DATA AND OBSERVATIONS:
Appearance of copper oxide (before heating):
Grayish-black, opaque, solid powder.
Observations during the process of heating:
Glowing orange, opaque solid.
Appearance of copper (after heating):
Brown, shiny, opaque, solid.
Trial 1
Mass of empty test tube
Mass of empty test tube (dry)
34.249g
Mass of test tube and copper oxide
Mass of test tube and copper oxide (dry)
36.111g
Mass of test tube and remaining copper
35.725g
34.217g
36.085g
CALCULATIONS:
1) Mass of copper remaining:
Trial 1:
35.725g - 34.249g = 1.476g
Trial 2:
35.725g - 34.217g = 1.508g
2) Mass of oxygen removed:
Trial 1:
36.111g - 35.725g = 0.386g
Trial 2:
36.085g - 35.725g = 0.360g
3) Moles of copper:
Trial 1:
Trial 2:
4) Moles of oxygen:
Trial 1:
Trial 2:
1.476g 1 mole = 0.02323 moles
63.55g
1.508g 1 mole = 0.02373 moles
63.55g
0.386g 1 mole = 0.0241 moles
16.0g
0.360g 1 mole = 0.0225 moles
16.0g
5) Ratio of copper to oxygen:
Trial 2
35.725g
Trial 1:
Cu = 0.02323 = 0.964
O
0.0241
Cu = 0.02373 = 1.05
O
0.0225
Trial 2:
6) Empirical formula of compound:
Trial 1:
CuO
Trial 2:
CuO
7) Percent error:
Trial 1:
Trial 2:
(0.964 - 1.000) x 100 = -3.60% error
1.000
(1.05 - 1.000) x 100 = 5.00% error
1.000
QUESTIONS:
1) Copper (II) Oxide. The ratio is 1:1 meaning there is the same amount of copper as
there is oxygen.
2) We assumed in trial 1 that the test tube was dry and that the only mass lost during the
heating process was oxygen. In fact, we also lost some water vapor to account for the
mass differences.
3) The remaining oxygen within the unreduced copper oxide would be assumed to be
copper and therefore given the atomic mass of copper when calculating the moles of
particles involved. This would cause the copper amount to seem much higher than it
actually is and result in a numerator heavy fraction that would calculate to a positive
error.
4) The water vapor accounts for the seemingly “phantom” oxygen. Due to the fact that
the test tube was not purged of water vapor prior to the experiment, some water vapor
was removed during heating and accounted for (in the mass numbers) as being oxygen.
5) Mass of mercury remaining:
Mass of oxygen removed:
Moles of mercury:
Moles of oxygen:
Ratio of mercury to oxygen:
4.207g
0.334g
0.0210 moles
0.0209 moles
1.00
Empirical formula for this oxide of mercury:
6) Fe:
2.635g 1 mole = 0.04718 moles
HgO
1
x3
3
55.85g
O:
1.006g 1 mole = 0.06288 moles
1.33
16.00g
The empirical formula for this oxide of iron is Fe3 O4
7) Choose a 100.g sample.
C:
89.8g 1 mole = 7.48 moles
1
H:
10.2g 1 mole = 10.2 moles
1.33
The empirical formula for propadiene is C3 H4
x3
4
x3
x3
3
4
CONCLUSION:
1) We were trying to demonstrate that all this mumbo jumbo we’ve been learning about
the formulas of chemical compounds actually works in real life. The biggest assumption
we made is that we could effectively and accurately break down a chemical compound
and determine how much elemental “stuff” it was composed of.
2) Some sources of error were uncontrollable. During our lab, which was constricted due
to the time of the period, we were not able to continually retest our “dry” test tube to
insure that it was truly dry. We were also unable to use a moisture absorbing packet
throughout our tests to absorb the water vapor and prevent it from recondensing on the
surface of our test tube.
3) Performing the experiment in an environment where water vapor had been completely
absorbed would have resulted in much more accurate data. If one was also able to deduce
when every last bit of copper oxide had been fully heated and when all of the oxygen was
completely removed the level of accuracy would have been much higher. Scales of higher
precision would have also aided in attaining results with less error.