The Chemical Formula for Methane, Ethane and Propane
Avogadro's Law
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Materials
Glass jacket
Slow eudiometer
H-base -PASSRod, stainless steel 1818,250 mm
Right angle clamp
Universal clamp
Gas bar
Rubber caps, pack of 20
Syringe 20 ml, Luer, 10 off
Cannula 0.45~13mm, Luer, 1 out of
Lab thermometer, -10...+150~C
Magnetic stirrer bars, 1 out of
Magnet, d= l 0 mm, l = 200 mm
Heating apparatus
Power regulator
Glass beaker, tall, 250 ml
Graduated vessel,l [,with handle
Digital barometer
Nomogram
High voltage supply unit, 0-10 kV
Connecting cable, 50 KV, 1000 mm
Steel cylinder oxygen, 2 1, filled
Reducing valve for oxygen
Table stand for 2 Isteel cylinders
Wrench for steel cylinders
Compressed gas, methane, 12 1
Compressed gas, ethane, 12 1
Space for notes
Compressed gas, propane, 7 1
Fine control valve
Hose clips, d = 8-12 mm, 2 pcs.
Funnel, do= 55 mm
Beads, 200 g
Spoon, special steel
Silicone rubber tubing, d = 7 mm
Sodium chloride, 500 g
Water, distilled, 5 1
Safety warning
The gases methane, ethane and propane are easily
flammable and form explosive mixtures with air.
All flames must be extinguished before gas samples
are taken.
Procedure
Prepare the gas bar by pouring so much water
through the funnel into the small gasometer that the
Erlenmeyer flask and the right-angled glass tube are
filled as air bubble-free as possible (Fig. 1). Excess
water runs out of the glass tube, so that the correct
amount of water required for filling adjusts itself
automatically.
Now connect the glass tube to the source of gas
(compressed gas container) with a length of silicone
tubing and start a slow flow of gas. The gas forces
the water out of the flask up into the funnel and is
Fig. 1
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P H W E series ol pmaim .Handbook Glarsisckelsynem.D PHYWE SYSTEME GMBH -0.37070 GWngen. Germany
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The Chemical Formula for Methane, Ethane and Propane
The Chemical Formula for Methane, Ethane and Propane
itself collected in the Erlenmeyer flask. When filling
is complete, close the end of the glass tube with a
rubber cap. Mount the eudiometer in the glass jacket
and fill the jacket with an aqueous salt solution
c(NaCI) = 3 molll (105 g NaCl in 600 ml water),
adding a few boiling stones (beads)(Fig. 2).
ture), ensuring that there is a permanent ignition
spark. Switch off the high voltage supply unit, tap the
movable plunger of the eudiometer gently a few
times and read off the volume shown by the edge of
the plunger. Calculate the hydrogen content of the
hydrocarbon from the difference in volumes. This
result allows you to construct a hypothetical compound and a corresponding equation for the reaction
with oxygen.
To test this hypothesis, carefully dry the inside of the
eudiometer and heat the salt solution which surrounds it up to boiling (approx. 103'C) by means of
the heating apparatus. At this temperature, switch on
the high voltage supply unit (maximum setting) and
inject the appropriate stoichiometric hydrocarbonloxygen mixture into the eudiometer. Read the
volume, convert it to standard conditions and check
your hypothesis.
Using the 20 ml syringe, draw 15 ml of oxygen and
2 ml of one of the pure hydrocarbons out of the small
gasometers. Switch on the high voltage supply unit
and set the voltage to the maximum value. Inject the
hydrocarbonloxygen mixture slowly through the rubber cap into the slow eudiometer (room tempera-
Fig. 2
I
Results
The injected gas mixture burns continuously in the
eudiometer at the permanent spark across the spark
gap. Water condenses on the colder eudiometer
surfaces. This is the reason why the volume between
the movable plunger and the fixed plunger with the
ignition device is smaller than that of the original gas
mixture. Table 1 shows some results obtained.
In the second part of the experiment, water cannot
condense on the hot eudiometer surfaces. The volume read is, after conversion to standard conditions,
in most cases larger than the injected gas volume.
Table 2 shows data obtained from measurements on
the combustion of stoichiometric mixtures in the cold
and and in the hot eudiometer. Conversion of the
volumes to standard conditions was made using the
nomogram 40440.00.
Table 1
Conclusions
The complete combustion of a hydrocarbon in an
excess of oxygen gives, acc. to:
Table 2
carbon dioxide and water (steam). Assuming that the
gases behave sufficiently ideally, we can determine
the chemical formula for the hydrocarbon used from
the decrease in volume found in the experiment:
According to Avogadro, when one part by volume (or
ml) of a gaseous hydrocarbon reacts with one part
34
12237
e P H W E SYSTEME GMBH 0.37070 Gainpen. Germany
PHVWE seder d publiiljon Handbwk Glass jack* system .
by volume of oxygen, they give one part by volume
of carbon dioxide:
C + 0,
so that with a "gaseousn C, acc. to
the initial volume of y + 1 parts by volume is reduced
by the volume of the hydrocarbon used toy parts by
volume.
The hydrogen contained in a hydrocarbon burns,
acc. to
to water. In the second part of the experiment, using
the hot eudiometer, the water is in vapour form and
reduces the decrease in volume resulting from the
reaction from 1 part by volume to 1 4 4 parts by
volume. Under the conditions used in the first part of
the experiment, however, water condensed on the
eudiometer surfaces, resulting in an additional decrease in volume of x/4 parts by volume.
, , the reaction
For a hydrocarbon of composition CH
can be described by the following equation:
CyH,
+ (y + x/4)
4
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~
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0,
+ y CO, + xf2 H20
In the first part of the experiment, this reaction leads
to a decrease in volume of A V = 1+x/4 parts by
volume per part by volume of the hydrocarbon used.
This allows us to determine the chemical formula for
obtained
methane from the results given in Table l,
in the first part of the experiment:
A decrease in volume V of 4 ml was found on the
combustion of 2 ml of methane. Correcting this to one
part by volume of hydrocarbon we have 412 = 1+x/4,
giving X = 4.
It is clear that the decrease in volume found comes
from the combination of the hydrocarbon volume
consumed and the oxygen volume which was converted to water. This allows us to calculate the theoretical volume of steam and the proportion of steam
to used gas. From this we have, when 2 ml 0, were
converted to water, a theoretical steam volume of
4 ml and a steam to gas proportion of 2, which
corresponds to 4 hydrogen atoms per molecule.
PHYWEse"s 01 pvblicstion .Handbaok Glass jecket rystem - 0 PHVWESYSTEME GMBH .D-37070 Muingen. Germany
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