Determination of the Percentage Oxygen in Air

Transcription

1 CHEM 121L General Chemistry Laboratory Revision 1.1 Determination of the Percentage Oxygen in Air In this laboratory exercise we will determine the percentage by volume of Oxygen in Air. We will do this by allowing the Oxygen in the Air to react with Iron in the form of Steel Wool. This will convert the gaseous Oxygen into solid Iron Oxide. This will cause the volume of the Air to decrease as it becomes devoid of Oxygen. Once the reaction is complete, the decrease in the Air s volume will be equal to the volume the gaseous Oxygen occupied in the Air. One of the earliest observations concerning combustion was that it stops when the supply of Air is removed from the combustible material. Various theories were put forth to explain this observation. The first of these theories to take hold was the Theory of Phlogiston. This theory, put forth in 1667 by Johann Joachim Becher, postulated; that bodies were composed of three earths - terra lapidea (vitreous), terra mercurialis (mercurial), terra pinguis (fatty). [Becher] considered combustible substances to be rich in terra pinguis, which was lost during burning,. Even metals that were calcinable contained some of the fatty earth. [These ideas were] later developed into an elaborate chemical system in which the term "phlogiston" replaced terra pinguis. The Development of Modern Chemistry Aaron J. Ihde

2 P a g e 2 According to this theory, during combustion, a hypothetical substance called Phlogiston is released. Air was believed to absorb the Phlogiston until a point of saturation. It is at this saturation point that combustion ceases, unless fresh Air is introduced. Material Dephlogisticated Material + Phlogisticated Air This theory was capable of explaining not only combustion and calcination, but was also capable of explaining the smelting of metal ores and respiration. Although inconvenient facts kept pointing to inconsistencies in the Theory of Phlogiston, it was not until the discovery of the gas Oxygen in the late 1700's that it was finally abandoned. Gaseous Oxygen was first produced by Carl Wilhelm Steele in 1774 when, as a result of the strong heating of the Red Calx of Mercury (Mercuric Oxide), the Calx decomposed into elemental Mercury and a gaseous substance now known as Oxygen. Antoine Lavoisier and Joseph Priestley were then able to show that Oxygen is a component of Air and is required for combustion and respiration. With these discoveries it was realized that Air supports combustion because of the presence of Oxygen. When Air is depleted of Oxygen, it no longer supports combustion. Material + Oxygen Oxide of the Material Although it seems as though the Theory of Phlogiston has merely been inverted, Oxygen is an isolatable substance, whereas Phlogiston is not. Additionally, all those inconvenient facts unexplainable by the Theory of Phlogiston suddenly become explainable by the presence and reactivity of Oxygen. Thus, as Ihde points out; "The phlogiston concept gradually disappeared as its elder adherents were removed from the scene by death." Lavoisier was able to show that Air that combined with elemental Mercury lost about 1/6 of its volume. In turn, he was able to regenerate this volume of gas by decomposing the Red Calx that formed. Thus, he reasoned, Air is approximately 1/6 Oxygen. We will do something similar in our determination of the Oxygen content of Air. We will combine Air with elemental Iron. In the presence of Water and an acid catalyst, the Oxygen in Air reacts with Iron to form a complex Hydrate: 2 Fe(s) + 3 O 2 (g) + x H 2 O Fe 2 O 3 xh 2 O(s) (Eq. 1) We must be careful not to use too much acid, however, because if the system is too acidic the Iron will react directly with the acid to form Hydrogen gas: Fe(s) + 2 H + (aq) Fe 2+ (aq) + H 2 (a) (Eq. 2) Thus, we will soak Steel Wool in a dilute aqueous solution of Acetic Acid and allow it to react with the Oxygen in Air. As it does this, the volume of the Air will contract. When complete, the

3 P a g e 3 volume lost by the Air will be equal to the volume Oxygen initially present. We will measure this volume lost by the Air by carrying out this reaction in an inverted test tube placed in a bath of Water. As the Oxygen is depleted from the Air and its volume contracts, Water will rush into the test tube to replace it. By measuring the volume of this Water, we will be measuring the volume of Oxygen in the Air. This method of measuring the volume of Oxygen in Air has one draw-back. The Air above the Water is Moist due to some Water Vapor that will be present in the gas. We can, however, correct for this by subtracting out the volume of the Air occupied by the Water Vapor. Thus, by applying an appropriate correction factor, we can determine the Volume Percentage Oxygen in Dry Air. We can then compare this determination with the accepted literature result.

4 P a g e 4 Pre-Lab Questions 1. Dry Air is reported to be 20.94% Oxygen. What is the Percentage Error in Lavoisier s determination? 2. Write a balanced chemical reaction for Scheele s method for producing Oxygen. 3. The word Oxygen is derived from the Greek word οξυγόνο. What is the meaning of this word and why was it selected as the name for Scheele s new gas?

