Part 1: Evaporation and Intermolecular Attractions

CHEM&152 Spring 2009 INTERMOLECULAR FORCES and SOLID-LIQUID TRANSITIONS Fill-in Name Pre-lab attached (p 10-11) Stamp Here Partner Lecture Instructor Parts 1 and 2 of this experiment may be done on separate days. Date Helpful info: Today you will be using a program called LoggerPro. It is a powerful program where data can be acquired using probes, and sensors. The data can then be manipulated. Below is a key for the shortcut menu displayed by LoggerPro. Some of these buttons will come in handy in this lab and in the future! http://www.phas.ubc.ca/~year1lab/p100/loggerpro/loggerpictures/toolbar.jpg Part 1: Evaporation and Intermolecular Attractions Purpose Investigate the relationship of dispersion forces and hydrogen bonding forces in intermolecular attractions. Discussion In this experiment, temperature probes covered with filter paper are placed in various liquids. Evaporation occurs when the probe is removed from the liquid's container (see figure 1). This evaporation is an endothermic process that results in a temperature decrease. The magnitude of a temperature decrease is, like viscosity and boiling temperature, related to the strength of intermolecular forces of attraction. In this experiment, you will study temperature changes caused by the evaporation of several liquids and relate the temperature changes to the strength of intermolecular forces of attraction. The temperature change is greater if more evaporation occurs (weaker intermolecular forces). You will use the results to predict, and then measure, the temperature change for several other liquids. You will encounter two types of organic compounds in this experiment alkanes and alcohols. The two alkanes are pentane, C 5 H 12, and hexane, C 6 H 14. In addition to carbon and hydrogen atoms, alcohols also contain the -OH functional group. Methanol, CH 3 OH, and ethanol, C 2 H 5 OH, are two of the alcohols that we will use in this experiment. You will examine the molecular structure of alkanes and alcohols for the presence and relative strength of two intermolecular forces hydrogen bonding and dispersion forces. Dispersion forces exist between any two molecules and generally increase as the molecular weight of the molecule increases. A hydrogen bond can occur when, in one molecule, a hydrogen atom is bonded directly to an N, O, or F atom (the donor) and that hydrogen is attracted to a lone pair on an N, O, or F atom in another molecule (the acceptor). Both a donor and an acceptor must be present for a hydrogen bond to occur. Exp#2 IMF Rev Sp09AEM Spring 2009 Page 1 of 11

FIGURE 1 Caution: In all experiments today keep liquids away from the computers and keyboards. Be very careful when stirring with direct-connect temperature probes. Clean them gently with water and wipe gently/blot with paper towels. Each temperature probe costs approx. $25. Materials **Please keep the organic solutions in the bottles** Computer methanol, CH 3 OH Procedure Serial Box Interface or ULI ethanol, C 2 H 5 OH Logger Pro 1-propanol, C 3 H 7 OH two Vernier Temperature Probes 1-butanol, C 4 H 9 OH 6 pieces of filter paper (2.5 cm X 2.5 cm) pentane, C 5 H 12 2 small rubber bands hexane, C 6 H 14 masking tape sodium thiosulfate pentahydrate 1. Obtain and wear goggles. Collect the necessary reagents and materials. 2. Prepare the computer for data collection by opening the Vernier Software and LoggerPro. You will need to then find the appropriate experiment using the pathway: File Open Experiments Chemistry with Computers 09 Evaporation. You may need to confirm the identity of the temperature probes that you are using in the Sensor Confirmation dialog box if it pops up when you open the program. You should see a temperature scale on the y-axis and time on the x-axis. 3. You will need to set the time scale for data collection. Under the Experiment menu, select Data Collection. Change the Length to 400 seconds and click OK. Then rescale the horizontal axis to 0 to 400 seconds by clicking once on the lowest number and changing it to 0, and clicking on the highest number and changing it to 400. The vertical axis of the graph will have temperature scaled from 5 to 30 C, change this to read 0 to 30 by clicking once on the minimum and changing it to 0. 4. Wrap Probe 1 and Probe 2 with square pieces of filter paper secured by small rubber bands as shown in Figure 1. Roll the filter paper around the probe tip in the shape of a cylinder. Hint: First slip the rubber band up on the probe, wrap the paper around the probe, and then finally slip the rubber band over the wrapped paper. The paper should be even with the probe end. 5. Stand Probe 1 in the ethanol container and Probe 2 in the 1-propanol container. Make sure the containers do not tip over. The liquids are in bottles that will be reused by different groups so be sure not to contaminate them. Exp#2 IMF Rev Sp09AEM Spring 2009 Page 2 of 11

