CHEM 122L

General Chemistry Laboratory

Revision 1.6

The Freezing Point Depression of t-Butanol

To learn about the Freezing Point Depression of Solvents.

To learn about Colligative Properties.

To learn about the determination of Molecular Weights.

In this laboratory exercise we will measure the Freezing Point Depression of t-Butanol when a

solute is added to it. We will use this data to determine the Molecular Weight of the solute. This

will then be compared to the solute’s known Molecular Weight to establish the reliability of this

methodology for Molecular Weight determinations.

The physical properties of any solvent are inherently altered by the presence of a solute. For

instance, the Boiling Point, Freezing Point, Density and Viscosity of Water, 100.0oC, 0.0

oC,

0.998220

g/mL and 1.00220

mPa sec respectively, will all change upon the addition of solutes.

As an example, the addition of Ethylene Glycol (CH2OHCH2OH) to Water changes these

properties relative to the amount (wt %) added:

wt% Freezing Pt. [oC] Density [g/mL] Viscosity [mPa sec]

0.0 0.00 0.99821 1.002

0.5 -0.15 0.9988 1.010

1.0 -0.30 0.9995 1.020

2.0 -0.61 1.0007 1.048

3.0 -0.92 1.0019 1.074

4.0 -1.24 1.0032 1.099

5.0 -1.58 1.0044 1.125

6.0 -1.91 1.0057 1.153

CRC Handbook of Chemistry and Physics

Focusing on the Freezing Point, we note the freezing point of the solution drops as the

concentration of Ethylene Glycol is increased. This is referred to as the Freezing Point

Depression of the solvent. This is why Ethylene Glycol works as an Antifreeze in your car’s

cooling system; it depresses the freezing point of Water so the coolant does not freeze during the

winter months.

The Freezing Point Depression is defined as:

Tf = Tfo - Tf (Eq. 1)

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where Tf is the freezing point of the solution and Tfo is the freezing point of the pure solvent. If

we convert the Concentration Scale from wt% to molality, the above data presents itself as:

molality [mol/kg] Tf [oC]

0.081 0.15

0.163 0.30

0.329 0.61

0.498 0.92

0.671 1.24

0.848 1.58

1.028 1.91

The conversion of concentration units is straight forward. For example, consider the conversion

of 6 wt% to molality Ethylene Glycol in Water. First assume we have 100g of solution. At 6%,

the solution is composed of 6g Ethylene Glycol and 94g of Water. Converting the mass Ethylene

Glycol to number of moles, we have:

# mole Eth. Gly. = 6 g / (62.07 g/mole) = 0.09667 mole Eth. Gly.

Converting the mass Water to kilograms, we have:

# kg Water = 94 g x (1 kg / 1000 g) = 0.094 kg

So, the molality of the solution is:

molality = = = 1.028 m

If we plot the Freezing Point Depression data in this guise, we find a linear relationship between

FP Depression and molality exists.

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In other words, the Freezing Point Depression is proportional to the solution’s molality:

Tf ~ m (Eq. 2)

The proportionality constant (slope of the line) is called the Freezing Point Depression Constant,

Kf. Thus, (Eq. 2) becomes:

Tf = Kf m (Eq. 3)

with Kf = 1.86oC/m for the Aqueous Ethylene Glycol case. Further, if we plot similar data for a

series of different solutes, Methanol (CH3OH) and Urea( NH2CONH2), we find they all fall

along the same line; slope = 1.86 oC/m.

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This means the Freezing Point Depression Constant Kf does not depend on the nature of the

solute. These observations concerning the Freezing Point Depression are generally true of all

solutions. The value of Kf does not depend on the nature of the solute, but it does depend on the

nature of the solvent. Experimentally determined values of Kf for various solvents are reported

below:

Solvent Kf [oC/m]

Water 1.86

Naphthalene 6.9

Benzene 4.90

Camphor 37.7

p-Dichlorobenzene 7.10

t-Butanol 9.1

Physical properties of a solution, like the Freezing Point Depression, that depend on the amount

of solute present, but not on the nature of the solute, are collectively referred to as Colligative

Properties; Colligative (from the Latin co-, together, and ligare, to bind) because they are bound

together in their common origin in the Laws of Thermodynamics. Two other common

Colligative Properties are the Boiling Point Elevation (Tb) and the Osmotic Pressure ().

