Calculating Molar Mass from
Freezing Point Depression
A simple Lab Experiment
on Colligative Properties
http://classweb.gmu.edu/jschorni/chem212/MolWgtFreezingPoint.ppt
Molecular Mass by Freezing Point Depression
Background
Vapor Pressure
 The melting and freezing points for a substance
are determined by the vapor pressure of the
solid and
liquid states.
 Vapor pressure is the pressure exerted by a
vapor in equilibrium with its liquid or solid
phase.
 Vapor pressure is determined by the ability of
particles at the liquid / solid surface to escape
into the vapor phase.
Molecular Mass by Freezing Point Depression
Boiling Point - Pure liquid
 The boiling point of a pure liquid is the temperature at
which the vapor pressure is equal to the pressure of its
environment.
 The “normal boiling point” is when the vapor pressure
is exactly 1 atmosphere.
Melting Point – Pure Solid
 The melting point of a substance is the temperature at
which the liquid and solid have the same vapor
pressure.
 The normal melting point is the temperature at which
the solid and liquid have the same vapor pressure and
the total pressure is 1 atmosphere.
Molecular Mass by Freezing Point Depression
Solutions
 When a solute (different substance than the solvent) is
dissolved in a pure nonvolatile solvent, the vapor
pressure of the resulting solution decreases.
 In a pure liquid – the solvent - all particles at the liquid
surface are the same.
 In a solution (mixture of solvent & solute particles), the
surface of the liquid is occupied by both solvent and
solute particles.
 Thus, there are fewer solvent particles in a solution to
enter the vapor phase resulting in a lower vapor
pressure (at all temperatures).
Molecular Mass by Freezing Point Depression
Boiling Point - Solutions
 A solution with its lower vapor pressure than the pure
solvent, must, therefore, be heated to a higher
temperature before the vapor pressure equals
atmospheric pressure.
 The solution, therefore, has a higher boiling point than
the pure solvent.
 Thus, adding a solute to a solvent increases the boiling
point of a solution.
Molecular Mass by Freezing Point Depression
Freezing Point - Solutions
 The freezing point of a solution is the temperature at
which the liquid solvent and the pure solid solute can
coexist at equilibrium simultaneously, that is, they have
the same vapor pressure.
 A solute dissolved in water at 0oC has a higher vapor
pressure than water, therefore, the solute will not
dissolve.
 The vapor pressure of a solid solute decreases faster
than the solvent as the temperature decreases.
 At some temperature lower than the melting point of
solvent, the vapor pressures of the solute and solvent
will become equal and the solute dissolves.
 Thus, adding a solute to a solvent decreases the
freezing point.
Phase
diagram
showing
the effect
of
nonvolatile
solute on
freezing
point and
boiling
point.
Molecular Mass by Freezing Point Depression
Colligative Properties
 Colligative properties of solutions are properties
that depend upon the concentration of solute
molecules or ions, but not upon the identity of the
solute.
 Colligative properties include freezing point
depression, boiling point elevation, vapor pressure
lowering, and osmotic pressure.
Molecular Mass by Freezing Point Depression
Molality (not to be confused with Molarity, moles/liter)
 The investigation of Colligative properties of solutions
requires the use of a particular concentration unit.
 Molality is defined as the number of moles of solute
dissolved in a Kilogram (not liter) of solvent.
moles solute
Molality (m ) 
kilogram solvent
 For each unique solvent, one mole of solute in a
kilogram of solvent lowers the freezing point a specified
amount.
 The freezing point change is directly proportional to the
amount of solute, not the identity of the solute.
 For the solvent water, one mole of solute will lower the
freezing point of a kilogram of water from 0oC to -1.86oC
Molecular Mass by Freezing Point Depression
 The constant of proportionality between the molality of a
solution and the change in freezing point is called the
“Molal Freezing Point Depression Constant, Kf.
 The molal freezing point depression constant is unique
for each different solvent.
Water
- 1.86 oC m-1 (units of “oC / molality”)
Camphor - 39.7 oC m-1
 The relationship:
Δ
f
 kf  m
where : Δ t f - freezing point change in oC
k f - Freezing Point Depression Constant (o C/mole)
m - molality (moles solute/kg solvent)
Freezing Point Depression
• The freezing point depression, Tf is equal to the
freezing point of the solvent minus the freezing point
of the solution.
• It is also proportional to the molal concentration of
the solute.
Tf = Kf m
• Where Kf, the freezing point depression constant,
depends only on the solvent.
Molecular Mass by Freezing Point Depression
The Experiment
1. Determine the molecular (molar) mass of Isopropyl
Alcohol, i.e., its molecular weight, by measuring its
freezing point depression in an aqueous solution.
