Yaworski 1
Michael Yaworski
Partner: Eric Kenny
SCH 3UI
15 May 2013
Lab Investigation 5.51 – Which Reagent Is Limiting and How Much Precipitate Is Formed?
Purpose:
The purpose of this lab was to test gravimetric stoichiometry by calculating the
theoretical yield of the precipitate formed and the actual yield of the precipitate formed in the
reaction of copper(II) sulfate with strontium chloride to find the percent yield.
Procedure:
1. 2.007 g of copper(II) sulfate pentahydrate, CuSO4 • 5H2O (s), was massed.
2. 1.994 g of strontium chloride hexahydrate, SrCl2 • 6H2O (s) was massed.
3. 35 mL of distilled water was measured and combined with the copper(II) sulfate
pentahydrate to form a solution in one beaker.
4. 37 mL of distilled water was measured and combined with the strontium chloride
hexahydrate to form a solution in separate beaker.
5. The two solutions aqueous copper(II) sulfate pentahydrate and aqueous strontium
chloride hexahydrate were combined in a beaker.
6. The filter paper was massed.
7. The mixture was filtered.
8. The filter paper was massed again.
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Observations:
Substances
copper(II) sulfate
pentahydrate
CuSO4 • 5H2O (s)
Properties
Mass (g)
Water Used
(mL)

light blue
2.007
N/A
strontium chloride
SrCl2 • 6H2O (s)

white
1.994
37
copper(II) sulfate solution
CuSO4 (aq)


vibrant blue
clear
N/A
35
strontium chloride
solution
SrCl2 (aq)


clear
colourless
N/A
N/A


white precipitate (before filtration)
white precipitate with green edges
precipitate + filter paper:
2.106
N/A
filter paper:
0.710
N/A
Mixture of Solutions
Filter Paper
(after filtration)


clear mixture
blue mixture

green coloured rim (after filtration)
Analysis:
The purpose of this lab was to test gravimetric stoichiometry by calculating the
theoretical yield of the precipitate formed and the actual yield of the precipitate formed in the
reaction of copper(II) sulfate with strontium chloride to find the percent yield.
To find the theoretical mass of the precipitate, the limiting reagent, if there was one, must
first have been identified.
The balanced chemical equation of the reaction was as follows:
CuSO4 (aq) + SrCl2 (aq) → CuCl2 (aq) + SrSO4 (s)
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The mass of copper(II) sulfate pentahydrate was observed and the molar mass of it was
calculated:
mCuSO4 • 5H2O = 2.007 g
MCuSO4 • 5H2O = MCu + MS + 4MO + 5 (2MH + MO)
MCuSO4 • 5H2O = 63.55 g/mol + 32.06 g/mol + 5 (2 • 1.01 g/mol + 16.00 g/mol)
MCuSO4 • 5H2O = 249.71 g/mol
The amount in moles of copper(II) sulfate pentahydrate was then calculated:
nCuSO4 • 5H2O =
nCuSO4 • 5H2O =
mCuSO4 • 5H2O
MCuSO4 • 5H2O
2.007 g
249.71 g/mol
← Amount in moles of copper(II) sulfate pentahydrate
available to consume all of strontium chloride hexahydrate.
The mole ratio of the reactants was then used to calculate the number in moles of the strontium
chloride hexahydrate required to consume all of the copper(II) sulfate pentahydrate:
nSrCl2 • 6H2O
1
=
nCuSO4 • 5H2O
1
nSrCl2 • 6H2O = nCuSO4 • 5H2O
nSrCl2 • 6H2O ≈ 8.037 • 10-3
← Approximate amount in moles of strontium chloride
hexahydrate required to consume all of copper(II) sulfate pentahydrate.
The amount in moles of strontium chloride hexahydrate available to consume all of the
copper(II) sulfate pentahydrate was then calculated in order to compare it to the amount required
to determine the limiting reagent.
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The mass of strontium chloride hexahydrate was observed and the molar mass of it was calculated:
mSrCl2 • 6H2O = 1.994 g
MSrCl2 • 6H2O = MSr + 2MCl + 6 (2MH + MO)
MSrCl2 • 6H2O = 87.62 g/mol + 2 (35.45 g/mol) + 6 (2 • 1.01 g/mol + 16.00 g/mol)
MSrCl2 • 6H2O = 266.64 g/mol
The amount in moles of strontium chloride hexahydrate was then calculated:
nSrCl2 • 6H2O =
mSrCl2 • 6H2O
MSrCl2 • 6H2O
nSrCl2 • 6H2O =
1.994 g
266.64 g/mol
nSrCl2 • 6H2O ≈ 7.748 • 10-3
← Approximate amount in moles of strontium chloride hexahydrate
available to consume all of copper(II) sulfate pentahydrate.
Therefore, because the amount of strontium chloride hexahydrate available to consume all of the
copper(II) sulfate pentahydrate was less than the amount required, strontium chloride
hexahydrate was the limiting reagent.
The molar mass of the precipitate, strontium sulfate, was then calculated:
MSrSO4 = MSr + MS + 4MO
MSrSO4 = 87.62 g/mol + 32.06 g/mol + 4 (16.00 g/mol)
MSrSO4 = 183.68 g/mol
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The mole ratio of the precipitate to the limiting reagent was then used to calculate the amount in
moles of the precipitate:
nSrSO4
nSrCl2 • 6H2O
=
1
1
nSrSO4 = nSrCl2 • 6H2O
nSrSO4 =
1.994 g
266.64 g/mol
The mass of the precipitate was then calculated based on the molar mass and number in moles of it:
mSrSO4 = nSrSO4 • MSrSO4
mSrSO4 =
1.994 g
• 183.68 g/mol
266.64 g/mol
mSrSO4 ≈ 1.374 g
Therefore, the theoretical mass of the precipitate was approximately 1.374 g.
The actual mass of the precipitate was then calculated based on observations:
mass of precipitate = mass of filter paper plus precipitate – mass of filter paper
mass of precipitate = 2.106 g – 0.710 g
mass of precipitate = 1.396 g
The percent yield of the precipitate was then calculated:
percent yield =
actual yield
x 100 %
theoretical yield
percent yield =
1.396 g
x 100 %
1.374 g
percent yield ≈ 101.6 %
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Therefore, because the percent yield is within ± 10% of 100%, according to the Nelson
textbook series, the experiment was successful in proving that gravimetric stoichiometry worked.
To note, however, the percent yield should have been less than 100%, as the actual mass
should have been less the theoretical mass: mass is always lost because of inevitable sources of
error, but mass can never be gained. A human error in the experiment was the solution was not
completely filtered through the filter paper, hence the green colour on the filter paper, and so the
massing of the filter paper included some solution, which increased the mass of the filter paper
and then the calculated mass of the precipitate. Other reasons for the percent yield being greater
than 100% were included in the sources of error.
Sources of Error:

The chemicals used were not completely pure which affected the theoretical calculations of the
mass of the precipitate, as well as the reaction that occurred between the two solutions. The
amount of purity of the chemical should have been accounted for in the theoretical calculations
because lower purity of the reactants resulted in a lower actual amount of the intended reactants;
the calculations were done assuming that the chemicals were 100% pure. That resulted in an
actual yield less than the theoretical yield. The purity of the chemicals could have also resulted in
other processes, such as oxidation, which was also not accounted for in the calculations.

Mass readings were not completely accurate in that they fluctuated constantly and it was
impossible to get a perfectly accurate mass reading on the reactant. Inaccurate mass readings
resulted in not perfectly calculated theoretical mass, which could have been more or less than it
should have been.
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
The transfer of the chemicals and solutions resulted in a loss of mass of the reactants and the
products. After the hydrated compounds were massed, they were moved to the beaker and some
mass could have been lost. More unavoidable, however, after the hydrated compounds were
mixed with water to form a solution, the solutions were then combined and a portion of the
solution was undoubtedly left in beaker it was being moved from, which also resulted in a loss of
mass of the reactants. Finally, when the mixture of solutions were moved from the beaker to the
filter paper, there was undoubtedly some of the mixture left in the beaker, which resulted in a loss
of mass of the product. All of the loss of mass of the reactants and products resulted in a smaller
actual yield than it should have been.

Getting all of the mixture out of the beaker when filtering it was impossible to do, especially
without dissolving the precipitate in the process. To get all, or most of, the precipitate out of the
beaker into the filter paper, excess amount of water would have to be used which would then
dissolve the precipitate. In an effort to avoid dissolving the precipitate in the beaker, enough
water was used to get a lot of the precipitate out of the beaker, but not all. That resulted in a loss
of product, as some precipitate remained in the beaker and was not accounted for in the actual
yield, so the actual yield was less than what it should have been.

After the mixture of the two solutions was filtered, the precipitate remained in the filter paper as a
white colour. However, after the precipitate was left to dry, the precipitate had a green tint on the
edges, just as the filter paper turned green, which indicated that there was partial solution left
unfiltered and remained in the filter paper with the precipitate. That error occurred because of the
effort to avoid dissolving the precipitate as well: the solution was not completely rinsed, with
water, through the filter paper. That error resulted in the observed mass of the precipitate being
greater that it actually was, as the solution was mixed with the precipitate still and was not
accounted for in the calculation. That increased the percent yield.