6. Prepare 2 pieces of masking tape, each about 10-cm long, to be used to tape the probes in position during step 7. 7. After the probes have been in the liquids for at least 30 seconds, begin data collection by clicking Collect. Monitor the temperature for 15 seconds to establish the initial temperature of each liquid. Without hitting stop, simultaneously remove the probes from the liquids and tape them so the probe tips extend 5 cm over the edge of the table top as shown in Figure 1. Make sure that the filter paper goes to the tip of the probe, but not past the tip. 8. When both temperatures have reached minimums and have begun to increase, click Stop to end data collection. Click the Statistics button,, then click OK to display a box for both probes. Record the maximum (T 1 ) and minimum (T 2 ) values for Temperature 1 (ethanol) and Temperature 2 (1 propanol) in the table. 9. For each liquid, subtract the minimum temperature from the maximum temperature to determine T, the temperature change during evaporation. 10. Roll the rubber band up the probe shaft and dispose of the filter paper in a beaker your instructor has placed in the hood. 11. Based on the T values you obtained for these two substances, plus information in the Pre -Lab exercise, predict the relative size of the T value for 1 -butanol. Compare its hydrogen-bonding capability and molecular weight to those of ethanol and 1-propanol. Record your predicted T, then explain how you arrived at this answer in the space provided. Do the same for pentane. It is not important that you predict the exact T value; simply estimate a logical value that is higher, lower, or between the previous T values. (e.g. > 4 C or < 8 C) 12. Test your predictions in Step 10 by repeating Steps 3-9 using 1-butanol for Probe 1 and pentane for Probe 2. 13. Based on the T values you have obtained for all four substances, plus information in the Pre - Lab exercise, predict the T values for methanol and hexane. Compare the hydrogen -bonding capability and molecular weight of methanol and hexane to those of the previous four liquids. Record your predicted T, then explain how you arrived at this answer in the space provided. 14. Test your prediction in Step 12 by repeating Steps 3-9, using methanol with Probe 1 and hexane with Probe 2. Exp#2 IMF Rev Sp09AEM Spring 2009 Page 3 of 11

Data Table Substance Tmax ( C) Tmin ( C) T ( C) (T max T min ) ethanol Explanation Predicted T ( C) 1-propanol 1-butanol pentane methanol hexane Questions 1. Take a moment to look at your T values. In the space below, organize your samples from smallest T to largest T for your alkanes, and again for your alcohols: Alkanes: < Alcohols: < < < 2. Discover what the T tells you about how readily a sample evaporates. Circle the appropriate response to each statement below. Large T values mean that the sample does does not evaporate readily. Small T values mean that the sample does does not evaporate readily. Exp#2 IMF Rev Sp09AEM Spring 2009 Page 4 of 11

3. Correlate your answers to how readily a sample evaporates to the strength of the intermolecular forces in that molecule. (again, circle the appropriate response to complete the statement.) If the sample evaporates readily, then the relative strength of the IMFs in the molecule is: (circle one) strong weak. If the sample does not evaporate, then the relative strength of the IMFs in the molecule is: (circle one) strong weak. 4. Looking at the formulas for your alkanes, what is the only difference between formulas? 5. What IMF(s) are present in the alkanes (list all present)? 6. Looking at the formulas for your alcohols, what is the only difference between the formulas? 7. What IMF(s) are present in the alcohols (list all present)? Exp#2 IMF Rev Sp09AEM Spring 2009 Page 5 of 11

Using your answers to the guided questions on page 5, answer the following questions: 1. Using Excel, plot a graph of T values of the four alcohols versus their respective molecular weights. Plot molecular weight on the x-axis and T on the y-axis. Label each point with the identity of the alcohol. From the graph, comments on the relationship between the strengths of intermolecular forces and a molecule s molecular weight: As molecular weight increases, the strength of a molecule s IMF. Explanation: 2. Two of the liquids, pentane and 1-butanol, had nearly the same molecular weights, but significantly different T values. Based on their intermolecular forces, explain why there was a difference in T values of these substances. Be specific. 3. Which of the alkanes studied has the stronger intermolecular forces of attraction? The weaker intermolecular forces? Explain using the results of this experiment. Include in your explanation comments about the strengths of their IMFs. 4. Which of the alcohols studied has the strongest intermolecular forces of attraction? The weakest intermolecular forces? Explain using the results of this experiment. Include in your explanation comments about the strengths of their IMFs. Exp#2 IMF Rev Sp09AEM Spring 2009 Page 6 of 11