5 P a g e 5 Procedure 1. Obtain two 20 x 150 test tubes without lips. Mark each tube, from the open end, using a wax pencil, every 0.5 cm for a total length of 6 cm. (You will use these markings to judge the rate of reaction by observing how fast Water fills the tube.) Arrange to clamp these tubes to a ring stand in an inverted position in a 1 L beaker filled with tap Water. 2. On an Analytical Balance, weigh out two ~0.5g pieces of fine Steel Wool (size 00). Do not compress the Wool. 3. In a fume hood, obtain 50mL Acetone, 50 ml of 1.0 M Acetic Acid and 50 ml 0.1 M Acetic Acid in 100 ml beakers. 4. Using forceps, in a fume hood, rinse one of the pieces of Wool in Acetone for about 30 seconds. This will remove any oil from the surface of the Steel Wool. Shake off the excess Acetone and drain it briefly on a paper towel. Do not compress the Steel Wool. 5. Transfer the Wool to the 1 M Acetic Acid. Agitate the Wool in the Acid for about 1 minute. This will help remove any Oxide coating from the Steel Wool and prepare the Wool s surface for reaction with Oxygen. Shake off the excess acid and drain the Wool briefly on a paper towel. Do not compress the Steel Wool. 6. Transfer the Wool to the 0.1 M Acetic Acid. Agitate the Wool in the Acid for about 30 seconds. Remove the Wool and shake it vigorously in the sink to remove as much of the acidic solution as possible. A small amount of this solution will remain adhered to the Wool s surface. This will provide the aqueous acidic solution required to catalyze the reaction between the Iron and the Oxygen. Do not compress the Steel Wool.

6 P a g e 6 7. Insert the Steel Wool into the bottom third of your test tube. Do not compress the Wool as you want as much surface area exposed to the Air as possible. 8. Invert the test tube and lower it into a water tank and clamp it into place. The mouth of the beaker must be below the level of the Water throughout the experiment. 9. At 5 minute intervals, note the level of the Water in the tube. Also, at some point during the reaction measure the temperature of the Water. We will take this temperature to be the temperature of the Air in the test tube. 10. Prepare a second sample of Wool for a second trial. Start the second trial while you are waiting for the first trial to be completed. 11. On a Top-Loading Balance, weigh a 250 ml beaker. 12. When the level of the Water in the test tube is no longer rising, wait 5 minutes longer and then adjust the height of the Water in the test tube such that it equals that of the Water in the beaker. (This will ensure the pressure of the gas in the tube is equal to that of the atmosphere. Think about why this is necessary in terms of Boyle s Law.) 13. Trap the Water in the test tube by firmly holding a large rubber stopper against the mouth of the tube while the mouth of the tube is still under Water. 14. Unclamp the test tube and remove it from the Water. Have your lab partner tap dry your hand, the stopper and the test tube. Quickly transfer the trapped Water to the weighed beaker. Re-weigh the beaker. 15. Using forceps remove the Steel Wool from the test tube and transfer it to the weighed beaker with the Water. Re-weigh the beaker and its contents. 16. Now, fill the test tube to the brim with Water. Add this to the beaker and re-weigh it. 17. Complete your second trial. 18. Measure the barometric pressure of the Atmosphere. When the level of the Water in the test tube is equal to the level of the Water in the beaker, the Atmospheric pressure is equal to the pressure of the gas in the tube.

7 P a g e 7 Data Analysis 1. Calculate the volume of Oxygen in the Moist Air from the mass of the Water trapped in the test tube. (Take the density of Water to be g/ml at Room Temperature.) Mass Water Trapped = Mass Step #14 Mass Step #11 2. Calculate the volume of the Steel Wool in the test tube from its initial mass. (Take the density of Steel Wool to be 7.70 g/ml.) Mass Steel Wool = Mass Step #2 3. Calculate the volume of the solution adhering to the Steel Wool. (Take the density of the adhering solution to be g/ml.) Mass Adhering Solution = Mass Step #15 - Mass Step #13 - Mass Step #2 4. Calculate the volume of the test tube from the mass of the Water that fills the test tube. (Take the density of Water to be g/ml at Room Temperature.) Mass Water in Full Tube = Mass Step #16 - Mass Step #15 5. Calculate the volume of the Air initially in the test tube. This will equal the volume of the test tube minus the volume of the Steel Wool and the volume of the adhering solution. Volume = Data Anal. Step #4 - Data Anal. Step #2 - Data Anal. Step #3 6. Calculate the percentage of Oxygen in the Moist Air. 7. Because the Air above the Water in the test tube contains some Water Vapor, we must correct this result to obtain the percentage Oxygen in Dry Air. Recall, the pressure of the gas in the test tube is equal to the barrometric pressure; P gas = P Barr. So, we need to subtract out the Vapor Pressure of Water P H2O and correct the % Oxygen in Moist Air according to: % Oxygen in Moist Air = % Oxygen in Dry Air x (P Barr - P H2O ) / P Barr Values for the Vapor Pressure of Water can be found in the Appendix. Calculate the percentage Oxygen in Dry Air. 8. Dry Air is reported to be 20.94% Oxygen. Calculate the percentage error for your determination.

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9 P a g e 9 Post Lab Questions 1. One of the inconvenient facts the Phlogistonists had to contend with was that metals were observed to increase in mass when they calcinate. Why was this fact inconvenient and how did the Phlogistonists explain it? 2. How would our results be affected if the solution adhering to the Steel Wool is too acidic and this resulted in the production of Hydrogen gas? 3. How many moles of Oxygen are contained in 10 ml of the pure gas at 25 o C and 1 atm pressure? How many grams of Iron is required to react with this much Oxygen?

10 P a g e 10 Appendix - Vapor Pressure and Density of Water Temp [ o C] Vapor Press [mmhg] Density [g/ml]