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Experiment
10
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Atomic
Weight of a Metal
Overview
In this experiment, you
used a gas buret to measure the volume of hydrogen gas that was
evolved when a small piece of magnesium or zinc was reacted with
acid. From the volume of hydrogen that was collected, you
should be able to calculate the number of moles of hydrogen
that was evolved, and from this (and the stoichiometry of the
reaction), the number of moles of Mg or Zn that must have
been present in the sample you took. Since you also weighed the
sample of Mg or Zn, you can calculate the atomic weight of the metal
you used (g/mole).
Data
Suppose the following data
were recorded for Trial 1 of the experiment (we won't bother with
data for the other Trials, but your own data table should be for all
three Trials).
|
|
Metal Used
|
Mg |
Mass beaker +
sample |
10.2011 g
|
Mass empty beaker
|
10.1243 g
|
Initial reading
of gas tube |
2.25 mL |
Final reading of
gas tube |
79.31 mL
|
Temperature of
hydrogen gas |
24oC
|
Difference in
water levels |
35.2 mm H2O
|
Temperature of
water in trough |
23oC
|
Density of water
at temp. of trough |
0.99751 g/mL (from
Appendix) |
Aqueous vapor
pressure at water temp. |
21.1 mm Hg (from
Appendix) |
Barometer reading
|
759.9 mm Hg (from
Instructor) |
Barometer
temperature |
24oC
(from Instructor) |
Barometer scale
correction |
3.00 mm Hg (from
Appendix) |
|
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Calculations
Page 106, Part II
1. Mass of Sample
The mass of metal sample
used is just the difference between the masses of the empty beaker
and the beaker with the sample in it.
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Mass of Sample =
10.2011 g - 10.1243 g = 0.0768 g |
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2.Volume of Gas Collected
The volume of hydrogen gas
just represents the difference in volume levels in the gas tube
before and after the reaction. For my data
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Volume of gas
collected = 79.31 mL - 2.25 mL = 77.06 mL |
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3. Corrected Barometric
Pressure
Remember that the apparent
height of the mercury column in a barometer is also affected by the
temperature. The Appendix to the lab manual lists corrections for
barometeric readings for different temperatures. For my data above,
the barometer was at 24oC and had a mercury height of
759.9 mm Hg. The correction listed in the Appendix for these
conditions is 3.0 mm Hg. So the correced barometric pressure is
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Corrected barometric
pressure = 759.9 mm Hg - 3.0 mm Hg = 756.9 mm Hg |
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4. Mercury Equivalent of
Difference in Water Levels
You'll remember that in
this experiment you had to measure in millimeters, with a ruler or
meterstick, the difference in height between the level of the water
in the gas tube and that in the trough or beaker. Since there was
liquid water inside the gas tube, the column of liquid water
contributes a portion of the total pressure inside the gas tube. The
total pressure inside the gas tube (which is equal to the
atmospheric pressure) is composed of: the pressure of the hydrogen
gas (what we want!), the pressure of water vapor which is mixed with
the hydrogen gas, and the pressure due to the column of liquid
water. For my data above, I found that the column of liquid water in
the gas tube had a height of 35.2 mm above the surface of the water
in the trough. Since we measure the pressure of gases in units of mm
Hg (torr), we can convert the pressure due to the 35.2 mm of water
into the equivalent level of Hg which would have the same pressure,
by using the densities of mercury (13.596 g/mL) and water (0.99751
g/mL at the temperature of my experiment.
5. Pressure of Hydrogen
gas
The total pressure inside
the gas tube (which is equal to the atmospheric pressure) is
composed of: the pressure of the hydrogen gas (what we want!), the
pressure of water vapor which is mixed with the hydrogen gas, and
the pressure due to the column of liquid water.
That is Ptotal
= Patmosphere = Phydrogen + Pwater vapor
+ Pliquid water
So if we subtract the
pressure of water vapor (21.1 mm Hg at the temperature of my
experiment) and the pressure due to the column of liquid water (2.58
mm Hg as calculated in Part 4 above) from the total corrected
barometric pressure (756.9 mm Hg from Part 3 above), this will give
us the pressure of the hydrogen gas itself.
