Experiment

9

 

Overview 

In this experiment, you were given an unknown sample which consisted partly of the substance potassium chlorate (KClO₃) and the remainder inert material.

 

When potassium chlorate is heated strongly (in the presence of manganese dioxide catalyst), it decomposes according to the following reaction:
 

As you can see, elemental oxygen gas is produced by the reaction. You measured the mass of oxygen produced from your sample, as well as the volume of oxygen (by displacement of water). From this information, you are able to calculate the molar volume of oxygen at STP (i.e., the volume occupied by one mole of O₂ at 0^oC and 1 atm pressure).

 

Data 

Suppose the following data had been recorded for this experiment (Page 67, Section I)
 

Calculations 

Page 67, Part II

1. Mass of Water

This is the mass of water collected in the beaker (which can be related to the volume of oxygen which was produced from the reaction of the sample of potassium chlorate in the test tube).
 

2. Volume of Water

In step1 above, you calculated the mass of water which was transferred from the round flask to the beaker. During the experiment, you measured the temperature of the water in the beaker (21^oC for my data above), and looked up in the Appendix of the lab manual the density of water at that temperature (0.99802 g/mL at 21^oC). Remember that density represents the relationship between the mass of a material and its volume.
 

3. Volume of Oxygen at Experimental Conditions

You will remember that before you could use the apparatus in this experiment for the reaction, you first had to fill the tube running from the round flask to the beaker with water. This was so that the first mL of oxygen produced would be able to push a mL of water from the flask into the beaker. That is, the apparatus was set up so that the volume of water collected would exactly equal the volume of oxygen generated by the chemical reaction. This assumes, of course, that all the rubber tubing connections in your apparatus were secure, and that you adjusted/equalized the pressures as directed in the lab manual.
 

4. Atmospheric Pressure

The atmospheric pressure while you were doing your experiment (which you need to have to do the rest of the calculations) was not the pressure read from the barometer. The mercury level visible in a barometer is due to two effects: the pressure of the atmosphere is the major contribution, but the mercury level is also a function of the temperature of the barometer. In your data table, you were given both the mercury level of the barometer (Part I f) and the temperature of the barometer (Part I g). In Part 1 h you were asked to look up in the Appendix the "barometer scale correction". The barometer scale correction corresponds to that portion of the mercury level that was due to the temperature.

 

For my data, using the Appendix of barometer scale corrections, the correction to be subtracted from the mercury level is 2.5 mm Hg. Since the table does not give all possible temperatures or all possible mercury levels, you should estimate the correction for your particular data (you will notice that the corrections do not vary much around normal lab conditions).
 

5. Pressure of oxygen alone.

In Part 4 directly above, we calculated the corrected atmospheric pressure. Because of the way the experiment was set up (through the step of equalizing pressures that you performed), this also represents the total pressure inside the round flask into which the oxygen was generated. However, the gas in the round flask was not pure oxygen gas. Because the oxygen was bubbled through water, the gas in the flask is a mixture of oxygen and water vapor. The saturation amount of water vapor needed to fill a flask (in terms of the partial pressure of water vapor) is a fixed quantity at a given temperature. In Part I i of the data, you were asked to look up the "aqueous vapor pressure" at the temperature of the oxygen in the Appendix of the lab manual. For 21^oC, the saturation aqueous vapor pressure is 18.6 mm Hg. This means that of the total of 757.7 mm Hg pressure in the round flask, 18.6 mm Hg is due to water vapor. The pressure of the oxygen gas itself is then given by subtraction.
 

6. Absolute temperature of oxygen.

The thermometers you use in lab measure temperatures in degrees Celsius. Since the Celsius temperature scale is a human invention, we cannot expect oxygen molecules to know anything about it!!! We need to calculate the temperature of the oxygen on the fundamental Absolute (Kelvin) temperature scale.
 

7. Volume of oxygen at STP

According to my data above, we collected 156.4 mL of oxygen at a pressure of 739.1 mm Hg at a temperature of 21^oC (294 K). Since the volumes of gases vary so much with temperature and pressure, it is common to reduce all gas volume measurements to what they would have been if the experiment had been performed at the chosen Standard Temperature and Pressure for gases (STP), which has been chosen to be a temperature of 0^oC (273 K) and a pressure of exactly 760. mm Hg (1.00 atm). Fill in the combined gas law given as follows:

• P₁ = 739.1 mm Hg
• V₁ = 156.4 mL
• T₁ = 294 K
• P₂ = 760 mm Hg
• V₂ = Volume at STP we want
• T₂ = 273 K

If you then solve for V₂, you come up with
 

8. Mass of oxygen produced.

At this point, we switch gears a little bit. For the last few calculations, we have been dealing with the volume of the oxygen. Now we switch to the mass of oxygen. This is calculated from the change in mass of the test tube before (Part Ic) and after (Part Id) heating. When the test tube was heated, the oxygen gas was liberated from the test tube.
 

