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How to Convert Molarity to Molality: A Step-by-Step Guide

In chemistry, molarity and molality both express a solution's concentration, but are measured differently.

  • To calculate molarity, you divide the moles of solute by the solution's liters. Molarity is designated by a capital "M".

  • To find molality, you divide the moles of solute by the kilograms of solvent. Molality is designated by a lowercase "m" or "b".

Scientists often express concentrations in molality when they publish because, unlike molarity, molality is not temperature-dependent. This independence makes it easier for scientists worldwide to reproduce the work.

Because molarity is volume-based and molality is mass-based, you can't convert directly between them without knowing the solution's density and the solvent's molar mass. In this article, you'll learn exactly how to do that!

🙋 Interested in learning more about molarity and molality? Check out our molarity 🇺🇸 and molality 🇺🇸 calculators.

When you know the molarity of a solution but need the molality, you cannot simply swap the units, because as we said above, molarity is based on volume (liters of solution), and molality is based on mass (kilograms of solvent). To link the two, you need two additional pieces of information:

There are two ways to write the conversion formula, depending on the units you're working with.

Standard formula

(Density in g/mL; molar mass in g/mol.)

b=1000×Mρ×1000−M×W\mathrm{b = \frac{1000\times M}{\rho \times 1000 - M \times W}}

where:

  • b\text{b} — Molality or molal concentration (mol/kg);
  • M\text{M} — Molarity or molar concentration (mol/L);
  • ρ\rho — Density of solution (g/mL); and
  • W\text{W} — Molar mass of solute (g/mol).

This version works directly with the most common laboratory units (g/mL for density and g/mol for molar mass), and includes the factor 1000 to convert grams to kilograms automatically.

Alternative formula

(Density in kg/L; molar mass in kg/mol.)

b=Mρ−M×W\mathrm{b = \frac{M}{\rho - M \times W}}

where:

  • b\text{b} — Molality or molal concentration (mol/kg);
  • M\text{M} — Molarity or molar concentration (mol/L);
  • ρ\rho — Density of solution (kg/L); and
  • W\text{W} — Molar mass of solute (kg/mol).

💡 A note on units

For this formula to work correctly, density must be expressed in kg/L and molar mass in kg/mol:

  • If your data is expressed in g/mL for density, the good news is that you don't have to change the number, because 1 g/mL=1 kg/L\mathrm{1\ g/mL=1\ kg/L}, so only the unit changes.
  • However, if your molar mass is expressed in g/mol (as is usually the case), you must convert it to kg/mol by dividing it by 1000 before using it in the equation.

Want to deepen your knowledge of solution concentrations? Check out our article on another concentration comparison: Normality vs. Molarity: What’s the Difference?.

Sometimes, you know a solution's molality but need to know its molarity. This is often the case when you prepare solutions based on mass (molality), but you must indicate the concentrations based on volume (molarity).

As with the molarity-to-molality formula, you can use two formulae depending on the units you are working with.

Standard formula

(Density in g/mL; molar mass in g/mol.)

M=b ×ρ×10001000+b×W\mathrm{M = \frac{b \times \rho \times 1000}{1000 + b \times W}}

where:

  • b\text{b} — Molality or molal concentration (mol/kg);
  • M\text{M} — Molarity or molar concentration (mol/L);
  • ρ\rho — Density of solution (g/mL); and
  • W\text{W} — Molar mass of solute (g/mol).

As with the formula for converting molarity to molality, this formula is ideal when density is expressed in g/mL and molar mass in g/mol, as is the case in most data tables and laboratory references. As you know by now, the factor 1000 in the formula automatically converts grams to kilograms.

Alternative formula

(Density in kg/L; molar mass in kg/mol.)

M=b ×ρ1+b×W\mathrm{M = \frac{b \times \rho}{1 + b \times W}}

where:

  • b\text{b} — Molality or molal concentration (mol/kg);
  • M\text{M} — Molarity or molar concentration (mol/L);
  • ρ\rho — Density of solution (kg/L); and
  • W\text{W} — Molar mass of solute (kg/mol).

💡 A note on the units

A quick reminder in case you missed it in the previous section. But, for this formula, density must be expressed in kg/L and molar mass in kg/mol:

  • If your density is given in g/mL, the numerical value stays precisely the same as 1 g/mL=1 kg/L\mathrm{1\ g/mL=1\ kg/L}, so only the unit changes.
  • However, you must convert molar mass g/mol to kg/mol by dividing by 1000 before using it.

Sometimes, the density conversion is not that simple, so don't hesitate to visit our density converter 🇺🇸.

Now that we have seen the formulae for converting molarity to molality and molality to molarity, let's take the time to understand what is actually happening in the calculations. We will go through two step-by-step examples to clarify the reasoning before checking our results with the formulae.

Example #1: Molarity to molality

Problem:
Determine the molality of a 2.0 M NaOH solution with a 1.20 g/mL density.

Solution:

  1. Assume you have 1 L of solution. This is a crucial step. The amount of solution is not given, but you need to have a specific quantity to do the calculations, and one liter is the best assumption for this problem.

  2. Find the total mass of the solution:

1 L ⁣× ⁣1.2 g/mL ⁣× ⁣1000 mL/L=1200 g\mathrm{1 L\! \times\! 1.2 g/mL\! \times\! 1000 mL/L = 1200 g}

This gives 1200 g of solution.

