You have learned from laboratory
experience that there is sometimes a difference between an experimental result and a
mathematical result. Using stoichiometry, we can mathematically determine the amount
of a product that should be formed during an experiment, yet we sometimes find that we
don't end up with exactly the right amount of product. We use sources of error to
explain this difference, and we have even done percent error calculations to calculate how
far off we are from the expect results. Percentage yield problems fall under this
type of problem. They simply allow us to calculate what percent of the expected
product we are able to account for by the end of our experiment. The formula that we
use is;
actual amount of product
percentage yield = ------------------------------------------- x 100
expected amount of product
Example 1. A student conducts a single displacement
reaction that produces 2.755 grams of copper. Mathematically he determines that
3.150 grams of copper should have been produced. Calculate the student's percentage
yield.
Solve:
actual amount of product: 2.755 g
expected amount of product: 3.150 g
actual amount of product
percentage yield = ------------------------------------------- x 100
expected amount of product
2.755g
percentage yield = --------------- x 100
3.150g
percentage yield = 87.4603174 %
percentage yield = 87.46 %
Example 2. A student completely reacts 5.00g of
magnesium with an excess of oxygen to produce magnesium oxide. Analysis reveals 8.10
g of magnesium oxide. What is the student's percentage yield?
Solve:
In this problem, you are given the actual amount of the
product, but you are not given the expected amount of the product. The second mass
shown, 5.00g, is the mass of one of the reactants. In order to determine the
expected amount of the product in this problem, you must begin with a mass-mass problem.
First, write a balanced chemical equation for the
reaction:
2Mg(s) + O2(g)
----> 2MgO(s)
Now, label the given and the unknown and
cross out the rest:
given
unknown
2Mg(s) + O2(g) ----> 2MgO(s)
5.00 g
X g
Change the mass given to moles by dividing
by the molar mass of Mg.
mass given: 5.00g
molar mass of Mg: 24.3 g
5.00g
Number of moles of Mg = -----------
24.3 g/mole
Number of moles of Mg = 0.206 moles
Compare the molar ratio between the given
and the unknown to determine the number of moles produced.
Coefficient of Mg; 2
Coefficient of MgO: 2
Number of moles of Mg: = 0.206 moles
Number of moles of MgO: = ?
number of moles of given
number of moles of unknown
-------------------------------- =
--------------------------------------
coefficient of given coefficient of unknown
0.206 moles X
moles
-------------- = -------------
2
2
Number of moles of MgO produced = 0.206
moles
Now, change the number of moles of MgO
produced to mass by multiplying by the molar mass of MgO.
# of moles of MgO = 0.206 moles
Molar mass of MgO = 40.3g/mole
mass = # of moles x molar mass
mass of MgO = 0.206 moles x 40.3 g/mole
mass of MgO = 8.30 g
Now, you are ready to solve the percentage
yield problem.
actual mass of MgO produced = 8.10 g
expected mass of MgO = 8.30 g
actual amount of product
percentage yield = ------------------------------------------- x 100
expected amount of product
8.10 g
percentage yield = ------------ x 100
8.30 g
percentage yield = 97.6 %
Now, practice what you have learned
with the worksheets and links below.
Please forward all questions, comments and criticisms to Gregory L. Curran.
© Copyright 2004 Fordham Preparatory School, All Rights Reserved.
Last Modified February 07, 2008 |
