For the experiment, we needed to do 6 different
calculations. The first one, was to calculate the pressure of the lab and
convert it to atm (atmospheres). The pressure of the lab was of 760.0 mmHg, and
to pass it to atmospheres, we only had to do a rule of three with atm and mmHg,
that in fact, 760 mmHg is 1 atm, so, finally, the pressure in atmospheres was 1
atm. But because we needed to make sure of the significant figures, and there
were 4 significant figures in 760.0, we have to put 4 significant figures at
the end so the result at the end will be 1.000 atm.
The second calculation was the volume of water
displaced during the evaporation in the experiment. When each one in its
experiment calculated the water displaced, we made an average between all
the data and finally we obtained that the volume of H2 displaced was 13.9 ml,
but for the calculations, we needed to pass it to litres, so we only needed to
divide 13.9 by 1000. So at the end the volume of water displaced was of 0.0139
l. At the beginning, the volume of H2 displaced had 3 significant figures
because it was 13.9 ml, so we have to make sure that at the end it has 3
significant figures, and in the final result of 0.0139 l, it has 3 significant
figures.
The third calculation was to calculate the
temperature. Each one in it experiment had to calculate the temperature of the
liquid at the end of the experiment, but we had to do an average between all
the data of the temperature of each experiment and at the end the average temperature
was of 23.6 ºC. But for the calculations, we needed to pass it
to Kelvin so we only needed to add to the original degree temperature 273, so
the temperature was of 297.6 K. We have to make sure that at the end it has the
same number of decimal places than at the beginning as we've summed, so we see
it and we confirm that it's correct as at the end it has the same number of
decimal places than at the beginning.
The fourth calculation was to calculate the moles of
Mg and of H2, that we only needed to calculate the moles of Mg that was by
dividing the average mass of Mg in grams of the experiment by the molar mass of
Mg that is 24.31g/mol (Webqc.org, 2015). So 0.0135 g (average grams of the
experiment)/24.31 g/mol (molar mass of Mg) is 0.000555 moles of Mg. Then, only
with the reaction equation (Mg + 2HCl --> MgCl2 + H2), with the mole ratio
between Mg and H2 (1:1), we could conclude that the moles of H2 were the same
than the moles of Mg, 0.000555 moles. At the beginning, the average grams of
the experiment had 3 significant figures because it was 0.0135 g, so we have to
make sure that at the end it has 3 significant figures as it is the less number
of significant figures of the calculations, and in the final result of 0.000555
moles of Mg and of H2, it has 3 significant figures.
The fifth calculation was to calculate R that is the
gas constant. That was easy done by using the equation to calculate the gas
constant (R=P*V/T*moles). So with the equation, when we substituted the data,
the final result was that the gas constant (R) was 0.0843 atm l / K mol. So
that was the final result. Then, to calculate the %error, we needed the
literature value that is 0.082 atm l / K mol and the experimental value that is
the final result of the experiment that is 0.0843 atm l / K mol. At the
beginning, the lowest number of significant figures of the calculations was 3,
so we have to make sure that at the end it has 3 significant figures, and in
the final result of 0.0843 atm l / K mol, it has 3 significant figures.
Finally, when we had all the data collected, only with
a specific equation, we finally calculated the %error. That equation is % error
= (literature value - experimental value) / literature value) * 100. So at the
end, when we substituted the data, the final result of % error is 2.8%. That
means that the %error was very small so we all did very well the experiment. At
the beginning, the lowest number of significant figures of the calculations was
2, so we have to make sure that at the end it has 2 significant figures, and
CONCLUSION
As we said before on the calculations, the gas
constant of the experiment was 0.0843 atm l / K mol, and the percentage error
was 2.8%. The percetage error in fact, is small and we suppose with it that the
experiment was very accurate as the percentage of error was very small.
Finally, we calculated the percentage error with the literature gas constant
(R), that is 0.082 atm l / K mol (Anne Marie Helmenstine, 2015), we've
investigated and we've found the gas constant literature value on the previous
web page that is specialized on chemistry and on the gas constant definition
and explanation.
EVALUATION OF THE METHOD
First of all, we founded an error at the first part of
the experiment, the pieces of magnesium reacted with the oxigen making on it's
surface a layer of magnesium oxyde. We notice this error when we saw that the
magnesium wasn`t bright and polished as a pure magnesium piece will be.
This is a random error because were aren´t sure that all the magnesium has the
same amount of oxidation. This is good example of random error because it is
caused by a unpredictable change in the equipment and condition of an
experiment.
In the next step of the process of the experiment we
founded another error, the balance were we measure the weight of the piece of
magnesium could be badly calibrated, that type of balance is very sensible and
a bad calibration could change completely the result of the experiment. This is
a huge error that we should try to solve at the start of the experiment by tare
the balance well. It is a systematic error because it is a problem in the
experimental set-up that results in the measured values always deviating from
the value in the same direction, that is, always higher or always lower
temperature.
Another error that we founded during the process of
the experiment was that the pressure inside the graduated test tube wasn´t the
same as the pressure that was outside in the lab (760,0 mmHg). To solve this
problem we should use a pressure sensor inside the tube to measure the pressure of the graduated
test tube, in this way, the results will be much more precise. This is another
example of random error because the pressure varies in the different graduated
test tube.
The final error that we founded was that the
temperature of the distilled water that we put inside the beaker varies with
the temperature of the gas inside the graduated test tube, the difference of
temperature affects the final result so it isn´t precise. This is another
example of a random error because the temperature of the water varies between
one beaker and another, the temperature of the acid is the same in all the
groups because we take the acid from the same beaker but the water came from
different squeeze bottles with different temperatures, to solve this
problem we should measure the temperature of the water and the temperature of the acid.
