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Name: Nguyen Minh Hanh

Student ID number: 4705201

Submission Date: September 28th

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)

29.4917

87.1452

29.7295

29.5927

Mass of container with Aliquot 1 (grams)

38.5528

97.4656

39.7494

38.9436

Mass of container with Aliquot 2 (grams)

46.7034

107.1981

48.9488

49.5667

Mass of container with Aliquot 3 (grams)

56.6007

117.1532

59.7491

59.5741

22.0

22.0

22.0

22.0

0.99780

0.99780

0.99780

0.99780

Temperature (OC) Density of Water (g/ mL)

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: 1023 Mass of empty container (grams)

50.2456

Mass of container with Aliquot 1 (grams)

56.8135

Mass of container with Aliquot 2 (grams)

63.3821

Mass of container with Aliquot 3 (grams)

69.9547

Temperature (OC)

21.5

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 liquid Temperature of Solvents (OC) Time (min)

Isopropyl Acetate

Hexane

Heptane

Methanol

Ethanol

23.1

24.0

20.3

22.1

25.1

24.7

21.4

0.5

21.9

19.0

9.4

17.8

13.8

19.4

9.6

1.0

20.8

15.6

4.7

15.1

8.7

16.2

5.9

1.5

20.1

13.6

3.5

13.7

6.6

14.6

4.6

2.0

19.4

12.3

3.0

13.0

5.6

13.9

4.2

2.5

19.1

11.6

3.0

12.7

5.3

13.4

4.1

3.0

18.7

11.1

3.1

12.5

5.3

13.2

4.1

3.5

18.6

10.8

4.0

12.5

5.1

12.9

4.4

4.0

18.3

10.8

6.9

12.5

5.1

12.7

5.3

4.5

18.2

10.8

11.0

12.5

5.1

12.7

7.8

5.0

18.2

10.8

12.2

5.3

12.7

11.5

5.5

18.2

10.8

12.2

5.3

12.7

6.0

18.1

10.8

12.2

5.5

12.7

Initial

Water

Unknown Code:1626

2. Calculate the volume 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)

38.5528 29.4917 = 9.0611

97.4656 87.1452 = 10.3204

39.7494 29.7295 = 10.0199

38.9436 29.5927 = 9.3509

Mass Aliquot 2 (grams)

46.7034 29.4917 = 17.2117

107.1981 87.1452 = 20.0529

48.948829.7295 = 19.2193

49.566729.5927 = 19.974

Mass Aliquot 3 (grams)

56.6007 29.4917 = 27.109

117.153287.1452 = 30.008

59.749129.7295 = 30.0196

59.574129.5927 = 29.9814

Calculated volume of Aliquot 1 (mL)

9.0611 / 0.99780= 9.0811

10.3204/ 0.99780= 10.3432

10.0199/ 0.99780 = 10.042

9.3509/ 0.99780= 9.3715

Calculated volume of Aliquot 2 (mL)

17.2117 / 0.99780= 17.2496

20.0529/ 0.99780= 20.0971

19.2193/ 0.99780 = 19.2617

19.974 / 0.99780= 20.018

Calculated volume of Aliquot 3 (mL)

27.109 / 0.99780= 27.1688

30.008/ 0.99780= 30.0742

30.0196/ 0.99780 = 30.0858

29.9814/ 0.99780= 30.0475

Error in volume of Aliquot 1 (mL)

9.0811 - 10 10.3432 -10 10.0420 - 10 = 0.9189 = 0.3432 = 0.042

9.3715 - 10 = 0.6285

Error in volume of Aliquot 2 (mL)

17.2496 20 = 2.7504

20.0971-20. 19.2617 - 20 = 0.0971 = 0.7383

20.0180 - 20 = 0.018

Error in volume of Aliquot 3 (ml)

