Molarity calculator
Find molarity, the grams of solute you need, or a dilution with M1V1 = M2V2.
- 🇺🇸 US federal & state rules
- IUPAC CIAAW atomic weights
- See sources
Molarity
1 M
mol/L
- Moles of solute
- 0.1 mol (100 mmol)
- Mass concentration
- 58.44 g/L
- Millimolar
- 1,000 mM
This is an example. Enter your numbers above and press the button to see your own result.
- Use the total volume of the finished solution, not the volume of water you started with.
Show the math
| Moles | n = mass ÷ molar mass = 5.844 ÷ 58.44 = 0.1 mol |
|---|---|
| Volume | 100 mL = 0.1 L |
| Molarity | M = n ÷ V = 0.1 ÷ 0.1 = 1 mol/L |
What to do next
How it works
Moles n = mass ÷ molar mass, and molarity M = n ÷ volume in liters. Grams needed = M × V × molar mass. For dilution the moles of solute stay the same, so M1 × V1 = M2 × V2.
The molar masses in the substance list add up the abridged standard atomic weights from IUPAC (CIAAW) and round to two decimals.
| Substance | Molar mass (g/mol) |
|---|---|
| Sodium chloride, NaCl | 58.44 |
| Sodium hydroxide, NaOH | 40 |
| Hydrochloric acid, HCl | 36.46 |
| Sulfuric acid, H₂SO₄ | 98.07 |
| Nitric acid, HNO₃ | 63.01 |
| Acetic acid, CH₃COOH | 60.05 |
| Ammonia, NH₃ | 17.03 |
| Potassium hydroxide, KOH | 56.11 |
| Potassium chloride, KCl | 74.55 |
| Calcium chloride, CaCl₂ | 110.98 |
| Calcium carbonate, CaCO₃ | 100.09 |
| Sodium bicarbonate, NaHCO₃ | 84.01 |
| Magnesium sulfate, MgSO₄ | 120.36 |
| Copper(II) sulfate pentahydrate, CuSO₄·5H₂O | 249.68 |
| Glucose, C₆H₁₂O₆ | 180.16 |
| Sucrose, C₁₂H₂₂O₁₁ | 342.3 |
| Ethanol, C₂H₅OH | 46.07 |
Things to keep in mind
The volume is the volume of the whole solution. Dissolving 58.44 g of NaCl in 1 L of water gives a little more than 1 L, so the concentration ends up slightly under 1 M.
Sources
Rules applied as of 2026-10-19. See the sources below for details.
- IUPAC CIAAW: Abridged standard atomic weightsChecked 2026-10-19
- IUPAC Gold Book: amount concentrationChecked 2026-10-19
Frequently asked questions
What is the molarity formula?
Molarity M = moles of solute ÷ liters of solution, and moles = grams ÷ molar mass. 5.844 g of NaCl (58.44 g/mol) in 100 mL is 0.1 mol ÷ 0.1 L = 1 M.
How many grams do I need for a 0.5 M solution?
Multiply molarity × liters × molar mass. For 250 mL of 0.5 M NaCl: 0.5 × 0.25 × 58.44 = 7.305 g.
How do I use M1V1 = M2V2?
Fill in three of the four values. To make 500 mL of 0.1 M from a 1 M stock, V1 = 0.1 × 500 ÷ 1 = 50 mL of stock, topped up to 500 mL.
What is the difference between molarity and molality?
Molarity divides by liters of solution. Molality divides by kilograms of solvent, so it does not change with temperature.