Experiment 5: Colligative Properties

Experiment 5 Colligative Properties GRADING RUBRIC EXPERIMENT 5 – LAB REPORT Worksheet (80 pts) Student name: ____ ________________________________ Lab partner name: TA name: ______ Day & time of the lab: ______2/28/22_____________________ The lab report packet for this experiment should consist of the following in THIS ORDER: The lab report packet for this experiment should consist of the following: 1. (10 pts) IN-lab notebook pages 1.1. Experimental procedure 1.2. Recorded experimental values and the file name (as saved on your USB drive) 1.3. Observations if any were noted 2. (70 pts) The WORKSHEET 3.1 (10 pts) Introduction 3.2 (45 pts) Results and discussion: Complete (full sentence) answers to the discussion question and properly formatted figures (DQ; Provide well- rounded answers composed with complete sentences for full credit. If you need to refer to any of the data or tables produced during the “Data Analysis” section, clearly indicate their location within your answers for the graders to find them readily.) 3.3 (10 pts) Post-Lab data analysis 3.4 (5 pts) References Complete this worksheet and turn in your digital lab packet to Canvas. Feel free to adjust space in this document as needed. The tables provided here are only an example of these may look like, feel free to replace the entire tables with your own creations using Excel, etc All tables and figures (graphs) must be accompanied by captions which should be positioned above tables and below figures. TURN IN the digital copy to Canvas on the due date by 11:59 pm. All electronic material must be completed by the due date to avoid late penalties. 1 (10 pts) IN-LAB notebook pages 2 3 Full credit for attaching all completed copy pages from your lab notebook. You must record everything: 4 ü detailed experimental procedure ü UNKNOWN NUMBER ü observations ü measurements recorded with all available significant figures and units as appropriate If something is deficient, inaccurate, or wrong, points will be deducted points. (10 pts) INTRODUCTION Introduction must be properly formatted and written with complete sentences in paragraph form, not in a numbered or bulleted list. For instructions on how to write an introduction correctly please reference the “Tips for Writing a Lab Report” handout on Canvas. (4 pts) Discuss the purpose and specific objectives of this experiment. This is similar to the first pre-lab question, but not identical (past tense, third person). (3 pts) Include and discuss at least one recent journal references (published within the last year or two) with research relevant to types of reactions or methods used in this lab. (3 pts) Include and discuss at least one historical perspective or application of the scientific principles or methods used in the lab. The purpose of this experiment was to identify two unknown electrolytes and two non electrolytes through monitoring the freezing point depressions. The ionic compound, or electrolyte, was either NaCl or KI, and the other nonelectrolyte compound was either fructose or sucrose. The results for all the freezing point depression were analyzed, corrected, and averaged before data was analyzed and compounds were identified. Additionally, error analysis were performed for the two compounds that were found during the experiment. One objective of this experiment were to measure the freezing point depressions fore pure water, ionic, and molecular compound in aqueous solutions. Another objective was to add custom linear trendlines in Logger Pro 3.14 to the freezing and cooling regions of the curves. Students were also able to identify unknown substances by calculating the molecular weight of the four unknown substances from the averaged normalized freezing points. The last