  1. Calculate the grams of the solute. 2.0M means 2.0 moles of sodium hydroxide per liter of solution, and sodium hydroxide's molar mass is 40.0 g/mol. So let's convert 2.0 moles to grams:
2.0 mol×40.0 g/mol=80.0 g\mathrm{2.0\ mol \times 40.0\ g/mol = 80.0\ g}

This gives 80.0 g of sodium hydroxide.

  1. Calculate the kilograms of the solvent:
1200 g⏟of solution−80.0 g⏟of solute=1120.0 g=1.120 kg⏟of solvent\begin{align*} \underbrace{\mathrm{1200\ g}}_\text{of solution} - \underbrace{\mathrm{80.0\ g}}_\text{of solute} &= \mathrm{1120.0\ g}\\ &= \underbrace{\mathrm{1.120\ kg}}_\text{of solvent} \end{align*}

This gives 1.120 kg of solvent.

  1. Calculate the molality:
2.0 moles of solute1.120 kg of solvent≈1.79 mol/kg\frac{\mathrm{2.0\ moles\ of\ solute}}{\mathrm{1.120 kg\ of\ solvent}} \approx \mathrm{1.79\ mol/kg}

Answer:
The molality is 1.79 moles per kg of solvent\mathrm{1.79\ moles\ per\ kg\ of\ solvent}.

Now that we have worked through the logic manually, let's check our result by using the two molarity to molality formulae directly.

b=1000×2.0 mol/L1.2 g/mL×1000−2.0 mol/L×40.0 g/mol=2.0 mol/L1.2 kg/L−2.0 mol/L×0.040 kg/mol≈1.79 mol/kg\footnotesize \begin{align*} \text{b} &= \mathrm{\frac{1000\times 2.0\ mol/L}{1.2\ g/mL \times 1000 - 2.0\ mol/L \times 40.0\ g/mol}}\\[1.2em] &= \mathrm{\frac{2.0\ mol/L}{1.2\ kg/L - 2.0\ mol/L \times 0.040\ kg/mol}}\\[1.2em] &\approx \mathrm{1.79\ mol/kg} \end{align*}

Unsurprisingly, we get the same result!

Example #2: Molality to molarity

Problem:
Find the molarity of a 2.5 molal (mol/kg) H2SO4 solution with a 1.54 g/mL density.

Solution:

  1. Assume you have 1 kg of solvent. This is an important step. The amount of solvent is not given, but you need to have a specific quantity to do the calculations.
  2. Find the number of moles of solute:
2.5 mol/kg×1 kg=2.5 mol\mathrm{2.5\ mol/kg \times 1 kg = 2.5\ mol}

This gives 2.5 moles of solute.

  1. Find the mass of the solute. The molar mass of sulfuric acid is 98.09 g/mol.
2.5 mol×98.09 g/mol≈245.23 g\mathrm{2.5\ mol \times 98.09\ g/mol \approx 245.23\ g}

This gives 245.23 g of solute.

  1. Find the total mass of the solution. Note that 1 kg of solvent equals 1000 g of solvent.
1000 g+245.23 g=1245.23 g\mathrm{1000 g + 245.23 g = 1245.23 g}
  1. Find the volume of the solution:
1245.23 g1.54 g/mL≈808.59 mL≈0.809 L\frac{\mathrm{1245.23 g}}{\mathrm{1.54\ g/mL}} \approx \mathrm{808.59 mL \approx 0.809 L}

This gives you about 0.809 L of solution.

  1. Calculate the molarity:
2.5 mol0.809 L≈3.09 mol/L\frac{\mathrm{2.5\ mol}}{\mathrm{0.809\ L}} \approx \mathrm{3.09\ mol/L}

Answer:
The molarity is about 3.09 moles per liter of solvent\mathrm{3.09\ moles\ per\ liter\ of\ solvent}.

As above, let's check our result by using the two molality to molarity formulae.

M=2.5 mol/kg ×1.54 g/mL×10001000+2.5 mol/kg×98.09 g/mol=2.5 mol/kg ×1.54 kg/L1+2.5 mol/kg×0.09809 kg/mol≈3.09 mol/L\footnotesize \begin{align*} \text{M} &= \mathrm{\frac{2.5\ mol/kg \times 1.54\ g/mL \times 1000}{1000 + 2.5\ mol/kg \times 98.09\ g/mol}}\\[1.2em] &= \mathrm{ \frac{2.5\ mol/kg \times 1.54\ kg/L}{1 + 2.5\ mol/kg \times 0.09809\ kg/mol}}\\[1.2em] &\approx \mathrm{3.09\ mol/L} \end{align*}

And... once again we get the same result!

In summary, converting between molarity and molality is a little more complex than it appears, as it requires knowledge of density and molar mass. But we are confident that with our step-by-step guide and a little practice, these conversions will become a quick and reliable addition to your chemistry toolkit.

Molality only concerns masses and therefore does not vary with temperature, whereas molarity concerns volumes, which vary with temperature. That is why molality is preferable to molarity.

No. Molarity is the amount of substance directly proportional to the number of moles of solute and inversely proportional to the volume of the solution in liters. Molality, on the other hand, is defined as the ratio between the number of moles of solute and the mass of the solvent in kilograms.

Yes. The lowercase letter m indicates molality, which is calculated from the number of moles of solute per kilogram of solvent. Molality can also be expressed with a lowercase b. A solution using these units is called a molal solution (for example, 0.1 m NaOH is a 0.1 molal solution of sodium hydroxide). The uppercase letter M indicates molarity, the number of moles of solute per liter of solution (not solvent).

This article was written by Claudia Herambourg and reviewed by Steven Wooding.