Molar Mass by Boiling Point Elevation PDF

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Lab Report Molar Mass...

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Carmen Belk Professor Brian Reid CHEM-1412.035 09 February 2021 Lab Report - Molar Mass by Boiling Point Elevation

PRE-LAB Questions 1. Define the term “colligative properties” and give examples. Colligative properties are the physical properties of a solution. Vapor pressure lower, boiling point higher, freezing point lower, and osmotic pressure. 2. What is molality?

3. Calculate the molality of a solution that contains 1.875 g of potassium chloride (KCl) in 175g water

4. How much will the boiling point of water change when 2.00 g of urea (60.1 g/mol) is

dissolved in 150 g of water?

5. A solution of 100 grams of brucine in 1 kg chloroform freezes at –64.69 °C. What is the molar mass of brucine?

Objective In this experiment, the boiling point elevation constant of Ethanol is calculated. This is then used to determine the molar mass of an unknown solute.

Introduction The solution used in this experiment is a mix of Ethanol and Urea. A solution is defined as a solute homogeneously dissolved in a solvent, with the solvent usually being of a greater quantity. The solute used is Urea and the solvent used is Ethanol. In the second part of this experiment, the solvent is still Ethanol, and the solute is an unknown substance. This experiment also deals with physical properties and how they affect the freezing and boiling points of solutions. The physical properties are divided into extensive properties which are mass and volume, intensive properties like density and concentration, and colligative properties which only apply to solutions. Colligative properties are things like vapor pressure

lowering, boiling point elevation, freezing point lowering, and osmotic pressure. They depend on the amount of solute dissolved into a solvent, regardless of the type of solute. When a solute is dissolved into a solvent, it creates bonds between the solute and solvent that are harder to break compared to bonds between just the solvent molecules. This means more energy is needed to break the bonds then boil out the liquid solvent, which is why the boiling point goes up for a solution compared to pure solvent. This also means the freezing point will be lower for the solution than pure solvent because it’s harder to freeze a mix of chemicals than it is to freeze a pure chemical.

Procedure 1. First prepare a 90 °C water bath by heating 400 mL of water in a 600 ml beaker. 2. The mass of Urea (60.06 g/mol) necessary to make 50 mL of 0.5 molal solution in Ethanol (d = 0.789 g/ml) is 1.18g. Measure out three Urea samples of the calculated mass. 3. Measure 50.00 mL of ethanol and place it into a clean and dry 8 x 1-inch test tube. Place a small capillary tube (about 3-4 inches long) in the ethanol, open end down. The test tube is then fitted with a two holed rubber stopper with thermometer (or temperature sensor) inserted so that the tip of the thermometer is about an inch below the surface of ethanol, and an 8 inch length of glass tubing. The test tube is then clamped and immersed in the hot water bath and heated gently. 4. Use this set up to determine the boiling point of pure ethanol. 5. Use the same ethanol and determine the elevated boiling point when 0.5molal, 1.0molal,

and 1.5molal urea is added (measured in step 2). 6. Construct and record your data in a table form. 7. Construct a graph of T; boiling point (y-axis) vs. m; molality (x-axis). Show the equation of the line and calculate the slope. Make sure to properly label the axes and title the graph. 8. Get an unknown from your lab instructor. To a new batch of 50.0 mL ethanol, add a mass of the unknown equal to the mass of 0.5molal of Urea (calculated in step 2). Use the above procedure to measure the boiling point elevation of the unknown/ethanol solution. Calculate the molar mass of the unknown. 9. Given the density of ethanol (0.789 g/cm 3 ), the mass of ethanol used may be found and hence the mass of unknown in 1 kg of solvent. From these data and the molality, the molecular weight of the unknown may be determined.

Results Include data tables, graphs, statistics, and calculations. Must show ALL work in calculations. Calculation for mass of Urea needed to make 50 mL of .5 molal solution:

1.44 g of unknown dissolved in 30 mL of ethanol:

Amount of urea (molar mass = 60.06 g/mol) required to make 0.5, 1.0 and 1.5 molal Urea solution in 30 mL of ethanol (density of ethanol = 0.789 g/mL):

Kb according to graph: 2.52

Kb according to calculation: 2.53 Kb ethanol accepted value: 1.22 Kb Percent error: 106.6%

Change in temperature of unknown solution: 2.00 C

Molar mass of unknown using experimental value: 76.60 g/mol

Molar mass of unknown using actual value: 37.11 g/mol

Discussion The procedure said to determine the mass of Urea necessary to make 50 mL of 0.5 molal solution in Ethanol, so this is the first calculation shown in the results section of the report. This

number was not used later on, it was just solved per the instructions and used as a practice problem. The data given to students is from Urea dissolved in 30 mL of Ethanol, so the calculations below the data tables reflect that. For the first part of the experiment, the amount of Urea required to make 0.5, 1.0 and 1.5 molal Urea solution in 30 mL of ethanol was calculated to be .7108g, 1.422g, and 2.13g, respectively. The density of ethanol was taken and multiplied with the milliliters of ethanol, then that number was divided by 1000 and multiplied by the respective molality. This was multiplied by the molar mass of Urea to get each answer. The boiling points of the three Urea concentrations plus the boiling point of the pure solvent came out to be 76.4, 77.6, 78.8, and 80.2 degrees Celsius. These values were put in excel and graphed. After plotting a trendline and finding the formula of the line, the Kb was found to be 2.52 (Kb equal to the m value in the formula y=mx+b). The values were also calculated by hand using the (change in)Tb=m*Kb formula. The result was a Kb of 2.53. The difference of .01 is probably because of rounding. The accepted value of Kb is 1.22 (Eastman & Rellefson, 307), but the value found from experimentation is 2.52. These two values have a percent error of 106.6%, so the Kb found through experimentation is about twice the amount it should be. This is most likely due to the measurements being off. The boiling point of pure ethanol measured is 76.4 Celsius, but the accepted boiling point is 78.4 Celsius (Eastman & Rellefson, 307). This alone is enough to throw the calculations off. This could be caused by a faulty visual temperature reading or a faulty thermometer, or measuring the BP at the wrong time, like before it was really “boiling”. The last part of the lab involved using both the measured and actual Kb value to calculate the molar mass of an unknown solvent. The measured boiling point of ethanol is 76.4 Celsius

and the boiling point of the ethanol plus unknown is 78.4 Celsius, which means the change in Tb is 2.0 degrees. The molar mass of the unknown calculated with the experimental Kb value is 76.60 g/mol. The molar mass calculated with the actual Kb value is 37.11 g/mol. This discrepancy is because of the differences in Kb value.

Conclusion This experiment had bad measurements that provided a Kb value for ethanol that was far greater than what was to be expected. This interfered with the results of the unknown solute’s molar mass.

Works Cited

Eastman. E.D. and Rollefson, G.K. Physical Chemistry 1947 ed. McGraw-Hill p. 307...