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Percentage Yield

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

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  1. Percentage Yield of a Chemical Reaction

  2. 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

  3. 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.

  4. 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

  5. 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

  6. 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

  7. 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.

  8. 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

  9. % = (actual/theoretical) * 100% = (1.7g/9.1g) * 100% = 19% (to two sf) Does this answer make sense?

  10. 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?

  11. 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

  12. p 262 PP 31 – 33 p 264 PP 34 – 37 Homework . . . there’s more . . .

  13. 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%.

  14. 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?

  15. 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

  16. Homework PP #38, 39, 40 on p 269 SR #1 – 4 on p 270 Get started on Ch 7 review problems.

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