1.76 mM. Multiply by 1,000 for mM and by 1,000,000 for µM.
How much peptide do I need for a given molarity?
Moles required equals molarity times volume in litres; mass equals moles times molecular weight. For 10 mL of 100 µM BPC-157: 1 × 10⁻⁴ mol/L × 0.010 L = 1 µmol, and 1 µmol × 1,419.55 g/mol = 1.42 mg. In practice it is easier and more accurate to dilute an existing stock using C1V1 = C2V2 than to weigh out 1.42 mg.
Are daltons the same as grams per mole?
Numerically, yes. A peptide with a molecular weight of 1,419.55 Da has a molar mass of 1,419.55 g/mol. That equivalence is what makes every mass-to-mole conversion a single division.
Why do two peptides at the same mg/mL have different molarities?
Because molarity counts molecules and mg/mL counts mass. One milligram of GHK-Cu at 403.93 Da is 2.476 µmol; one milligram of semaglutide at 4,113.58 Da is 0.243 µmol — a 10-fold difference in molecules for the same mass. Any comparison intended to be about receptor occupancy has to be run in molar units.
What is the net peptide content correction?
Synthetic peptides are isolated as salts, so a 5 mg vial contains 5 mg of powder including counter-ions and residual water, of which typically 70–90% is peptide. At 85% net peptide content, a 5 mg vial holds 4.25 mg of peptide, and reconstituting with 2 mL gives 1.50 mM rather than the 1.76 mM the label implies — a 15% systematic error in every derived figure.
How do I hit an exact concentration despite the salt?
Work backwards from the corrected peptide mass. To reach exactly 2.5 mg/mL of peptide from a vial holding 4.25 mg of peptide, add 4.25 ÷ 2.5 = 1.70 mL of diluent rather than 2 mL. The certificate of analysis has to state net peptide content for this to be possible.
Does GHK-Cu's molecular weight include the copper?
Yes. The catalogued 403.93 Da is the copper(II) complex. Computing from the free glycyl-histidyl-lysine tripeptide alone gives about 340.4 Da and an error of roughly 19% in every molarity calculation derived from it.
Does adding 2 mL of diluent give exactly 2 mL of solution?
Slightly more, because the dissolved solid occupies volume. For a few milligrams of peptide the displacement is negligible against 2 mL. It stops being negligible for large-mass vials — above roughly 50 mg — where final volume should be measured rather than assumed.

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