16.16: Calculating Molar Mass

We know that we can put antifreeze into a radiator and keep an engine from freezing up. By knowing how cold it will get and how much water is in the radiator, we can determine how much antifreeze to add to achieve our desired freezing point depression. We can do this because we know what the antifreeze is. Can we switch things around and get some information about the properties of the antifreeze (such as its molecular weight) from the freezing point decrease? It turns out that we can do this fairly easily and accurately.

Calculating Molar Mass

In the laboratory, freezing point or boiling point data can be used to determine the molar mass of an unknown solute. Since we know the relationship between a decrease in freezing point and the concentration of solute, if we dissolve a known mass of our unknown solute into a known amount of solvent, we can calculate the molar mass of the solute. The $$K_f$$ or $$K_b$$ of the solvent must be known. We also need to know if the solute is an electrolyte or a nonelectrolyte. If the solvent is an electrolyte, you would need to know the number of ions produced when it dissociates.

Example 16.16.1

$$38.7 \: \text{g}$$ of a nonelectrolyte is dissolved into $$218 \: \text{g}$$ of water. The freezing point of the solution is measured to be $$-5.53^\text{o} \text{C}$$. Calculate the molar mass of the solute.

Solution:

Step 1: List the known quantities and plan the problem.

Known

• $$\Delta T_f = -5.53^\text{o} \text{C}$$
• Mass $$\ce{H_2O} = 218 \: \text{g} = 0.218 \: \text{kg}$$
• Mass solute $$= 38.7 \: \text{g}$$
• $$K_f \left( \ce{H_2O} \right) = -1.86^\text{o} \text{C}/\textit{m}$$

Unknown

• Molar mass solute $$= ? \: \text{g/mol}$$

Use the freezing point depression $$\left( \Delta T_f \right)$$ to calculate the molality of the solution. Then use the molality equation to calculate the moles of solute. Then divide the grams of solute by the moles to determine the molar mass.

Step 2: Solve.

\begin{align} \textit{m} = \frac{\Delta T_f}{K_f} &= \frac{-5.53^\text{o} \text{C}}{-1.86^\text{o} \text{C}/\textit{m}} = 2.97 \: \textit{m} \\ \text{mol solute} &= \textit{m} \times \text{kg} \: \ce{H_2O} = 2.97 \: \textit{m} \times 0.218 \: \text{kg} = 0.648 \: \text{mol} \\ \frac{38.7 \: \text{g}}{0.648 \: \text{mol}} &= 59.7 \: \text{g/mol} \end{align}

The molar mass of the unknown solute is $$59.7 \: \text{g/mol}$$. Knowing the molar mass is an important step in determining the identity of an unknown. A similar problem could be done with the change in boiling point.