Part 2: A Study of Solid-Liquid Transitions Purpose Study the relationship of temperature to the transition of matter between the liquid and solid states. Discussion Every compound has certain physical and chemical properties that are unique to that compound. Two physical properties that are commonly recorded for a compound are boiling point and freezing or melting point. Today, you will study the latter. The melting and freezing points for a compound occur at the same temperature. This temperature is the point where the compound's solid and liquid states exist in equilibrium with each other with no change in temperature (at constant pressure). If the compound is turning to a liquid we talk about a melting point, and if it is turning solid we call it the freezing point. freezing point Liquid Solid Liquid Solid melting point As long as both the solid and liquid states are existing together, the temperature remains constant. However, as soon as one of the states disappears, the temperature will change. All solids absorb heat as they melt. The heat of fusion is the amount of heat required to melt a unit quantity of substance (usually 1 g or 1 mole) with no change in temperature. For water this value is 6.01 kj/mole or 79 cal/g. We know that at the melting point no change in temperature occurs. If the solid is absorbing heat and no rise in temperature is occurring, how then is this heat energy being used? In order to understand this we need to first discuss the crystal properties of solids. Most solid substances are crystalline. This means that in the solid state the atoms, ions or molecules of the compound are arranged in a definite three-dimensional order. They are held in this order by intermolecular forces holding the molecules (or atoms/ions) together. When we begin to heat a crystalline solid, the molecules (or atoms/ions) take up the heat energy until they finally gain enough energy to overcome the attractive forces holding them in the crystal structure. When a molecule gains sufficient energy it breaks free of the crystal and becomes part of the liquid state. If the heating is continued, eventually all the molecules in the solid will gain enough energy to break free and the solid turns to liquid. At this point, any additional heat energy will raise the temperature of the liquid. Conversely, if no additional heat is added to the liquid it will begin to crystallize. As it crystallizes, the liquid gives off heat energy. This heat energy keeps the temperature constant (i.e. at the freezing point) until all the liquid crystallizes. Then the temperature begins to drop. Sometimes it is possible to cool a liquid using an ice bath to a temperature that is below its normal freezing point. A liquid existing in this state is said to be supercooled. This state is not stable and crystallization will occur if the solution is stirred or if a "seed" crystal is added. The seed crystal induces crystallization by providing a pattern of the crystal structure for the liquid compound to line up with. As soon as crystallization begins, heat energy is given off and the temperature rises to the normal freezing point and remains there until all the liquid has solidified. Exp#2 IMF Rev Sp09AEM Spring 2009 Page 7 of 11

Procedure In the experiment today you will measure the freezing point of a liquid and the freezing point of a supercooled liquid using the same temperature probes you used in Part 1. One partner should start at the Experimental Set-Up while the other does the Computer Set-Up. Computer Set-Up 1. Verify that the computer is still set up for Experiment 9. 2. Under the Experiment menu open Data Collection. Choose the Collection tab. Change Length to 30 minutes and collect at a sampling speed of 5 per minute. Press Done. 3. Change the temperature scale on the y-axis to read 0 to 100 C by clicking on the minimum and changing it to 0 and then the maximum and changing it to 100. You should change the x-axis to read from 0 to 30 minutes. Experimental Set-Up 1. Obtain two test tubes from the lab bench which each contain about 10 g of hydrated sodium thiosulfate crystals (Na 2 S 2 O 3 5 H 2 O). 2. Fill a 250 ml beaker about 1/2 full with cold (use 3-5 ice cubes) water for use later. On a small piece of weighing paper, obtain two seed crystals of sodium thiosulfate pentahydrate. 3. Do this part at your lab bench, not next to the computers. Disconnect the temperature probes from the interface and insert them into the test tubes. Heat each test tube in a hot water bath, stirring gently with the temperature probes until the crystals in both tubes are melted. If the sodium thiosulfate hydrate begins to crystallize, you must reheat the test tube. Be careful to keep the wires from the probes away from the hot plate or the burner. You may use a clamp on the ring stand to suspend the wires. Be careful melting the samples, monitor them closely. No crystals should be present once melted, however, if the crystals are heated too hot, dehydration occurs and the melting point changes. Do not contaminate the materials with water or other substances. The next few steps must be done smoothly and quickly in order, MAKE SURE YOU READ AHEAD AND ARE PREPARED! 4. Clamp the test tubes to a ring stand. Move the ring stand with the test tubes clamped onto it, to the computer and reconnect each temperature probe. The temperature of both should be about 60 C. If the liquid begins to crystallize you will need to reheat it gently. You may do this with the temperature probe in the test tube by unplugging it from the interface box. Be careful not to heat the wires. 5. Press Collect to begin to collect data. 6. After about 30 seconds, raise the beaker of cold tap water around one of the test tubes (#2). This will be your supercooled sodium thiosulfate pentahydrate. 7. When the temperature of the supercooled liquid reaches about 30 C (do not stir the supercooling tube or you may cause crystallization), remove the beaker of water first, then drop a seed crystal into the supercooled liquid and begin stirring both test tubes (gently of course). NOTE: Do not attempt to move the thermometer when it is frozen in the sodium thiosulfate pentahydrate. As the material solidifies, you may not be able to stir very much. Exp#2 IMF Rev Sp09AEM Spring 2009 Page 8 of 11

