Molarity Calculator
Compute the molarity of a solution, the mass of solute needed for a target concentration, or the volume to make from a given amount — all from M = n/V with n = mass/MW.
Molarity
0.1 mol/L
- Moles of solute
- 0.1 mol
- Solute mass
- 5.844 g
- Molar mass
- 58.44 g/mol
- Volume
- 1 L
- Molarity
- 0.1 mol/L
Molarity M = n / V, where n is moles of solute and V is solution volume in litres. Moles are obtained from mass and molar mass by n = mass / MW. Rearrange to get mass = M × V × MW or V = (mass / MW) / M.
How to use this calculator
Pick what you want to solve for at the top — molarity, solute mass, or solution volume — then fill in the other three quantities. Mass is in grams, molar mass in grams per mole, volume in litres, and molarity in mol/L (equivalently M). If you only have volume in millilitres, divide by 1000 first (250 mL = 0.25 L). The calculator returns the missing quantity as the primary result and shows the full breakdown — moles, mass, molar mass, volume and molarity — so you can copy any of them straight into a lab notebook. Molar masses for common solutes: NaCl 58.44, glucose 180.16, sucrose 342.30, water 18.02, NaOH 40.00, HCl 36.46, KCl 74.55. If you have a chemical formula but no molar mass, use the Molecular Weight Calculator first.
How the calculation works
Molarity is the amount of solute, in moles, per litre of solution: M = n / V. The unit is mol/L, almost always written M (so 0.10 mol/L = 0.10 M). Moles are obtained from the mass of solute by dividing by molar mass: n = m / MW, where MW is in grams per mole. Substituting gives M = m / (MW · V), and rearranging gives the other two forms you need at the bench: m = M · V · MW and V = m / (MW · M). The molar mass MW is fixed by the chemical formula — for an element it is the atomic mass in g/mol, for a compound it is the sum of the atomic masses of its constituent atoms. Molarity is the most common concentration unit in chemistry because it links directly to stoichiometry: one mole of any substance contains the same number of formula units (Avogadro’s number, 6.022 × 10²³), so a balanced equation reads straight off as a ratio of molarities × volumes.
Worked example
How would you prepare 125 mL of a 1.32 M sodium sulfate (Na₂SO₄, MW 142.04 g/mol) solution? Choose "Solve for solute mass", enter molarity 1.32, molar mass 142.04, volume 0.125. The calculator returns mass = 1.32 × 0.125 × 142.04 = 23.4 g. So you weigh out 23.4 g of Na₂SO₄, dissolve in a little water in a 125 mL volumetric flask, and top up to the line. Reverse the problem and check: enter mass 23.4 g, MW 142.04, volume 0.125 L and solve for molarity — you get 1.318 M, matching the textbook (Brown, LeMay, Bursten et al., Sample Exercise 4.13).
Frequently asked questions
What is molarity?
Molarity is the concentration of a solute expressed as moles of solute per litre of solution. The symbol is M (capital), the unit is mol/L, and the formula is M = n / V. A 1.0 M glucose solution contains one mole of glucose (180.16 g) dissolved and diluted to one litre of total solution — not one litre of water added to the glucose. That distinction matters because the solute itself takes up volume; molarity is always measured against the final solution volume, which is why volumetric flasks (calibrated to a single line for a single volume) are the standard tool for making molar solutions.
What is the difference between molarity and molality?
Molarity (M) is moles of solute per litre of solution; molality (m, lowercase) is moles of solute per kilogram of solvent. Two practical differences flow from this. First, molality uses mass instead of volume, so it does not change with temperature — solution volume expands when heated but solvent mass does not. Second, molality counts only the solvent, while molarity counts the whole solution including the solute. Molarity is more convenient in the lab because volume is easier to measure than mass, but for colligative-property work (freezing-point depression, boiling-point elevation, osmotic pressure) molality is the standard.
How do I prepare a molar solution from a solid solute?
Calculate the mass needed: m = M × V × MW. Weigh that mass on a balance accurate to at least 0.1 % of the value (an analytical balance for milligrams). Transfer the solid into a volumetric flask of the target volume — not a beaker or a graduated cylinder, which are too imprecise. Add a portion of the solvent and swirl until the solute fully dissolves; some solids dissolve slowly or release heat, so wait until the flask returns to room temperature before the final step. Top up with solvent to the calibration mark on the flask neck, holding the flask at eye level. Stopper and invert several times to mix. The result is a solution of known molarity at the calibration temperature (usually 20 °C).
What does "dilute a 6 M stock to 0.5 M" mean in practice?
Use the dilution equation M₁V₁ = M₂V₂, where M₁ and V₁ are the stock molarity and volume taken, and M₂ and V₂ are the diluted molarity and final volume. To make 250 mL of 0.5 M from 6 M stock: V₁ = (0.5 × 0.25) / 6 = 0.0208 L = 20.8 mL of stock, then top up to 250 mL with solvent in a volumetric flask. The moles of solute are conserved — you are spreading the same n across a larger V, so M falls. This calculator focuses on the forward problem (m, V, M relationships); for dilution-only work you can apply the same formula by hand in one line, or use a dedicated dilution calculator.
How do I find the molar mass of a compound?
Sum the atomic masses (from the periodic table, in g/mol) of every atom in the chemical formula. For NaCl: 22.99 (Na) + 35.45 (Cl) = 58.44 g/mol. For glucose C₆H₁₂O₆: 6 × 12.01 + 12 × 1.008 + 6 × 16.00 = 72.06 + 12.10 + 96.00 = 180.16 g/mol. For hydrates, include the water of crystallisation — CuSO₄·5H₂O is 159.61 + 5 × 18.02 = 249.71 g/mol, and that is the mass you weigh out. Use the Molecular Weight Calculator if you would rather paste the formula than add it up by hand.
Does temperature affect molarity?
Yes, slightly. Molarity is defined on volume, and water expands by about 0.02 % per °C around room temperature, so a 1.000 M solution at 20 °C becomes about 0.998 M at 30 °C — a 0.2 % drop. This is negligible for routine work but matters in precise analytical chemistry and at temperature extremes. For temperature-independent concentration use molality (mol solute per kg solvent) or mass fraction; for a defined preparation temperature, always quote the molarity together with the temperature ("0.100 M at 20 °C"). Volumetric glassware is calibrated at 20 °C by convention.