Transcript Percent yield

Chapter 9 Chemical
Quantities in Reactions
9.3
Limiting Reactants
On a space shuttle, the LiOH in the canisters
removes CO2 from the air.
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Theoretical, Actual, and
Percent Yield
Theoretical yield
• The maximum amount of product calculated using
the balanced equation.
Actual yield
• The amount of product obtained when the
reaction takes place.
Percent yield
• The ratio of actual yield to theoretical yield.
Percent yield =
actual yield (g)
x 100
theoretical yield (g)
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Calculating Percent Yield
Suppose you prepare cookie dough to make 5
dozen cookies. The phone rings and you answer.
While talking, a sheet of 12 cookies burns and you
have to throw them out. The rest of the cookies are
okay. What is the percent yield of edible cookies?
Theoretical yield 60 cookies possible
Actual yield
48 cookies to eat
Percent yield
48 cookies x 100% = 80.% yield
60 cookies
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Calculations for Percent Yield
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Learning Check
Without proper ventilation and limited oxygen, the
reaction of carbon and oxygen produces carbon
monoxide.
2C(g)
+
O2(g)
2CO(g)
What is the percent yield if 40.0 g of CO is
produced when 30.0 g of O2 is used?
1) 25.0%
2) 75.0%
3) 76.2%
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Solution
STEP 1 Write the given and needed quantities.
Given 40.0 g of CO produced (actual)
30.0 g of O2 used
Need percent yield of CO
STEP 2 Write a plan to calculate the theoretical yield
and the percent yield.
g of O2
moles of
moles of
g of CO
O2
CO
(theoretical)
Percent yield of CO = g of CO (actual)
x 100%
g of CO (theoretical)
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Solution (continued)
STEP 3 Write the molar masses and the mole-mole
factors from the balanced equation.
1 mol of O2 = 32.00 g of O2
1 mol O2 and 32.00 g O2
32.00 g O2
1 mol O2
1 mol of O2 = 2 mol of CO
1 mol O2 and 2 mol CO
2 mol CO
1 mol O2
1 mol of CO = 28.01 g of CO
1 mol CO and 28.01 g CO
28.01 g CO
1 mol CO
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Solution (continued)
STEP 4 Calculate the percent yield ratio by dividing the
actual yield (given) by the theoretical yield x 100%.
30.0 g O2 x 1 mol O2 x 2 mol CO x 28.01 g CO
32.00 g O2 1 mol O2
1 mol CO
= 52.5 g of CO (theoretical)
Percent yield:
40.0 g CO (actual)
x 100 = 76.2% yield (3)
52.5 g CO (theoretical)
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Learning Check
When N2 and 5.00 g of H2 are mixed, the
reaction produces 16.0 g of NH3. What is the
percent yield for the reaction?
N2(g) + 3H2(g)
2NH3(g)
1) 31.3 %
2) 56.7 %
3) 80.0 %
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Solution
STEP 1 Write the given and needed quantities.
Given 16.0 g of NH3 produced (actual)
5.00 g of H2
Need percent yield of NH3
STEP 2 Write a plan to calculate the theoretical yield
and the percent yield.
g of H2
moles of H2
moles of NH3
g of NH3
(theoretical)
Percent yield = g of NH3(actual)
x 100%
g of NH3(theoretical)
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Solution (continued)
STEP 3 Write the molar masses and the molemole factors from the balanced equation.
1 mol of H2 = 2.016 g of H2
1 mol H2 and 2.016 g H2
2.016 g H2
1 mol H2
3 mol of H2 = 2 mol of NH3
3 mol H2 and 2 mol NH3
2 mol NH3
3 mol H2
1 mol of NH3 = 17.03 g of NH3
1 mol NH3 and 17.03 g NH3
17.03 g NH3
1 mol NH3
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Solution (continued)
STEP 4 Calculate the percent yield ratio by
dividing the actual yield (given) by the
theoretical yield x 100%.
5.00 g H2 x 1 mol H2 x 2 mol NH3 x 17.03 g NH3
2.016 g H2 3 mol H2
1 mol NH3
= 28.2 g of NH3 (theoretical)
Percent yield = 16.0 g NH3 x 100 = 56.7 % (2)
28.2 g NH3
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