Laboratory 3 Freezing Point Depression Colligative properties of solutions depend only on the concentration of the solution. Freezing point depression is an example of a colligative property. Drivers on icy streets depend on it. Salt placed on the road in the winter, dissolves into the snow or ice and lowers the freezing point of water, preferably below the road’s temperature. Pure water freezes at 0o C. If one mole of dissolved particles is added to one kilogram of water it dissolves at -1.86 oC. This gives the freezing point constant for water, kfp = 1.86oC/m, where m is molality of the solution. Molality is defined as the moles of solute per kilogram of solvent. When salt is dissolved in water, two ions are present in solution, Na+ and Cl-, so the number of particles is two, and if one mole of sodium chloride is dissolved in one kilogram of water the freezing point of the water drops from 0 oC to -3.72 oC. If one mole of sugar is dissolved in one kilogram of water the freezing point will be -1.86 oC because sugar does not dissociate and gives only one particle in solution with water. So it is clear why salt is used instead of sugar on icy roads. This freezing point depression constant is only good for water. Each solvent has a different kf. The general equation for freezing point depression is: ΔTfp = i kfp m (1) where ΔTfp is the freezing point change, kfp is the freezing point depression constant and m is the molality of the solution, and i is the van Hoff't factor which tells the number of particles in solution. For NaCl this is 2. For sugar this is 1. Molality is as the unit of concentration instead of molarity. Molarity has the units moles solute/liter solution. Liquids tend to change volume when the temperature changes therefore molality, which is a temperature independent unit is used. In this experiment first we will determine the freezing point of the pure material, stearic acid. Then we will determine the freezing point depression constant, kfp, of this material using a known substance, and finally we will determine the molar mass of an unknown substance, by its freezing point depression of stearic acid. The normal freezing point is determined by removing heat from a substance. In theory when heat is removed from a liquid the liquid will decrease in temperature until it reaches the freezing point. The liquid then stays at the freezing point until it becomes a solid, and then the temperature of the solid decreases. The freezing point is determined to be the constant temperature or the flat part of the cooling curve. In practice when you remove heat from a liquid, 21 there is a fast decrease in temperature for the liquid, followed by a much flatter, but still sloped section where freezing occurs. The experimental cooling curve for stearic acid is given in Figure 1. Cooling Curve 95 90 85 Temperature (Celsius) y = -0.105x + 89.378 80 Freezing Point is the intersection of the two lines 75 70 y = -0.0085x + 66.2 65 60 55 0 100 200 300 400 500 Time (seconds) For a pure substance typically a very nice cooling curve is obtained. Solutions may exhibit more complicated cooling curves. An example of cooling curve for a solution is shown below: 22 600 Cooling Curve 2 (Stearic Acid with Palmitic Acid) 85 80 y = -0.11x + 80.667 Temperature (Celisus) 75 Freezing point is the intersection of the two lines 70 65 y = -0.0089x + 63.694 60 55 0 100 200 300 400 500 600 Time (seconds) Because it is difficult to obtain a good cooling curve for the solutions, be sure that they are well mixed and 100% dissolved, and measure them very carefully. For both of the cooling curves given above, two straight lines have been fit to the data, one for the portion where the liquid is cooling and one for the portion where freezing occurred. A consistent method for determining the freezing point is to use the temperature at which these lines intersect, which is indicated by the vertical arrow on both plots. The minor grid lines have also been included to make it easy to read the temperature. Procedure: The freezing point of stearic acid 1. Prepare a hot water bath in a 250 mL beaker. Place it on the hot plate and begin to heat it. Take a large test tube and add 3 grams of stearic acid. Measure the mass exactly. 2. Using your test-tube holder, put the test tube into the hot water bath. Put your thermometer into the test tube, not the water bath. 3. When the stearic acid has completely melted, remove it from the hot water bath and insert the test-tube into a block of Styrofoam, which has a hole for the test tube in it. 4. Record the temperature every 30 seconds until the stearic acid has completely solidified. 23 700 Repeat the procedure using the same sample of stearic acid. Plot the cooling curve for stearic acid, temperature (y axis) versus time for both cooling curves. You can plot the data using EXCEL, but however you plot the data you must read the value of the freezing point to one decimal point. Directions for making a plot with EXCEL First start the program, in a windows system usually you use a double click. The program opens to a fresh sheet. To move between sheets, click on the tabs on the bottom of the page. The first step is to input some of your data. The easiest method to keep everything straight is to use different sheets for your different cooling curves. Use the first sheet to analyze your freezing point of pure stearic acid. In the first column enter the time. Just type in the number and hit return. In the second column, type the temperature. Now plot the line. To make a graph in Excel 97, I have found the