experiment to calculate oxygen in air

UNIT 6 Determination of the Percent Oxygen in Air – INORGANIC CHEMISTRY PRACTICALS

Share this:

Unit 6 determination of the percent oxygen in air.

1.0   Introduction

2.0   Objectives

• Calculations

4.0  Conclusion

5.0  Summary

6.0  Tutor Marked Assignments

• UNIT 5 Determination of hardness of water using complexometric titration - INORGANIC CHEMISTRY PRACTICALS
• Chemistry Form 1 Notes : AIR OXYGEN AND COMBUSTION

7.0  References/Further Reading

1.0 Introduction

Air is a homogeneous mixture of gases such as nitrogen, oxygen, argon, and trace amounts of other elemental gases and carbon dioxide. The amounts of each gas can be measured both by weight and volume to determine the percent composition. In this experiment, the students will measure gas volume s using gas measuring burets. Since gases are very sensitive to changes in temperature and pressure, the students should carefully note atmospheric pressure, laboratory and water temperatures.

2.0 Objectives

At the end of this unit , should be able to perform an experiment is to determine the percentage by volume of oxygen in air.

• Main Content

Consider for a moment the air that you breathe. Since the time of the ancient Greek philosophers, people have realized that air is critical to life, though with little understanding of why.

We now know that the most common gases in air are nitrogen (78%), oxygen (about 21%), and argon (almost 1%). Other molecules are present in the atmosphere as well, but in very small quantities.

In this laboratory experiment, you will perform a procedure to verify the oxygen content of

Anyone who has witnessed rust on a car, bicycle or barbed wire fence knows that this reaction occurs spontaneously, though the rate can be very slow. To hasten the process and complete the data collection in one laboratory period, we will first “activate” the iron by washing it with acetic acid. It is believed that a small amount of acid catalyzes the reaction, though the mechanism is not well understood. On the other hand, an excess of acid could interfere with the results by reacting with the iron itself, to form hydrogen gas.

The experimental set-up is shown in the figure As the oxygen in air reacts with iron to form solid iron(III) oxide, the volume of the trapped air should decrease and water will enter the test tube. This change in volume is equal to the volume of oxygen consumed in the reaction.

Assuming that the length of the test tube is proportional to its volume and that the change in the length of the column of air in the test tube is due only to the removal of oxygen, the percentage of oxygen can be determined by calculating the change in the volume of air in the test tube.

To ensure that all oxygen is completely reacted, iron will be present in excess. A second question that you will attempt to answer in this experiment is what is the optimal quantity of iron to be used? To find the answer to this question, it will be necessary to perform a series of experiments. Rather than doing these all yourself, you will pool data with others in the class.

3.2 Procedure

• Fill a 15-cm test tube completely with water. Pour the water into a 100-mL beaker and weigh. Record the temperature of the water.
• Measure the length in millimeters of the test tube. Measure to the point halfway between the end and beginning of the rounded end. Attach a plastic metric ruler with two rubber bands so that the metric length begins at the lip of the tube. The rubber bands should be placed around the bottom half of the test tube, leaving your view of the top half unobstructed.
• Fill a 400-mL beaker about ¾ full with water.
• Obtain a small piece of steel wool from the front bench. Measure and record the mass.

Steps 5 – 7 should be done quickly while working in a fume hood. You will need the weighed piece of steel wool, forceps, and the test tube used in step 1.

• Holding the steel wool with forceps, rinse thoroughly with acetone. (This will remove any oils from the surface of the steel wool.) Shake off excess acetone in the acetone waste bucket.
• Soak the steel wool in a 50:50 vinegar/water mixture for 1 minute, making sure that all of the steel wool is under the surface of the solution. Remove the steel wool and shake off excess solution in the acetic acid waste bucket.
• Pull apart the steel wool to increase the surface area and insert it into the bottom of the test tube. Push the steel wool loosely into the test tube with a glass stirring rod.
• Working back at your station, cover the end of the test tube with your finger and quickly invert the test tube assembly into the beaker of water, as shown in figure 1, removing your finger once the opening of the test tube is under water. If necessary, adjust the 0.0 mm ruler mark to the water level inside the tube. Record the time.

Figure 1. Experimental set-up

• After 5 minutes, move the test tube so that the water level inside the test tube is equal to the water level inside the beaker. You will find this easiest to accomplish by holding the ruler against the side of the beaker. Measure and record the height of the water in the test tube and then rest it on the bottom again.
• Measure and record the height of the water in the test tube every five minutes using the procedure in step 9 until the water level stops changing. Take two or three readings at the final constant level .
• Remove the wire from the test tube, record its color, discard it and clean the test tube.
• Repeat steps 3 – 11 with a fresh piece of steel wool.

3.3 Calculations

Steps 1 – 3 may be done while collecting data (above) and must be completed before leaving lab.

• For each trial, prepare an Excel table and record the water level (mm), time (minutes) and percent change in the water level reading.
• Graph the percent change in the water level reading versus time. (Curves from both trials may be recorded on the same graph.
• Record your value for the mass of iron used, the percent volume of oxygen in air and the time it took to reach constant volume on the class summary table in lab. Before leaving, copy the class information into your lab notebook.
• Calculate the class average and standard deviation for the percent volume of oxygen.
• 1.0 g of steel wool?
• the heaviest piece of steel wool?

