Molality Formula Calculator

Molality ($b$, often written $m$) expresses concentration per mass of solvent:

$b = \frac{n_{\text{solute}}}{m_{\text{solvent}}}$

where $n_{\text{solute}}$ is the amount of dissolved solute in moles and $m_{\text{solvent}}$ is the mass of the solvent alone, in kilograms. The unit is mol/kg, written m and read "molal": a 0.400 m solution contains 0.400 mol of solute per kilogram of solvent.

Since moles come from mass and molar mass, the expanded form is

$b = \frac{m_{\text{solute}} / M_{\text{solute}}}{m_{\text{solvent}}}$

with masses in grams and kilograms respectively and $M$ the molar mass in g/mol.

Mass does not expand when a solution is warmed, so molality is preferred whenever temperature changes — which is why the colligative-property laws, such as freezing-point depression $\Delta T_f = K_f\,b\,i$, are written in molality rather than molarity.

Step-by-Step

1. Identify solute and solvent. The solvent is the component present in excess, usually water.
2. Convert the solute mass to moles: $n = m/M$, taking $M$ from the chemical formula.
3. Get the solvent mass in kilograms. If the problem quotes the mass of solution, subtract the solute first: $m_{\text{solvent}} = m_{\text{solution}} - m_{\text{solute}}$. Then divide grams by 1000.
4. Divide: $b = n / m_{\text{solvent}}$.
5. Round to the fewest significant figures among the measured inputs.

Converting Molarity to Molality

This needs the solution density $\rho$. Take exactly 1 L of solution: it weighs $1000\rho$ grams and contains $c$ moles, i.e. $cM$ grams, of solute. The solvent is the difference, so

$b = \frac{1000\,c}{1000\rho - cM}$

with $\rho$ in g/mL, $c$ in mol/L, and $M$ in g/mol. Density is the piece most often missing: molarity alone can never be converted to molality.

Sanity Check

For a dilute aqueous solution ($\rho \approx 1.00$ g/mL, very little solute) the two concentrations are numerically close, $b \approx c$. Treat that as a check on your arithmetic, never as a shortcut.

• Dividing by the mass of the solution — the denominator is solvent only. When a problem gives the solution mass, subtract the solute before dividing.
• Leaving the solvent mass in grams — the formula demands kilograms; skipping the factor of 1000 inflates the answer a thousandfold.
• Confusing molality with molarity — molal is mol/kg and molar is mol/L, and the symbols m and M differ only in case. Check which denominator the question describes.
• Using the wrong molar mass — for a hydrate such as $\mathrm{CuSO_4 \cdot 5H_2O}$, the water of crystallisation counts toward the mass of the sample.
• Dropping the van 't Hoff factor — molality counts formula units, so $\Delta T_f = K_f\,b\,i$ needs $i$ ($i \approx 2$ for NaCl, which yields two ions).
• Assuming the density is 1.00 g/mL — that holds only for near-pure water, and a molarity conversion is very sensitive to it.