objective was to compare the measured freezing points to theoretical freezing points after the unknown compound has been correctly identified Historical In an investigation of the freezing point depression of water in the presence or absence of cholesterol was measured through using NMR, nuclear magnetic resonance. Moreover, this experiment was conducted because it is well known that cholesterol plays a massive role within the cell membrane as it increases fluidity, but there are two types of water molecules within the cell membrane: bound and free water. Bound water molecules are bound to the heads of phospholipids near the surface of the bilayer. Moreover, bound water molecules are able to resist freezing and stay very mobile. Thus, this experiment was conducted to understand bound water molecule’s freezing point depressions within a cell membrane. In order to conduct this experiment, different temperatures were used to obtain results from a series of 2H quadrupole coupling NMR spectra. With that being said, the results show that with an increase of cholesterol, the freezing point decreases. This is similar to the colligative property experiment as when 5 salt was added to the ice bath the temperature lowered. Therefore, there is a similarity between cholesterol lowering the freezing point within the water in the cell membrane and salt lowering the freezing point of ice. There is an application aspect in a sense that the salt in the ice bath lowers the freezing point of water similar to how the cholesterol in the cell membrane lowers the freezing depression of water. Recent Co-precipitation synthesis of Bismuth Manganite was conducted through the assistance of an ice bath. In order to do this, precursor solutions were prepared through dissolving 0.1M of Bi(NO3).5 H2O and also 0.2 M of MnSO4.H2O in 50mL of distilled water. Additionally, 5 mL of HNO3 was added to the bismuth nitrate solution and it was mixed thoroughly. This is similar to the methodology that was conducted for colligative properties because after the molecular solid and ionic solid was added with the water, it was stirred. Additionally, 5 mL of the bismuth nitrate solution is collected with a volumetric pipette and placed in an ice bath. This similar to when the glass vial with the solution was placed in the ice bath. Another similarity with the synthesis of bismuth manganite is the stirring of the solution after compounds are added with water. Therefore, there are similarities in methodology. (45 pts) RESULTS and DISCUSSION Provide well-rounded answers composed with complete sentences for full credit. If you need to refer to any of the data or tables produced during the “Data Analysis” section, clearly indicate their location within your answers for the graders to find them readily. Results (8 pts) Include all relevant graphs and tables as specified in the Lab Manual. All included figures and tables must be cited by name in the main text. 6 Figure 1: A time versus temperature graph for Deionized Water with a cooling and freezing trendline and experimental freezing point. Figure 2: A time versus temperature graph for an unknown ionic solid with a cooling and freezing trendline and experimental freezing point. 