Lab tool
Molarity calculator
The relationship is mass = concentration × volume × molecular weight. Fill in any three fields and the fourth is calculated for you.
The short answer
Molarity is moles of solute per litre of solution: concentration (mol/L) = mass (g) ÷ [molecular weight (g/mol) × volume (L)].Rearranged for the value usually wanted at the bench, mass = concentration × volume × molecular weight.
Two shortcuts follow from it. One millilitre of a 1 mM solution needs a number of micrograms equal to the molecular weight — 1,419.55 µg for BPC-157. And a 1 mg/mL stock is 1,000,000 ÷ molecular weight in µM, so 1 mg/mL of BPC-157 is 704.4 µM.
These are measurement calculations for laboratory work. The materials referenced are supplied as laboratory research material, not as drugs, supplements or consumer products.
mass (g) = concentration (mol/L) × volume (L) × molecular weight (g/mol)
Shown on every product page — e.g. BPC-157 is 1419.55 g/mol.
Solving for: Mass
Result
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Leave one field blank and it will be calculated from the other three.
The formula and its rearrangements
- Molar concentration
- C (mol/L) = m (g) ÷ [ MW (g/mol) × V (L) ]
- Mass
- m (g) = C (mol/L) × V (L) × MW (g/mol)
- Volume
- V (L) = m (g) ÷ [ C (mol/L) × MW (g/mol) ]
- Molecular weight
- MW (g/mol) = m (g) ÷ [ C (mol/L) × V (L) ]
- mg/mL to micromolar
- C (µM) = 1,000,000 × mg/mL ÷ MW (g/mol)
- Micrograms for 1 mL of a 1 mM solution
- m (µg) = MW (g/mol)
Every unit prefix is a power of ten, so the arithmetic is the same whichever row of the selectors you use — mM with mL behaves exactly like M with L. If a prefix is the part that keeps going wrong, the unit converter isolates just that step.
Worked example: 1 mL of 1 mM BPC-157
- Molecular weight. BPC-157 is 1419.55 g/mol, from the specification block on its product page.
- Target. 1 mM in 1 mL, i.e. C = 0.001 mol/L and V = 0.001 L.
- Multiply out. m = 0.001 × 0.001 × 1419.55 = 1.41955 × 10⁻³ g, which is 1.42 mg or 1,419.55 µg.
- Sanity-check the shortcut. The microgram figure came out equal to the molecular weight, as it always does for 1 mL of 1 mM. If your answer is not within rounding of the molecular weight, a prefix is wrong.
- Read it the other way. That same vial content dissolved to 1 mg/mL would be 1,000,000 ÷ 1419.55 = 704.4 µM.
Reference table: peptide molecular weights
| Peptide | MW (g/mol) | µg for 1 mL of 1 mM | 1 mg/mL stock is |
|---|---|---|---|
| KPV | 342.43 | 342.43 | 2,920.3 µM |
| GHK-Cu (Copper Peptide) | 403.93 | 403.93 | 2,475.7 µM |
| Ipamorelin | 711.85 | 711.85 | 1,404.8 µM |
| TB-500 (Thymosin Beta-4 Fragment) | 889.02 | 889.02 | 1,124.8 µM |
| PT-141 (Bremelanotide) | 1,025.16 | 1,025.16 | 975.5 µM |
| BPC-157 | 1,419.55 | 1,419.55 | 704.4 µM |
| MOTS-c | 2,174.55 | 2,174.55 | 459.9 µM |
| Semaglutide | 4,113.58 | 4,113.58 | 243.1 µM |
| Tirzepatide | 4,813.45 | 4,813.45 | 207.8 µM |
| Retatrutide | 4,731.3 | 4,731.3 | 211.4 µM |
Molecular weights are read from the catalog specification for each listing, so they match the figure on the lot-matched certificate of analysis. Note how far apart the molar concentrations are at the same 1 mg/mL: a tripeptide and an acylated 39-mer differ by more than a factor of ten.
Common questions
What is molarity?
Molarity is the amount of substance dissolved per litre of solution, measured in moles per litre (mol/L, written M). One mole is 6.022 × 10²³ molecules, so molarity counts molecules rather than mass. Two peptides at the same mg/mL are at very different molarities if their molecular weights differ.
How do you calculate molarity from mass?
Divide the mass by the molecular weight to get moles, then divide by the volume in litres: concentration (mol/L) = mass (g) ÷ (molecular weight (g/mol) × volume (L)). Rearranged for mass it reads mass (g) = concentration (mol/L) × volume (L) × molecular weight (g/mol).
How do you convert mg/mL to molarity?
A concentration of 1 mg/mL is 1 g/L, so its molarity in mol/L is 1 ÷ the molecular weight. In micromolar that is 1,000,000 ÷ molecular weight. A 1 mg/mL BPC-157 stock, at 1419.55 g/mol, is therefore 704.4 µM.
How much peptide is needed for 1 mL of a 1 mM solution?
A number of micrograms equal to the molecular weight. 1 mM × 1 mL = 1 micromole, and one micromole weighs the molecular weight in micrograms. So 1 mL of 1 mM BPC-157 (1419.55 g/mol) needs 1419.55 µg, which is 1.42 mg.
Where do you find a peptide molecular weight?
It is on the lot-matched certificate of analysis and in the specification block of every product page in this catalog, alongside the molecular formula and CAS number. Use the free-acid or salt mass that matches the material you actually weighed — a trifluoroacetate or acetate salt is heavier than the free peptide.