TAHA Engineering Toolsby TAHA Knowledge Lab

Chemical Engineering

Molarity Calculator

Molarity from mass, molar mass and volume — or the mass or volume needed for a chosen molarity, with purity.

Inputs

Calculate from
A volumetric flask filled with solution up to a mark on its neck labelled V. A small pile labelled m of solute sits beside it with an arrow into the flask.mark: final volume Vm (solute)c = n ÷ Vn = m ÷ M
A mass m of solute (molar mass M) is dissolved and made up to the mark: the final solution volume V. Molarity c = n ÷ V with n = m ÷ M. Diagram not to scale.

From the substance’s data sheet or calculated from standard atomic weights. 1 to 2000 g/mol.

The total volume after dissolving and making up to the mark — not the volume of solvent added.

100% for a pure substance. A lower purity means more must be weighed for the same amount of solute.

Result

Molarity
1
Amount of solute
0.1
Mass concentration of the solute
58.44
Working
n = m × purity ÷ M = 5.844 g × 1 ÷ 58.44 g/mol = 0.1 mol c = n ÷ V = 0.1 mol ÷ 0.1 L = 1 mol/L

The solution has a molarity of 1 mol/L (0.1 mol of solute).

Educational estimate — check the assumptions below.

About this calculator

Molarity (molar concentration, c) is the amount of solute in moles per litre of solution, written mol/L or M. It links the mass you weigh to the number of molecules in a volume, through the solute’s molar mass. Use this calculator to find the molarity of a solution, the mass needed for a chosen molarity, or the final volume for a mass you have.

This is a calculation tool only. It does not give instructions for handling chemicals; follow the safety data sheet and your laboratory’s procedures.

Formula

n = m × P ÷ M    c = n ÷ V    m = c × M × V ÷ P

m = mass weighed (g), P = purity as a fraction, M = molar mass (g/mol), n = amount (mol), V = final solution volume (L), c = molarity (mol/L).

Worked example

  • 5.844 g of sodium chloride (M = 58.44 g/mol) made up to 100 mL: n = 5.844 ÷ 58.44 = 0.1000 mol → c = 0.1000 ÷ 0.100 L = 1.000 mol/L.
  • For 250 mL of 0.5 mol/L with the same salt at 99% purity: m = 0.5 × 58.44 × 0.250 ÷ 0.99 = 7.379 g.

Assumptions and limitations

  • V is the final volume of the solution, measured after the solute has dissolved. Molarity changes slightly with temperature because the volume does.
  • The purity entered is the mass fraction of the solute in the material weighed. Water of crystallisation must be included in the molar mass used (for example a hydrate).
  • The solute is assumed to dissolve completely.
  • A calculation tool only, not chemical handling advice.

References

  • IUPAC, Quantities, Units and Symbols in Physical Chemistry (the “Green Book”): amount concentration c = n ÷ V.