Molecule guides
Moles, Molarity and mg/mL: The Arithmetic Behind Peptide Concentrations
Three equations, one division each — plus the salt-content correction that quietly inflates every figure derived from a vial label.
Three relationships tie molecular weight, moles and molarity together — moles = mass ÷ MW, molarity being moles ÷ litres, and C₁V₁ = C₂V₂ for dilutions — and since molecular weight expressed in daltons is numerically the same as grams per mole, each conversion amounts to one division. A vial label and a syringe deal in mg/mL; receptor binding, enzyme kinetics and cell-culture protocols are written in molarity. Moving between the two is the arithmetic that keeps two peptides "at the same concentration" from sitting at wildly different molecule counts. Below, every conversion is worked through using real catalogued masses, the net-peptide-content correction that nearly everyone leaves out is covered, and a µmol-per-mg reference table closes the piece.
The three equations
- Getting moles from mass: mol = mass (g) ÷ MW (g/mol). Put into the units a peptide laboratory actually works in, µmol = mg ÷ MW × 1,000, or equally µmol per mg = 1,000 ÷ MW.
- Getting molarity from concentration: because x mg/mL equals x g/L, molarity (mol/L) = (mg/mL) ÷ MW. Multiply the result by 1,000 for mM or by 1,000,000 for µM.
- Dilution: C₁V₁ = C₂V₂, using any consistent pair of units. Rearranged for the stock volume needed, V₁ = (C₂ × V₂) ÷ C₁.
That is the whole toolkit. All three are built into the molarity calculator, but being able to do the arithmetic by hand is worth it, because that is how a calculator fed the wrong units gets caught.
Worked example 1: from vial to molarity
Consider a 5 mg BPC-157 vial with a molecular weight of 1,419.55 Da.
- Reconstitution. 5 mg into 2 mL of diluent → 5 ÷ 2 = 2.5 mg/mL.
- Conversion. 2.5 mg/mL is 2.5 g/L; dividing by 1,419.55 g/mol gives 0.001761 mol/L, or 1.76 mM.
- In µM, should the protocol use those units: 1,761 µM.
- A check. For BPC-157, µmol per mg is 1,000 ÷ 1,419.55 = 0.704, so the 5 mg vial holds 3.52 µmol, and 3.52 µmol dissolved in 0.002 L is 1,761 µM — the same answer arrived at differently.
Worked example 2: from molarity to mass
Suppose a cell-culture protocol asks for 10 mL of a 100 µM working solution.
- Moles needed. 100 µM is 1 × 10⁻⁴ mol/L, and in 0.010 L that comes to 1 × 10⁻⁶ mol, or 1 µmol.
- Mass needed. 1 µmol × 1,419.55 g/mol = 1,419.55 µg, that is 1.42 mg.
- Alternatively, make it by dilution. Starting from the 1.76 mM stock above, V₁ = (100 µM × 10 mL) ÷ 1,761 µM = 0.568 mL of stock, brought up to 10 mL.
Nearly always the dilution route wins: pipetting 0.568 mL is easier to do accurately than weighing 1.42 mg, and it consumes less material.
Worked example 3: equal mass is not equal moles
Preventing exactly this failure is why the conversion exists.
- GHK-Cu at 403.93 Da: 1,000 ÷ 403.93 means 1 mg is 2.476 µmol.
- Semaglutide at 4,113.58 Da: 1,000 ÷ 4,113.58 means 1 mg is 0.243 µmol.
Milligram for milligram, the copper tripeptide delivers 10.2 times as many molecules as semaglutide does. Comparing two peptides in mg/mL therefore compares different numbers of things wherever receptor occupancy is the point. The same difficulty sits inside any blend mixed at a 1:1 mass ratio, which is why blends against vials of single peptides matters for stoichiometric work.
Worked example 4: correcting for net peptide content
Peptides made synthetically are isolated as salts, so a vial marked 5 mg holds 5 mg of powder — peptide together with counter-ion, residual water and solvent. Net peptide content usually falls between 70 and 90%, and the certificate should state it.
- Without correction. 5 mg in 2 mL → 2.5 mg/mL → 1.761 mM, as calculated above.
- With correction at 85%. Real peptide = 5 × 0.85 = 4.25 mg; in 2 mL that is 2.125 mg/mL, and dividing by 1,419.55 gives 1.497 mM.
- The size of the error. (1.761 − 1.497) ÷ 1.761 = 15% too high, carried forward into every subsequent dilution and every concentration reported.
