# You have learned from laboratory experience that there is

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```					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                            Xg

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 %

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