Percent Yield
• In theory, everything is perfect. People do what is right, even
if it hurts them, simply because that is what is right. If a
person works hard toward a goal, one day he will achieve it. If
I buy expensive, energy efficient light bulbs that are supposed
to last for 5 years, I will only have to buy light bulbs once
every 5 years.
• Unfortunately, we do not live in a theoretical world. In our
world, things go wrong. And as we’ve seen with many other
things, as goes the world, goes chemistry.
• Therefore, our perfectly balanced chemical equations that say
if we put in so many grams/moles of X we will get back so
many grams/moles of Y are victims of our imperfect world
also. In the real world, we will only get a percentage of the Y
out that the chemical equation predicted for us.
Percent Yield…continued
• Chemical Equation produces → Theoretical (perfect) Yield
• Real World produces → Actual (imperfect) Yield
• The actual yield can never be greater than the theoretical yield.
• The actual Yield can never be more than 100% of the theoretical yield.
• The discrepancy is not due to an incorrect equation. It is due to
imperfections in the process such as:
•
•
•
•
Reactants or products leak out, especially when they are gases.
The reactants are not 100% pure.
Some product is lost when it is purified.
There are also many chemical reasons, including:
o The products decompose back into reactants (as with the ammonia process).
o The products react to form different substances.
o Some of the reactants react in ways other than the one shown in the equation.
– These are called side reactions.
• The reaction occurs very slowly. This is especially true of reactions
involving organic substances.
Steps to Solve Percent Yield Problems
1.
2.
Identify the Limiting Reactant in the reaction.
Calculate the Theoretical Yield of the substance in question.
• Identify what you know and what you do not know and write it down.
• Convert any masses given to moles.
•
Mass of substance (g)
= Number of Moles
Molar Mass of substance (g/mol)
• Determine the Mole Ratio needed.
• Coefficient of unknown substance in BALANCED chemical equation
Coefficient of Limiting Reactant in BALANCED chemical equation
• # mol unknown molecule
# mol Limiting Reactant
• # mol comes from BALANCED chemical equation NOT the problem itself.
• Use the mole ratio to determine the amount of product the known reactant will
produce.
• Known Moles of Limiting Reactant X Mole Ratio = Moles of Unknown
• Convert moles of product to grams.
• # of Moles Calculated X Total Molar Mass (g/mol) = Theoretical Mass of substance (g)
3.
Determine % Yield
•
Mass of Actual Yield (g) X 100 = % Yield
Mass of Theoretical Yield (g)
Percent Yield Practice
We
1. Dichlorine monoxide, Cl2O, is sometimes used
as a powerful chlorinating agent in research. It
can be produced by passing chlorine gas over
heated mercury (II) oxide according to the
following equation: HgO + Cl2 → HgCl2 + Cl2O
What is the % yield, if the quantity of reactants
is sufficient to produce 0.86g of Cl2O but only
0.71g is obtained?
You
Percent Yield Practice …continued
2. Acetylene, C2H2, can be used as an industrial starting
material for the production of many organic
compounds. Sometimes, it is first brominated to form
1,1,2,2-tetrabromoethane, CHBr2CHBr2, which can then
be reacted in many different ways to make other
substances. The equation for the bromination of
acetylene follows: C2H2 + 2Br2 → CHBr2CHBr2
If 72.0g of C2H2 reacts with excess bromine and 729g of
the product is recovered, what is the % yield of the
reaction?
Percent Error Notes
Percent error is a calculation of how efficient your
lab technique is. Just like % Yield, it compares your
Theoretical yield to your Actual yield. However, the
focus is on what was not produced instead of what
was produced.
Steps to Solve % error Problems:
1. Identify your Actual and Theoretical Yields
2. Mass of Theoretical Yield(g) - Mass of Actual Yield (g) X 100 =
Mass of Theoretical Yield (g)
Percent Error Practice
We
1. Dichlorine monoxide, Cl2O, is sometimes used
as a powerful chlorinating agent in research. It
can be produced by passing chlorine gas over
heated mercury (II) oxide according to the
following equation: HgO + Cl2 → HgCl2 + Cl2O
What is the % yield, if the quantity of reactants
is sufficient to produce 0.86g of Cl2O but only
0.71g is obtained?
You
Percent Error Practice …continued
2. Acetylene, C2H2, can be used as an industrial starting
material for the production of many organic
compounds. Sometimes, it is first brominated to form
1,1,2,2-tetrabromoethane, CHBr2CHBr2, which can then
be reacted in many different ways to make other
substances. The equation for the bromination of
acetylene follows: C2H2 + 2Br2 → CHBr2CHBr2
If 72.0g of C2H2 reacts with excess bromine and 729g of
the product is recovered, what is the % yield of the
reaction?
Homework
1. Modern Chemistry Book page 321 questions
22-30
2. Modern Chemistry Book page 318 Practice
Questions 1 and 2
3. Whoosh Bottle Questions
4. Al Leftovers Lab Part II