Experiment 2: Analysis of a KClO3 Mixture and the

Molar Volume of Oxygen

Prepared by:

Janine V. SameloNevah Rizza L.Sevilla

Edessa Joy R. SumagaysayPauline Bianca R. Alfonso

Jean Annerie M. HernandezBSChem 3

Objectives:

• To determine the volume occupied by one mole of gas at 273K and 760 torr

• To determine the percent purity of a sample by mass of a chemical decomposition reaction

MethodologyWeigh a 200mm Pyrex test tube. Weigh approximately 1 g

of a KClO3 mixture to the nearest milligram (0.001 g).

Assemble the apparatus, clamp the pyrex test tube to the ring stand and connect the gas delivery tube to the pheumatic trough. Fill the trough about 2/3 full of water. Connect the gas delivery tube to the Pyrex test

tube.

Fill the 250 –florence flask)with water to the rim, slide a glass plate over the rim and invert the flask over the gas outlet in

the pneumatic trough.

Remove the glass plate. Support the receiving flask with a ring and clamp. Check to make sure no bubbles have entered

the receiving flask.

Heat the test until O2 gas is generated. When no further evolution of oxygen is observed, disconnect the gas delivery tube

from the pyrex test tube first and then allow it to cool.

Bring the pressure inside the receiving flask to atmospheric pressure by raising or lowering the flask until the water levels are

the same inside and out.

Record the temperature of the water in the trough and the barometric pressure. Get the vapor pressure of water from any

book at that particular temperature.

Dry the outside of the florence flask and weigh the flask and its contents (including the glass plate) to the nearest 0.1 gram.

Fill the flask to the rim with water, slide the glass plate over the rim and again weigh the flask and its contents.

Weigh the cool 200-mm pyrex test tube to the nearest milligrams. Repeat the experiment if time allows.

2KClO3(s) --> 2KCl(s) + 3O2(g)

Data:  DATA

Weight of test tube and contents before heating

42.10 g

Weight of test tube and contents after heating

41.78 g

Weight of empty test tube 41.10 g

Weight of flask after reaction 184.58 g

Weight of flask full of water 438.00 g

Volume of oxygen collected 253.42 mL

Barometer reading 999.95 mb

Temperature of oxygen 30.2 oC

Vapor pressure of water at the temperature of the oxygen

31.8 mmHg

Calculations

• Pressure of “dry” O2 torr:

Patm = PO2 + PH2O

PO2 = Patm - PH2O

Po2 = 718.22418 torr

• Volume of O2 at STP

• Weight of O2 produced

Wt = wt of test the before rxn – wt of test tube after rxnWt= 42.10 g – 41.78 gWt= 0.32 g

• Moles O2 produced

• Molar volume of O2

=

• Moles KClO3 decomposed

• Weight of KClO3 in initial mixture

• Percent KClO3 in initial mixture

Discussion:• The mixture of KClO3 and MnO2 was decomposed by

heating to generate the oxygen gas which displaces the water from the Florence flask. The flask was raised and lowered repeatedly until the water levels are the same inside and out to obtain the atmospheric pressure, which, according to Dalton's Law of Partial Pressures, equal to the sum of the partial pressures of oxygen and water vapor. The barometric pressure of the atmosphere was 999.95 mbs, and the vapor pressure of the water inside the flask at 30.2oC was 31.8 mmHg. Getting their difference, the pressure of the gas collected is 718.22418 torr.

Discussion• Using Boyle’s Law and Charle’s Law the volume of the wet

oxygen collected was corrected.

• The mass of O2 produced after the reaction was determined by subtracting the mass of the test tube containing the sample before and after the reaction which was then converted to the number of moles produced by dividing it to the molar mass of oxygen. The molar volume of the oxygen was calculated by dividing its corrected volume by the moles of oxygen. With this, the computed molar volume of O2 from the experiment is L/mol.

Discussion

• The number of moles of KClO3 reacted was determined using the stoichiometry of the reaction where 2 moles of KClO3 react with 3 moles of O2 which was then multiplied to the molar mass of KClO3 to get its mass and then finally multiplying to 100 to get the percent KClO3 in the original mixture.

Conclusion:• Based on the experimental result, the molar volume

of the gas collected is L/mol. The molar volume of any gas at STP ("Standard temperature and pressure" of 273.15 K and 1 atm) is 22.414 L which gives a -3.91% difference between them.

• The percent KClO3 in the original mixture is 81.67 %.• Errors in the experiment could be accounted from

the small bubbles inside the flask while it was submerged in the water bath and there could be a little amount other gases from the atmosphere dissolved in the water used.

References

 • answers.yahoo.com/question/index?

qid=20071129131545AAOL3Yra• http

://chemmovies.unl.edu/chemistry/smallscale/SS031.html

• http://wwwchem.csustan.edu/chem1102/molarvol.htm

Post Lab Questions

• 1.) A mixture of KClO3 and MnO2 weighing 1.75 g was heated to 300oC until decomposition was complete. After cooling, the mixture weighed 1.11 grams. A.) Write a balanced equation for the decomposition. B.) How many grams and moles of O2 were produced in the decomposition? C.) How many moles and grams of KClO3 were present in the original mixture? D.) Calculate the percent KClO3 in the original sample.

2KClO3 ======== 2KCl + 3O2

Δ• MnO2 is just a catalyst of the decomposition of

KClO3.

Mass of O2 reacted = 1.75 g- 1.11 g

= 0.64 g

.

2.) If an air bubble accidentally enters the gas- collecting flask, how would this affect the reported number of moles of KClO3 decomposed? Explain.

• Oxygen is generated by the decomposition of the KClO3 (with MnO2 as catalyst). The air in the flask is heated and expands. This increased volume of gas, bubbles out of the system and is collected in the florence flask. If an air bubble accidentally enters the flask, the reported number of moles of decomposed will also increase because the number of moles decomposed is directly proportional to the number of moles of generated oxygen based on the reaction stoichiometry.

3.) Suppose that the water inside the gas-collecting flask was higher than the water level outside the flask. If no attempt was made to make the two water levels the same, how would this error in technique affect your report of moles of oxygen gas collected? Explain.

• In this situation, the pressure inside the flask is smaller than the pressure outside. If no attempt was made to make the two water levels the same, the partial pressure of O2 would be larger and since pressure is directly proportional to the number of moles, the calculated moles of oxygen would be greater than the actual moles of oxygen generated in the reaction.