Experiment 4

Freezing Point Depression

Objective:  To observe the effect of concentration on the freezing point of a solution and to utilize this information to determine the molecular weight of a solute.

Background:

      At a given pressure, the temperature at which a specific pure substance undergoes a particular phase change (freezing, boiling, sublimation, etc.) is always the same. For example, pure water at 1 bar pressure will always freeze at 0oC and boil at 100oC. However, if impurities (solutes) are added to the pure solvent in the liquid phase, the temperatures at which phase changes from the liquid occur will be different than those for the pure solvent. In particular the freezing point of a solution formed by mixing a pure liquid with a small amount of solute will decrease and the boiling point of the solution will increase. These effects are referred to as freezing point depression and boiling point elevation and both are part of a larger class of related phenomena called colligative properties - other examples include the lowering of the vapor pressure of a pure liquid by the addition of solutes and the phenomena of osmotic pressure, which is important in keeping blood cells from collapsing. In our everyday lives, the phenomenon of freezing point depression explains why road ice will melt when salt is poured on it (the freezing point is lowered by the addition of the impurity) and why we put ethylene glycol (antifreeze) in our automobile radiators to protect our cooling systems from freezing in extreme cold. Physically, freezing point depression and boiling point elevation occur because a solution consisting of a solvent (the formerly pure liquid) with a small amount of solute added is thermodynamically more stable than the pure liquid itself - primarily due to the fact that the addition of the solute increases the entropy of the solution. This additional stability increases the range of temperatures at which the liquid is stable - lowering the freezing point and increasing the boiling point. In this lab you will be exploring how the freezing point depression of water depends upon the amount and nature of the added solute.

Safety

1.  Do not allow dry ice to touch bare skin -- it can cause severe frostbite.

2.  You must wear goggles at all times while performing this laboratory.

3.  Never ingest chemicals or put any object (pens?) into your mouth in the lab.  Chemicals in the lab may be contaminated or poisonous.

 

Prelaboratory Exercise

Think back to the lab in which you plotted the freezing curve of water.  (See Freezing of Water and Solutions) 

What did the freezing curve of water look like?

Why did the freezing curve look like that?

Water is the coolant used in automobile radiators because it can absorb large amounts of thermal energy (heat) to help keep the engine cool. Water typically freezes at 0° C (32° F) and at temperatures below 0° C water is a solid.  A solid coolant in a car will not perform its intended function; and since water expands as it freezes it could expand and damage the radiator and parts of the engine.

Given the potential problems above, why is water used as a coolant in automobile engines – even in cold climates?

How do we keep the water in the radiator from freezing?

When you have your car “winterized” you can test how low the temperature of the radiator coolant can go before you have a problem.

 How is the temperature at which the coolant will freeze determined?

The prelab for this week is to think about and answer the above questions.

   IMPORTANT: You cannot participate in the lab without a pre-lab journal entry.  So bring a copy of your completed pre-lab to the laboratory.

Explore the website for this experiment.

  Click here to receive a pre-laboratory template that you can use to write up your observations as a Word document.  Print the Word document and take that hard-copy with you to the laboratory, but be sure to save the document to a 3.5” floppy disk!  You may turn in the printed copy, but save an electronic copy of all of your work until the end of the semester.

Laboratory Experiment (Part 1)

Key questions for group: 15 minutes 

·  How do your explanations in the pre-laboratory compare with one another?

 ·  Describe the explanations that you made.  Discuss until you reach consensus on the observations that go on the blackboard.

 Class discussion/ Instruction:  5 – 10 minutes

 ·   Groups share their explanations with other groups in the laboratory.

 Investigation

For this investigation you will use the Serial Box Interface and LoggerPro software to record the cooling curve and to determine the freezing point of a water/alcohol mixture.  Each group will be assigned a different alcohol to use in the investigation; your group will be assigned one of the following: 
Alcohol Formula Molecular Weight (g/mol) Freezing Point
(°C)
Methanol CH3OH 32.04 -93.9
Ethanol CH3CH2OH 46.07 -117.3
1-Propanol (n-propyl alcohol) CH3CH2CH2OH 60.11 -126.5
2-Propanol (isopropyl alcohol) (CH3)2CHOH 60.11 -89.5
1-Butanol 
(n-butyl alcohol)
CH3CH2CH2CH2OH   74.12 -89.5

Consider your experience in previous experiments -- discuss how your group will determine the freezing point of the alcohol/water mixture, and how that freezing point differs from the normal freezing points of pure water and the pure alcohol.  

 ·  What materials might you use?