7 Figure 3. Time versus temperature graph for an unknown molecular solid with a cooling and freezing trendline and experimental freezing point Discussion Questions 1. (3 pts) While setting up the ice-salt bath, if the bath was warmer than the desired temperature you were instructed to add more salt rather than more ice. What is the reasoning behind that? The temperature that was trying to be achieved was one between -10°C and -15°C which is lower than the freezing point of water. Salt, however, can lower the freezing point of water as when salt is added the delta G decreased because the entropy is increasing. Therefore, adding more ice is inefficient because in order to make the reaction more spontaneous, salt must be added to increase the entropy which lowers delta G. Furthermore, adding more ice can only make the temperature reach 0°C while the salt can make it go below 0°C. 2. (5 pts) Discuss the calculations performed in Steps 1-3 of the Data Analysis. Using their average normalized freezing points, what were the calculated molecular weights of Unknowns 1 and 2? The calculation performed in steps 1-2 of the data analysis are divided into three steps. The first step is to use the average values from unknown 1 in the formula 𝜟𝑻𝒇 = 𝒌𝒇 ⋅𝒎⋅𝒊 . The 𝜟𝑻𝒇 is found through subtracting 0°C from the average values given. The kf value is a constant given which is 1.86 °C /m and i, the vant hoff factor is 2, because both KI and NaCl have a vant hoff of 8 2. Moreover, these values are plugged into the equation, leaving the only variable to solve for being m, the molality. Therefore, the first step is to solve for the molality. The second step to find the molecular weight is to find the number of moles, given molality. The formula to find this is simply m = mol solute/ kg solvent. The kg solvent is 0.02 kg because 20 mL of solvent was used in the experiment, which 0.02 kg when multiplied by the density of water 1 g/mol. The previously calculated molality is then set equal to the moles of solute (unknown) divided by the kg of solvent, which then produces the moles of solute. The third step was to convert the moles of solute into molecular weight which is done through dividing grams of the unknown by the moles. After doing this the molecular weight will be calculated: 200 g/mol for unknown 1 and 100 g/mol for unknown 2. 𝜟𝑻𝒇 = 𝒌𝒇 ⋅ 𝒎 ⋅ 𝒊 Equation 1: 𝜟𝑻𝒇 = 𝒌𝒇 ⋅𝒎⋅𝒊 Kf = 1.86 c/m 𝜟𝑻𝒇 = change in freezing temperarture m = molality i = vant hoff factor (2) 𝜟𝑻𝒇 = 0.0 °C – (-1.7°C) = 1.7 °C 1.7 °C = 1.86 °C c/m x m x 2 1.7 °C / (3.72 c/m) m = .46 mol/kg Equation 2: m = (mol of solute) / (kg of solvent) Density of water = 1 g/mol Sample solvent = 20 mL 20 mL x 1 g/mol = 0.02 kg Mol of solute = .46 mol/kg .46 mol/kg = (mol of solute) / (0.02 kg) Mol of solute = .009 mol Equation 3 : (1.5 g) / (moles of solute) Mol of solute : .009 mol Molecular weight (g/mol) = 1.5 g / .009 mol Molecular weight of unknown 1 + 166.67 g/mol Molecular weight : 200 g/mol 9 3. (2 pts) Based on the known molecular weights of NaCl and KI, identify Unknowns 1 and 2. The known molecular weight of KI is 166 g/mol, and the molecular weight of unknown 1 is 200 g/mol meaning unknown 1 is KI. Additionally, the known molecular weight of NaCl is 58.44 g/mol and the molecular weight of unknown 2 is 50 g/mol meaning unknown 2 is NaCl. 4. (4 pts) Based on your assigned unknown from Part B (which has now been identified), use Equations 1 and 2 to calculate the “theoretical” freezing point of your Part B sample using the exact amount (in grams) that you recorded in Step 10 of the Procedure. Discuss this calculation. Embed your properly formatted equations into your lab write-up. The assigned unknown for Part B was unknown 2, and in order to calculate the ‘theoretical’ freezing point two equations must be used: m = mol solute/ kg and 𝜟𝑻𝒇 = 𝒌𝒇 ⋅𝒎⋅𝒊. Equation 2, m = mol solute/ kg solvent is used to find the molality which can be used for equation 1. In equation 2, volume is known (0.02 kg) because density of water (1 g/mL) is multiplied by the volume 20mL to get .02 kg. Additionally, the weighed amount of ionic solid from the experiment is used, which is 1.494 g. Using both the mass of the solvent and mass of solid, .02 kg, equation 2 can be used to find the molality which is simply diving the grams of solid over the kg of