Transcript Freezing-point depression

Exp 12A: A Molar Mass from Freezing-Point
Depression
Purpose
1. Determine the freezing-point depression in a solution
when a solute is added to a pure solvent
2.
Use the freezing-point depression of the solution to
determine the molar mass of the solute
3.
Detemine the molecular formula of the solute based on
the given empirical formula and the calculated molar
mass
•
See also: Silberberg, Chapter 13, pp. 515-520, and
Sample problems 13.7 and 13.8
Exp 12A: A Molar Mass from Freezing-Point
Depression
• The melting point is the temperature at which a solid
changes to a liquid
• The freezing point is the temperature at which a liquid
changes to a solid
• The freezing point of pure water is 0°C
• The melting point can be depressed by adding a solute
such as a salt.
– Use of ordinary salt (sodium chloride, NaCl) on icy roads in the
winter helps to melt the ice from the roads by lowering the melting
point of the ice.
– Making ice-cream by using a water-salt mixture to lower the
freezing point of the solution
• A solution (solvent + solute) typically has a measurably
lower melting point than the pure solvent
Exp 12A: A Molar Mass from Freezing-Point
Depression
• The freezing point depression ΔTf is a colligative
property of the solution
– properties that depend on the concentration of solute
molecules or ions, but not upon the identity of the solute
– they include freezing point depression, boiling point
elevation, vapor pressure lowering, and osmotic pressure
• Example
– ethylene glycol (C2H6O2) in automobile cooling systems
– ethylene glycol (antifreeze) protects against freezing by
lowering the freezing point and permits a higher operating
temperature by raising the boiling point
Exp 12A: A Molar Mass from Freezing-Point
Depression
Coolant
Pure Water
Freezing Point
Boiling Point
0°C/32°F
100°C/212°F
50/50
C2H6O2/H2O
-37°C/-35°F
70/30
C2H6O2/H2O
-55°C/-67°F
106°C/223°F
113°C/235°F
Freezing Point Depression and Boiling Point
Elevation of Some Solvents
Solvent
Formula
Melting
Point (°C)
Boiling
Point (°C)
Kf(°C/m)
Kb(°C/m)
Water
H2O
0.000
100.000
1.858
0.521
Acetic acid
HC2H3O2
16.60
118.5
3.59
3.08
Benzene
C6H6
5.455
80.2
5.065
2.61
Camphor
C10H16O
179.5
...
40
...
Carbon disulfide
CS2
...
46.3
...
2.40
Cyclohexane
C6H12
6.55
80.74
20.5
2.79
Ethanol
C2H5OH
...
78.3
...
1.07
Exp 12A: A Molar Mass from Freezing-Point
Depression
• For dilute solutions freezing-point depression is found to
be proportional to the molal concentration cm of the
solution:
• ΔTf = Kf cm
where Kf is called the freezing-point-depression constant
• Kf (cyclohexane) = 20.5oC/m
• Molality (m) equals the number of moles of a given
substance per kilogram of solvent (mol/kg)
• it does not change with the temperature, as it deals with
the mass of solvent rather than the volume of solution
 molality of a solution is always constant irrespective of the
physical conditions like temperature and pressure
Exp 12A: A Molar Mass from Freezing-Point
Depression - Experimental
Part A: Measuring the freezing point of cyclohexane
• Pipet exactly 20.0 ml cyclohexane in a large test tube. Close
immediately
• Calculate mass of cyclohexane
– d = 0.779 g/ml
– 20.0 ml = 20.0 ml x 0.779 g/ml = 15.58 g = 15.6 g
• Record volume and calculated mass on your worksheet
Exp 12A: A Molar Mass from Freezing-Point
Depression - Experimental
Part A: Measuring the freezing point of cyclohexane
• Place the test tube with cylcohexane in a beaker with
crushed ice and water
• Stir gently, but constantly
• When the temperature gets close to the freezing point of
cyclohexane (melting point = 6.6oC), take temperature
readings every 15 sec. This will take 10-15 min
• Record the temperature change to the nearest 0.1oC. Stop
the experiment once all cyclohexane is frozen and the
temperature decreases further.
• Allow cyclohexane to melt completely.
