Molar Volume of Oxygen

Lab #1 24/01/22
Analysis of 𝐾𝐶𝐿𝑂3𝑎𝑛𝑑 𝑡ℎ𝑒 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑀𝑜𝑙𝑎𝑟 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑂𝑥𝑦𝑔𝑒𝑛 𝑎𝑡 𝑆𝑇𝑃

Purpose:
The purpose of this experiment is to experimentally determine the molar volume of oxygen
under STP conditions and compare it to the actual value of the volume of an ideal gas at STP.
Additionally, we will attempt to calculate the mass of 𝐾𝐶𝐿𝑂3 when given an unknown sample.
Materials:
1. clamps
2. bunsen burner kit
3. 500ml beaker
4. Digital Scale
5. Stand and clamp with three holders
6. 500 ml Florence Flask with a stopper fitted with two rubber tubing
7. Pinch Clamp
8. Boiling tube with a stopper fitted with a rubber tubing
9. Stand and Clamp
10. weighing container
Chemical materials :
𝐾𝐶𝐿𝑂3 ( potassium chlorate)

𝑀𝑁𝑂2 ( magnesium dioxide)
Procedure:
1. All the materials were gathered
2. The florence flask was filled with water and set under the already set up apparatus
3. The pressure of the flask was equalized to atmospheric pressure by blowing in the tube
until all the flow of water was steady and there were no air bubbles. The end of tube C
was blown until there was water in the capillary tube flowing into the beaker. When there
was a constant water flow, the end of the capillary tube was clamped with a pinch clamp.
4. The boiling tube and beaker were cleaned and dried

5. The following was weighed and recorded in the data table
a. the empty boiling tube
b. 0.5g – 0.55g of Potassium Chlorate (KClO3) in the boiling tube
c. A Spatula’s tip full of MnO 2 was placed into the container containing the
Potassium Chlorate.
d. The boiling tube was weighed again.
e. A clean and dry 500ml beaker
6. The mouth of the boiling tube was connected with the end of tube C and clamped with a
three fingered clamp to the retort stand. The boiling tube was positioned at about 45
7. The Bunsen Burner Kit was set up and the flame was lit with the flint.
8. The Bunsen flame was placed under the Boiling tube, making sure it is not touching the
bottom of Boiling tube
a. Once the flame was set under the test tube and the pressure in the Florence flask
increased, ( around 4-5 bubbles) and the valve that leads to the beaker was opened
to collect the water.
b. The flame was moved at a back and to motion under the boiling tube tube.
9. When there is no further production of bubbles in the conical flask and the tubing is
removed from the beaker so that no more water enters the beaker.
a. The stopper is removed from the boiling tube. (This prevented backflow of the
water into the boiling tube)
b. The tube with contents and the beaker with the water was weighed. The
temperature of the water was measured and recorded
10. All the materials were washed and the apparatus was stored away.
Observations, Results, and Data
Results
Trial Trial 1(g)
Weight of empty weighing container (g) 30.38g
Weight of weighing container + KClO3 (g) 30.92g
Weight of weighing container + KClO3 + MnO4 (g) 31.14g
Weight of test tube and contents before heating(g) 31.14g
Weight of test tube and contents after heating (g) 30.88g
Weight of empty beaker n/a
Volume of oxygen collected (ml) 182ml
Weight of beaker + water collected n/a
Temperature of oxygen (K) 273.15K
Calculations:
1. Calculate the number of moles present in the mixture
𝑚𝑜𝑙 𝐾𝐶𝐿𝑂3 = 𝑚𝑜𝑙 𝑂2 * (2 𝑚𝑜𝑙 𝐾𝐶𝐿𝑂3/ 3 𝑚𝑜𝑙 𝑂2)

𝑚𝑜𝑙 𝑂2 = 𝑚𝑎𝑠𝑠/ 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 = (0.26g / 32 g mol-1) = 8.126 X 10-3 mol
mol 𝐾𝐶𝐿𝑂3 = 8.126 X 10-3 * (⅔) = 5.417 X10 -3 mol
2. Calculate the mass percentage of 𝐾𝐶𝐿𝑂3 present in the mixture
𝑚𝑎𝑠𝑠 % = (𝑚𝑎𝑠𝑠𝐾𝐶𝐿𝑂 / 𝑡𝑜𝑡𝑎𝑙 𝑚𝑎𝑠𝑠) * 100

3
𝑚𝑎𝑠𝑠 𝐾𝐶𝐿𝑂3 = 𝑚𝑜𝑙 𝐾𝐶𝐿𝑂3 * 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝐾𝐶𝐿𝑂3

𝑚𝑎𝑠𝑠 𝐾𝐶𝐿𝑂3 = 5.417 X10 -3 mol * 122.55 g/mol = 6.639 X 10-1 g
−1
𝑚𝑎𝑠𝑠 % = ( 6. 639 𝑋 10 𝑔 / .76 g ) * 100 = 87. 35 %
3. Calculate the Molar Volume of Oxygen
𝑚𝑜𝑙𝑎𝑟 𝑉 = 𝑉𝑂 / 𝑚𝑜𝑙 𝑂
2 2
molar mass = (15.999 * 2) = 32 g/ mol

