Chemistry Vant Hoff Factor, BP, FP etc

Chemistry 1C Freezing Point Depression

Larson/Daley 1 May 23, 2008 FPDepression.doc

Freezing Point Depression, the van’t Hoff Factor, and Molar Mass Objectives

• To understand colligative properties. • To find the freezing point depression of a solution. • To determine the van't Hoff factor for acetic acid dissolved in cyclohexane. • To deduce the nature of acetic acid dissolved in cyclohexane. • To find the molar mass of an unknown organic compound.

Discussion – Freezing Point Depression The vapor pressure of a solution containing a non-volatile solute is lower than that of the pure solvent. Consequently, the boiling point of a solution is higher. The freezing point of a solvent is also affected by the solute; the freezing point of a solution is lower than that of the pure solvent. The freezing point of a solution, Tf, is

Tf = T°f – ∆Tf

Where T°f is the freezing point of the pure solvent and ∆Tf is the freezing point depression. Freezing point depression, like boiling point elevation, is termed a colligative property. Such properties depend only on the concentration of the solute particles in the solution, and not on their nature.

For dilute solutions, ∆Tf = mKf = T°f – Tf (1)

Where m is the molality of the solution and Kf is the molal freezing point depression constant, a property of the solvent. Values of Kf for various solvents are given in table I.

Table I Consider solutions of electrolytes, such as NaCl, dissolved in water; in this case the solute dissociates to form ions in the solution. Thus there are more solute particles in solution than for the same molality of a non-electrolyte, such as sucrose or methanol, dissolved in water. For dilute solutions of NaCl(aq), the solution contains essentially completely separated Na+(aq) and Cl–(aq) ions. As a

result, the molality of the solute particles (ions in this case) is twice what it would be for a non-electrolyte, and ∆Tf is twice as large. On the other hand, if the solute particles associate in solution to form aggregates then the number of solute particles (aggregates in this case) is reduced, causing ∆Tf to be smaller than if the solute did not associate. The degree of dissociation or association of the solute particles is given by the van't Hoff factor, i. This factor is the ratio of the number of moles of solute particles in solution to the number of moles of solute dissolved.

i = moles solute particles/moles solute dissolved If the solute exists as single molecules (neither dissociating or associating) in solution, then i = 1

If the solute dissociates in solution, then i > 1

If the solute associates in solution, then i < 1

The value of the van't Hoff factor depends upon: the nature of the solute AND solvent and the concentration of the solute in the solution. We can determine the van't Hoff factor experimentally for a given system, that is for a given solute, solvent and concentration of solute, by comparing the experimentally measured freezing point depression to the calculated freezing point depression given by equation (1)

!

i =!Tf(measured)

!Tf(calculated) (2)

Solvent Normal Freezing Point Pure Solvent, T°f (°C)

Kf (°C/m)

Water 0.00 1.86 Benzene 5.50 5.12 Camphor 179.75 39.7

Cyclohexane ? 20.0



An alternative form of equation (1) that explicitly takes into account dissociation or association of the solute is:

∆Tf = imKf (3) (m is the molality without taking dissociation or association into consideration) Equation 3 can also be used to determine the van't Hoff factor for a given system, once the freezing point depression is measured.

In this experiment, we will focus our attention on the effect that the natures of the solute and solvent have on the value of a van't Hoff factor; we will not investigate concentration effects. Specifically, we will investigate the behavior of acetic acid dissolved in a non-polar solvent, cyclohexane. The questions we will address include:

Does acetic acid dissociate, associate, or exist as single molecules when dissolved in cyclohexane? Can we explain the behavior of acetic acid in cyclohexane?

Discussion – Molar Mass from Freezing Point Depression Data The freezing point depression of a solution can also be used to determine the molar mass of the solute. Using equation (3) we can solve for the molality of a solution (if i is known) once ∆Tf has been measured:

!

m =!Tf

iKf

(4)

From the molality and known mass of solvent, the moles of solute can be found. From the moles of solute and known mass of solute the molar mass of the solute can be determined. In this experiment, you are going to determine the molar mass of an unknown solid organic compound dissolved in cyclohexane.

