Experiment 13 PDF

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practical lab report for EXPERIMENT 13: COLLIGATIVE PROPERTIES
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EXPERIMENT 13: COLLIGATIVE PROPERTIES

Table of Contents 1.1.1. Prelab 1.1.2. Purpose 1.1.3. Introduction 1.1.4 Safety Information 1.1.5 Experimental Raw Data and Observations 1.1.6 In-Lab 1.1.7 Results and Discussion 1.1.8 Lab Report Date: March 22nd, 2017

References: ScienceLab, compiler. 2003 May 21. Stearic acid MSDS [Internet]. Houston, Texas: Sciencelab; cited 2007 January 31. Available from: http://www.sciencelab.com/msds.php?msdsId=9927609 McCarthy, S. M., & Gordon-Wylie, S. W. (2005). A greener approach for measuring colligative properties. Journal of Chemical Education, 82( 1), 116. University of North Carolina at Chapel Hill Chemistry Department. C2016. General Chemistry 102L Lab Manual: Spring 2017. Plymouth (MI): Macmillan Learning Curriculum Solutions. pp. 23-29.

Purpose: The purpose of this lab is to measure the freezing-point depression in order to determine the quantitative molar mass of an unknown solute. This lab will demonstrate the effect of a solute on the freezing point of the solvent. Introduction: This lab uses different fatty acids to demonstrate colligative properties. Colligative properties, including vapor pressure depression, boiling point elevation, freezing point depression, and osmotic pressure, are properties that describe the behavior of a

solution. They are not dependent on the identity or type of solute, but rather they depend on the total quantity or concentration of the solutes. The major component of a solution is the solvent and the minor components are the solutes. This experiment focuses on the freezing depression, which refers to when the freezing point of a solution is lower than the freezing point of the pure liquid solvent. This is directly proportional to the molality (moles of solute per kilogram of solvent) given by the equation: ΔTf  = Kfm  ΔTf refers to the change in freezing point temperature, Kf is the molal-freezing point depression  constant for the solvent, and m  is the molal concentration of the solution. In ionic solutes or strong acids and bases, the molal concentration of the solution is the sum of the molality of all the dissociated ions. In this experiment, long-chain fatty acids are used. These are solids at room temperature and are weak covalent acids, so dissociation is not a factor. The molar mass of a solute can be calculated using the freezing point of a solution, the masses of solute and solvent, and the molal freezing-point-depression constant.

Experimental Process and Measurements Fill a 600 mL beaker about ¾ full with tap water and begin heating on a hot plate. Obtain 2 test tubes: one 18x150 mm and one 25x150. Insulte the smaller tube by wrapping it in paper towel and placing it in the larger tube. Remove the smaller tube and weigh it, recording the mass in the data table. Ad about 8 grams of stearic acid by filling the tube about ¾ full. Weigh the filled tube and record the mass. Set the sample in the hot water bath until the solid has melted. Once the sample has melted, place the thermometer in the test tube and continue heating until the temperature reaches 85 degrees celsius. Remove the tube from the hot water bath and place it back in the insulation apparatus. Carefully stir the sample with the thermometer and record the temperature every 30 seconds until the change is less than 0.2 degrees celsius per reading for 2 minutes. Repeat for an additional trial. Weight out about 0.7 grams of the unknown to the nearest 0.001 g and record the mass in your data table. Add the unknown to stearic acid in the test tube. Set the tube in the hot water bath and wait until the solution has melted. Continue heating until the solution reaches 85 degrees celsius. Place the tube in the insulation apparatus and stir the solution until the temperature changes less than 0.2 degrees celsius per reading for 2 minutes. Repeat for trials II and III. Use the same unknown for all three trials. For trial

II, a second portion of the unknown will be added. Therefore, the solution will contain about twice as much, and trial three will contain about three times as much. Find the literature values of molar mass, freezing point, and Kf for stearic acid. Generate individual graphs of Temperature vs. Time for each trial of the pure stearic acid. Determine the freezing point from the inflection point of the graph. Determine the mean freezing point and calculate the percent error. For the unknown solute trials, generate a scatter plot containing Temperature vs. Time data for each trial of the unknown solution. From the inflection points in the data sets, determine all three freezing points. Determine the change in freezing point using the mean experimental freezing point from part A. Calculate the molality of each trial and use this to determine the molar mass for each. Calculate the mean molar mass of the unknown and the CV for the three trials. Use the table to identify your unknown, and calculate the percent error for the molar mass. Experimental Report Information This lab includes work with hot water, so in order to minimize risk, this should be handled carefully with proper heat protectant equipment. Goggles, lab coats, and gloves should be worn at all times to prevent contact with the organic liquids. The information on hazards and precautions for used materials and the links for msds sheets are provided below: ●

Stearic acid: Slightly hazardous in case of skin contact (irritant), of eye contact (irritant), of ingestion, of inhalation.

