Transcript Percent Yield - Conejo Valley Unified School District

Percent Yield
Theoretical Yield
 The maximum amount of product that can be
produced from a given amount of reactant
 Theoretical yield is calculated using
stoichiometry
Actual Yield
 The amount of product actually produced
when the chemical reaction is carried out in
an experiment
Percent Yield
 A comparison of the actual and theoretical yield
 In general, the higher the yield, the better the
results are from the experiment.
% Yield = actual yield (experiment)
theoretical yield (calculation)
× 100
Steps
1. Identify what is given in the problem.
A.
B.
One product and one reactant (go to step 2)
One product and two reactants (go to step 3)
2. The product given is the actual yield, calculate
the theoretical yield using stoichiometry and
the reactant given
3. The product given is the actual yield, calculate
the limiting reactant and that becomes your
theoretical yield
4. Calculate the percent yield using the equation
Determine the theoretical yield of Ag2CrO4 if
0.500 g of AgNO3 is used to react with K2CrO4.
Also if 0.455 g of Ag2CrO4 is obtained from an
experiment, calculate the percent yield.
2 AgNO3 + 1 K2CrO4
0.500g AgNO3
1 mol AgNO3
1 Ag2CrO4 + 2 KNO3
1 mol AgCrO4
169.88 g AgNO3 2 mol AgNO3
331.74 g AgCrO4
1 mol AgCrO4
= 0.488 g AgCrO4
% Yield = Actual
×100 =
Theoretical
0.455 g AgCrO4
0.488 g AgCrO4
×100 = 93.2%
CCl4 was prepared by reacting 100.0 g of CS2
and 100.0 g Cl2. Calculate the theoretical yield
and percent yield if 65.0 g of CCl4 was obtained
from the reaction.
1 CS2 + 3 Cl2
100.0g CS2 1 mol CS2
76.15 g CS2
100.0g Cl2 1 mol Cl2
70.90 g Cl2
1 CCl4 + 1 S2Cl2
1 mol CCl4
153.8 g CCl4
1 mol CS2
1 mol CCl4
1 mol CCl4
153.8 g Ccl4
3 mol Cl2
1 mol CCl4
% Yield = Actual
×100 =
Theoretical
= 202.0 g CCl4
= 72.31 g CCl4
65.0 g CCl4
72.31 g CCl4
L.R. = theoretical
×100 = 89.9%
Silver bromide (AgBr) was prepared by reacting 200.0
g of magnesium bromide and 100.0 g of silver nitrate.
Calculate the theoretical and percent yield if 100.0 g of
silver bromide was obtained from the reaction.
1 MgBr2 + 2 AgNO3
200.0g MgBr2 1 mol MgBr2
2 mol AgBr
184.11 g MgBr2
100.0g AgNO3 1 mol AgNO3
1 Mg(NO3)2 + 2 AgBr
187.77 g AgBr
1 mol MgBr2 1 mol AgBr
2 mol AgBr
187.77 g AgBr
169.88 g AgNO3 2 mol AgNO3 1 mol AgBr
% Yield = Actual
×100 =
Theoretical
=408.0g AgBr
= 110.5 g AgBr
L.R. = theoretical
100.0 g AgBr
×100 = 90.5%
110.5 g AgBr