Determination of an Unknown Liquid Lab PDF

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report for lab 1...

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Instructions: Please complete the following fields, corresponding to Experiment 1 in Chemistry 1000H. If you have any questions, please contact the laboratory coordinator.

Name: Jorja Gallivan

Student ID number: 0668356

Partner Name: Holly Hodgson

Experiment Date: Friday September 27th, 2019

CHEM1000H – LAB REPORT 1

1. Raw data: (all masses need to have 4 decimal places)

A. Determination of Glassware Accuracy Table 1: Data to determine the uncertainty in glassware using the density of water 50 mL Beaker 50.0 mL Graduated Cylinder

10.00 mL volumetric pipette

50.00 mL Burette

Mass of empty container (grams)

30.0978

87.3916

N/A, 50mL N/A, 50mL beaker used beaker used when weighed when weighed

Mass of container with Aliquot 1 (grams)

38.3309

96.8318

38.5080

40.5510

Mass of container with Aliquot 2 (grams)

47.7706

106.2239

48.3746

49.4086

Mass of container with Aliquot 3 (grams)

58.0832

116.3747

58.3294

59.4016

Temperature (OC)

21.0

22.0

21.5

22.5

Density of Water (g/ mL)

0.8233

0.9440

0.8410

1.0453

Note: Aliquot 2 contains both 10 mL additions, Aliquot 3 contains all 3 10 mL additions

B. Density of an Unknown Sample Table 2: Data to determine the density of an unknown sample Code for unknown sample: 10A Mass of empty container (grams)

57.809

Mass of container with Aliquot 1 (grams)

59.5564

Mass of container with Aliquot 2 (grams)

68.1847

Mass of container with Aliquot 3 (grams)

76.8297

Temperature (OC)

22.5

Density (g/mL)

0.1747

Note: Aliquot 2 contains both 10 mL additions, Aliquot 3 contains all 3 10 mL additions

C. Evaporative Properties of Solvents Table 3: Time vs Temperature data for the evaporation of liquids Temperature of Solvents (OC) Time (min)

Water

Isopropyl Acetate

Hexane

Heptane

Methanol

Ethanol

Unknown Code: 10A

Initial

19.9

19.8

20.3

20.5

20.6

18.8

21.1

0.5

18.2

11.1

8.7

14.9

14.3

7.1

19.5

1.0

17.3

9.6

4.2

13.4

12.6

5.0

16.8

1.5

16.7

9.0

2.9

12.2

11.9

3.6

13.1

2.0

15.6

8.5

4.6

13.2

11.0

3.5

11.1

2.5

14.9

8.1

6.2

13.3

10.5

4.4

10.0

3.0

14.4

8.4

7.7

13.1

10.5

5.0

9.3

3.5

14.4

8.5

8.7

12.9

10.5

5.6

9.1

4

14.4

8.9

8.7

12.9

10.5

5.9

8.8

4.5

14.4

9.2

8.7

12.9

10.5

5.9

8.9

5

14.4

9.2

8.7

12.9

10.5

5.9

9.4

2. Determine the theoretical mass of each aliquot, as well as the error associated with each piece of glassware from Part A, using the density of water at your recorded temperature (see appendix Table A) and your mass measurements. ANSWER: Show a complete sample for each calculation (see note from “How To” document) Table 4: Uncertainty in glassware using theoretical mass values 50 mL Beaker 50.0 mL Graduated Cylinder

10.00 mL volumetric pipette

50.00 mL Burette

Mass Aliquot 1 (grams)

38.3309 - 30.0978 =8.2331

96.8318 -87.3916 =9.4402

38.5080 -30.0978 =8.4102

40.5510 -30.0978 =10.4532

Mass Aliquot 2 (grams)

47.7706 -30.0978 =17.6728

106.2239 -87.3916 =18.8323

48.3746 -30.0978 =18.2768

49.4086 -30.0978 =19.3108

Mass Aliquot 3 (grams)

58.0832 -30.0978 =27.9854

116.3747 -87.3916 =28.9831

58.3294 -30.0978 =28.2316

59.4016 -30.0978 =29.3038

Theoretical mass of Aliquot 1 (grams)

0.99802*10 0.99780*10 =9.9802 =9.9780

0.99791*10 =9.9791

0.997685*10 =9.97685

Theoretical mass of Aliquot 2 (grams)

0.99802*20 0.99780*20 =19.9604 =19.9560

0.99791*20 =19.9582

0.997685*20 =19.9537

Theoretical mass of Aliquot 3 (grams)

0.99802*30 0.99780*30 =29.7624 =29.9340

0.99791*30 =29.9373

0.997685*30 =29.93055

Error in mass of Aliquot 1 (grams)