It should be noted the data presented above is for relatively dilute solutions; concentrations

below ~1.0 molal. As a solution becomes more concentrated, not only is the Freezing Point

Depression no longer proportional to the molality, but it will begin to depend on the nature of the

solute. The above data extended to ~30 molal demonstrate this deviation for Ethylene Glycol

and Urea.

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CRC Handbook of Chemistry and Physics

Thus, the Freezing Point Depression relationship of (Eq. 3) is a limiting law; true only in the limit

of zero concentration:

Tf = Kf m (Eq. 4)

Physically, the solute acts to disrupt the formation of the solvent’s crystal structure, causing the

freezing point to decrease. At sufficiently low concentrations, it doesn’t matter what the nature

of the solute is so long as it can disrupt crystal formation. As the solute’s concentration

increases, crystallization of the solvent becomes more difficult and the freezing point of the

solution decreases further. However, if the solution becomes sufficiently concentrated, solute

molecules begin to interact with each other, as well as with the solvent molecules, in ways that

cause the freezing point to deviate from the ideal prediction of (Eq. 3). In fact, Colligative

Property data is useful in measuring these solution non-idealities.

In addition to being useful for measuring non-idealities in solutions, Colligative Property data

can also be useful in determining the Molecular Weight of a solute. If we measure the Freezing

Point Depression due to the presence of a solute of unknown Molecular Weight, we can

determine the molality of the solute. And, if we know the mass of the solvent used to prepare the

solution, we can determine the number of moles of solute present. Finally, if we know the mass

of the solute used, we can calculate the solute’s Molecular Weight using the definition of the

Molecular Weight:

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Molecular Weight = (Eq. 7)

As an example, consider a solution prepared from 1.00g of an unknown solute in 50.0g of the

solvent Benzene. The Freezing Point Depression is measured to be 1.81oC. We can determine the

molality of the solute using (Eq. 3):

m = Tf / Kf = 1.81oC /( 4.90

oC/m) = 0.369 m

This then allows us to calculate the number of moles solute:

# moles = molality x #kg solvent = 0.369m x 0.0500kg = 0.0185 mole

Now we apply (Eq. 7) to determine the Molecular Weight:

MW = 1.00g / 0.0185 mole = 54.0 g/mole

For small molecules, Colligative Property data is usually not sufficiently accurate to be useful for

Molecular Weight determinations. (A Mass Spectrometers offers a very, very accurate and

precise alternative.) However, for high molecular weight polymers (proteins, nucleic acids,

carbohydrates, synthetic organic polymers), Colligative Property data can be useful for

Molecular Weight determinations.

Finally, as a practical point, the Freezing Point of the solvent, and the solution, can each be

determined by measuring their Cooling Curves; the temperature of the system followed over

time when the system is immersed in a cold bath. For the pure solvent, the cooling of the liquid

will be relatively linear. Following a brief period of supercooling, which occurs because the

system is cooling faster than the molecules can lock into position in a crystalline structure, the

temperature of the system will remain relatively constant while the freezing process occurs.

Then, the temperature of the system will again decline as the solid cools. The Freezing Point Tfo

is the temperature at which the constancy occurs.

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This process is modified slightly for the solution case. As crystals of the solvent begin to form,

the solution becomes more concentrated, and the freezing point is depressed further. Hence, the

region of the Cooling Curve representing the freezing of the solvent will not be constant, but

instead will fall with time. The Freezing Point Tf is extrapolated from this data.

In this laboratory exercise, we will measure the Freezing Point Depression of a t-Butanol

solution:

when one of the following solutes is added:

Ethanol (CH3CH2OH)

Propanol (CH3CH2CH2OH)

Isoamyl Alcohol (CH3CH(CH3)CH2CH2OH)

Each of these solutes, as well as the solvent t-Butanol, is an Organic Alcohol; an organic

compound containing the –OH functionality.

t-Butanol is industrially important for its solvent power. It is used for the removal of water from

substances, in the extraction of drugs, in the manufacture of perfumes (particularly as an

important raw material for the preparation of artificial musk), in the recrystallization of

chemicals, and as a chemical intermediate (e.g., in the manufacture of t-butyl chloride and of t-

butyl phenol). It is an authorized denaturant for ethyl alcohol. It has been patented for use as a

gasoline antiknock agent.

