Experiment 14 Lab Report Chem Molar Mass of of a Solid PDF

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Lab Report Molar Mass of a Solid...

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Molar Mass of a Solid

Owayne Haughton Lab Partner: Gary DeLancer, Fazia Rahman Course: Chemistry 1310 Instructor Name: Maysoon Saleh Lab Assistant: Michaela Johnson Date Experiment was performed: 2/8/18

Abstract: The purposes of this experiment were to observe and measure the effect of a solute on the freezing point of a solvent and to determine the molar mass of the unknown solute labeled “Eagles”. The colligative property of freezing point depression was the theory at work behind what was observed in the experiment trials. The unknown molar mass was calculated to be on average 96.65g/mol. The freezing point of the solvent was lowered by the addition of the nonvolatile solute to the Cyclohexane solvent. 1

Introduction: The purposes of this experiment were to find the molar mass of the unknown solute using the freezing point depression as the colligative property and measure the effect of the solute on the freezing point of the solution. The instruments used in this experiment included multiple sized beakers, a thermometer, a digital thermometer, a computer, a test tube, and a clamp. The 600ml beaker was used to hold and insulate the 400ml beaker containing the ice water bath. The thermometer was used to determine the freezing point of the solution. The digital thermometer and computer were used to collect the data of the freezing point change. The test tube was used to contain the cyclohexane and unknown solute. And the clamp was used to keep the test tube stable throughout the experiment. The clean 250ml beaker and test tube are weighed. Twelve ml of cyclohexane is then added to the test tube. The test tube (containing the cyclohexane only) is then submerged into the ice water bath to reach its freezing point, from cooling curve. With use of the software named LoggerPro on the computer, the data is plotted on a graph to show the temperature vs time needed to obtain the cooling curve for cyclohexane. In the second part of the experiment the mass of the cyclohexane is calculated by subtracting the mass of the beaker and test tube with the mass of the beaker, test tube, and cyclohexane. About .1g of unknown solute is then added to the test tube containing the cyclohexane. To determine the freezing point change between the freezing point of cyclohexane compared to that of cyclohexane + unknown solute, the test tube containing both is then submerged into the ice water bath. The lowest temperature that the solution reaches is then the freezing point after .1g of solute is added. That process is repeated as another .1g of solute is added to see how much the freezing point will drop again. This experiment is performed in order to obtain the data needed to calculate the molar mass of the unknown solute through finding the moles of the solute using the following equation  2

(∆Tf = kf x molality). The factors that affect the results of this experiment include not adding more ice to the ice water bath as the ice melts away. This is because the ice water needs to stay as close to its freezing point as it can so that the cyclohexane can get as close to its freezing point as it can. Other factors that may affect the results include not adding the exact amount of unknown solute weighed. If the solute weighed does not match up with the exact amount of solute transferred into the test tube, then the results obtained will be inaccurate. The hypotheses of this experiment are that the solute added should lower the freezing point of

the solution. As the equation states added will result in the

(∆ T f =k f x

moles of solute ) . The more grams of solute kg of solvent

∆ T f to become larger.

Materials and Methods: Please refer to page 189 of Laboratory Manual for Principles of General Chemistry by J.A. Beran. One deviation from the original experiment involved the amount of trials preformed. For the sake of time there were only two trials of this experiment preformed instead of three. There were no other deviations to this experiment.

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Data: A. Freezing Point of Cyclohexane Mass of beaker, test tube (g) Freezing point, from cooling curve (℃)

109.65g 7.1 ℃

B. Freezing Point of Cyclohexane plus “Eagles” Solute 1. Mass of beaker, test tube, cyclohexane (g) 2. Mass of cyclohexane (g) 3. Tared mass of added solute (g) 4. Freezing point, from cooling curve (℃)

Trial 1 117.89g 8.24g .1055g 4.8 ℃

Trial 2 117.89 8.24g .1020g 4.1 ℃

Calculations 1. Kf for cyclohexane (℃ x kg/mol) 2. Freezing-point change (℃) 3. Mass of cyclohexane in solution (kg) 4. Moles of solute, total (mol) 5. Mass of solute in solution, total (g) 6. Molar mass of solute (g/mol) 7.Average molar mass of solute (g/mol) 8. Standard deviation of molar mass 9. Relative Standard Deviation of MM

20 2.3 .00824 kg .00095 mol .1055 g 111.05 (g/mol) 96.65 (g/mol) 20.36 21%

20 3 .00824 kg .00124 mol .1020 g 82.26 (g/mol)

Calculations: (For Trial 1) 2. Freezing point of pure solvent – freezing point of solution (solvent + unknown solute) = 7.1 ℃ - 4.8 ℃ = 2.3 ℃

3. Mass of Cyclohexane (g) * 1 (kg) / 1000 (g) = Mass of Cyclohexane (kg = 8.24 g * 1 kg / 1000 g = 0.00824 kg Cyclohexane

4. Moles of solute = molality (mol/kg) * Mass of Cyclohexane (kg) = 4

(2.3 ℃* 0.00824 kg)/ 20 ℃ = 0.00095 mol

6. Molar mass of the solute = mass of the solute in solution / moles of the solute = 0.1055 (g) / 0.00095 = 111.05

7. Average molar mass of solute = (molar mass from trial 1) + (molar mass from trial 2) = (111.05 + 82.26) / 2 = 96.65 (g/mol)

