# I heard someone tell me a formula to convert molarity to molality: M= m(density of solution - 0.001 x M x Mb) where M-molarity, m-molality, Mb-molecular weight of solute. Could you please tell me if the formula is right, stating an example too? Thanks

Jul 5, 2015

Yes, that formula is correct.

#### Explanation:

You don't actually need to memorize a formula that takes you from molarity to molality, all you really need to know is what molairty and molality mean.

A solution's molarity is defined as the moles of solute per liters of solution.

$C = \frac{n}{V} _ \text{sol}$

On the other hand, a solution's molality is defined as moles of solute per kilograms of solvent.

$b = \frac{n}{m} _ \text{sol}$

So, let's try to determine how you can use one to get to the other. Let's say that you're dealing with a solution of known density and volume in which you dissolve a known mass of solute, ${m}_{\text{solute}}$.

The solution's molarity will be

$C = {n}_{\text{solute"/V_"sol}}$

The solution's molality will be

b = n_"solute"/(m_"solvent" * 0.001)

You multiply the mass of the solvent by 0.001 to go from grams to kilograms, which is what the equation for molality needs.

Notice that both equations feature the number of moles of solute, ${n}_{\text{solute}}$, which means that you can write

$C \cdot {V}_{\text{sol" = n_"solute" = b * m_"solvent}} \cdot 0.001$, or

$C \cdot {V}_{\text{sol" = b * m_"solvent}} \cdot 0.001$

The mass of the solvent can be written as the mass of the solution minus the mass of the solute

${m}_{\text{solvent" = m_"sol" - m_"solute}}$

C * V_"sol" = b * (m_"sol" - m_"solute") * 0.001

The mass of the solution can be written using the solution's volume and density

$\rho = {m}_{\text{sol"/V_"sol" => m_"sol" = rho * V_"sol}}$

This means that you have

C * V_"sol" = b * (rho * V_"sol" - m_"solute") * 0.001

The mass of the solute can be expressed using its number of moles and molar mass

${m}_{\text{solute}} = n \cdot {M}_{M}$

C * V_"sol" = b * (rho * V_"sol" - n * M_M) * 0.001

Divide both sides of the equation by ${V}_{\text{sol}}$ to get

$C \cdot \cancel{{V}_{\text{sol")/cancel(V_"sol") = b * (rho * cancel(V_"sol")/cancel(V_"sol") - (n * M_M)/V_"sol}}} \cdot 0.001$

But $\frac{n}{V} _ \text{sol}$ is actually the solution's molarity, so you have

$C = b \cdot \left(\rho - C \cdot {M}_{M}\right) \cdot 0.001$

Now, in order to get the units to match, you actually need to have the density expressed in grams per liter, as opposed to grams per mL.

Do get around this problem, multiply the density by 1000 to get

$C = b \cdot \left(\rho \cdot 1000 - C \cdot {M}_{M}\right) \cdot 0.001$,

which is equivalent to

$\textcolor{g r e e n}{C = b \cdot \left(\rho - 0.001 \cdot C \cdot {M}_{M}\right)}$

Do a quick unit's test to see if the result makes sense

"moles"/"L" = "moles"/"kg" * ("g"/"mL" - 0.001 * cancel("moles")/"L" * "g"/cancel("mol"))

cancel("moles")/"L" = cancel("moles")/"kg" * (g/"mL" - underbrace(0.001 * "g"/"L")_(color(blue)(="g"/"mL")))

$\frac{1}{\text{L" = 1/"kg" * "g"/"mL" <=> 1/"L" = 1/(0.001 * cancel("g")) * cancel("g")/"mL}}$

Finally,

$\frac{1}{\text{L" = 1/underbrace(0.001 * "mL")_(color(blue)("=L")) = 1/"L}}$

Here's a quick example. Let's say that you have a 1-L solution of glucose (${M}_{M}$ is 180.16 g/mol) that has a molarity of 0.25 M and a density of 1.014 g/mL. Use this formula to determine its molality.

According to the formula, you'd have

$b = \frac{C}{\rho - 0.001 \cdot C \cdot {M}_{M}} = \frac{0.25}{1.014 - 0.001 \cdot 0.25 \cdot 180.16}$

$b = \text{0.258 molal}$

This is the same result you'd get using the classic approach.

${m}_{\text{water" = m_"solution" - m_"glucose}}$

${m}_{\text{water" = rho * V_"sol}} - n \cdot {M}_{M}$

m_"water" = 1.014"g"/cancel("mL") * 1000cancel("mL") - 0.25"moles" * 180.16"g"/cancel("mol")

${m}_{\text{water" = "1014 g" - "45.04 g" = "968.96 g}}$

Therefore,

$b = \frac{n}{m} _ \text{water" = "0.25 moles"/(968.96 * 10^(-3)"kg") = "0.258 molal}$