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Stoichiometry and the Math of Equations Part 4: Percent Yield 1

Stoichiometry and the Math of Equations Part 4: Percent Yield 1

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Page 1: Stoichiometry and the Math of Equations Part 4: Percent Yield 1

Stoichiometry and the Math of Equations

Part 4: Percent Yield

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Page 2: Stoichiometry and the Math of Equations Part 4: Percent Yield 1

Objectives

• Explain that the final answer to an excess-limiting problem is the theoretical amount of product made

• Use this knowledge and experimental data to calculate the percentage yield of a reaction

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Page 3: Stoichiometry and the Math of Equations Part 4: Percent Yield 1

Percent Yield• The final answer to ANY excess-limiting

problem is the theoretical yield of product- the amount of product predicted from the amounts of the reactants given – Represents the maximum amount of

product that can be made• Reality:

– The theoretical yield is rarely obtained in a real laboratory setting

– Side reactions consume some of the reactants or products

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Page 4: Stoichiometry and the Math of Equations Part 4: Percent Yield 1

Percent Yield• Percent yield - the actual yield, the

amount actually obtained, is often compared to the theoretical yield as a percentage

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Page 5: Stoichiometry and the Math of Equations Part 4: Percent Yield 1

Percent Yield• 68.5 g of CO is reacted with 8.60 g of H2

gas to produce methanol (CH3OH). If 55.7 g of methanol is actually produced, calculate the % yield of the methanol

• The formula for calculating percent yield is:

– Actual = What you got (experimental)– Theoretical = What you should have

gotten in theory (expected)5

Page 6: Stoichiometry and the Math of Equations Part 4: Percent Yield 1

68.5 g of CO is reacted with 8.60 g of H2 gas to produce methanol (CH3OH). If 55.7 g of

methanol is actually produced, calculate the % yield of the methanol

First – find the theoretical yield (do an excess-limiting problem)

68.5 g 8.60 g ? g CO + 2 H2 → CH3OH

1 mole 2 mole 1 mole

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Page 7: Stoichiometry and the Math of Equations Part 4: Percent Yield 1

68.5 g of CO is reacted with 8.60 g of H2 gas to produce methanol (CH3OH). If 55.7 g of

methanol is actually produced, calculate the % yield of the methanol

Find out which one is limiting: 68.5g CO x1 mol CO x 2 mol H2 x 2.02g H2

1 28.0g CO 1 mol CO 1 mol H2

= 9.88 g H2 so H2 is limiting

8.60g H2x 1 mol H2 x 1 mol CH3OH x 32.0g CH3OH =

1 2.02g H2 2 mol H2 1 mol CH3OH

= 68.1 g CH3OH

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Page 8: Stoichiometry and the Math of Equations Part 4: Percent Yield 1

68.5 g of CO is reacted with 8.60 g of H2 gas to produce methanol (CH3OH). If 55.7 g of methanol is actually produced, calculate the % yield of the

methanol

• Next, find the percent yield:What we got (given in problem)

What we should have gotten

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Page 9: Stoichiometry and the Math of Equations Part 4: Percent Yield 1

Objectives

• Explain that the final answer to an excess-limiting problem is the theoretical amount of product made

• Use this knowledge and experimental data to calculate the percentage yield of a reaction

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