Molality Calculator

Find Molality from Solute and Solvent

Molality Formula

Formula

\[ b = \frac{n_{\text{solute}}}{m_{\text{solvent (kg)}}} \quad \text{where} \quad n_{\text{solute}} = \frac{m_{\text{solute (g)}}}{M} \]

Where:

\(b\) — molality of the solution (mol/kg)
\(n_{\text{solute}}\) — amount of solute in moles (mol)
\(m_{\text{solvent}}\) — mass of the solvent only, in kilograms (kg)
\(m_{\text{solute}}\) — mass of the dissolved substance in grams (g)
\(M\) — molar mass of the solute (g/mol)

Molality is defined as the number of moles of solute dissolved in one kilogram of solvent. If the solute amount is already known in moles, the first equation is used directly; if only the mass is known, it is first converted to moles with the second equation.

Molality – Calculation Example

Example: 58.44 g of sodium chloride (NaCl, molar mass 58.44 g/mol) is dissolved in 500 g of water. Find the molality of the solution.

\[ n = \frac{58.44 \text{ g}}{58.44 \text{ g/mol}} = 1 \text{ mol} \]

\[ m_{\text{solvent}} = 500 \text{ g} = 0.5 \text{ kg} \]

\[ b = \frac{1 \text{ mol}}{0.5 \text{ kg}} = 2 \text{ mol/kg} \]

Result: the solution has a molality of 2 mol/kg (a 2 molal, or 2 m, NaCl solution).

How to Use the Molality Calculator

Choose the Input Method: select "Solute mass + molar mass" if you know the mass of the dissolved substance, or "Solute amount in moles" if you already have the amount in moles.
If you chose the mass method, enter the Solute Mass and pick its unit (g, mg, kg, lb or oz), then enter the Molar Mass of the substance in g/mol. For common solutes, use values such as 58.44 for NaCl or 180.16 for glucose.
If you chose the mole method, enter the Solute Amount (Moles) directly.
Enter the Solvent Mass and select its unit. This must be the mass of the solvent alone (for example, the water), not the total mass of the solution.
Read the results: Molality in mol/kg, along with the calculated Moles of Solute and the Solvent Mass converted to kilograms.

What Is Molality?

Molality (symbol \(b\), sometimes written \(m\)) is a measure of the concentration of a solution: it tells you how many moles of solute are dissolved in each kilogram of solvent. Unlike molarity, which is expressed per liter of solution, molality is based purely on the mass of the solvent. This makes it independent of temperature and pressure, because mass does not change when a liquid expands or contracts.

A 1 mol/kg solution (often called "1 molal" or "1 m") contains one mole of solute per kilogram of solvent. In the example above, 1 mole of NaCl in 0.5 kg of water gives 2 mol/kg — twice as concentrated as a 1 m solution.

Molality is especially important in fields where temperature varies. It appears in the equations for boiling point elevation and freezing point depression (colligative properties), in cryoscopic and ebullioscopic measurements, and in thermodynamic relations involving solvent activity. For laboratory work at constant temperature, however, molarity or mass percent is often more convenient; see the C1V1 = C2V2 dilution calculator for concentration-based calculations.

Because molality is defined per kilogram of solvent, the same amounts of solute and solvent always give the same molality, no matter the density of the liquid or the temperature of the lab. To go the other way — from a weighed sample to a number of particles — you may also find the atoms to moles converter useful.

Assumptions, Limitations and Tips

Solvent mass, not solution mass. The most common mistake is entering the total mass of the solution. If you only know the solution mass, subtract the solute mass first: \(m_{\text{solvent}} = m_{\text{solution}} - m_{\text{solute}}\).
Molar mass must be accurate. The moles of solute depend directly on the molar mass you enter. For hydrated salts (e.g. CuSO₄·5H₂O) include the water of crystallization in the molar mass.
Units are converted automatically. Masses can be entered in grams, milligrams, kilograms, pounds or ounces; the calculator converts everything to mol/kg internally.
Molality vs. molarity. Do not confuse the two: a 1 m (molal) solution and a 1 M (molar) solution generally contain different amounts of solute. Molality uses solvent mass; molarity uses the final volume of the solution.
Real solutions. Molality is a purely stoichiometric quantity. It says nothing about whether the solute dissociates (NaCl forms ions in water) or how the particles behave; for colligative-property calculations, account for the van 't Hoff factor separately.
Positive values only. The calculator requires positive solute and solvent amounts; enter zero masses to see an error message rather than a meaningless result.

Frequently Asked Questions

How do I calculate molality step by step by hand?

First convert the solute mass to moles by dividing by the molar mass: \( n = m / M \). Then convert the solvent mass to kilograms (divide grams by 1000). Finally divide: \( b = n / m_{\text{solvent}} \). For example, 36 g of glucose (M = 180.16 g/mol) in 250 g of water gives \( n = 0.2 \) mol, \( m = 0.25 \) kg, and \( b = 0.8 \) mol/kg.

How do I convert molality to molarity?

You need the density of the final solution. First find the solution mass: \( m_{\text{solution}} = m_{\text{solvent}} + m_{\text{solute}} \). Divide by density to get the volume in liters, then \( c = n / V \). Without density, the conversion is impossible, because molality is based on solvent mass while molarity is based on solution volume. For dilute aqueous solutions the two values are close, since 1 kg of water occupies about 1 L.

How do I convert molality to mass percent or mole fraction?

A 1 mol/kg solution means 1 mole of solute per 1000 g of solvent. Mass percent: \( \% = 100 \cdot m_{\text{solute}} / (m_{\text{solute}} + m_{\text{solvent}}) \). For mole fraction, convert solvent mass to moles using its molar mass (18.015 g/mol for water), then \( x_{\text{solute}} = n_{\text{solute}} / (n_{\text{solute}} + n_{\text{solvent}}) \). Example: 1 mol NaCl in 1 kg water gives \( x \approx 1/56.6 \approx 0.0177 \).

What molar mass should I enter for hydrated salts?

Use the molar mass of the full hydrate formula, including the water of crystallization. For CuSO₄·5H₂O that is 249.68 g/mol, not the 159.61 g/mol of anhydrous CuSO₄. Weighing out hydrate crystals means the crystal water is part of your measured mass. The water of crystallization also adds to the total solvent in rigorous work, though in most textbook problems it is ignored.

Why is molality used instead of molarity for boiling and freezing point calculations?

Molality is temperature-independent because mass does not change when a liquid expands or contracts, while volume — and therefore molarity — does. Colligative property equations such as \( \Delta T_b = i \cdot K_b \cdot b \) and \( \Delta T_f = i \cdot K_f \cdot b \) rely on a concentration that stays constant as the temperature rises or falls during the measurement, which is exactly what mol/kg provides.

When is 1 molal approximately equal to 1 molar?

For very dilute aqueous solutions, where the solute adds negligible volume and the solution density is close to 1 kg/L. In that limit, 1 kg of solvent produces roughly 1 L of solution, so mol/kg and mol/L give nearly the same number. As concentration rises, the two diverge: dissolving a lot of solute changes the volume and density, so a 1 m solution can be noticeably more concentrated than a 1 M one.

Do I need to account for the van 't Hoff factor when calculating molality?

No — molality itself is purely stoichiometric and does not depend on dissociation. The van 't Hoff factor \( i \) only enters when you use molality for colligative properties, because NaCl splits into two ions (\( i \approx 2 \)) and MgCl₂ into three (\( i \approx 3 \)). A 1 mol/kg NaCl solution still has molality 1; it is the effective particle concentration that is roughly doubled.