Molality & Normality Calculator
Calculate molality from solvent mass or normality from equivalents and n-factor.
Choose the right mode
Molality and normality both describe concentration, but they answer different questions. Molality compares moles of solute with kilograms of solvent. Normality compares reactive equivalents with liters of solution, so its value depends on the reaction role of the solute.
| Mode | First value | Second value | Result |
|---|---|---|---|
| Molality | moles solute | kg solvent | mol/kg |
| Normality from equivalents | equivalents | liters solution | N |
| Normality from n-factor | molarity | n-factor | N |
Formulas used
Worked examples
Molality: 0.50 mol solute in 0.250 kg solvent gives 0.50 / 0.250 = 2.00 mol/kg.
Normality from equivalents: 0.100 equivalents in 0.500 L gives 0.100 / 0.500 = 0.200 N.
Normality from n-factor: 0.250 M H2SO4 with n-factor 2 in an acid-base setup gives 0.250 x 2 = 0.500 N.
Molality vs molarity
Molality uses solvent mass, so it does not change just because the solution expands or contracts with temperature. Molarity uses final solution volume, so it is more convenient for volumetric glassware but can shift when volume changes. This is why freezing-point depression and boiling-point elevation problems often use molality.
Normality needs context
Normality is reaction-specific. HCl usually has n-factor 1 for acid-base neutralization. H2SO4 is often treated as n-factor 2 when both acidic hydrogens react. In redox chemistry, n-factor comes from electrons transferred. If your class gives a reaction, choose the n-factor from that reaction rather than from the formula alone.
Where this calculator is useful
- Freezing-point depression and boiling-point elevation problems.
- Acid-base normality practice and titration setup checks.
- Comparing molarity, molality, and normality in solution chemistry.
- Preparing quick examples for concentration units beyond molarity.
- Checking equivalent-based calculations before lab reports.
Common mistakes
- Using kg of solution instead of kg of solvent for molality.
- Using liters of solvent instead of liters of final solution for normality.
- Treating n-factor as fixed when the reaction changes.
- Entering grams where the formula requires moles or equivalents.
- Mixing molality units, mol/kg, with molarity units, mol/L.
Rounding and result checking
Molality is reported as mol/kg and normality as N or eq/L. If the denominator is less than 1, the result can be larger than the numerator, which is normal. For example, 0.50 mol in 0.250 kg solvent gives 2.00 mol/kg. Keep enough digits through the division or multiplication, then round the final concentration to the measured input with the fewest significant figures.
Related Chemistry Tools
FAQs
What is molality?
Molality (m) is a concentration unit defined as moles of solute per kilogram of solvent: m = moles solute ÷ mass solvent (kg). Units are mol/kg, often abbreviated 'm' (lowercase, distinct from M for molarity). The defining feature is that molality is based on the mass of solvent, not the volume of solution. Mass does not change with temperature, so molality is temperature-independent — a property that matters when working with colligative properties like boiling-point elevation (ΔTb = Kb·m) and freezing-point depression (ΔTf = Kf·m), which are themselves temperature-driven phenomena. Example: 0.5 mol of NaCl dissolved in 1.0 kg of water gives 0.5 / 1.0 = 0.5 m. Tip: when the problem mentions colligative properties, reach for molality; when it mentions volumetric titrations or stock-solution dilutions, molarity is usually more convenient.
How to calculate molality?
Use m = (moles of solute) ÷ (mass of solvent in kg). Procedure: convert solute mass to moles via moles = mass ÷ molar mass; convert solvent mass to kg (divide grams by 1000); divide moles by solvent mass. Worked example: dissolve 9.80 g of H2SO4 (M = 98.08 g/mol) in 250 g of water. Moles H2SO4 = 9.80 ÷ 98.08 = 0.0999 mol; solvent mass = 0.250 kg; molality = 0.0999 ÷ 0.250 = 0.400 m. Watch the wording: the denominator is solvent only, not total solution. If a problem gives 'mass of solution', subtract the solute mass before computing molality.
What is the difference between molarity and molality?
Molarity (M) is moles of solute per litre of solution. Molality (m) is moles of solute per kilogram of solvent. Practical differences: (1) The denominator is solution volume for M and solvent mass for m. (2) Molarity depends on temperature (volumes expand and contract); molality does not (mass is invariant). (3) Lab convenience favors molarity for stock solutions and titrations because volumes are quicker to measure than masses. Theoretical work — colligative properties, freezing-point depression constants, boiling-point elevation — uses molality because the equations require a temperature-independent concentration. For dilute aqueous solutions at room temperature, M ≈ m because 1 L of dilute solution weighs about 1 kg of water. The two diverge for concentrated solutions or for solvents whose density differs from 1 g/mL.
How to calculate molality from molarity?
Use m = (1000 × M) ÷ [(1000 × d) − (M × Mw)], where M is molarity (mol/L), d is solution density (g/mL), and Mw is the solute's molar mass (g/mol). Derivation: take 1.000 L of solution; its mass is 1000 × d g; the solute mass is M × Mw g; subtracting gives the solvent mass in g; dividing M by that mass (in kg) gives molality. Worked example: 2.00 M H2SO4 with density 1.10 g/mL. Mass of 1 L = 1100 g; mass of solute = 2.00 × 98.08 = 196.2 g; mass of solvent = 1100 − 196.2 = 903.8 g = 0.9038 kg; molality = 2.00 / 0.9038 = 2.21 m. Sanity check: molality is slightly larger than molarity for dense solutions because the solvent mass is less than the solution mass; the gap grows with concentration and density.
Does molality change with temperature?
No — molality is independent of temperature. It is defined using the mass of solvent, and mass does not vary with temperature. Molarity, in contrast, depends on solution volume, which expands when heated and contracts when cooled, so the molarity of a given solution decreases slightly as temperature rises. This is the practical reason colligative-property equations (ΔTb = Kb · m for boiling-point elevation, ΔTf = Kf · m for freezing-point depression, π = MRT for osmotic pressure when used in dilute aqueous solutions) are written in molality, mole fraction, or mole-based forms — quantities that don't drift with temperature. For room-temperature stock solutions the molarity-temperature dependence is small and usually ignored, but it matters for high-precision work and for solutions taken to extreme temperatures.