27.1688 30 = 2.8312

30.0742 30 = 0.0742

30.0858 - 30 =0.0858

30.0475 - 30 =0.0475

Average Error

(0.9189 + 2.7504 + 2.8312)/3= 2.1668

(0.3432 + 0.0971 + 0.0742)/3= 0.1715

(0.042 + 0.7383 + 0.0858)/3= 0.2887

(0.6285 + 0.018 + 0.0475)/3 = 0.2313

Standard Deviation

1.0815

0.1491

0.3900

0.3443

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

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: No, the pipette is not the best piece of glassware to use for measuring 10ml of the unknown sample because it does not have the least average amount number and it does not have the least amount of standard deviation from its average error. However the 50.0 mL Graduated Cylinder has. Therefore, the pipette is not the best equipment.

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: 1626 Mass of Aliquot 1 (grams)

56.8135 - 50.2456 = 6.5679

Mass of Aliquot 2 (grams)

63.3821 - 50.2456 = 13.1365

Mass of Aliquot 3 (grams)

69.9547 - 50.2456 = 19.7091

Density of Aliquot 1 (g/mL)

6.5679 / 10 = 0.65679

Density of Aliquot 2 (g/mL)

13.1365 / 20 = 0.65679

Density of Aliquot 3 (g/mL)

19.7091 / 30 = 0.65697

Average density (g/mL)

(0.65679+ 0.65679+ 0.65697)/3= 0.65695

Standard deviation

0

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

5. Plot your Time vs Temperature data from Table 3 (using graphing format provided in Blackboard “How To” document and appendix II of the lab manual) for your samples. (One plot for each of the three solvent pairs and one plot for your unknown, 4 graphs total). 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:

TEMPERATURE (DEGREES CELSIUS)

WATER TEMPERATURE VS. TIME Water 23.1

25

Linear (Water 23.1)

20

15

10

5

0 0

1

2

3

4

TIME (MINUTES)

5

6

7

Isopropyl Acetate Temperature Vs. Time Temperature (degrees celsius)

20

Isopropyl Acetate 24

18 16

Linear (Isopropyl Acetate 24)

14 12 10 8 6 4 2 0 0

1

2

3

4

5

6

7

6

7

Time ( minutes )

TEMPERATURE (DEGREES CELSIUS)

HEXANE TEMPERATURE VS. TIME Hexane 20.3

Linear (Hexane 20.3)

12

10 8 6 4 2 0 0

1

2

3

4

TIME (MINUTES)

5

UNKNOWN SUBSTANCE 1626 TEMPERATURE VS. TIME Unknown Code:1626 21.4

Linear (Unknown Code:1626 21.4)

TEMPERATURE (CELSIUS)

14 12 10 8 6 4 2 0 0

1

2

3

4

5

6

TIME (MINUTES)

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: Water has the bonding bonds as the intermolecular forces, so a high heat is a requirement to convert the liquid water to the gas form. The graph of isopropyl acetate is steeper since it lacks hydrogen bonding. This means it needs lower heat than water to evaporate, so water evaporates slower than isopropyl acetate. When we compare the graph of hexane and isopropyl acetate, the curve of hexane is a sharper decline in temperature. Hexane needs colder heat to evaporate because it has a longer carbon chain which is dispersion forces. The unknown substance has same appearance of the the curve of hexane but it is wider because the heat requires to evaporate is hotter than hexane.

7

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: Using the Table B in the appendix and evaporation and evaporative properties, the unknown sample 1626 is hexane because it has the average density same with hexane which is 0.65g/mL. Besides that, the temperature which is required to evaporate the unknown sample has a very small errors compared to the evaporative temperature of hexane 8. Report the unknown solvent code for the sample portrayed in the lab video ANSWER: The unknown solvent code for the sample portrayed in the lab video is 1023

References: Cite any/all ‘outside’ materials used to complete this report. Blackboard CHEM1000H

Tro, N. J., Fridgen, T. D., & Shaw, L. (2020). Real Gases:The Effects of Size and Intermolecular Forces . In Chemistry: a molecular approach (3rd ed., pp. 181–184). Pearson Canada....