easiest method is to make sure your x values are in one column and your y values are in the next column. Using the mouse highlight your data, x and y for your standard curve. Then click on the chart wizard icon, it looks like a bar graph. A series of 4 dialog boxes open to make the chart. Dialog Box 1: First the chart wizard asks what type of graph you want. Click on xy plot, on the right side of the window it asks what type of xy plot you want. Click on the style of graph you like. For this graph you will sketch your own line so just use plot points without a line. Dialog Box 2: Now the chart wizard asks if the data you want plotted is A1:B5. If the range is what you want hit the next button, otherwise edit the range of data, and then hit the next button. Dialog Box 3: gives you the opportunity to add legends or labels if you like. So you will be able to read the temperature values accurately, click on gridlines and check the box for minor gridlines for the y-axis. When you are happy with your graph, click on next. Dialog Box 4: gives you a choice of placing the chart on the page with the data or on a new page. Decide which you want, typically a new page is neater. Click finish, and a graph of time versus temperature appears. You should see two regions with two different slopes. The intersection of these two lines gives the freezing point of pure stearic acid. Take the average of the two freezing points to get the freezing point of pure stearic acid. If your two values differ by more than 2o C take another cooling curve. You can compare this value with a literature value from either the Aldrich catalogue or the CRC Handbook. 24 Part 2 Freezing point depression constant of stearic acid Weigh about 0.5 g of palmitic acid. Record the mass to the nearest milligram. Add this to your test tube containing stearic acid. Calculate the molality of the solution. Mix the solution thoroughly. Using the same procedure as above, take a cooling curve for this solution. Using your test tube holder put the test tube into the waterbath. As the solution dissolves use a stirring rod to mix it thoroughly. Remove the liquid solution from the hot water bath and place it in the insulated block. Again record the time and temperature every thirty seconds until the solution is completely solid. Thaw the solution and repeat the procedure. The freezing point of the solution is the intersection of the two lines with different slopes. Plot both cooling curves and determine the freezing point of the solution. Average the two measurements to get the freezing point; remember it should be lower than that of pure stearic. Once again if your two freezing points differ by more than 2o C, you should repeat the measurement one more time, or if your average freezing point is higher than that of pure stearic acid. Now we can use equation 1 to determine the freezing point depression constant for stearic acid. Determine the molality of the solution, and then the freezing point depression constant. Discard the solution in the appropriate waste container. Part 3 The molar mass of an unknown compound Using a clean the test tube. Add 3 grams of pure stearic acid. Record the number of your unknown in your lab notebook. Add about 0.5 g of your unknown, recording the mass to the nearest milligram. Determine the freezing point of your unknown solution by again recording two cooling curves. ΔTfp = the freezing point of pure pure stearic acid minus the freezing point of the unknown solution. Using the average of your two freezing points determine ΔTfp. The molality of your solution, m = ΔTfp / kfp , where kfp was determined in part 2. Determine the molality of your unknown solution. The molality is moles solute/kg solvent. Determine the moles of unknown solute in the solution. This is the number of moles present in your test tube. The molar mass of your unknown is grams/mole. Divide the recorded mass of your unknown by moles unknown to determine the molar mass. Hints for organizing data and performing calculations: You could set up one big data table with all the data you need to collect. Pure Stearic Acid MassInitial stearic acid = Palmitic acid Masspalmititc acid added = 25 Unknown MassUnknown Run 1 Time Run 2 Temp Time Run 1 Temp Time Run 2 Temp Time Run 1 Temp Time Run 2 Temp Time Temp Plot both pure runs on the same graph using different colors or x's and o's. You want to do this while taking the data, or immediately after so you can see if there is any problem. Determine the freezing points and average the two runs. Plot your other runs in the same fashion. In your notebook write stearic acid acid with Palmitic acid with Unknown Tf = Here put the values with the correct significant figures in each column. For palmitic acid and your unknown: i = 1, neither will dissociate in stearic acid ΔTf = the difference of the freezing point of each solution and pure stearic molality for palmitic solution is: moles palmitic acid/kg stearic acid = mpalmitic acid and now for stearic acid kfp , = ΔTfp palmitic acid /mpalmitic acid Using this kfp, determine the molality of your unknown solution. munknown = ΔTfp unknown / kfp To get the moles of unknown from the molality of the unknown solution remember m=moles solute/kg solvent. Just multiple the molality by the kg of stearic acid used. From this molar mass unknown, MM, is: MM = massunknown / molesunknown Question: Based on the information in the introduction how much NaCl in grams is needed to lower the freezing point of 1.2 liters of water 2.5oC? How many grams of CaCl2 would be needed? Safety and Disposal: The stearic acid, palmitic acid and unknown solutions should be placed in the waste container provided. To get the last bit out of the test tubes, rinse with hot isopropanol. 