Mechanics (PreLab Questions) 1. A cooling curve would be the inverse of the warming curve, with the curve starting at a high temperature and decreasing over time. Heating curves show the process of sublimation while cooling curves show the process of deposition. 2. We prevent supercooling by

Results Observations In lab The steric acid in boiling water appears to shrink in solid form down the tube.. Goes from powder to a crystalized appearance.. bubbles come off of solid inside the tube.Solid disappears and liquid is

clear in glass.As we stired and measured the temp of the liquid in the jacket insulated tube, it solidified. As soon as it was put back in water in wet back to clear liquid form. Same results for second trial.

Unknown used for (3) trials is: UNKNOWN 3 We predict with the next trial the freezing point will drop one degree After adding unknown and mixing to cool,.... the resulting liquid is cloudy

Source of error- minimal amounts of unknown each trial stuck to side of test tube and did not completely dissolve in solution Also, amounts that were weighed in beaker were not completely transferred to the tube.

Without unknown completely solid when cooled.. after trial three of unknown, the solution is a thick liquid, foggy substance

Table 1: Masses of Substances

Mass of stearic acid (g) Code for unknown:

8.03 Unknown 3

Mass of unknown, first addition(g)

.777

Mass of unknown, second addition (g)

.719

Total mass of solute, second addition (g)

1.50

Mass of unknown, third addition (g)

.742

Total mass of solute, third addition (g)

2.24

Mass of stearic acid was found by weighing the test tube and tarring the scale to find the mass of just the stearic acid.

Figure 1: Temperature vs. Time for Stearic acid with no Unknown

This figure shows the temperature vs. time plot for the solution of only stearic acid. The freezing point is found where the cooling curve levels off. For trial 1, it occurred at 69.2 ℃ and for trial 2 it occurred at 69.3 ℃.

Table 2: Freezing Point Calculations of Stearic Acid Trial 1 Freezing Point of Stearic Acid (℃)

Trial 2 69.2

Average Freezing Point (℃)

69.3

69.25

Reference value for freezing point (℃)

69.4 (msds stearic acid, pg. 3) 0.22 %

Percent Error (%) Reference value for Kf for stearic acid( *kg/mol)

4.50 ( J ournal of Chemical Education, pg 116)

Figure 2: Temperature vs. Time for Three Trials with Unknown This figure shows the temperature vs. time plot for the three trials conducted adding more unknown each time. The freezing point is found where the cooling curve levels off. For trial 1, it occurred at 68.0 ℃, for trial 2 it occurred at 66.0 ℃, and for trial 3, it occurred at 63.5 ℃.

Trial 1

Trial 2

Trial 3

Total mass of solute (g)

.777

1.50

2.24

Freezing point of

68.0

66.0

63.5

solution (

)

Delta T (

)

Molality (moles/Kg) Moles of unknown solute (moles) Calculated molar mass (g/mol)

1.3

3.3

5.8

0.289

0.733

1.30

0.00232

.00589

.0104

335

255

215

Mean molar mass (moles)

268

Standard deviation (moles)

61.1

RSD (%)

22.8

Identity of unknown

Palmitic Acid

Molar mass literature value (g/mol)

256.42

Molar mass percent error (%)

Sample Calculations Percent Error in the Freezing Point of Stearic Acid c alculated freezing point : 69.25 ℃ l iterature value freezing point : 69.40 ℃ % error = |measured − a ccepted|/accepted x 100 % error = |69.25 − 69.4|/69.4 x 100 = 0.22%

Delta Tf f reezing point of stearic acid − freezing point trial 1 69.25 ℃ − 68.0 ℃ = 1.25 ℃

4.52

f reezing point of stearic acid − freezing point trial 2 69.25 ℃ − 66.0 ℃ = 3.25 ℃ f reezing point of stearic acid − freezing point trial 3 69.25 ℃ − 63.5 ℃ = 5.75 ℃ Molality of Unknown Fatty Acid ΔT f = K fm F or T rial 1 : 1.25 ℃ = (4.50 ℃ * k g/mol)(m) m = 1.25 ℃ / 4.50 ℃ * k g/mol m = 0.278 mol/kg Molar Mass of Unknown m olality = moles solute / kg solvent F or T rial 1 : m ass of stearic acid (solvent) = 0.00803 kg m oles solute = molality * k g solvent = 0.287 mol/kg * 0.00803 kg moles solute = 0.00223 moles mass of solute = 0.777 g molar mass = grams/moles = 0.777 g/0.00223 moles RSD% of Experimental Molar Mass RSD % = SD /mean * 100 S D = 54.7 mean = 271.7 RSD% = (54.7/271.7) * 100 = 20.1% Molar Mass % Error calculated mean molar mass = 271.7 accepted molar mass = 280.45 % error = |measured − a ccepted|/accepted x 100 % error = |271.7 − 280.45|/280.45 x 100 = 3.12%