9.9802 -8.2331 =1.7471

9.9780 -9.4402 =0.5378

9.9791 -8.4102 =1.5689

9.97685 -10.4532 =0.47635

Error in mass of Aliquot 2 (grams)

19.9604 -17.6728 =2.2876

19.9560 -18.8323 =1.1237

19.9582 -18.2768 =1.6814

19.96537 -19.3108 =0.65457

Error in mass of Aliquot 3 (grams)

29.7624 -27.9854 =1.777

29.9340 -28.9831 =0.9509

29.9373 -28.2316 =1.7057

29.93055 -29.3038 =0.62675

Average Error

1.93723

0.8708

1.652

0.58589

Standard Deviation

0.2480

0.24580

0.0601

0.07828

**error in mass is shown as an absolute value Note: Aliquot 2 contains both 10 mL additions, Aliquot 3 contains all 3 10 mL additions

3. Was the pipette the ‘best’ piece of glassware to use for measuring 10.00 ml of your unknown sample, and if not which would have been better? Explain, based on the accuracy of each piece of glassware. ANSWER: According to our measurement data, the 50.0mL burette was the best piece of glassware for accuracy, the 50.0mL graduated cylinder was the second best, the 10.00mL pipette was the third best and the 50.0mL beaker was the least accurate. However, this may be due to human error within the experiment because in theory the pipette is the most accurate piece of glassware. This is because the beaker and graduated cylinder have larger increments making them less accurate and harder to get a very precise answer, and the burette was difficult to close the stopcock at the exact proper measurement.

4. Determine the density of your unknown sample. ANSWER: Show a complete sample for each calculation (see note from “How To” document) Table 5: Density of an unknown sample with standard deviation Code for unknown sample: 10A Mass of Aliquot 1 (grams)

59.5546 -57.809 = 1.7456

Mass of Aliquot 2 (grams)

68.1847 -57.809 =10.3757

Mass of Aliquot 3 (grams)

76.8297 -57.809 =19.0207

Density of Aliquot 1 (g/mL)

1.7456/10 =0.17456

Density of Aliquot 2 (g/mL)

10.3757/20 =0.518785

Density of Aliquot 3 (g/mL)

19.0207/30 =0.634

Average density (g/mL)

0.442448

Standard deviation

0.195178

Note: Aliquot 2 contains both 10 mL additions, Aliquot 3 contains all 3 10 mL additions

5. Plot your Time vs Temperature data form Table 3 (using graphing format provided in Blackboard “How To” document and appendix II of the lab manual) for each of your samples. You can plot the graphs manually using the paper template provided with experiment 1 in Blackboard or you can use an appropriate graphing program to complete them electronically. *See Part C- Note with procedure

Using this information, compare the evaporative rate to the intermolecular forces present and the possible effect of structure and size in each of the known solvents. ANSWER: With the prior knowledge of intermolecular forces, is it known that molecules with weaker forces evaporate at a faster rate, by looking and comparing the evaporation rate graphs it is easy to determine which solvents have stronger intermolecular forces which have weaker. According to our data, ethanol had the fastest evaporation rate therefore meaning it has the weakest intermolecular forces. It is closely followed by hexane and isopropyl acetate indicating that this solvents have weaker forces as well. The strongest of the forces is hydrogen bonds, therefore making sense that water was the slowest evaporating. After water, our next slowest was our unknown which tells us about its forces making it easier to identify. Next, the slowest were heptane and methanol indicating they as well had stronger forces than the isopropyl acetate, hexane, and ethanol.

6. Compare the evaporative rates and properties of the known solvents to your unknown sample and comment on the intermolecular forces present in each case. ANSWER: When comparing the temperatures for the evaporative rates of the known solvents, it was evident that our unknown sample had a much slower and less drastic decrease in temperature than all of the known substances except for water which had the slowest rate. This is due to the fact that water consists of hydrogen bonding, the strongest intermolecular force, meaning it will evaporate the slowest which is evident in our data. Since our unknown was only faster than water, this suggest that our unknown has fairly strong intermolecular forces.

7. Based on the density (using Table B in the appendix) and evaporative properties, what is the identity of your unknown sample? (don’t forget to include the code in your response) ANSWER: Based on the density of our unknown sample, it was concluded that there was some form of error made within our experiment. The density did not match any of the given densities within the appendix list. However, during experimentation it was easily recognized

that our unknown sample had a familiar scent, one very similar to nail polish remover. With prior knowledge we were aware that nail polish remover is composed of acetone, one of the solvents on the list. Our error in density was 0.347552g/mL, however, in conclusion with other observations consisting of smell as well as intermolecular/evaporation properties, we came to the conclusion that our unknown sample is in fact acetone. Unknown sample code : 10A

References: none ❑ Textbook- section or page number ❑ Blackboard- Chemistry 1000H ❑ Other (specify)...