Isopropyl Alcohol (C3H7O)
2. For this experiment it is assumed that the vapor
pressure of isopropyl alcohol (the solute) at 0oC is
negligible.
3. This experiment is carried out in a water, ice, salt (NaCl)
mixture, which itself is an example of freezing point
lowering, where the freezing point of water is lowered to
around -10oC.
Molecular Mass by Freezing Point Depression
The Apparatus
 Test Tube fitted with a rubber stopper
into which is inserted a thermometer.
 Thermometer is held in place by the
rubber stopper and the wire is free to
move up and down for stirring.
 The cold bath, into which the test tube
is inserted, is a large beaker containing
the mixture of water, ice, salt.
 The water, ice, salt mixture needs to be
at a temperature of at least -10.0oC.
 *** Do not use the thermometer to stir
the solution.
Molecular Mass by Freezing Point Depression
The Procedure – Part I
1. In a large beaker place equal amounts of ice and salt,
and a small amount of water.
2. Stir mixture.
3. Adjust salt content so that the temperature of the
solution is at least -10oC.
4. Rinse the freezing point apparatus – test tube – with
distilled water.
5. Place approximately 25.0 mL of distilled water,
measured to the nearest 0.1 mL, in the test tube.
6. Stopper the test tube with rubber stopper holding the
thermometer and wire stirrer.
7. Clamp test tube in position so that the contents are
below the water level in the beaker.
Molecular Mass by Freezing Point Depression
The Procedure (Con’t)
8.
Move the stirrer up and down constantly to
agitate the water.
9.
Record at least 5 temperature readings at 30
second intervals or until stirring the freezing
mixture become difficult.
10. Record the freezing the point of distilled water.
11. Extract the test tube from the water/ice bath and
allow the freezing mixture to thaw (use warmth of
your hand to facilitate the thawing).
12. Repeat the freezing procedure for a second
freezing point value.
13. Calculate the mean freezing point of distilled
water.
Molecular Mass by Freezing Point Depression
The Procedure – Part II
1.
Use the buret to dispense 6-7 milliliters of the
isopropyl alcohol to a small test tube, recording
the exact initial and final volumes of the buret.
Note: Burets can and must be read to the nearest
0.01 mL.
2.
Transfer the contents of the small test tube to the
freezing point apparatus containing the thawed
distilled water used in the first part of the
experiment.
3.
Continuously agitate the mixture with the stirrer.
4.
Record the temperature every 30 seconds
Molecular Mass by Freezing Point Depression
5. Continue the agitation until the temperature
readings have been constant for five (5)
consecutive readings or until stirring the frozen
mixture becomes difficult.
6. Record the freezing point of the solution.
7. Thaw the mixture and repeat the freezing process.
8. Calculate the mean freezing point.
Molecular Mass by Freezing Point Depression
Supercooled Solutions
1. If the water/alcohol mixture does not appear to
freeze at a constant temperature, the mixture may
be Supercooled.
2. A Supercooled liquid freezes slightly below its
theoretical freezing point.
3. To obtain the freezing point, construct a graph of
temperature (y-axis) vs. time (x-axis).
4. The resulting plot should show a curve with two
relatively straight segments.
5. The intersection of these two straight lines is the
freezing point of the solution.
To find the
Freezing point
temperatures
when
supercooling
occurs,
Ignore the dip
in the line
graph and
find the
intersection
point between
the two lines
and then
extrapolate to
the
temperature
axis.
Molecular Mass by Freezing Point Depression
Calculations
1. The solvent in this experiment is water.
Density = 1.0 g/mL
2. The solute in this experiment is Isopropyl Alcohol
Density = 0.785 g/mL
3. The Molal Freezing Point Depression Constant, Kf,
for water is 1.86 oC/mole
4. From the average experimental melting point of ice
(distilled water) and the average freezing point for
the alcohol/water mixture calculate the freezing
point depression, oC, for this experiment.
5. From the Freezing Point Depression Expression
calculate the molality of the water/alcohol solution.
tf = kf  m
Molecular Mass by Freezing Point Depression
5. Calculate the number of moles of solute (isopropyl
alcohol) in the solution.
molality
= moles solute / kilogram solvent
 moles solute = molality  kilograms
6. Molecular (Molar) Mass, or Molecular Weight of Solute
Solute
= Isopropyl Alcohol
MWt
= grams solute / moles solute
grams solute = Vol solute  Density
 MWt
= (Vol solute  Density) / mole
7. Percent Error
% Error 
[Accepted Value - Observed Value]
 100
Accepted Value
Accepted Value = 60.1 g/mole