EXP 13 LAB Report - A Carbonate Analysis; Molar Volume of Carbon Dioxide PDF

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A Carbonate Analysis; Molar Volume of Carbon Dioxide...

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Chiara De Jesus Lab Partners: Diego A., Ronald R. November 6, 2020 EXP 13 A Carbonate Analysis

A.

Calculation of Mass of KClO3 Sample for Analysis Mass of KClO3 for ~40 mL O2

1 2 3

C.

Unknown Sample No. ___GOLD____ Mass of sample (g) Mass of generator + sample before reaction (g) Instructor's approval of apparatus

Trial 1 0.302 42.030

Trial 2 0.301 22.463

Determination of Volume, Temperature, and Pressure of the Carbon Dioxide Gas

1

Initial volume of water in "O2-collecting" graduated cylinder (mL)

0.00

0.00

2

Final volume of water in "O2-collecting" graduated cylinder (mL)

71.10

3 4 5

Volume of O2 gas collected (L) Temperature of water (°C) Barometric pressure (torr)

7.11E-02 21.00 760.98

75.40 7.54E02 23.00 761.00

6

Vapor Pressure of H2O at 21°C (torr)

18.70

21.00

7

Pressure of "dry" O2 gas (torr)

742.28

740.00

Calculations Pressure of dry O2 gas = Barometric pressure – vapor pressure Trial 1 = 760.98 – 18.70 = 742.28 torr Trial 2 = 761 – 21 = 740 torr

D

Mass of Oxygen Evolved

1

Mass of generator + sample after reaction (g)

22.036

22.364

2

Mass loss of generator = mass O2 evolved (g)

19.994

0.099

742.28 6.44E02 6.25E01

740.00 6.81E02 3.09E03

0.10

22.02

Molar volume of O2 Gas 1

Pressure of "dry" O2 gas (torr)

2

Volume of O2 gas at STP (L)

3

Moles of O2 gas evolved (mol)

4

Molar volume of O2 gas at STP (L/mol)

5

Average molar volume of O2 gas at STP (L/mol)

11.06

Calculations Mass loss of generator = mass O 2 evolved Mass of generator + same before reaction – mass of generator +sample after reaction Trial 1 = 42.030 – 22.036 = 19.994 Trial 2 = 22.463 – 22.364 = 0.099 Volume of O2 gas at STP = VO2 (at STP) = VO2 expt. * (PO2 EXPT/760 torr) *(273K/TO2 expt) Trial 1 = 0.0711*(742.28/760)*(273 K/21 + 273.15 K) = 0.0644 Trial 2 = 0.0754*(740/760)*(273 K/21 + 273.15 K) = 0.0681 Moles of O2 gas evolved Trial 1 = 19.994 g (1 mol/32 g) = 0.625 mol

Trial 2 = 0.099 g (1 mol/32 g) = 0.00309 Molar volume O2 gas at STP Discussion/Conclusion This experiment required the application of Dalton’s law of partial particles, a combination of Boyle’s and Charles’ laws, and stoichiometry to determine the percent of an anhydrous salt in a mixture and the molar volume of the gas evolved. We had the unknown sample, Gold, and the data of this experiment concluded that the average molar volume of O 2 gas at STP (standard temperature and pressure) was 11.06 L/mol. This was found by first calculating the volume of O2 gas at STP with the equation: VO2 (at STP) = VO2 expt. * (PO2 EXPT/760 torr) *(273K/TO2 expt) and getting 0.0644 L for Trial 1 and 0.0681 L for Trial 2, respectively. Next, the moles of O2 gas evolved was calculated by taking the mass of O2 evolved from each trial and multiplying it by 1 mol/the molar mass of O 2, resulting in getting 0.625 moles for Trial 1 and 0.00309 moles for Trial 2. To find the molar volume of O 2 gas at STP for each volume, we divided the volume of O2 evolved by the moles of O2 gas evolved and received 0.10 for Trial 1 and 22.02 for Trial 2. Taking those two values and adding them together, then dividing it by 2, we got 11.06 L/mol as the average molar volume. Errors to note would be KCl3 is used and the output will be O2 not CO2....