4.0 Conclusion

Report your values for the percent oxygen in air, as well as the class mean and standard deviation. Comment on the reproducibility of the results. How does your value compare to the class value and the accepted value (20.833%)? Explain the comparison by reference to errors in your procedure and any assumptions that were made. (When thinking about assumptions, recall what you know about the behaviors of gases, including solubility, partial pressures and the temperature- volume relationship.)

For this experiment, we also want to determine the optimal amount of steel wool that

should be used. Is there a minimum value, below which the time frame becomes too long? Is there a maximum value beyond which there is no time advantage noticed? What about the volume occupied by the steel wool itself? What percentage of the total volume does it occupy and at what point does it become significant? Consider both of these factor s when trying to determine an optimal mass of iron to use for this procedure.

Discuss these results in terms of collision theory. Why did rates change as the mass of iron was increased? How would they have been impacted if the steel wool was in a tight wad, rather than spread out? What was the role of the acetic acid and how does a catalyst function?

5.0 Summary

In this unit you have been able to perform an experiment to determine the percentage by volume of oxygen in air.

6.0 Tutor Marked Assignments (TMA)

• In step 9 you equalized the water level s inside the test tube and beaker. What was the purpose of this step?
• If you had done this experiment at the top of Mt. Everest, would the results be the same? Explain your answer.

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Accept Our Privacy Terms.*

I have read and accepted the Privacy Policy *

WAEC SYLLABUS FOR COMMERCE 2021/2022 (WASSCE)

Waec syllabus for office practice 2021/2022 (wassce), related posts.

Physics Form 1 Notes : CHAPTER FOUR – PRESSURE

Business Studies Form 4 Notes : MONEY AND BANKING

TRANSPORT AND COMMUNICATION QUESTIONS AND ANSWERS

METAL WORK KCSE PAST PAPERS QUESTIONS AND ANSWERS PDF

There are also other gases present in even smaller amounts - like neon and krypton, for example.

But there is another gas whose percentage varies according to the conditions. That is water vapour. In hot humid places on Earth that can be as much as 4%.

We tend to simplify these figures for general use:

Nitrogen makes up about 4/5 of the atmosphere.

Oxygen makes up about 1/5 of the atmosphere.

Finding the percentage of oxygen in the air

The classic method of doing this is to pass 100 cm 3 of air backwards and forwards over heated copper packed tightly into a glass or silica tube. A silica tube looks like glass, but is much more resistant to melting.

The air is pushed backwards and forwards across the heated copper until there is no further reduction in volume. The apparatus is allowed to cool because hot air expands, and the volume remaining is measured.

Some of the copper in the tube turns black as it turns to copper(II) oxide.

2Cu + O 2     2CuO

It is important that there is an excess of unchanged copper in the tube at the end so that you can be sure that the reaction stopped because the oxygen was all used up.

The volume left is 79 cm 3 , showing that air contains 21% oxygen.

Just out of interest:   This experiment is almost always described without any criticism, and people are always pleased that it seems to produce such a good result. However, there are two potential errors in it.

The first is that, however tightly you pack the copper into the tube, there will always be a certain amount of unmeasured air between the bits of copper, in the rubber tubing and in the tubes at the end of the syringes. If this totalled, say, 5 cm 3 , the volume would have decreased by an extra 1 cm 3 because of its oxygen content.

The other problem is that the accepted value of about 21% for oxygen depends on it being dry air. In most parts of the world, it isn't going to be dry air - there could be up to 4% of water vapour in the air. The percentage of oxygen in damp air will be below 21% because of the space taken up by water vapour, and so the volume will decrease less than you might expect.

As it happens, these errors work in opposite directions, and more or less cancel each other out.

Why am I telling you all this? Because if you want to do science in the future, you have to be critical and not just accept everything you are told!

If you are doing an exam, do not comment on this!

An alternative method using the rusting of iron

When iron rusts, it combines with oxygen and water to produce hydrated iron(III) oxide - that's iron(III) oxide with water attached to it, Fe 2 O 3 .xH 2 O. "x" shows a variable amount of water.

Obviously, if iron rusts, it must be removing oxygen from the air. This simple experiment uses this to find the percentage of oxygen in the air.

The iron wool is placed in a measuring cylinder and wetted by adding some water.

Most of the water is poured off, but it is best to leave a little bit in the cylinder so that when you invert it over a beaker of water, the water level in the cylinder falls within the cylinder markings.

You make a note of the water level in the cylinder (showing the volume of gas contained in it), and then leave it for a few days for the iron to rust.

The water level obviously rises as the oxygen is used up. Make a note of the new level.

Made up results!

Volume of air at start = 9.7 cm 3 Volume left at end = 7.7 cm 3 Volume of oxygen = 9.7 - 7.7 cm 3 = 2.0 cm 3 Percentage of oxygen = (2.0 / 9.7) x 100 % = 21% (to 2 significant figures)

Note:   If you do this, you will almost certainly use steel wool rather than iron wool. This is iron which has a very small amount of carbon in it. It behaves chemically as if it were pure iron.

If you were being fussy, you might argue that we haven't allowed for the space taken up in the measuring cylinder by the iron. In fact, this is going to be pretty small. You don't need much iron for the small amount of oxygen in the tube to react with. And, because it is used as a loose wool, most of its volume is air anyway.

Where would you like to go now?

To the air and oxygen menu . . .

To the Chemistry 14-16 menu . . .

To Chemguide Main Menu . . .

© Jim Clark 2020