CHE 115 LAB #8
Name: Matt Wilhelm
Date 11/8/2011
Experiment 8: Coordination Chemistry
Purpose
In this experiment a coordination compound was made by reacting copper ions with a common
ligand, ammonia, and the percent yield of the reaction was determined. A sample of copper
sulfate pentahydrate was obtained and massed. This sample was dissolved in water and then
reacted with ammonia and 95% ethanol. After the reaction, using a vacuum filtration system the
solid product was collected on a vacuum filter. During this process the substrate was washed
with ethanol/ammonia and 95% ethanol to completely dry solid. Once the product was dried
completely the mass of the product was recorded. This value was then used to calculate the
percent yield, theoretical yield and actual yield of the reaction. This process utilizes the ionic
properties of the solid being used in the reaction as well as the chemical and physical properties
of coordination compounds to obtain the results portrayed in the experiment.
Procedure
Procedure is listed in CHE 115 Lab General and Analytical Chemistry III Laboratory Manual
Fall Quarter 2011, DePaul University Experiment 8. There were no deviations from the
procedure listed in the Lab Manual.
Data and Results
Table 1, Raw Data
Mass of CuSO4 ٠ 5H2O (g)
Mass of beaker (g)
Mass of beaker & product (g)
Mass of product (g)
Molar mass [Cu(NH3)4]SO4٠ 5H2O (g/mol)
Percent yield [Cu(NH3)4]SO4٠ 5H2O
1.553 g
31.48 (g)
32.5204 (g)
1.0404 (g)
245.772 (g/mol)
68%
*The above data was collected and calculated during the experiment. The percent yield of the product was
68%.
Experimental Description
The synthetic process observed during the experiment was the dissolution of CuSO4٠ 5H2O and
the synthesis of [Cu(NH3)4]SO4٠ 5H2O. This process began by massing out a sample of copper
(CuSO4 ٠ 5H2O), for this experiment the mass used was 1.558 grams. This copper sample was
then dissolved in approximately 5.0 mL of water. Once this solid was dissolved 5 mL of 15 M
ammonia was added to the solution and covered with a watch glass so that no evaporation will
impact the results. The solution was light purple after adding the ammonia. Next and under a
hood, 8 mL of 95% ethanol was added to the solution and stirred with a glass rod. The solution
turned dark purple at this point. Once the solution was mixed properly it was cooled in an ice
bath for 10 minutes with a watch glass covering it. During the process the solution is able to
react and the solid forming is able to precipitate. During this process the vacuum filtration
system was set up. The filter paper was moistened so that no precipitate will get through the filter
and into the filtration beaker.
Once the solution had cooled for 10 minutes, the solution was slowly poured onto the filter paper
set up in the filtration device. The filtration device began sucking the liquid and aqueous material
through the filter while the filter paper collected the solid. The color of the filtrate was dark
bluish/purple and the solid forming on the filter paper appeared to be a light bluish/purple color.
The color of the filter paper was similar but lighter than the initial copper sample (CuSO4 ٠
5H2O). As the product was finishing filtering it was then washed with 10 mL of 1:1
ethanol/ammonia solution. Using a Pasteur pipet small portions of this wash solution were added
to top of filter paper. After the 10 mL of the ethanol/ammonia solution was finished the product
was washed once again with 4.5 mL of 95% ethanol. During this process the solid forming is
able to be isolated as the wash substances are stripping any unnecessary ions from the surface of
the solid. The solid was then dried for an additional 10 minutes.
Using a spatula the solid collected on the beaker was scraped into a pre-weighed beaker. The
initial and final beaker weights can be seen in Table 1. The final weight of the substance
collected in the beaker was established by using a mass by difference method shown below in the
calculations section. Using the final weight of the substance the theoretical yield, percent yield
and actual yield of the solid formed [Cu(NH3)4]SO4٠ 5H2O were calculated.
Calculations
Theoretical Yield
To calculate the theoretical yield of [Cu(NH3)4]SO4٠ 5H2O from the starting amount of CuSO4 ٠
5H2O the mass of the sample of CuSO4 ٠ 5H2O must be divided by the molecular mass of
CuSO4 ٠ 5H2O. This value is the moles of theoretical yield and must then by multiplied by the
molecular weight of the product, this will reveal the theoretical yield.
1.5528 g CuSO4 ٠ 5H2O (g) / 249.7 g/mol = 0.006218662 moles theoretical
0.006218662 moles theoretical x 245.772 g/mol = 1.5284 g theoretical yield
Actual Yield
The actual yield was found using a mass by difference technique. The mass of the beaker was
obtained. The product was the placed into the beaker and weighed again. The initial weight of
the beaker without the product was subtracted from the weight of the beaker and the product.
[Mass beaker + Product] – [Mass beaker] = Actual Yield (g)
32.520 g – 31.480 g = 1.040 g
Percent Yield
The percent yield of the product was established by using the theoretical yield and the actual
yield obtained above. The actual yield was divided by the theoretical yield. This value was then
multiplied by 100% and the resulting value is the percent yield.
[Actual Yield g] / [Theoretical Yield g] = Percent Yield %
1.0404 g / 1.52837 g = 0.680 x 100% = 68%
Discussion
Based upon the color of the filtrate it is not possible to obtain a 100% yield on this reaction. For
starters, obtaining 100% yield in any reaction performed by hand or imperfect machines is not
likely as there are many ways in which failure to achieve 100% yield can happen. With respect to
the color of the filtrate, it doesn’t appear likely to obtain 100% yield. The color of the initial
substrate was a definitive bright light blue color. The color of the filtrate was a darker blue and
the color of the solid collected on the filter was a lighter blue. With this being said, the color on
the filter and the color of the filtrate had similar appearances in places. The color of the filtrate is
obviously part of the compound that is not forming the solid and is definitively different from the
color of the solid forming on the filter. The color of the substance on the filter is more similar to
the initial compound used although there appears to be components of each substance both in the
filter and filtrate. With this being said it appears that some of the solid was able to get through
the filter and some of the liquid or aqueous material was dried to the filter. This means it is
unlikely to get a 100% yield as there is may be solid lost in the beaker through the filter or
imbedded on the filter itself. In addition, it is also possible that some of the solution evaporated
during the initial mixing phase even though a watch glass was used.