We will use this Freezing Point Depression data to determine the Molecular Weight of the solute.

This can then be compared with the Molecular Weight determined from the chemical formula to

judge the usefulness of this methodology in determining Molecular Weights of small

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compounds. Keep in mind, just because a measurement is theoretically possible does not mean it

is experimentally practical. Further, just because a measurement is not useful for one calculation

does not mean it is not useful for another.

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Pre-Lab Questions

1. The following Cooling Curve data was obtained when ParadichloroBenzene was added to

liquid Naphthalene:

Time [min] Temp. [oC]

0.0 59.7

0.5 58.0

1.0 56.5

1.5 54.8

2.0 53.4

2.5 52.1

3.0 50.9

3.5 49.5

4.0 48.4

4.5 48.6

5.0 48.7

5.5 48.6

6.0 48.5

6.5 48.4

7.0 48.3

7.5 48.2

8.0 48.1

Use this data to prepare a Cooling Curve. You should use a software package such as

Excel to graph the data. Add appropriate “Trendlines” to determine the Freezing Point of

this solution.

(Hint: Break the data into three regions; cooling of the liquid, supercooling and freezing of

the solution. Add “Trendlines” for the first and third regions. These two “Trendlines”

intersect at the Freezing Point. So, set the two equations equal to each other and solve for

x. Insert the value of x into either equation and solve for the Freezing Point.)

2. The addition of 7.131g of an unknown compound to 45.05g of p-Dichlorobenzene lowers

the freezing point from 53.5oC to 49.7

oC. What is the molecular weight of the compound?

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Procedure

1. Assemble the freezing point apparatus as pictured below:

The Water bath, an 800 mL beaker, should be cooled with a small amount of Ice. Make

sure the Ice-Water bath is cooled to very close to 0oC.

2. Use a clean, dry 150 mL beaker to support the large test tube.

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Weigh this assembly on a Top-Loading Balance to a precision of 0.01g. Add about 25g

(~31 mL) of t-Butyl Alcohol, and reweigh the assembly. (The container of t-Butanol is

located in the fume hood where it is being warmed in a Water bath to liquify the alcohol.

When obtaining your sample be sure the outside of the container is thoroughly dry as even

minute amounts of Water in the t-Butanol will contaminate the alcohol such that it is

unusable.)

!Good Practice – Always use the same balance for repeat weighings!

3. Separately fill another 800 mL beaker with Water, place it on a Hot Plate, place the

solvent-containing test tube, with the looped stirrer and thermometer, in the Water bath and

heat the system until all the solvent has melted and has been heated to ~40oC.

4. Remove the test tube from the hot Water bath and place it in the Ice-Water bath. The Ice-

Bath should be stirred occasionally during the course of your measurements.

5. Measure the temperature every 10 sec until solidification is complete. During this time the

solvent should be continuously stirred using the looped stirrer. Do not dispose of the

solvent when you are done obtaining the Cooling Curve. You will use this same

solvent for the next part of the experiment.

6. Obtain a clean, dry 10 mL graduated cylinder with cork. Add ~1.5g of a solute chosen

from those listed in the discussion to the graduated cylinder and cork the cylinder. Weigh

the graduated cylinder with solute on a Top-Loading Balance to a precision of 0.01g.

7. Again, heat the solvent in the large test tube until it is melted and again at ~40oC.

8. Quantitatively transfer the solute from the graduated cylinder to the large test tube

containing the warmed solvent. Be certain none of the solute adheres to the walls of the

test tube. If some does, roll the test tube so that the solvent contacts and dissolves the

solute. Immediately re-cork the “empty” cylinder. Re-weigh the graduated cylinder. The

difference in weights of the “full” and “empty” cylinders gives the mass of the solute used.