solvent. After getting molality, equation 1 can be used which is used to find the freezing point. In equation 1, Kf is given 1.86 ° C/m, molality was previously calculated, and the vant hoff factor is 2. With this information, the values plugged in will give the freezing point depression. Theoretical Freezing Equation 2: m = mol of solute/ kg solvent volume: 20 mL - 20 mL x 1 g/mL = 20 g/ 1g/mL = 20 g/ 1000 = .02 kg Known molecular weight : 55.84 g/mol Moles of solute : (1.494g) / (58.44 g/mol) = .02556 mol m = 0.02556 mol/ .02 kg m = 1.28 mol/kg Equation 1: 𝜟𝑻𝒇 = 𝒌𝒇 ⋅𝒎⋅𝒊 Kf = 1.86 °C/m m = 1.28 mol/kg i=2 𝜟𝑻𝒇 = (1.86 °C/m) ( 1.28 mol/kg) (2) = 4.76 ° C = 𝜟𝑻𝒇 = 5.0 °C 10 Freezing point of water = 0.0 °C 0 °C – 4.0°C = -5.0 °C 5. (2 pts) How does your “theoretical” freezing point from Discussion Question 4 compare to the “corrected” freezing point that you calculated in Step 22 of the Procedure? Use the following equation to determine the % error of your results: theoretical value − measured value 3 × 100 % theoretical value The theoretical freezing point from discussion question 4 compares to the corrected freezing point by having the theoretical freezing point being warmer than the corrected as the theoretical is -4.0 °C while the corrected freezing point is -8.4°C. The percent error was 110% because using the equation given: % error = 3 (#$.&)#(#(..)) % error = B #$.& B × 100 % = 68% 6. (5 pts) Discuss the calculations performed in Steps 4-6 of the Data Analysis. Using their average normalized freezing points, what were the calculated molecular weights of Unknowns 3 and 4? The molecular weight of unknown 3 is 100 g/mol and 300 g/mol for unknown 4. In order to do this the average values from unknown 4 in equation 1 𝜟𝑻𝒇 = 𝒌𝒇 ⋅𝒎⋅𝒊 . The 𝜟𝑻𝒇 is found through subtracting 0°C from the average values given. The kf value is a constant given which is 1.86 °C /m and i, the vant hoff factor is 2, because both sucrose and fructose have a vant hoff of 1. Moreover, these values are plugged into the equation, leaving the only variable to solve for being m, the molality. Therefore, the first step is to solve for the molality. In equation 1, all the values are there except “m”, which means kf, i, and 𝜟𝑻𝒇 are plugged into the equation to find “m". The second step to find the molecular weight is to find the number of moles, given molality. The formula to find this is simply m = mol solute/ kg solvent. The kg solvent is 0.02 kg because 20 mL of solvent was used in the experiment, which 0.02 kg when multiplied by the density of water 1 g/mol. The previously calculated molality is then set equal to the moles of solute (unknown) divided by the kg of solvent, which then produces the moles of solute. The third step was to convert the moles of solute into molecular weight which is done through dividing grams of the unknown by the moles. 11 7. (2 pts) Based on the known molecular weights of fructose and sucrose, identify Unknowns 3 and 4. The known molecular weight of fructose is 180.16 g/mol and sucrose is 342.3 g/mol. This means that by comparing the molecular weights from discussion question 6, the molecular weight was 100 g/mol for unknown 3 and 300 g/mol for unknown 4. Therefore, unknown 3 is fructose and unknown 4 is sucrose. 