26 Measurement of Physical Properties In any measurement it is important to know the precision of the measurement and also its accuracy. All physical measurements should be made as precisely and accurately as possible. Maximizing the precision of a measurement is accomplished by using the most precise equipment available, and using it properly. If possible it is also wise to compare your result with either a known or theoretical value. Significant Figures When calculating a result from more than one measurement it important to retain the uncertainty information from all the measurements. There is an entire field of mathematics devoted to this topic. In this course we use the relatively simple method of significant figures. A summary of the rules with examples: Addition and subtraction: line up the numbers to be added or subtracted; the answer is truncated to the decimal place of the least precise number. Ex. 12.1 + 2.345 = 14.4 15.678 - 2.2 = 13.5 (notice I rounded up) Multiplication and Division: Significant Figures in the answer are equal to the number of significant figures in the least precise number. 15.6 x 2.1 = 31 16.789 ⎟ 25.67432 = 0.65392 25.1 x 3.00 = 75.0 Note zeroes before another number as in 0.65392 do not count. In the middle and the end they count. Laboratory Notebooks In this course you will be required to keep a laboratory notebook. A good laboratory notebook is an accurate record of everything, which occurred in the lab. In patent disputes a good lab book versus an inaccurate lab book can mean millions of dollars. In this course it may mean hundreds of points. Before the lab you will be required to prepare a lab report outline to be completed during the lab session. Each lab report will contain: 7 Title and Purpose 1. Procedure and Observations 2. Data and Calculations 3. Results and Conclusions 4. Answers to questions in manual An example of a lab report is shown below: 1. Title: Density of liquid and a solid. Purpose to measure the density of a liquid and an unknown solid. 2. Procedure: Observations Part 1 Liquid Unknown # 5 smells like gasoline Weigh an empty 10.0 mL volumetric flask Fill with unknown liquid. Weigh filled volumetric flask Mass of empty flask = 12.032 grams Mass of full flask = 18.685 grams Part 2 Solid Part 2 Unknown #12 Shiny orange color Fill a graduated cylinder with about 25 mL of water Measure precisely volume of water. Volume of water = 24.83 mL Volume of water + metal = 28.53 mL Mass of Dry metal = 46.409 grams weigh dry solid sample Data and Calculations: Part 1 Liquid: Density = Mass / volume Mass of liquid = mass of liquid + flask - mass of flask = 18.685 grams - 12.032 grams = 6.652 grams. Density = 6.652 grams / 10.000 mL = 0.6652 g/mL Part 2 solid volume of solid = volume of solid + water - volume water = 28.53 mL - 24.83 mL = 3.70 mL density = 46.409 g/3.70 mL = 12.5 g/mL Results and Conclusions: The density of the liquid was determined to be 0.6652 g/mL by comparison with the density table in the CRC it appears the sample could be hexane, which has a density of 0.660 g/mL The density of solid was 12.5 grams / mL. The solid looked like copper, but the density of copper from the CRC is: 8.94 g/mL, which is significantly less than my unknown sample. Therefore although the sample looks like copper it must be something else. 8 In this example, the data is recorded in the section with the observations, and the procedure is recorded in one column and the observations are recorded in an adjoining column. This allows you to record your observations with the correct section of the procedure. In some experiments, the type and volume of data is better recorded in a table. In this case it should follow the procedure section. You should still leave room in the procedure section for observations. One of the objectives of this course is for students to learn how to determine what data they need to collect, and how to organize it. For some experiments explicit instructions for organizing the data and calculations will be given, but for other experiments you will need to determine this for yourself before class. In the case of repetitive calculations tables are necessary. A spreadsheet such as Excel can be used, and instructions are included for the Reaction Rate experiment. All your calculations must follow the rules for significant figures and every value must have a correct unit. A spreadsheet or calculator will not determine the correct number of significant figures; it is up to you. When determining the results and conclusions, there are some things to keep in mind. The results should relate back to the purpose. Address directly if the purpose was fulfilled. If the result is a number clearly restate what it is and the unit for the number. If possible compare your result with a literature value. If you received no result or an unexpected result, give some scientific explanation of this. Human error is not a good explanation, because the experiment or section, which was in error, should be repeated. Thoroughness is important but it is not necessary to write everything you know about density or volume etc. To be ready to use all the lab time efficiently, before lab class you should have completed the purpose, procedure and arranged the data table or written down what you need to measure. Lab Instructors may have additional report requirements. 9
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