1. Percentage Yield of a Chemical Reaction

2. 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

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 NH 3 are produced from the reaction of 7.5 g N 2 with sufficient H 2, according to: N 2 (g) + 3 H 2 (g)  2 NH 3 (g) Steps to solve this problem: 1.Determine theoretical yield using stoich. in units of grams. 2. Calculate % yield.

8. Theoretical yield is 1 mol N 2 (g):2 NH 3 (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 NH 3 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, CaCO 3, thermally decomposes to produce CaO and CO 2 according to CaCO 3 (s)  CaO(s) + CO 2 (g) If the reaction proceeds with a 92.5% yield, what volume, at SATP, of CO 2 can be expected if 12.4 g CaCO 3 is heated?

11. 1 mol CaCO 3 (s):1 mol CO 2 (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 FeS 2. When a 13.9 g sample of impure iron pyrite is heated in the presence of oxygen, O 2, 8.02 g of Fe 2 O 3 is produced according to: 4 FeS 2 (s) + 11 O 2 (g)  2 Fe 2 O 3 (s) + 8 SO 2 (g) What is the % purity of the iron pyrite sample?

15. 4 FeS 2 (s) + 11 O 2 (g)  2 Fe 2 O 3 (s) + 8 SO 2 (g) 4 mol FeS 2 :2 mol Fe 2 O 3 8.02 g ↓/159.7 g/mol 0.100 mol  (X4/2)  0.0502 mol ↓*120.0 g/mol 12.0 g FeS 2 This represents the mass of pure FeS 2. % 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.