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Phydrogen
= 756.9 mm Hg - 21.1 mm Hg - 2.58 mm Hg = 733.2 mm Hg
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6. Absolute temperature
of Hydrogen gas
The Celsius temperature
scale is a human invention, and we can't expect hydrogen gas
molecules to know about it. The temperature of my hydrogen sample
was 24oC, which has to be converted to Absolute (Kelvin)
degrees:
7. Moles of Hydrogen
Produced
All of the above
calculations have been leading us to being able to figure the amount
of hydrogen generated using the ideal gas law (PV = nRT).
Fill in the following data and solve for n -- the number of
moles of hydrogen:
- P = 733.2 mm Hg (733.2 torr)
- V = 77.06 mL (0.07706 L)
- R = 62.358 L torr/mol K
- T = 297 K
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8. Moles of Metal
For magnesium, the
balanced chemical equation is Mg + 2HCl ®
Mg Cl2 + H2
Since the coefficients of
magnesium metal and hydrogen gas are the same in the balanced
chemical equation, if 0.00305 moles of hydrogen gas was produced,
then the sample of magnesium reacted also must have consisted of
0.00305 moles.
9. Experimental
Gram-Atomic Weight (Molar Mass) of Metal
The gram atomic weight of
a metal represents the mass of 1 mole of the metal expressed in
grams. More recently, this has been termed the "molar mass" of the
metal. In any case, we want the number of "grams per mole" for the
metal. In Part 1 of the calculations, we determined that the sample
of magnesium weighed 0.0768 g. In Part 8 of the calculations, we
determined that the sample of magnesium consisted of 0.00305 moles.
Putting these together gives the molar mass
Parts 10, 11, 12
By now I'm sure you're an
expert at calculating means, deviations, and the average deviation
for a set of measurements. If you've forgotten, see the help on
Experiments 1, 2, 4, or 9.
Parts 13 and 14
The actual gram atomic
mass (molar mass) of magnesium is 24.31 g. In Part 14, you calculate
how far off your answer is on a percentage basis. For my data, the %
error is
Questions
We can't solve the problems for you, but
here are some hints as to how to go about the task.
- This question is essentially the same
as the calculations above, only going backwards. Plus, you don't
have to worry about all the corrections to the pressure, since the
pressure given (720 torr) can be assumed to be the pressure due to
the hydrogen alone. Use the ideal gas equation (PV = nRT)
with the given data to figure out n (which will be the number
of moles of hydrogen gas that is present in the sample). Then use
the balanced chemical equation to figure out how many moles of Al
metal would have to react to produce the calculated amount of
hydrogen (make sure the equation is balanced: this is not a
1:1 reaction). Finally, use the atomic mass of Al to figure out the
mass of Al for the calculated number of moles. The answer is 0.045
g, but you'll have to show a complete solution to receive credit on
your report.
- First figure out how many moles of K
is present in 0.0379 g K using the molar mass of K. Then, use the
balanced chemical equation to figure out how many moles of hydrogen
would result from this amount of K reacting with HCl. Finally, use
the ideal gas equation (PV = nRT) filling in this number of
moles, along with the conditions represented by "STP".
- This one is a killer!! You are really
going to have to use your skills in algebra to solve it. Approach
the problem this way: let x represent the mass of Mg in the
sample and let y represent the mass of Zn. We know that (x
+ y) = 0.1000 g. From the volume of hydrogen present at STP,
we can calculate the number of moles of hydrogen that the mixture
generated using the ideal gas equation. Since both Mg and Zn produce
hydrogen on a 1:1 stoichiometric basis (from the balanced chemical
equations), the number of moles of hydrogen produced is equal to the
total number of moles of Mg and Zn combined. How are
the number of moles of Mg and Zn related to their molar masses? In
terms of the x and y variables we have defined,
the number of moles of Mg is (x/24.31) and the number of
moles of Zn is (y/65.38). The rest of the calculation is left
to you......!!!
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