8. Standard deviation of molar mass = = √(96.65 – 111.05)2 + (96.65 – 82.26)2 = 20.36MM

9. Relative standard deviation = standard deviation / average mean * 100 = 20.36 / 96.65 * 100 = 21%

Discussion: The purposes of this experiment were to find the molar mass of the unknown solute using the freezing point depression as the colligative property and measure the effect of the solute on the freezing point of the solution. The instruments used in this experiment included multiple sized beakers, a thermometer, a digital thermometer, a computer, a test tube, and a clamp. The 600ml beaker was used to hold and insulate the 400ml beaker containing the ice water bath. The thermometer was used to determine the freezing point of the solution. The digital thermometer and computer were used to collect the data of the freezing point change. The test tube was used

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to contain the cyclohexane and unknown solute. And the clamp was used to keep the test tube stable throughout the experiment. The clean 250ml beaker and test tube are weighed to be 109.65 g. Twelve ml of cyclohexane is then added to the test tube. The test tube (containing the cyclohexane only) is then submerged into the ice water bath to reach its freezing point, from cooling curve. With use of the software named LoggerPro on the computer, the data is plotted on a graph to show the temperature vs time needed to obtain the cooling curve for cyclohexane. The freezing point of the cyclohexane was found to be 7.1 ℃. In the second part of the experiment the mass of the cyclohexane is calculated by subtracting the mass of the beaker and test tube (109.65 g) with the mass of the beaker, test tube, and cyclohexane (117.89 g) and the result was a mass of (8.24 g) of cyclohexane. About .1055g of unknown solute was then added to the test tube containing the cyclohexane. To determine the freezing point change between the freezing point of cyclohexane compared to that of cyclohexane + unknown solute, the test tube containing both was then submerged into the ice water bath. The lowest temperature that the solution reached was 4.8 ℃ and therefore that was the freezing point of the solution as the solute was added. That process is repeated as another .1020g of solute is added to see how much the freezing point will drop again. The freezing point dropped again to 4.1 ℃. This information was then used to calculate the molar mass of the solute. This experiment is performed in order to obtain the data needed to calculate the molar mass of the unknown solute through finding the moles of the solute using the following equation (∆Tf = kf x molality). 1.

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Figure 1 shows the freezing point, from cooling curve as the freezing point (for the cyclohexane only) reaches 7.1 ℃ . The sudden drop in temperature as the test tube was submerged into the ice bath is a result of adding more ice to the ice bath.

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2.

Figure two shows the plotted data from the first trial of the experiment. This is the trial where the freezing points drops from 7.1 ℃ and reaches 4.8 ℃ . The sudden spike up in the graph which is a rise in temperature must have been due to taking the test tube out of the bath as it was conceived that the freezing point had been reached. The points plotted after 4.8 ℃ can be ignored.

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3.

Figure 3 shows the data obtained in the second trial of the experiment as the freezing point went from 4.9 ℃ to 4.1 ℃ . This graph shows how the temperature stabilizes at about 4.1 which is a clear sign that the freezing point was reached at 4.1 ℃. Conclusion:

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Based on the Laboratory Manual for Principles of General Chemistry the results were as expected. The experiment was preformed exactly as the manual says and the results were similar to the pre-lab question results in the manual. The fact that the temperature dropped as the solute was added means that the equation (∆Tf = kf x molality) is true and that the lab was performed correctly. Some errors that may have altered the results of the experiment include not maintain the freezing point of the ice bath by adding more ice when the ice melts away. Also, most but not all of the solute was added into the cyclohexane. Some was stuck on the walls and did not mix with the cyclohexane which must have thrown of the results a bit.

Laboratory Questions 1. Part A.3. Some of the cyclohexane solvent vaporized during the temperature versus time measurement. Will this loss of cyclohexane result in its freezing point being recorded as too high, too low, or unaffected? Explain. Too high, because the more solvent that is added results in the lower freezing point. If some of the solute is vaporized then there is less solvent in the solution and therefore a higher temperature. 2. Part A.3. The digital thermometer is miscalibrated by +0.15ºC over its entire range. If the same thermometer is used in Part B.2, will the reported moles of solute in the solution be too high, too low, or unaffected? Explain. Too low, because when multiplying the freezing point with the mass of the cyclohexane then dividing by the kf of hexane the result will be slightly smaller.

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3. Part B.1. Some of the solid solute adheres to the side of the test tube during the freezing point determination of the solution in Part B.2. As a result of the oversight, will the reported molar mass of the solute be too high, too low, or unaffected? Explain. Too low, because the mass/mol ratio will be lower than it should because not all of the mass was actually adhered for. 4. Part B.2. Some of the cyclohexane solvent vaporized during the temperature versus time measurement. Will this loss of cyclohexane result in the freezing point of the solution being recorded as too high, too low, or unaffected? Explain. Too high, because the more solvent that is added results in the lower freezing point. If some of the solute is vaporized then there is less solvent in the solution and therefore a higher temperature.

5. Part B.2. The solute dissociates slightly in the solvent. How will the slight dissociation affect the reported molar mass of the solute—too high, too low, or unaffected? Explain. * Slight dissociation will cause increase in van haff factor and therefore molar mass will be too high. 6. Part B.3, Figure 14.3. The temperature versus time data plot (Figure 14.3) shows no change in temperature at the freezing point for a pure solvent; however, the temperature at the freezing point for a solution steadily decreases until the solution has completely solidified. Account for this decreasing temperature.

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The remaining solution continues to become more and more concentrated and have a progressive lower freezing point as the pure solvent solidifies out of the solution. 7. Part C.1. Interpretation of the data plots consistently shows that the freezing points of three solutions are too high. As a result of this “misreading of the data,” will the reported molar mass of the solute be too high, too low, or unaffected? Too low because the higher the freezing-point change the lower the molar mass will be.

Works Cited Beran, J.A., (2008). Page 287-298 Laboratory Manual for Principles of General Chemistry

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