Percentage Yield

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# Percentage Yield - PowerPoint PPT Presentation

Percentage Yield. of a Chemical Reaction. Let’s look at your last Chemistry Test. You scored 32/40. What’s your % grade? (32/40) * 100% = 80% What is the “theoretical” grade on this test? (theoretical = highest possible grade) 40 What was your “actual” grade? 32.

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### Percentage Yield

of a Chemical Reaction

Let’s look at yourlast Chemistry Test

(32/40) * 100% = 80%

What is the “theoretical” grade on this test?

40

32

What has this to do with Chemistry?

Theoretical yield of a chemical reaction is

predicted by

stoichiometry.

The amount of product obtained by the chemist is the

actual yield.

Actual yields are often less than theoretical yields due to
• competing (side) reactions
• loss of product due to poor lab technique
• chemical equilibrium (See y’all next year!)
• impure reactants
• an impure or contaminated product
• a solid product that hasn’t been

sufficiently dried

Calculating Percentage Yield

% yield = actual yield * 100%

theoretical yield

Use same units for actual yield as for the

theoretical yield.

ie. grams/grams; mol/mol, etc

Sample Problem 1

Determine the % yield when 1.7 g of NH3 are produced from the reaction of 7.5 g N2 with sufficient H2, according to:

N2(g) + 3 H2(g)  2 NH3(g)

Steps to solve this problem:

• Determine theoretical yield using stoich. in units of

grams.

2. Calculate % yield.

Theoretical yield is

1 mol N2(g) : 2 NH3(g)

7.5 g

↓(/28.0 g/mol)

0.27 mol(x 2/1) 0.54 mol

↓x 17.0 g/mol

9.1 g NH3 is the

theoretical yield

% = (actual/theoretical) * 100%

= (1.7g/9.1g) * 100%

= 19% (to two sf)

Sample Problem 2

Calcium carbonate, CaCO3, thermally decomposes to produce

CaO and CO2 according to

CaCO3(s)  CaO(s) + CO2(g)

If the reaction proceeds with a 92.5% yield, what volume, at SATP, of CO2 can be expected if 12.4 g CaCO3 is heated?

1 mol CaCO3(s) : 1 mol CO2(g)

12.4 g

↓/ 100.1 g.mol

0.123 mol(x 1/1)0.123 mol

↓ x 24.0 L/mol

2.97 L theoretical

↓ x 0.925 yield

2.75 L actual yield at SATP

p 262 PP 31 – 33

p 264 PP 34 – 37

Homework

. . . there’s more . . .

Percentage Purity

Impure reactants are often the cause of less than 100% yield.

For example, the reactant

you massed is only 70% pure.

What will this do to the % yield?

Yield will be 70%.

Sample Problem 3

Iron pyrite (fool’s gold) has the formula FeS2.

When a 13.9 g sample of impure iron pyrite is heated in the presence of oxygen, O2, 8.02 g of Fe2O3 is produced according to:

4 FeS2(s) + 11 O2(g)2 Fe2O3(s) + 8 SO2(g)

What is the % purity of the iron pyrite sample?

4 FeS2(s) + 11 O2(g)2 Fe2O3(s) + 8 SO2(g)

4 mol FeS2 : 2 mol Fe2O3

8.02 g

↓/159.7 g/mol

0.100 mol(X4/2)0.0502 mol

↓*120.0 g/mol

12.0 g FeS2 This represents the mass

of pure FeS2.

% purity = (12.0 g/13.9 g) * 100%

= 86.3% is the purity of iron pyrite

Homework

PP #38, 39, 40 on p 269

SR #1 – 4 on p 270

Get started on Ch 7 review problems.