Chemistry calculator
Molarity Calculator
Calculate the molarity of a solution from moles of solute, or from mass and molar mass, with the volume in liters or milliliters.
Molarity
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What is molarity?
Molarity is a measure of concentration. It tells you how many moles of solute are present in one liter of solution.
Molarity formula using moles
M is molarity in mol/L, n is the number of moles of solute, and V is the solution volume in liters.
Example using moles
Suppose a solution contains 0.5 moles of solute in 2 liters of solution.
The solution therefore has a molarity of 0.25 mol/L, or 0.25 M.
Frequently asked questions
What units are used for molarity?
Molarity is typically expressed as moles per liter, written mol/L or simply M.
Do I need to convert milliliters to liters?
Yes. The standard molarity formula uses liters. Calquo automatically converts milliliters to liters when you select mL.
How do I calculate moles from grams?
Divide the mass of the solute in grams by its molar mass in grams per mole.
What is molar mass?
Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol).
Is molarity based on solution volume or solvent volume?
Molarity uses the total final volume of the solution, not just the volume of the solvent before the solute is added.
How the calculator works
The calculator has two modes. In "I know moles" you enter the amount of solute in moles and the solution volume. In "I know mass" you enter the mass of solute in grams together with its molar mass in g/mol, and the calculator converts that to moles first. Either way, the volume can be typed in liters or milliliters; milliliters are divided by 1,000 so the result is always in moles per liter (M).
Under the main result you can see the moles of solute and the volume in liters that were actually used, which makes it easy to check each step against a textbook solution. Results are rounded to four decimal places. The volume and molar mass must be greater than zero, and the moles or mass cannot be negative.
Molarity formula
M is molarity in mol/L, n is the number of moles of solute, and V is the volume of the whole solution in liters. The second line converts a mass to moles, and the third combines the two steps into one. Because V must be in liters, a volume given in milliliters has to be divided by 1,000 before it goes into the formula.
Worked example: 5.85 g of sodium chloride (molar mass 58.44 g/mol) dissolved to a final volume of 250 mL. Moles: 5.85 ÷ 58.44 = 0.1001 mol. Volume: 250 mL = 0.250 L. Molarity: 0.1001 ÷ 0.250 = 0.4004 M, which a chemist would report as 0.400 M. A simpler case: 0.5 mol of glucose in 2 L of solution is 0.5 ÷ 2 = 0.25 M.
Finding the molar mass
Molar mass is the mass of one mole of a substance in grams, and it is the sum of the atomic masses of every atom in the formula. Sodium chloride is 22.99 + 35.45 = 58.44 g/mol. Glucose, C₆H₁₂O₆, is 6 × 12.011 + 12 × 1.008 + 6 × 15.999 = 180.16 g/mol. For hydrated salts include the water: copper(II) sulfate pentahydrate, CuSO₄·5H₂O, is 159.61 + 5 × 18.015 = 249.68 g/mol.
Using the anhydrous molar mass when you weighed out a hydrate is a classic error that makes the calculated molarity too high. Check the label on the reagent bottle, which usually prints the molar mass of the exact form you are holding.
Molarity compared with other concentration units
Molarity is the most common concentration unit in chemistry, but it is not the only one, and each is used for a reason:
- Molarity (M, mol/L) – moles of solute per liter of solution; changes slightly with temperature because the volume expands
- Molality (m, mol/kg) – moles of solute per kilogram of solvent; independent of temperature, used for freezing-point and boiling-point calculations
- Normality (N, eq/L) – equivalents per liter; depends on the reaction, common in acid-base and redox titrations
- Mass percent (% w/w) – grams of solute per 100 g of solution
- Parts per million (ppm) – milligrams of solute per liter for dilute aqueous solutions
Preparing a solution and common mistakes
To make a solution of a target molarity, reverse the formula: mass needed = M × V × molar mass. For 500 mL of 0.1 M NaCl, that is 0.1 × 0.5 × 58.44 = 2.922 g. Dissolve the solid in somewhat less water than the final volume, then add water up to the mark in a volumetric flask. The volume in the definition is the final volume of the solution, not the volume of water you started with, so do not simply add 2.922 g to 500 mL of water.
Other mistakes to watch for: forgetting to convert milliliters to liters (an answer 1,000 times too small), reading molar mass from the wrong compound, and mixing up solute and solvent. If you already have a concentrated stock and want to make a weaker solution, a dilution calculation using C1V1 = C2V2 is quicker than weighing anything.
Frequently asked questions
What is molarity?
Molarity is the number of moles of solute dissolved in one liter of solution. It is written as M or mol/L, so a 2 M solution contains 2 moles of solute in every liter.
How do you calculate molarity from grams?
Divide the mass by the molar mass to get moles, then divide by the volume in liters. For 10 g of NaOH (40.00 g/mol) in 2 L: 10 ÷ 40 = 0.25 mol, and 0.25 ÷ 2 = 0.125 M.
What is the molarity of 2 moles in 500 mL?
Convert the volume to liters first: 500 mL = 0.5 L. Then 2 ÷ 0.5 = 4 M.
What is the difference between molarity and molality?
Molarity is moles of solute per liter of solution; molality is moles of solute per kilogram of solvent. Molality does not change with temperature, while molarity does because liquids expand when heated.
How many grams do I need for 1 L of a 1 M solution?
One molar mass's worth of the compound. For sodium chloride that is 58.44 g, for glucose 180.16 g, and for sodium hydroxide 40.00 g, each dissolved and made up to a final volume of exactly 1 L.
Does temperature affect molarity?
Slightly. The number of moles stays the same, but the solution's volume expands as it warms, so the molarity falls a little. For precise work, prepare and measure solutions at the same temperature or use molality instead.