Molar Mass Determination by Freezing Point Depression
Molar mass determination by freezing point depression is a widely used laboratory technique for finding the molecular weight of an unknown compound. The method relies on a fundamental colligative property: adding a dissolved solute lowers the freezing point of a solvent. So by measuring this temperature change and applying a known cryoscopic constant, chemists can calculate the number of moles present and, from that, the molar mass of the solute. The approach is especially valuable for organic compounds, biological molecules, polymers, and other substances that are difficult to analyze using mass spectrometry And that's really what it comes down to..
What Is Freezing Point Depression?
Freezing point depression is the decrease in the freezing temperature of a solvent caused by the presence of a dissolved solute. The magnitude of the change depends on the number of dissolved particles, rather than their chemical identity. This makes freezing point depression a colligative property.
For a dilute ideal solution, the relationship is:
[ \Delta T_f = iK_fb ]
In this equation:
- (\Delta T_f) is the freezing point depression in degrees Celsius or kelvin.
- (i) is the van’t Hoff factor.
- (K_f) is the cryoscopic constant of the solvent.
- (b) is the molality of the solution in moles of solute per kilogram of solvent.
The observed freezing point depression is calculated from
[ \Delta T_f = T_{f,\text{pure}}-T_{f,\text{solution}} ]
Because a temperature difference has the same magnitude in degrees Celsius and kelvin, (\Delta T_f) may be recorded in either unit But it adds up..
Determining the Molar Mass
Once the freezing points of the pure solvent and solution have been measured, the molality of the solution can be found from
[ b=\frac{\Delta T_f}{iK_f} ]
For a non-electrolyte that does not associate or dissociate in the solvent, the van’t Hoff factor is approximately 1, so
[ b=\frac{\Delta T_f}{K_f} ]
The number of moles of solute is then calculated using the mass of solvent:
[ n_{\text{solute}}=b,m_{\text{solvent}} ]
where (m_{\text{solvent}}) is expressed in kilograms. Finally, the molar mass is obtained from
[ M=\frac{m_{\text{solute}}}{n_{\text{solute}}} ]
or, when the solute mass is expressed in grams and the solvent mass in grams,
[ M=\frac{1000K_fm_{\text{solute}}} {\Delta T_fm_{\text{solvent}}} ]
for a non-electrolyte. More generally, the van’t Hoff factor must be included:
[ M=\frac{1000iK_fm_{\text{solute}}} {\Delta T_fm_{\text{solvent}}} ]
Experimental Procedure
A typical freezing-point-depression determination follows these steps:
- Select an appropriate solvent. The unknown compound should dissolve readily, remain chemically unchanged, and not react with the solvent. The solvent should also have a conveniently measurable freezing point and a known cryoscopic constant.
- Measure the pure-solvent freezing point. A sample of the solvent is cooled slowly while being stirred. The temperature at which the first stable crystals form is recorded after correcting for any supercooling.
- Prepare the solution. A measured mass of the unknown solute is dissolved in a known mass of solvent.
- Measure the solution freezing point. The solution is cooled under the same conditions used for the pure solvent, and the onset of ice or solvent crystal formation is recorded.
- Calculate the depression and molar mass. The difference between the two freezing points is inserted into the appropriate equation.
Using several solution concentrations can improve reliability. A plot of (\Delta T_f) against molality should be approximately linear, and the molar mass is calculated from its slope:
[ \text{slope}=\frac{\Delta T_f}{b}=iK_f ]
Example Calculation
Suppose (5.0\ \text{g}) of water. Worth adding: 00\ \text{g}) of a non-electrolyte is dissolved in (100. 00^\circ\text{C}), while the solution freezes at (-1.On top of that, the pure water freezes at (0. 86^\circ\text{C}) And that's really what it comes down to..
[ K_f=1.86\ \text{K kg mol}^{-1} ]
The freezing-point depression is
[ \Delta T_f=0.00-(-1.86)=1.86\ \text{K} ]
So, the molality is
[ b=\frac{1.86}{1.86}=1.00\ \text{mol kg}^{-1} ]
The (0.1000\ \text{kg}) of water contains