8. (4 pts) Based on your assigned unknown from Part C (which has now been identified), use Equations 1 and 2 to calculate the “theoretical” freezing point of your Part C sample using the exact amount (in grams) that you recorded in Step 17 of the Procedure. Discuss this calculation. Embed your properly formatted equations into your lab write-up. Based on the assigned unknown, unknown 4 which is sucrose, from Part C, the theoretical freezing point is -1.0 °C. In order to calculate this, equation 1 and 2, m = mol solute/ kg and 𝜟𝑻𝒇 = 𝒌𝒇 ⋅𝒎⋅𝒊, must be used. Equation 2, m = mol solute/ kg solvent is used for finding the molality which can be used for equation 1. In equation 2, volume is known (0.02 kg) as the density of water (1 g/mL) is multiplied by the volume 20mL to get .02 kg. Additionally, the weighed amount of ionic solid from the experiment is used, which is 4.988 g. Both the mass of the solvent and mass of solid, .02 kg, is used in equation 2 can be used to find the molality which is simply diving the grams of solid over the kg of solvent. After getting molality, equation 1 can be used which is used to find the freezing point. In equation 1, Kf is given 1.86 ° C/m, molality was previously calculated, and the vant hoff factor is 1. With this information, the values plugged in will give the freezing point depression. Theoretical Freezing Equation 2: m = mol of solute/ kg solvent volume: 20 mL - 20 mL x 1 g/mL = 20 g/ 1g/mL = 20 g/ 1000 = .02 kg Known molecular weight : 342.3 g/mol Moles of solute : (4.988g) m = mol of solute/ kg solvent - > (4.988g) / (342.3 g/mol) = .01457 mol m = 0.01457 mol/ .02 kg m = .73 mol/kg Equation 1: 𝜟𝑻𝒇 = 𝒌𝒇 ⋅𝒎⋅𝒊 Kf = 1.86 °C/m 12 m = .73 mol/kg i=1 𝜟𝑻𝒇 = (1.86 °C/m) ( .73 mol/kg) (1) = 1.36 = 𝜟𝑻𝒇 = 1.0 °C Freezing point of water = 0.0 °C 0 °C – 1.0°C = -1.0 °C 9. (2 pts) How does your “theoretical” freezing point from Discussion Question 7 compare to the “corrected” freezing point that you calculated in Step 22 of the Procedure? Using the equation above, determine the % error of your results. The theoretical freezing point from discussion question 7 compares to the corrected freezing point by having the theoretical freezing point being warmer than the corrected as the theoretical is -3.0 °C while the corrected freezing point is -1.5°C. The percent error was 50% because using the equation given: (#*.&)#(#*.$) % error = B #*.& B × 100% = 50% 10. (4 pts) Unlike a pure solvent, when cooling a solution to its freezing point, a process called supercooling can occur. Define supercooling. Did you observe supercooling in any of your trials from Parts A-C? If so, give a plausible explanation as to why this might have occurred experimentally. Supercooling is the process where the temperature of a liquid or gas goes below its freezing point without it becoming a solid. As a result, when plotting the data, it was observed that there was a supercooling because there was in an increase in temperature, which occurs when latent heat is given off when the liquid is in the process of changing from liquid to solid. This occurred during Part A and C. This could have occurred because of uneven stirring of the solution. 13 11. (4 pts) Discuss all observed or reasonably predictable experimental errors. What impact did/would those errors have on your experimental data? (be specific) This supercooling that occurred in Part A and C could have occurred because the temperature of the solution is not consistent through the solution which could occurred experimentally due to uneven stirring of the liquid. Additionally, another experimental error could have been the temperature probe picking up the supercooling temperatures rather than the overall temperature of the solution, which would cause a percentage of the error when calculating the freezing point depression in the data. Thus, an error in the temperature probe picking up the supercooling temperatures could have resulted in error. Furthermore, an uncalibrated analytical balance can produce wrong mass values in the ionic and molecular weights for the unknown substances. (10 pts) POST-lab DATA ANALYSIS PART B: The Freezing Point of an Unknown Ionic Solid – NaCl or KI 1. (2 pts) Using the average normalized freezing points (emailed to you from your TA) of both Unknowns 1 and 2, use Equation 1 to calculate the concentrations (in molality) of each unknown. The freezing point depression constant 𝒌𝒇 of water is 1.86 °C/m. 