Discussion In Experiment 13, we compared the freezing point of stearic acid with that of solutions of stearic acid and an increasing amount of unknown fatty acid in order to find the molar mass and identity of the unknown. The first part of the lab measured the freezing point of stearic acid with no unknown. The graph produced from the data,

Figure 1, shows an exponential curve. The freezing point occurs where the graph levels off and the temperatures stop changing over time. I found an average freezing point of 69.25 ℃. The literature value for the freezing point of stearic acid is 69.4 ℃ (msds stearic acid, pg. 3) . Using this, I found a percent error of 0.22% (Table 1). Part 2 of the lab required running trials with increasing amounts of an unknown fatty acid. For each trial, we added about 0.8 grams of the unknown. Thus, the second trial contained twice as much unknown as the first, and the third contained three times as much. We compared the freezing points of each of these trials and determined that the freezing point dropped about 2 ℃ between each trial. As seen in Figure 2, trials 1-3 with the unknown produced exponential scatter plots, with freezing points of 68 ℃, 66 ℃, and 63.5 ℃ respectively. Using these values, we calculated the change in freezing point temperature and used ΔTf = Kfm to find the molality of each trial. I found the literature value of Kf to be 4.50 ℃*kg/mol (Journal of Chemical Education, pg 116). From this molality and our original mass of stearic acid, the solvent, we calculated a mean molar mass of the fatty acid. I found a mean molar mass of 271.7 g/mol with a standard deviation of 54.71 and a RSD% of 20.1%. Using Table 13-1 in the lab manual, I determined the unknown to be Linoleic Acid with a molar mass of 280.45 g/mol. I calculated a percent error of 3.12%. These values are seen in Table 2. By using calculated percent errors, I am confident in the accuracy of our results because the percent errors of our experimental values and the literature values were all within 5%. However, though I believe our results are accurate, the precision is not as high for the calculated molar masses, because the standard deviation was 54.71 with an RSD% of 20.1. This is relatively high and leads me to believe potential sources of error occurred. Possible sources of error could include inconsistent time measurements or inaccurate temperature readings. These could therefore lead to incorrect calculations. In addition, minimal amounts of the solutions solidified on the side of the test tube and were not touched by the thermometer. This caused loss of mass and uneven distribution of the unknown within the stearic acid. Thus I believe this is the most inherent source of experimental uncertainty. This would cause our molar mass to be lower than the accepted value. This loss of mass occurred in all three trials, however, the exact mass lost was inconsistent. Therefore, this primarily affected precision, as seen in the relatively high RSD%. The temperature could be read to one degree of uncertainty and the mass to 3 significant figures. This degree of uncertainty can potentially lead to a minimal decrease in accuracy and precision. However, with the values we received, I believe our results met reasonable amounts of accuracy and precision. Conclusion

This lab compared the freezing point of stearic acid with that of solutions containing an increasing amount of unknown. The freezing-point depression was used to calculate the molality and molar mass in order to determine the identity of the unknown. We calculated a mean molar mass of 271.7 g/mol and identified the unknown as linoleic acid. Post Lab Questions 1. Colligative properties are properties that depend on the concentration of the solute but not the identity of solute. In this experiment, we calculated the colligative property of freezing point depression. This is a colligative property because it is directly dependent on the concentration of the solute. We saw in this lab that as the concentration of the unknown increased, the freezing point depression decreased. This is different than something such as density because density considers the grams of the solute, which depends on the identity. In addition, colligative properties are only applied to solutions, and properties such as density and vapor pressure can be applied to any substance. 2. Adding chemical impurities to a pure solvent results in depression of the freezing point because an increase in particles disrupts the intermolecular forces. Therefore, less energy is needed to form them, decreasing the freezing point. Salt on the roads after snowfall acts as a solute that lowers the freezing point of the water. Thus, it prevents the melted snow from refreezing. 3. Super cooling refers the cooling of a liquid below its freezing point without a change in state. In order to prevent supercooling in our experiment, we used the thermometer to constantly stir the solution. This increased the entropy of the system and helped the solution to solidify. Based on our observations, we were able to prevent supercooling because we observed solidification in the test tube for all of our trials. 4. Equilibrium cooling means the slope of the curve is at a constant rate of zero. According to Figures 1 and 2, and the rest of the trials, each trial reached equilibrium cooling. If this had not occurred, the plot would not have a straight line and would raise and lower inconsistently when trying to reach equilibrium....