1. Experiment 2:
Analysis of a KClO3 Mixture
and the Molar Volume of
Oxygen
Prepared by:
Janine V. Samelo
Nevah Rizza L.Sevilla
Edessa Joy R. Sumagaysay
Pauline Bianca R. Alfonso
Jean Annerie M. Hernandez
BSChem 3

2. 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

3. Methodology
Weigh 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.

4. 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.

5. 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.

6. 2KClO3(s) --> 2KCl(s) + 3O2(g)

7. 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

8. Calculations
• Pressure of “dry” O2 torr:
Patm = PO2 + PH2O
PO2 = Patm - PH2O
Po2 = 718.22418 torr

9. • Volume of O2 at STP
• Weight of O2 produced
Wt = wt of test the before rxn – wt of test tube after
rxn
Wt= 42.10 g – 41.78 g
Wt= 0.32 g

10. • Moles O2 produced
• Molar volume of O2
=
• Moles KClO3 decomposed

11. • Weight of KClO3 in initial mixture
• Percent KClO3 in initial mixture

12. 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.

13. 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.

14. 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.

15. 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 percentKClO3 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.

16. References
• answers.yahoo.com/question/index?qid=2007
1129131545AAOL3Yra
• http://chemmovies.unl.edu/chemistry/smallsc
ale/SS031.html
• http://wwwchem.csustan.edu/chem1102/mol
arvol.htm

17. 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.

18. 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

19. .

20. 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.

21. 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.