𝜟𝑻𝒇 = 𝒌𝒇 ⋅ 𝒎 ⋅ 𝒊 Equation 1: 𝜟𝑻𝒇 = 𝒌𝒇 ⋅𝒎⋅𝒊 Kf = 1.86 c/m 𝜟𝑻𝒇 = change in freezing temperature m = molality i = vant hoff factor (2) 𝜟𝑻𝒇 = 0.0 °C – (-1.7°C) = 1.7 °C 1.7 °C = 1.86 °C c/m x m x 2 1.7 °C / (3.72 c/m) m for unknown 1 = .46 mol/kg Equation 1: 𝜟𝑻𝒇 = 𝒌𝒇 ⋅𝒎⋅𝒊 Kf = 1.86 c/m 𝜟𝑻𝒇 = change in freezing temperature m = molality i = vant hoff factor (2) 14 𝜟𝑻𝒇 = 0.0 °C – (-5.2°C) = 5.2 °C 5.2 °C = 1.86 °C c/m x m x 2 5.2 °C / (3.72 c/m) m for unknown 2 = 1.40 mol/kg 2. (1 pts) Use Equation 2 to convert each concentration from Step 1 to number of moles for Unknowns 1 and 2. Equation 2: m = (mol of solute) / (kg of solvent) Density of water = 1 g/mol Sample solvent = 20 mL 20 mL x 1 g/mol = 0.02 kg .46 mol/kg = mol of solute / 0.02 kg Mol of solute for unknown 1 = .009 mol Equation 2: m = (mol of solute) / (kg of solvent) Density of water = 1 g/mol Sample solvent = 20 mL 20 mL x 1 g/mol = 0.02 kg 1.40 = mol of solute/ 0.02 kg Mol of solute for unknown 2 = .03 mol 3. (2 pts) Since the freezing point values in Step 1 were normalized to correspond to 1.5 g of unknown, determine the molecular weights of Unknowns 1 and 2 assuming a 1.5 g sample of each. Equation 3 : (1.5 g) / (moles of solute) Mol of solute : .009 mol Molecular weight (g/mol) = 1.5 g / .009 mol Molecular weight of unknown1 : 166.67 g/mol Molecular weight unknown 1: 200 g/mol Equation 3 : (1.5 g) / (moles of solute) Mol of solute : .03 mol Molecular weight (g/mol) = 1.5 g / .03 mol Molecular weight of unknown 2: 50 g/mol Molecular weight unknown 2: 50 g/mol 15 PART C: The Freezing Point of an Unknown Molecular Solid – Fructose or Sucrose 4. (2 pts) Using the average normalized freezing points (emailed to you from your TA) of both Unknowns 3 and 4, use Equation 1 to calculate the concentrations (in molality) of each unknown. 𝜟𝑻𝒇 = 𝒌𝒇 ⋅ 𝒎 ⋅ 𝒊 Equation 1: 𝜟𝑻𝒇 = 𝒌𝒇 ⋅𝒎⋅𝒊 Kf = 1.86 c/m 𝜟𝑻𝒇 = change in freezing temperature m = molality i = vant hoff factor (1) 𝜟𝑻𝒇 = 0.0 °C – (-3.3°C) = 3.3 °C 3.3 °C = 1.86 °C c/m x m x 1 3.3 °C / (1.86 c/m) m for unknown 3 = 1.77 mol/kg Equation 1: 𝜟𝑻𝒇 = 𝒌𝒇 ⋅𝒎⋅𝒊 Kf = 1.86 c/m 𝜟𝑻𝒇 = change in freezing temperarture m = molality i = vant hoff factor (1) 𝜟𝑻𝒇 = 0.0 °C – (-1.9°C) = 1.9 °C 1.9 °C = 1.86 °C c/m x m x 1 1.9°C / (1.86 c/m) m for unknown 4 = 1.02 mol/kg 5. (1 pts) Use Equation 2 to convert each concentration from Step 4 to number of moles for Unknowns 3 and 4. Equation 2: m = (mol of solute) / (kg of solvent) Density of water = 1 g/mol Sample solvent = 20 mL 20 mL x 1 g/mol = 0.02 kg 16 1.77 mol/kg = (mol of solute) / (0.02 kg) Mol of solute for unknown 3 = .04 mol Equation 2: m = (mol of solute) / (kg of solvent) Density of water = 1 g/mol Sample solvent = 20 mL 20 mL x 1 g/mol = 0.02 kg 1.02 mol/kg = (mol of solute) / (0.02 kg) Mol of solute for unknown 4 = .02 mol 6. (2 pts) Since the freezing point values in Step 4 were normalized to correspond to 5.0 g of unknown, determine the molecular weights of Unknowns 3 and 4 assuming a 5.0 g sample of each. Equation 3 : (5.0 g) / (moles of solute) Mol of solute : .04 mol Molecular weight (g/mol) = 5.0 g / .04 mol Molecular weight of unknown1 : 125 g/mol Molecular weight unknown 3: 100 g/mol Equation 3 : (5.0 g) / (moles of solute) Mol of solute : .02 mol Molecular weight (g/mol) = 5.0 g / .02 mol Molecular weight of unknown 2: 250 g/mol Molecular weight unknown 4: 300 g/mol (5 pts) REFERENCES You must include at least two outside references, not including the ones used for the lab manual. K. Annamalai et al 2021 ECS J. Solid State Sci. Technol. 10 061001 Langmuir 2008, 24, 23, 13598–13604 Publication Date:November 7, 2008 https://doi.org/10.1021/la8023698 17