𝑚𝑎𝑠𝑠 𝑂2 = 31. 14𝑔 / 30. 08𝑔 = 0.26 g
𝑚𝑜𝑙 𝑂2 = 𝑚𝑎𝑠𝑠/ 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 = (0.26g / 32 g mol-1) = 8.126 X 10-3

−3
𝑚𝑜𝑙𝑎𝑟 𝑉 𝑂2 = . 182 𝐿 / 8. 126 𝑋 10 mol = 22.39 mol /L
Discussion:
The main purpose of this lab was to compare the molar volume of oxygen gas obtained
under STP conditions to the molar volume predicted by the ideal gas law. The ideal gas law can
be summarized by the equation: 𝑃𝑉 = 𝑛𝑅𝑇. The ideal gas law makes the assumption that there
is little to no intermolecular forces between atoms, in other words, the molecules are too far
away to experience an attraction. This lack of attraction means that the collisions between
molecules are elastic, i.e. the KE in the system remains constant before and after the reaction.
This means that all the changes in heat are accounted for in the change in Temperature. Along
with this, the fact that the molecules are far apart means that the volume of the molecules is
negligible compared to the volume of the gas. The (molar volume) is specifically the volume
of a gas per mole of the gas under STP conditions, which is defined as having a temperature of
273.15 K and a pressure of 1atm or 760 torr. This equation for molar can be found by
rearranging the ideal gas law equation : = 𝑅𝑇/𝑃.
Using this equation, the molar volume of an ideal gas under STP has been determined as 22.414
L.
This same concept to experimentally determine the molar volume volume of 𝑂2. To do
this a decomposition reaction is set up using 𝐾𝐶𝐿𝑂3 , with 𝑀𝑁𝑂2 as the catalyst. After the
decomposition reaction has occurred the mass lost from the 𝐾𝐶𝐿𝑂3 mixture will be the mass of
𝑂2 present in the reaction. This mass will then be used to calculate the moles of 𝑂2 present. This
experiment will also yield the volume of 𝑂2 because the pressure created by the reaction will
force the water out of the florence flask and into a separate beaker. Using these values gained
from the experiment the can be calculated, and since the conditions are STP, the molar
volume will be the volume of water collected divided by the moles of oxygen calculated.
Another goal of this experiment, was to determine the mass percentage of 𝐾𝐶𝐿𝑂3 used.
This had to be determined because the original mixture of 𝐾𝐶𝐿𝑂3 was impure. To do this,
stoichiometry had to be applied to first find the number of moles.
2 KClO3 (s) 2 KCl (s) + 3 O2 (g)

By balancing the reaction, it is shown that for every 2 mol of 𝐾𝐶𝐿𝑂3 there is 3 mol of 𝑂2. By
using this, we can find the mass of 𝐾𝐶𝐿𝑂3 used in the reaction, and in turn the mass percentage.
Conclusion:
This experiment was successful because both the purposes were fulfilled. The molar
volume of 𝑂2 calculated was 22.39 L which is only 0.01 L off from the 22.4 L predicted by the
ideal gas law. Similarly, using the methods mentioned in the previous paragraph, the mass
percentage of 𝐾𝐶𝐿𝑂3 present in the mixture was found to be 87.35%.
For the molar volume, the slight deviation from the predicted value can be accounted for
due to the fact that the STP conditions might have been slightly off during the experiment.

Molar Volume of Oxygen

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Lab #1 24/01/22

Analysis of 𝐾𝐶𝐿𝑂3𝑎𝑛𝑑 𝑡ℎ𝑒 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑀𝑜𝑙𝑎𝑟 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑂𝑥𝑦𝑔𝑒𝑛 𝑎𝑡 𝑆𝑇𝑃

𝐾𝐶𝐿𝑂3 ( potassium chlorate)

1. All the materials were gathered

4. The boiling tube and beaker were cleaned and dried

1. Calculate the number of moles present in the mixture

𝑚𝑜𝑙 𝐾𝐶𝐿𝑂3 = 𝑚𝑜𝑙 𝑂2 * (2 𝑚𝑜𝑙 𝐾𝐶𝐿𝑂3/ 3 𝑚𝑜𝑙 𝑂2)

2. Calculate the mass percentage of 𝐾𝐶𝐿𝑂3 present in the mixture

𝑚𝑎𝑠𝑠 % = (𝑚𝑎𝑠𝑠𝐾𝐶𝐿𝑂 / 𝑡𝑜𝑡𝑎𝑙 𝑚𝑎𝑠𝑠) * 100

𝑚𝑎𝑠𝑠 𝐾𝐶𝐿𝑂3 = 𝑚𝑜𝑙 𝐾𝐶𝐿𝑂3 * 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝐾𝐶𝐿𝑂3

molar mass = (15.999 * 2) = 32 g/ mol

𝑚𝑜𝑙 𝑂2 = 𝑚𝑎𝑠𝑠/ 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 = (0.26g / 32 g mol-1) = 8.126 X 10-3

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

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