- Putting it right. Correct the calculation, or else reconstitute with the corrected volume: to land on exactly 2.5 mg/mL of peptide from a vial holding 4.25 mg, add 4.25 ÷ 2.5 = 1.70 mL rather than 2 mL.
In qualitative work the correction seldom alters a conclusion. Where an IC₅₀, an EC₅₀ or a binding constant is being reported, leaving it out introduces a systematic error of 15–30%. Where to find the number is set out under third-party testing explained.
Reference table: µmol per milligram
When no molecular weight appears on the label, it can be calculated from the sequence: add up the average residue masses and add 18.02 Da, as described in how to read a sequence of amino acids.
Unit traps worth committing to memory
- 1 mg/mL equals 1 g/L. Inverting this by a factor of 1,000 is the single most frequent mistake.
- 1 mg equals 1,000 mcg. Labels are written in mg while working concentrations often appear in mcg/mL, which is what the unit converter is for.
- Each step between mM, µM and nM is a factor of 1,000. Cell-culture work usually operates in µM or nM, whereas a freshly reconstituted vial usually sits in mM.
- A dalton and a gram per mole carry the same number: a peptide of 1,419.55 Da is 1,419.55 g/mol.
- Added volume is not final volume. Putting 2 mL of diluent onto a lyophilized cake yields a little over 2 mL of solution; at peptide masses of a few milligrams the displacement is negligible, but it ceases to be so beyond roughly 50 mg per vial.
- Complexes such as copper peptides. The 403.93 Da quoted for GHK-Cu counts the copper ion; calculating from the free tripeptide sequence alone yields 340.4 Da and an error of 19%.
For the other half of the problem — picking a diluent volume in mg/mL and per-unit terms so that routine draws fall on readable syringe graduations — see the reconstitution arithmetic explained and the reconstitution calculator.
Questions
What converts mg/mL into molarity?
Since x mg/mL is the same as x g/L, molarity in mol/L is just (mg/mL) divided by the molecular weight in g/mol. Take BPC-157 at 2.5 mg/mL against a molecular weight of 1,419.55: 2.5 ÷ 1,419.55 = 0.001761 mol/L, that is 1.76 mM. To reach mM multiply by 1,000; for µM multiply by 1,000,000.
How much peptide does a given molarity require?
The moles you need are molarity multiplied by volume in litres, and the mass is those moles multiplied by 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. Diluting an existing stock via C1V1 = C2V2 is both easier and more accurate than trying to weigh 1.42 mg.
Is a dalton the same as a gram per mole?
Numerically it is. A peptide of molecular weight 1,419.55 Da has a molar mass of 1,419.55 g/mol. It is that equivalence which reduces every mass-to-mole conversion to one division.
Why do two peptides at identical mg/mL have different molarities?
Molarity counts molecules while mg/mL counts mass. A milligram of GHK-Cu at 403.93 Da amounts to 2.476 µmol, whereas a milligram of semaglutide at 4,113.58 Da amounts to only 0.243 µmol — ten times fewer molecules for the same weight. Anything meant to say something about receptor occupancy must therefore be run in molar units.
What is the net peptide content correction?
Because synthetic peptides are isolated as salts, a 5 mg vial holds 5 mg of powder inclusive of counter-ions and residual water, of which usually 70–90% is peptide. At 85%, that vial contains 4.25 mg of peptide, so reconstituting in 2 mL produces 1.50 mM rather than the 1.76 mM the label suggests — 15% of systematic error baked into every figure that follows.
How can an exact concentration be reached despite the salt?
Work back from the corrected peptide mass. Getting exactly 2.5 mg/mL of peptide out of a vial containing 4.25 mg means adding 4.25 ÷ 2.5 = 1.70 mL of diluent, not 2 mL. None of this is possible unless the certificate of analysis states net peptide content.
Does the molecular weight of GHK-Cu include the copper?
It does. The 403.93 Da in the catalogue refers to the copper(II) complex. Calculating instead from the free glycyl-histidyl-lysine tripeptide gives roughly 340.4 Da, which puts about 19% of error into every molarity derived from it.
Does adding 2 mL of diluent produce exactly 2 mL of solution?
A fraction more, since the dissolved solid takes up space. With only a few milligrams of peptide that displacement disappears against 2 mL. It ceases to be negligible for large fills — above roughly 50 mg — where the final volume ought to be measured rather than assumed.