Materials LabPro with link cable Ice TI Graphing Calculator Cyclohexane (50 mL per group of two students) Vernier Stainless steel temperature probe Glacial acetic acid. (0.2 mL per group of two students) (1) 600-mL beaker 200-µL automatic pipet with disposable tips

(two or more per lab, locate in hoods) (2) 250-mL beakers Waste container labeled “Cyclohexane”, “Acetic acid”. (1) large test tube Unknown solid organic compound(s).

(2 bottles of each unknown. ≈ 0.3 g per group of two students)

Procedure Part 1: Freezing Point of the Pure Solvent, Cyclohexane 1. Weigh a large, dry test tube supported-in a 250-mL beaker. (Test tube MUST BE CLEAN!!!) Remember

that to obtain the mass of the test tube, you can tare (zero) the balance with the beaker alone on it before placing the test tube in the beaker.

2. Put about 15-mL of cyclohexane in the test tube. 3. Put the test tube in a 600-mL beaker with enough room temperature water to fill the beaker about half

way. 4. Prepare a LabPro for temperature data collection. See the LabPro Quick Start Guide for details. 5. Plug the temperature probe into Channel 1 of the LabPro System. If you are using an older temperature

probe you may need to use a DIN-BTA adapter to connect the probe to the LabPro. Use the link cable to connect the LabPro System to the TI Graphing Calculator. Firmly press in the cable ends.

6. Place the temperature probe in the cyclohexane. Set up the calculator and LabPro for data collection. Use TIME GRAPH data entry with 5 seconds between

samples and 60 samples. • Begin to record the temperature by pressing [START] from the MAIN MENU. A temperature graph is

displayed on the calculator screen. 7. Begin cooling the water bath to ice temperature by adding enough ice to almost fill the beaker. Stir the

cyclohexane with the temperature probe and also move the test tube around in the water bath. Quick



cooling while simultaneously stirring with the temperature probe and moving the test tube around in the water bath tends to produce better cooling curves.

8. Make a note of the temperature and approximate time when the first appearance of cyclohexane crystals occurs.

9. When data collection stops a graph of temperature vs. time is displayed on the calculator screen. As you move the cursor right or left, the time (X) and temperature (Y) values of each data point are displayed below the graph. As a pure liquid freezes, the temperature remains constant. Be aware that even slight contamination due to unclean glassware can cause the temperature to slowly decrease instead of remaining constant as freezing occurs. Examine the graph to find the freezing point of cyclohexane. The freezing point can be found where the slope of the cooling curve breaks and flattens out. Record the freezing point (to nearest 0.1°C). Compare this value to the value you recorded when crystals were first observed. The two values should be very close.

10. Press [ENTER] to return to the MAIN MENU. 11. Remove the cyclohexane from the water bath and allow it to melt. Do not discard the cyclohexane, you

will be using it for Part 2 of this experiment.

Part 2: Freezing Point of the Acetic Acid Solution 1. Dry the outside of the test tube and reweigh the test tube with cyclohexane while being supported-in

a 250-mL beaker. 2. IN THE HOOD, use the automatic pipet provided to drip 200 µL (0.2 mL) of glacial (pure) acetic acid into

the cyclohexane, avoiding the side of the test tube. BE CAREFUL NOT TO TOUCH THE CYCLOHEXANE with the pipet tip! Drip the acetic acid into the cyclohexane by holding the tip just above the surface. (If the tip contacts the cyclohexane, finish delivery, eject the tip into the waste bottle and replace with a new tip for the next student.)

3. Reweigh the test tube with cyclohexane and acetic acid while being supported-in a 250-mL beaker. 4. After weighing, stir the solution well! (Why wait to stir until after weighing?) 5. Repeat the freezing process as in Part 1 for this mixture. 6. Unlike pure cyclohexane, cooling a mixture of acetic acid and cyclohexane results in a gradual drop in

temperature during the time period when freezing takes place. To determine the freezing point of the acetic acid-cyclohexane solution, you need to determine the temperature at which the mixture first started to freeze. Examine the data points to locate the freezing point of the solution, as shown in Figure 1. It may be best to locate the freezing point by transferring the data file into Graphical Analysis, using one of the computers located in the lab room, and then examining the graph using the interpolate mode of graphical analysis. (Your instructor can assist you with the transfer.) Record the freezing point (to nearest 0.1°C). Compare this value to the value you recorded when crystals were first observed. The two values should be very close.

7. Discard your solution into the appropriate waste beaker.

Figure 1: Determining the freezing point of a solution.

