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16.16: Calculating Molar Mass

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    53864
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    Putting antifreeze into a radiator will keep an engine from freezing. By knowing how cold the weather will get and how much water is in the radiator, one can determine how much antifreeze to add to achieve a specific desired freezing point depression. This is made possible by knowing what antifreeze is. Can things be switched around, so as to get some information about the properties of the antifreeze (such as its molecular weight) from the freezing point decrease? As it turns out, this can be done fairly easily and accurately.

    Figure \(\PageIndex{1}\): Changes in temperature.

    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, we would need to know the number of ions produced when it dissociates.

    Example \(\PageIndex{1}\)

    \(38.7 \: \text{g}\) 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

    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.

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

    Step 3: Think about your result.

    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.

    Summary


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