EXPERIMENT 8
DETERMINATION OF THE PERCENTAGE OF OXYGEN IN THE AIR
To measure the amount of oxygen in air we will use the same principle employed by the
pneumatic chemists--a sample of air will be trapped in a test tube along with something that
uses up all of the oxygen in the tube, allowing the water to rise in the tube. Rather than use a
candle or a mouse, we will employ the reaction of oxygen with iron in the form of steel wool.
When conditions are arranged properly, oxygen reacts rapidly and completely with the iron, as
described by the following (unbalanced) reaction (the balancing of this reaction will be left as
an exercise for you):
Fe + O2---> Fe2O3
This reaction is more complex than just the direct combination of oxygen with iron. It also
requires the presence of water and is accelerated by acids. However, the solution in contact
with the iron must not be allowed to become too acidic; otherwise some hydrogen will form by
the reaction
Fe + 2 H+ —> Fe2+ + H2
EXPERIMENT:
Avoid breathing the acetone vapor or spilling it on your skin. Acetone is a flammable solvent.
There must be no open flames in the hood.
Attach a strip of masking tape (on which to mark the water level) to two large culture tubes
(lipless test tube). Attach a set of clamps to your ring stand so that the test tubes can be
mounted, inverted, in a 1-L beaker filled nearly to the brim with tap water. Weigh two 1.0-g
portions of fine (size 00) steel wool. Do not compress the material. Obtain 100 ml each of
acetone, l.0 M acetic acid and 0.1 M acetic acid. To save materials, please arrange with a
nearby student to share your acetone and 1.0 M acetic acid, as these materials may be used by
two students to clean their steel wool pieces. However, each student should obtain their own
0.1 M acetic acid.
Using forceps, rinse a piece of steel wool in acetone for about 30 seconds to remove any oily
material from the surface of the steel wool. Shake off the excess acetone, drain the steel wool
briefly on a paper towel, and transfer it to the 1.0 M acetic acid solution. With your forceps,
agitate the steel wool occasionally for about a minute. Then shake off the excess, drain on a
paper towel briefly, and put the steel wool in the beaker containing 0.1 M acetic acid, agitating
it for about 30 s. Using forceps, remove the steel wool and shake it vigorously to remove as
much solution as possible. Then insert the steel wool in the 20- x 150-mm culture tube,
pushing it to the bottom half of the tube. Do not compress the steel wool. It should be spread
over most of the bottom half of the test tube.
Immediately invert the test tube and carefully lower it into the beaker of water and clamp it in
position. The mouth of the test tube must be below the water level throughout the experiment.
The initial volume of air is assumed to be the total volume of the test tube (minus the volume
of steel wool and adhering solution, which will be determined later). Rinse the forceps in tap
water and dry. At 5- to 10- min intervals, mark
the rising water level on the masking tape. Using
a second test tube, prepare the remaining piece of
steel wool and carry out a duplicate run. While
you are waiting for the reaction to be completed,
weigh a clean dry 250-mL beaker to the nearest
0.1 g. Record its mass. When no further change
in water level can be detected (usually 20 to 30
min are required), wait 5 min longer, then adjust
the height of the test tube so that the water levels
inside and outside the tube are the same. (When
the levels are the same, the pressure inside the
tube will be equal to the atmospheric pressure.)
Now trap the water that has risen in the tube by
pressing a rubber stopper firmly against the
mouth of the tube. (The stopper should be larger
than the mouth of the tube so that it does not
enter the tube.) Ask a laboratory neighbor to
unclamp the tube while you are holding the
stopper against the mouth of the tube. Carefully
transfer the water you have trapped into the
previously weighed 250-mL beaker. Reweigh and
record the mass of the beaker plus water. The
volume of this water corresponds to the volume
of oxygen that has reacted with the steel wool.
With forceps, remove the steel wool and put it in
the same previously weighed beaker containing
the water. Reweigh and record the mass. Finally,
fill the empty test tube to the brim with water and
add it to the same beaker and reweigh and record
the mass again. The initial mass of the empty beaker and the subsequent three weighings will
provide data from which you can calculate the total volume of the test tube, correcting for the
volume occupied by the steel wool and adhering acetic acid solution.
Alternative method: Instead of attaching tape to the side, you may use a ruler submerged in the
water to follow the progress of the rising water. You will need to record the rulers
measurements at 5-10 minute intervals
Name __________________
Date___________________
DETERMINATION OF THE PERCENTAGE OF OXYGEN IN THE AIR
(WORKSHEET)
DATA:
Density of water = 0.997 g/mL
Density of steel wool = 7.70 g/mL
Run #
Mass of
Steel
wool
Mass of
Beaker
Mass of Beaker +
Water
Mass of Beaker +
Water + Wet Steel
Wool
Volume of
Test Tube
Mass of
Water to Fill
Test Tube
CALCULATIONS:
Volume of oxygen in moist air: (mass of beaker and water - mass of beaker)/density of water
Volume of steel wool: (mass of Fe/density of Fe)
Volume of adhering solution: ((mass of beaker, water and steel wool - mass of beaker and
water - mass of Fe) / density of water)
Volume of test tube: ((mass of water, beaker, steel wool and water to fill test tube - mass of
water, beaker and steel wool) / density of water)
Volume of air in tube (volume of test tube - volume of steel wool - volume of adhering
solution)
Percent Oxygen in the Air (volume of oxygen / volume of air in tube) x 100
Results
Run #
Percent Oxygen
Avg =
Conclusions
Sources of Error
PROBLEMS
1. Consider the reactions given on the first page of this handout. How many grams of iron are
required to convert 80 mL of oxygen at STP?
2a. Calculate the total pressure in the flask after all of the oxygen has been removed from the
air in the flask. Assume that the temperature is 25̊C, the initial barometric pressure is 760 torr
and that air is 21% oxygen by volume.
2b.Actually the air inside the flask was in contact with water. Water vapor accounted for 23.6
torr of the total 760 torr of "wet" air inside the flask. If the O2 was removed from this air,
would the pressure be higher or lower than in 2a?