·  How will you set up the experiment? (Procedures)

·  What data will you collect?

·  How will you determine the confidence that you have in your results?

·  How will you communicate your results to other groups?

Sharing of Results from Part 1

Each team shares their results from Part 1 with the class.  Record the class data in your notebook.  Discuss trends seen in the data from Part 1.

 ·  What trends do you see in the data?

·  How can you account for the trends?  What might be the cause of the trends?

·  How could you test your assumptions?   How could you use the trends to investigate the molecular weight of an unknown? 

Laboratory Experiment (Part 2)

 Investigation

In Part 2 you will determine the molecular weight of an unknown alcohol by a method that utilizes freezing point depression.  In Part 1 you determined the normal freezing point of water and the freezing points with several alcohols mixed into the water.  You will use that information to calculate the change in the freezing point temperature which can then be used to determine the molecular weight of the unknown alcohol.

Follow the same procedures that were used in Part 1 with your unknown alcohol (assigned by your TA) and use that data to calculate the molecular weight of the alcohol.  (How to calculate molecular weight is in Related Materials.)

Procedures

Your TA will assign your team one of five alcohols to use.

(Open Experiment 2 in LoggerPro and adjust time and temperature ranges as appropriate.)

1.  Weigh out 10.0 grams of the alcohol that you were assigned.  Pour the sample into a clean, dry 250-mL Erlenmeyer flask.

2.  Weigh out 100.0 grams of distilled water and pour it into the Erlenmeyer flask containing the alcohol.  Mix the liquids.

3.   Prepare a dry-ice bath by filling a 600-mL beaker 2/3-full with dry-ice chunks.  CAUTION:  Be careful handling the dry ice as it can cause severe burns when it comes in contact with unprotected skin.

4.   Set up the Direct-Connect Temperature Probe and the LoggerPro software for collection of temperature data.  Each run should take less than 10 minutes.

5.   A pre-drilled and split rubber stopper will be used to hold the Temperature probe in your 50-mL Erlenmeyer flask.  Your TA will demonstrate how to insert the probe body sideways into the stopper.  Never push the probe tip through the hole in the stopper, because this causes the Teflon sleeve around the probe to leak, which corrodes the probe and destroys the electronics.

Data Collection. NOTE:  You will need to run the procedures first with pure water to determine a baseline freezing temperature.

6.   Pour approximately 30 mL distilled water into a 50-mL Erlenmeyer flask.  

7.  Place the rubber stopper holding the Temperature probe into the mouth of the flask so that the tip of the probe is in the liquid and the opening of the flask is stoppered.

8.   Place the flask in the dry-ice bath, making certain that the level of the dry ice is above the level of the water in the flask.  CAUTION:  Do not touch the dry ice with bare hands.

9.  Begin collecting data.  Make certain that you continually swirl the water in the flask to maintain a uniform temperature in your solution.

10.  After the freezing point has been reached and is relatively stable, you may stop the data collection.

11.  Repeat steps 7-10 with your 30 mL alcohol/water solution in the flask.

Related Materials

Hopefully, the trend you observed in Part 1 was that as the molecular weights of the alcohols increase, there is less effect on the temperature change.  This is because the magnitude of colligative properties is based on the number of particles in solution.  Equal masses of alcohols of different molecular weights means that there are fewer molecules of the heavier molecule.  For example: because of its lower molecular weight,10 grams of methanol (32.04 g/mole) would be a greater number of molecules than 10 grams of 1-butanol (74.12 g/mole).

The equation relating the freezing point depression and concentration of a solution is:

DTf  =  Km

where DTf  is the change in freezing point (the difference in the freezing point of a pure solvent and the freezing point of a solution using that solvent),  Kis a constant for the solvent, and m is the concentration of the solution in molal units (moles of solute per kilogram of solvent).  We can rearrange the equation to determine the molal concentration:

 m   =  DTf / K  =  moles solute /  kg solvent          The Kf of water is 1.86 ( ºC/m)

(VERY IMPORTANT m = molality = (moles of solute)/(mass solvent in kg)

You must keep the mass of the solvent in kg throughout the calculation, because K is in units of  ºC/molal) 

From the concentration (molality) and the weight (in kg) of solvent used, determine the number of moles of solute in your sample.

Knowing the weight of the solute you used (X g), and having determined the number of moles (Y moles) of solute from the freezing point depression determination  and calculation above, you can calculate the molecular weight of the solute:

Molecular Weight  =  X g / Y moles   =  (X/Y) g / mole

(X g and Y moles are different representations of the same quantity of the compound, and are therefore related.)