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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

Percentage Yield

of a Chemical Reaction

let s look at your last chemistry test
Let’s look at yourlast 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

what has this to do with chemistry
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.

slide4
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
slide5
Actual yields can also be greater than theoretical yields due to
  • an impure or contaminated product
  • a solid product that hasn’t been

sufficiently dried

calculating percentage yield
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
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.

slide8
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

slide9
% = (actual/theoretical) * 100%

= (1.7g/9.1g) * 100%

= 19% (to two sf)

Does this answer make sense?

sample problem 2
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?

slide11
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

slide12
p 262 PP 31 – 33

p 264 PP 34 – 37

Homework

. . . there’s more . . .

percentage purity
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
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 fes 2 s 11 o 2 g 2 fe 2 o 3 s 8 so 2 g
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
Homework

PP #38, 39, 40 on p 269

SR #1 – 4 on p 270

Get started on Ch 7 review problems.