Molarity (symbol M, units mol/L, also written mol·dm⁻³) is the amount of solute dissolved per litre of solution. It is the workhorse concentration unit in chemistry: titrations, reaction stoichiometry, buffer preparation and dilutions are all set up in molarity. This calculator solves the molarity relationship for any one of four quantities — the concentration itself, the amount of solute in moles, the mass to weigh out, or the volume of solution to make up — and it bridges grams and moles for you through the molar mass, so you can go straight from a balance reading to a concentration without a separate step.
How it works
The single relationship behind every mode is:
M = n / V
where n is the amount of solute in moles and V is the volume of solution in litres. Amount of substance connects to a mass on the balance through the molar mass Mr (g/mol):
n = mass / Mr, so M = mass / (Mr · V)
Choose what to solve for and the tool rearranges the equation, then shows the rearranged form filled in with your numbers:
- Molarity = mass ÷ (Mr × V), or n ÷ V if you enter moles directly
- Moles = M × V
- Mass = M × V × Mr
- Volume = n ÷ M
You can supply the molar mass two ways. Type a chemical formula — NaCl, Ca(OH)2, KMnO4, even hydrates like CuSO4(H2O)5 — and a recursive-descent parser totals the standard atomic weights (IUPAC 2021 conventional values) to give Mr. Or switch to manual entry and type a molar mass directly. Volumes accept litres, millilitres or microlitres; masses accept grams or milligrams; amounts accept moles or millimoles, all converted internally to a consistent base before the arithmetic runs.
Worked example
You need 250 mL of a 0.100 M sodium chloride solution. NaCl has a molar mass of 58.44 g/mol.
- Convert the volume to litres: V = 250 mL ÷ 1000 = 0.250 L
- Find the moles required: n = M × V = 0.100 mol/L × 0.250 L = 0.0250 mol
- Convert moles to a mass to weigh out: mass = n × Mr = 0.0250 mol × 58.44 g/mol = 1.461 g
Weigh 1.461 g of NaCl, dissolve it in a little water, then make the solution up to exactly 250 mL in a volumetric flask. Working the same example backwards: if you actually dissolved 1.461 g in 250 mL, the molarity is 1.461 ÷ (58.44 × 0.250) = 0.100 mol/L, confirming the result.
| Solve for | Rearranged formula |
|---|---|
| Molarity | mass ÷ (Mr × V) |
| Moles | M × V |
| Mass | M × V × Mr |
| Volume | n ÷ M |
A quick reference: a 1 M solution holds one mole of solute per litre. For NaCl that is 58.44 g made up to 1 L; for glucose (Mr ≈ 180.16 g/mol) it is about 180.16 g per litre. Every figure here is computed in your browser and nothing is uploaded or stored.