8. When the temperature of the liquid in test tube #1 (non-supercooled liquid) falls to 51 C, drop in a seed crystal of sodium thiosulfate to"seed" the liquid, thus aiding crystallization and preventing supercooling. Continue stirring (gently of course). 9. Continue collecting data for the time listed on the graph (30 minutes); until the temperature drops to ~ 40 C in both test tubes, or until the cooling curves level out and remain about level for 5 minutes 10. Stop the collection process. Print your graph of the data and attach it to the end of the experiment. (File Print Graph) Be sure to label the graph and indicate the identity of each curve. (by hand is ok) Analyze Data A. To determine the freezing temperature of the sodium thiosulfate pentahydrate, you need to determine the mean (or average) temperature in the portion of graph with nearly constant temperature. Move the mouse pointer to the beginning of the graph s flat part for the non-supercoooled substance. Press the mouse button and hold it down as you drag across the flat part of the curve, selecting only the points in the plateau. Click on the Statistics button,. Select the correct temperature probe for the nonsupercooled curve. The mean temperature value for the selected data is listed in the statistics box on the graph. Record this value as the freezing temperature of the non-supercoooled sodium thiosulfate pentahydrate below. Click on the upper-left corner of the statistics box to remove it from the graph. Analyze Data B. Repeat this process for the supercoooled sodium thiosulfate pentahydrate, Record below. 11. When you are finished, heat the solids gently to remove the temperature probes. The solid should be left in the test tube and returned to the reagent table to be used by the next group. Rinse the temperature probes with water and dry them carefully to remove any remaining solid. Questions: 1. Melting/freezing point the of non-supercooled liquid is C 2. Melting/freezing point of the supercooled liquid is C 3. Are the freezing/melting points of the non-supercooled and supercooled liquids approximately the same? Comment on any differences between the two values. 4. When the seed crystal was added to the liquid, record your visual observations of what happened to the solution in the test tube: 5. A. Did the temperature of the supercooled liquid rise or fall upon addition of a seed crystal? B. Was this an exo- or endo-thermic process? 6. What would the final temperature of both test tubes be if we let them sit for 8 hours? Exp#2 IMF Rev Sp09AEM Spring 2009 Page 9 of 11

pre-lab stamp here 1. Complete the table: Intermolecular Forces and Solid-Liquid Transitions Substance Formula Structural Formula (condensed structural formula is ok) Molecular Weight (g/mol) Hydrogen Bond (Yes or No) ethanol C 2 H 5 OH 1-propanol C 3 H 7 OH 1-butanol C 4 H 9 OH pentane C 5 H 12 methanol CH 3 OH hexane C 6 H 14 2. Which will evaporate faster, 1-butanol or pentane? Explain your choice. 3. State the following as either exo- or endo-thermic processes: Melting Vaporization Sublimation Freezing Condensation Deposition turn over for the last pre-lab question... Exp#2 IMF Rev Sp09AEM Spring 2009 Page 10 of 11

4. A thermometer is placed in a test tube of chipped ice at -5.0 C. The temperature is recorded at the time intervals shown below until room temperature is reached. Plot the data on graph paper (or using a graphing program like excel). Include a title, labeled axes, and write an explanation for what phases are present/what is occurring for both flat portions of the curve. Please write the explanation ON the plot above the horizontal regions. Plot time on the x-axis! Time (min) Temp ( C) Time (min) Temp ( C) Time (min) Temp ( C) 0-5.0 25 0.0 60 19.0 2-2.5 30 1.5 65 20.0 4-1.0 35 4.0 70 20.0 6 0.0 40 8.0 75 20.0 10 0.0 45 11.5 80 20.0 15 0.0 50 15.0 20 0.0 55 17.5 Exp#2 IMF Rev Sp09AEM Spring 2009 Page 11 of 11