9. Moles of oxygen

The number of moles of anything is based on the mass and the molar mass. Remember that it was O₂ that was generated in this experiment, so the molar mass we need is 32.00 g.
 

10. Molar Volume of Oxygen at STP

The "Molar Volume" of oxygen means the volume that 1.00 mole of oxygen would have at STP (0^oC, 1 atm). From my data above, we found that 6.641 X 10^-3 moles of O₂ had a volume of 0.141 Liters. Since the "molar volume" represents a ratio (Liters/mole), using my data we can calculate a value:
 

11. Theoretical Value of the Molar Volume

Here we are to calculate a theoretical value for 1.00 mol of a gas at STP, so that we can compare with our experimental value of 21.3 L/mol for O₂. You should use the Ideal Gas Law, PV = nRT, using the following values:

• P = 1.00 atm
• V = what we are looking for
• n = 1.00 mole
• R = 0.08206 L atm/mol K
• T = 273 K

Solving this leads to
 

12. % Error in Molar Volume

This represents the % difference between our experimental value (21.3 L/mole) and the theoretical value (22.4 L/mole) and shows how carefully we did the experiment.
 

This means we were about 5% off from the true value. The minus sign indicates that our answer was lower than the true value.

13. Moles KClO₃ in sample

Now we're switching gears again. We have calculated how much oxygen was generated, and used that to determine the volume of 1 mole of O₂ at STP. But the amount of oxygen generated can also tell us about the original unknown sample we began with. Remember that the sample consisted partly of KClO₃, which when heated, produced the oxygen. Based on the amount of oxygen we obtained, we can calculate back to find out what fraction of the original sample must have been KClO₃ (to have been able to generate the determined amount of oxygen).

The balanced chemical equation for the reaction is
 

This equation says that every two moles of KClO₃ will produce three moles of O₂. So if our experiment produced 6.664 X 10^-3 moles of O₂, the number of moles of KClO₃ that must have reacted to produces is two-thirds of this.
 

So 4.427 X 10^-3 moles of KClO₃ must have been present in the sample we heated.

14. Mass of KClO₃ in sample

In the step above, we calculated that the original sample must have contained 4.427 × 10^-3 moles of KClO₃ to have generated the amount of oxygen we collected. The molar mass of KClO₃ is 122.6 g. So the mass of the 4.427 × 10^-3 moles of KClO₃ is given by:
 

So the original sample must have contained 0.5428 g KClO₃ in order to have produced the amount of oxygen we collected.

15. Percent KClO₃ in sample

The unknown sample you were given consisted only partly of KClO₃. There was also some inert material that did not react when heated. We calculated above that the sample must have contained 0.5428 g of KClO₃ in order to have produced the amount of oxygen gas we collected. If we compare this mass of KClO₃ to the total mass of the original sample before heating, we can calculate the %KClO₃.

 

The mass of the original sample is calculated from Parts Ib (the mass of the empty test tube) and Ic (the mass of the test tube with the sample before it has been heated).
 

And then the percent KClO₃ (0.5428 g KClO₃ out of a total sample mass of 0.9352 g) is
 

Questions 

1. This question is a lot easier than the calculations we've done above, because we don't have to worry about the correction for water vapor. Just figure out how many moles of KClO₃ is present in the 1.235 g sample (this is pure KClO₃, not a mixture as in the experiment). Then use the coefficients of the balanced chemical equation to figure out how many moles of O₂ could be produced from this number of moles of KClO₃. Finally, use the fact that one mole of oxygen would occupy a volume of 22.4 L at STP to figure out the volume of the oxygen in this question at STP. The answer is 0.338 L (338 mL), but you will have to show the calculations to receive credit on your lab report.
2. First figure out how many moles of water is present in the 0.11 g (the molar mass of water is 18.02 g). Then use the ideal gas equation (PV = nRT) to figure out what volume this amount of water would occupy at 800^oC. Since the pressure is not specified in the question, you can assume that the pressure is 760 mm (1.00 atm) which is approximately the pressure under which this experiment was performed (or you could use the actual pressure of your own experiment). For a pressure of 760 mm Hg, the volume calculated would be 0.537 L (537 mL). You will have to show how to get this answer to receive credit on your lab report.
3. In Question 2, the volume of water vapor calculated was 537 mL. The error is thus just 537 ml (540 mL to 2 significant figures).  In percentage terms, this is a huge error

You can see from this why it was so important to have a completely dry test tube for the reaction. This is why we gave you a new, clean test tube rather than have you try to wash out and dry one from your locker.