Molality from Molarity Calculator
Enter molarity, solution density, and solute molar mass to find molality.
Result
How it is calculated
m = 1000 × M ÷ (1000 × ρ − M × MW), where M is the molarity in mol/L, ρ is the density of the whole solution in g/mL and MW is the solute's molar mass in g/mol. It works on one litre of solution: that litre weighs 1000 × ρ grams, the solute in it weighs M × MW grams, and the difference is the solvent mass. Dividing the moles of solute by the solvent mass in kg gives molality. All three inputs must be above zero, and if the solute mass would equal or exceed the solution mass (denominator of zero or less) there is no result. Results are shown to four decimals. The conversion is valid only at the temperature at which the density was measured.
Worked example
A 1 mol/L NaCl solution with density 1.04 g/mL and molar mass 58.44 g/mol. One litre of solution weighs 1000 × 1.04 = 1040 g. The salt in it weighs 1 × 58.44 = 58.44 g, so the water is 1040 − 58.44 = 981.56 g. Molality = 1000 × 1 ÷ 981.56 = 1.0188 mol/kg.
Common mistakes
- Using the density of the pure solvent (1.000 for water) instead of the density of the solution; the answer then only matches for very dilute solutions.
- Entering the density in kg/m³ (1040) or g/L; the tool expects g/mL, and 1040 would find no result.
- Treating molarity and molality as equal. They are close for dilute water solutions (0.1 M gives 0.1006 m with density 1.0) but diverge as concentration rises.
- Using the wrong molar mass, such as the anhydrous value for a hydrate that was weighed out; use the mass of the form that is dissolved.
- Using a density measured at a different temperature than the molarity was prepared at.
About this calculator
Molarity (mol solute per liter of solution) and molality (mol solute per kilogram of solvent) look similar but measure different things — molarity depends on solution volume, which shifts with temperature, while molality depends on solvent mass, which does not, making it the preferred unit for precise work like freezing-point depression or boiling-point elevation.
Converting between them requires the solution's density (ρ, in g/mL) and the solute's molar mass (MW, in g/mol): for one liter of solution, the total mass is 1000·ρ grams, the mass of solute is M·MW grams, and subtracting gives the solvent's mass — so molality m = 1000·M / (1000·ρ − M·MW).
Analytical and physical chemists use this conversion whenever a solution's molarity is known from preparation but a temperature-independent concentration is needed for colligative-property calculations or precise thermodynamic work.
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Comments (1)
The density to enter is that of the finished solution at its own concentration, not of pure water, since the formula needs the solution's mass per litre. With 1 M NaCl at 1.05 g/mL the answer is about 1.0085 mol/kg. If you get no result, the density is too low for that molarity: the solute alone would outweigh the solution. Which solute are you converting?