Lauric Acid

Freezing point depression: Lauric Acid Introduction: The purpose of this experiment is determine the freezing point depression constant for Lauric Acid by examining how solutes affect the melting/freezing point of a substance. The freezing point of a given substance is the temperature at which it converts from a liquid to solid. The impurities may get in the way of the normal molecular interactions. This causes more energy to be taken out of the system and so the temperature must lower. The solute present in the Lauric acid is benzoic acid. By using LabPro we will determine the freezing point of Lauric acid and the freezing point depression constant for the solution of Lauric acid and benzoic acid. Procedure: First, we set up a water bath consisting of a 400 mL beaker of water at about 60-70 °C. Place a test tube of pure Lauric acid n the beaker of warm water and allow it to melt. Once the Lauric acid is completely melted, place the temperature probe inside the test tube, place the test tube into a room temperature water bath, and begin collecting data in LoggerPro. As the Lauric acid begins to cool, the solution should be continuously stirred with the temperature probe as the data is collected. Repeat this experiment once more and we average the data from these two. We repeated this experiment with 0.750g BA/8.00 g LA and 1.50g BA/ 8.00 g LA. Once we obtain the data from each procedure, we analyze the freezing points of each. Equations: ΔT = Kf m Kf : freezing point depression constant ΔT: Lowering of the freezing point m: molality Observation/ Data Collection: The following graphs show the temperature change in the sample: 1. 8g Lauric Acid with 0.750g Benzoic acid. 2. 8g Lauric Acid with 1.50g Benzoic acid. Calculations: Discussion/Conclusion: The freezing point of the solution with a higher concentration of solute is lower than the solution with a lesser concentration of solute. In our experiment, the solution of Lauric acid and Benzoic acid with a concentration of 1.50g BA/8.00 g LA had a freezing point of 4.16 °C/m, and 0.750g BA/8.00 g LA had a freezing point of 4.68 °C/m. Errors could have originated if the temperature probe is not cleaned well enough. Also, error may have been occurred if stirring wasn’t continuous. The average freezing point depression constant was found to be 4.42 °C/m. This experiment was successful overall in teaching us how to determine the freezing point of substances with the presence of solutes. Worksheet: Freezing Point Depression 1. What is molality? Why is molality used in the calculation of the freezing point constant instead of molarity? Molality is defined as the number of moles of solute divided by the madd of the solvent. 2. Hydrogen bonds are (longer/shorter) than normal between H2O molecules in the liquid state. 3. As we try to determine the freezing point constant in this lab, what do we use as the solute? Banzoic acid 4. Define super – cooling? It is the process of lowering the temperature of a liquid below its freezing point without it becoming solid. 5. Draw a graph of time versus temperature representative of what you should obtain when freezing lauric acid. Label a) freezing point and b) super-cooling 6. You add 36 g of KCl (MW= 74.6 g/mol) to a beaker of 0.82 L of a substance (d = 1.5 g/mL) and allow it to dissolve. What is the molality of this solution? No of moles = 36g KCl x 1 mol/74.6g/mol = 0.48mol Molality = 0.48mol/0.82kg = 0.59mol/kg 7. Consider the solution in the previous problem. The normal melting point of the substance is 0.4 0C. Give the Kf of the solution if the freezing point of the solution (with the KCl) was −8.2 0C. Molality: 0.59mol/kg TF = -8.2 - 0.4 = -8.6 0C T/M = -8.6/0.59 = -14.58 0C/m 8. A student dissolves 60g of an unknown nonionizing solute into 1.50 kg of H2O. He experimentally determines that the freezing point is changed by 3 0C. What is the MW of the unknown solute? The Kf of water is 1.86 K/m KF = 1.86 K/mol T/M = KF 3/M = 1.86 K/m M = 1.61 mol/kg M = mol/1.50 = 1.61 mol/kg Mol = 2.42 mol 60g x 1/x = 2.42mol 24.79g = MW