• Repeat the freezing point determination
• Plot the data, determine the freezing point in each case,
and calculate the mean freezing point
Exp 12A: A Molar Mass from Freezing-Point
Depression - Experimental
Part B: Measuring the freezing point of solutions
• Weigh 0.24 - 0.25 g of the solute on weighing paper (sample
1). Determine the mass exactly (record in 4 decimals)
• Weigh 0.10 - 0.11 g of solute on a 2nd piece of weighing
paper (sample 2). Weigh in 4 decimals
• Transfer sample 1 to the solvent in freezing point depression
apparatus
• Stir until the solute is completely dissolved. All of the solute
must be dissolved
• Stir gently but constantly and cool the solution as before
• Determine the temperature as described under (A) until the
entire solution is frozen
• Remove the apparatus and allow the frozen solution to melt
completely.
• Graph your data!
• How do your data look?
Exp 12A: A Molar Mass from Freezing-Point
Depression - Experimental
Part B: Measuring the freezing point of solutions
• Add sample 2 to the solution containing sample 1 and repeat
the freezing process
• Plot data for both experiments
• Calculate the change in freezing point, DTf
• Calculate the molal mass of the solute from each DTf. Mind
the significant figures
• Obtain the mean molar mass
• Share data with the rest of the class
• Calculate the molecular formula of the solute from the grand
average and the empirical formula (C3H2Cl)
Exp 12A: A Molar Mass from Freezing-Point
Depression - Calculations
Prelab Problem #3
- a 0.2436-g sample of an unknown substance was dissolved in
20.0 mL cyclohexane.
- The density of cyclohxane = 0.779 g/mL.
- The freezing point depression was 2.5oC.
- Kf = 20.5oC/m
- Calculate the molar mass of the unknown substance.
Exp 12A: A Molar Mass from Freezing-Point
Depression - Calculations
Prelab Problem #3
Solution
• ΔTf = Kf cm = 2.5oC
• cm = ΔTf / Kf = 2.5oC / 20.5oC/m = 0.12 m = 0.12 mol solute/kg
solvent
• Convert molality to mols of solute
 cm = 0.12 m = 0.12 mol/kg =
= (0.12 mol solute/kg solvent) x (20.0 mL solvent x 0.779
g/mL) x (1 kg solvent/1000 g solvent) = 1.9 x 10-3 mol
solute
• molar mass = 0.2436 g solute/1.9 x 10-3 mol solute
= 1.3 x 102 g/mol
Exp 12A: A Molar Mass from Freezing-Point
Depression - Calculations
Freezing Point of Cyclohexane
• Volume of cyclohexane _____ mL
• Mass of cylohexane _____ mL x 0.779 g/mL = ____ g =
____ kg
• Mean freezing point ____ o C
Sample 1 (adding 1st amount of solute)
• Freezing point ____ o C
• Change DTf. ____ o C
• Molal concentration cm = DTf./Kf = ____ m (mol/kg)
• Molar concentration Mm = DTf./Kf (mol/kg) x volume (mL) x
density (g/mL) x 1 kg/103 g = _____ mol
• Molar mass = ___ g solute / ___ mol = g/mol
Exp 12A: A Molar Mass from Freezing-Point
Depression - Calculations
Sample 2 (adding 2st amount of solute)
• Freezing point ____ o C
• Change DTf. ____ o C
• Molal concentration cm = DTf./Kf = ____ m (mol/kg)
• Molar concentration Mm = DTf./Kf (mol/kg) x volume (mL) x
density (g/mL) x 1 kg/103 g = _____ mol
• Molar mass = ___ g solute / ___ mol = ____ g/mol
Exp 12A: A Molar Mass from Freezing-Point
Depression - Calculations
Determining molecular formula
• Empirical formula for the solute: C3H2Cl
• Formula weight = (3 x 12.01) + (2 x 1.008) + 35.45 = 73.50
g/mol
• Molar mass = _____ g/mol
• Formula units = (_____ g/mol)/73.50 g/mol = ___ (needs to
be (close to) a whole number)
• Molecular formula = (C3H2Cl) x (whole number)
This Thursday
• Exp. 12A: Postlab
• Exp. 13: Rate of an Iodine Clock Reaction
– Prelab preparations and
– Prelab assignment
Phase Diagram of Solvent and Solution
Exp 12A: A Molar Mass from Freezing- Point
Depression
Molarity
• molarity (M) equals the number of moles of a given
substance per liter of solution (mol/L)
• measurement of the absolute number of particles in a
solution, irrespective of their weight and volume
Molality
• molality (m) equals the number of moles of a given
substance per kilogram of solvent (mol/kg)
• it does not change with the temperature, as it deals with
the mass of solvent rather than the volume of solution
 molality of a solution is always constant irrespective of the
physical conditions like temperature and pressure