9. Stir the solvent-solute mixture in the large test tube and repeat the determination of the

Cooling Curve.

10. In order to obtain a second determination of the Freezing Point Depression of the solvent,

re-heat the solution and repeat the measurement of the Cooling Curve.

11. Repeat the determination of the Cooling Curve for a third trial.

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Data Analysis

1. Prepare Cooling Curves for the Solvent (t-Butanol) and the Solution; you should have four

curves. Use the same procedure at that used in Problem #1 of the Pre-Lab assignment to

determine the freezing points for each case. Lastly, report Tfo for t-Butanol and the

Average Tf for the Solution.

2. Determine the Freezing Point Depression.

3. Calculate the molality of the Solution.

4. Determine the Molecular Weight of the Solute from the Freezing Point Depression data.

5. Compare this to the Molecular Weight determined from the Chemical Formula of the

Solute. Calculate the Percentage Error in your FP MW determination.

6. Comment on which measurement or procedure has the largest error. How might you

improve this measurement?

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Post Lab Questions

1. Calculate the Freezing Point Depression which will result when 0.50g of each of the

following compounds is dissolved in 30.00g of Water:

Cmpd 1: MW = 100.0 g/mol

Cmpd 2: MW = 105.0 g/mol

Comment on the advisability of using the Freezing Point Depression measurement in

determining the Molecular Weight of each of these compounds.

2. In our discussion, we considered the Freezing Point Depression of Ethylene Glycol

(CH2OHCH2OH) in Water. Draw the Lewis Structure of Ethylene Glycol and explain why

it will form a solution with Water.

3. When adding Ethylene Glycol to your cooling system, it is usually mixed 50:50 by volume.

This will prepare a 52.7 wt% solution. Convert this concentration to molality and predict

the Freezing Point Depression based on (Eq. 3). The measured value, as reported by Dennis

R. Cordray, Lisa R. Kaplan, Peter M. Woyciesjes and Theodore F. Kozak in Fluid Phase

Equilibria 117 (1996) 146 is 38.3oC. What is the Percentage Error introduced by assuming

the solution is dilute?

4. When salting ice covered roads, why is it less desirable to use a salt such as Na3PO4 than it

is to use NaCl? (Two Reasons.)

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Addendum

Solutions versus Colloidal Systems

The above procedure for Molecular Weight determination relies on the fact that the solute forms

a true solution when dissolved in the solvent. This is not always the case. In many instances, the

solute will instead combine with the solvent to form a colloidal system.

Colloids are systems in which one phase is dispersed into another. Examples include fog, milk,

shaving cream, and Jello. At the macroscopic level a colloid such as milk appears homogenous.

However, upon microscopic examination, milk will really appear as extremely small droplets of

liquid fat dispersed in liquid Water.

Most colloids, such as milk, are opaque or cloudy. However, in some cases the system appears

transparent to the naked eye and therefore appears as though it is a solution. But because the

dispersed colloidal particles are at least as large as the wavelengths of visible light, a beam of

light passing through the system is scattered. This scattering phenomenon is known as the

Tyndall Effect. The Tyndall Effect is what is observed when a beam of light passes through dust

or fog dispersed in Air. This effect is one method for distinguishing between true solutions,

where the solutes are dissolved as individual molecules, and which do not exhibit a Tyndall

Effect, and colloidal systems which merely appear as though they are solutions.

Because Colligative Properties depend only on the number of “particles” dissolved in the

solvent, molecular weights derived from Freezing Point Depression data of colloidal systems are

really the average molar masses of the colloidal particles comprising the systems.

In this short exercise, we will examine two apparent “solutions”, one of which is a true solution

and the other of which is a colloid. We will identify which exhibits a Tyndall Effect, and is

therefore a colloid, and which does not.

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Procedure

1. Examine the two “solutions” provided by your laboratory instructor. One is an aqueous

mixture of Ferric Hydroxide; Fe(OH)3. The other is an aqueous mixture of organic dyes.

Can you tell which the true solution is and which is the colloid?

2. In a darkened room, pass a beam of light from a laser “pointer” through each system.

Which exhibits a Tyndall Effect and which does not? Identify the colloid.