Time

Tem

pera

ture Freezing Point



Part 3: Freezing Point of a Cyclohexane Solution Containing an Unknown Organic Compound 1. Weigh a large, dry test tube supported-in a 250-mL beaker. (Test tube MUST BE CLEAN!!!) 2. Wipe-clean the temperature probe. 3. Add approximately 13 mL (≈ 10 g) of cyclohexane to the test tube. 4. Weigh the test tube with cyclohexane while being supported-in a 250-mL beaker. 5. Determine the mass of cyclohexane added. 6. Find the solid organic unknown in the lab. Record the unknown ID code, mass of unknown needed per

gram of cyclohexane, and the van’t Hoff factor for the unknown (all given on the unknown label). 7. From the label information, calculate the mass of unknown to add to your cyclohexane. 8. Add the calculated mass of organic unknown (to within ±0.01 g) to your cyclohexane. 9. Stir the mixture with the temperature probe until the unknown solid has completely dissolved. 10. Repeat the freezing process as in Part 1 for this mixture. 11. Examine the graph and record the freezing point (to nearest 0.1°C) as in Part 2. 12. Discard your solution into the appropriate waste beaker.



Report Sheet Name Partner

Part 1 – Freezing Point of Cyclohexane Freezing point of pure cyclohexane

Part 2 – Freezing Point of Acetic Acid in Cyclohexane Mass of test tube Mass of test tube and cyclohexane Mass of test tube, cyclohexane and acetic acid Freezing point of mixture

Part 3 – Freezing Point of Unknown Solid in Cyclohexane Mass of test tube Mass of test tube and cyclohexane Mass of test tube, cyclohexane and unknown Unknown ID code (on unknown label) Mass unknown per gram cyclohexane (on unknown label) van’t Hoff factor for unknown (on unknown label) Freezing point of unknown mixture

Calculations Part 2 – Acetic Acid Solution From the data recorded: 1) Calculate the experimental freezing point depression of the cyclohexane-acetic acid solution.

a) Calculate the experimental van't Hoff factor for acetic acid in cyclohexane.



2) Using the van’t Hoff factor calculated above:

a) Does the acetic acid ionize or associate in cyclohexane? How do you know?

b) Determine either the % ionization of acetic acid (if i is >1) or how many acetic acid molecules are associated per solute particle formed (if i is <1) in the cyclohexane.

c) Explain why acetic acid behaves this way in cyclohexane. Use Lewis Structures and give a written explanation in terms of intermolecular forces.

Part 3 - Unknown Solid Molar Mass 1. From the data recorded, calculate the molar mass of the unknown solid. Show all your work below in

an orderly fashion. Label all calculated quantities.



Follow-up Questions 1. Why does the temperature continually decrease as a solution freezes instead of maintaining a constant

temperature as a pure liquid would do?

2. When making ice cream, the temperature of the ingredients is decreased to below the freezing point of pure water by using a surrounding ice bath that contains a large amount of table salt, NaCl.

a) Assuming that it dissolves completely and an ideal value for the van't Hoff factor, what mass of NaCl is needed to lower the freezing point of 5.5 kg of water to –5.5°C?

b) Making the same assumptions, calculate the mass of CaCl2 needed.

3. Conifer, a sugar (a non-electrolyte that does not associate in water) derivative found in conifers such as fir trees, has a composition of 56.13% C, 6.48% H and 37.39% O by mass. A 2.216 g sample is dissolved in 48.68 g of water and the solution is found to have a boiling point of 100.068°C. What is the molecular formula of conifer? (Kb for water is 0.51°C/m.)



4. Hydrogen chloride (HCl) is soluble in both water and in benzene (C6H6).

benzene a) For a 0.01 m HCl(aq) solution, the freezing point depression is about 0.04°C. Determine the van't Hoff

factor for HCl in water. i) Is this van't Hoff factor what you would expect for HCl dissolved in water? Explain why or why

not.

b) For a 0.01 m HCl dissolved in benzene solution, the freezing point depression is about 0.05°C. Determine the van't Hoff factor for HCl in benzene.

c) Compare the van’t Hoff factor for HCl when water is the solvent to the van’t Hoff factor for HCl when benzene is the solvent. Suggest a reason in terms of intermolecular forces for any difference between the two values.

Documents

Chemistry Vant Hoff Factor, BP, FP etc