1. Percent Yield

2. Definitions The theoretical yield is the maximum amount of product that could be made from the reactants. The actual yield is the amount of product that actually forms when the reactions happens. The percent yield is a ratio of the actual yield to the theoretical yield, expressed as a percent.

3. Equation % yield = actual yield x 100 theoretical yield

4. A real life problem A baker is following a recipe that is supposed to make 3 dozen cookies. She only gets 33 cookies when she is done. What is the % yield for her recipe?

5. Cookies % yield = actual yield x 100 theoretical yield

6. Cookies % yield = actual yield x 100 theoretical yield 33 X 100 = 91.66666667 = 92% 36

7. CaCO 3  CaO + CO 2 You have 24.8 g of CaCO 3. Determine the theoretical yield of CaO and then the % yield of this reaction if the actual yield was 12.0 g of CaO was obtained. 4 steps! 1. Change grams of reactant to moles of reactant. 2. Convert moles of reactant to moles of product. 3. Convert moles of product to grams of product. 4. Calculate the % yield.

8. CaCO 3  CaO + CO 2 You have 24.8 g of CaCO 3. Determine the theoretical yield of CaO and then the % yield of this reaction if the actual yield was 12.0 g of CaO was obtained. 1. Change grams of reactant to moles of reactant.

9. CaCO 3  CaO + CO 2 You have 24.8 g of CaCO 3. Determine the theoretical yield of CaO and then the % yield of this reaction if the actual yield was 12.0 g of CaO was obtained. 1. Change grams of reactant to moles of reactant. 24.8 g x 1 mol = 0.247786903 mol CaCO 3 1 100.086 g

10. CaCO 3  CaO + CO 2 You have 24.8 g of CaCO 3. Determine the theoretical yield of CaO and then the % yield of this reaction if the actual yield was 12.0 g of CaO was obtained. 2. Convert moles of reactant to moles of product.

11. CaCO 3  CaO + CO 2 You have 24.8 g of CaCO 3. Determine the theoretical yield of CaO and then the % yield of this reaction if the actual yield was 12.0 g of CaO was obtained. 2. Convert moles of reactant to moles of product. 0.247786903 mol CaCO 3 x 1 mol CaO = 1 1 mol CaCO 3 =0.247786903 mol CaO

12. CaCO 3  CaO + CO 2 You have 24.8 g of CaCO 3. Determine the theoretical yield of CaO and then the % yield of this reaction if the actual yield was 12.0 g of CaO was obtained. 3. Convert moles of product to grams of product.

13. CaCO 3  CaO + CO 2 You have 24.8 g of CaCO 3. Determine the theoretical yield of CaO and then the % yield of this reaction if the actual yield was 12.0 g of CaO was obtained. 3. Convert moles of product to grams of product. 0.247786903 mol CaO x 56.077 g CaO 1 1 mol CaO = 13.89514616 g CaO rounded to 13.9 g CaO. This is your theoretical yield!

14. CaCO 3  CaO + CO 2 You have 24.8 g of CaCO 3. Determine the theoretical yield of CaO and then the % yield of this reaction if the actual yield was 12.0 g of CaO was obtained. 4. Calculate the % yield.

15. CaCO 3  CaO + CO 2 You have 24.8 g of CaCO 3. Determine the theoretical yield of CaO and then the % yield of this reaction if the actual yield was 12.0 g of CaO was obtained. 4. Calculate the % yield. 12.0g x 100 13.9g = 86.3309353 rounded to 86.3%

16. Fe 2 O 3 + 3CO  2Fe + 3CO 2 84.8 g of Fe 2 O 3 is available in this reaction. 50.0 grams of Fe are made. What is the % yield? 1. Change grams of reactant to moles of reactant. 2. Convert moles of reactant to moles of product. 3. Convert moles of product to grams of product. 4. Calculate the % yield.

17. Fe 2 O 3 + 3CO  2Fe + 3CO 2 84.8 g of Fe 2 O 3 is available in this reaction. 50.0 grams of Fe are made. What is the % yield? 1. Change grams of reactant to moles of reactant.

18. Fe 2 O 3 + 3CO  2Fe + 3CO 2 84.8 g of Fe 2 O 3 is available in this reaction. 50.0 grams of Fe are made. What is the % yield? 1. Change grams of reactant to moles of reactant. 84.8 g Fe 2 O 3 x 1 mol Fe 2 O 3 1 159.687 g Fe 2 O 3 = 0.531038845 mol Fe 2 O 3

19. Fe 2 O 3 + 3CO  2Fe + 3CO 2 84.8 g of Fe 2 O 3 is available in this reaction. 50.0 grams of Fe are made. What is the % yield? 2. Convert moles of reactant to moles of product.

20. Fe 2 O 3 + 3CO  2Fe + 3CO 2 84.8 g of Fe 2 O 3 is available in this reaction. 50.0 grams of Fe are made. What is the % yield? 2. Convert moles of reactant to moles of product. 0.531038845 mol Fe 2 O 3 x 2 mol Fe 1 1 mol Fe 2 O 3 = 1.062077689 mol Fe

21. Fe 2 O 3 + 3CO  2Fe + 3CO 2 84.8 g of Fe 2 O 3 is available in this reaction. 50.0 grams of Fe are made. What is the % yield? 3. Convert moles of product to grams of product.

22. Fe 2 O 3 + 3CO  2Fe + 3CO 2 84.8 g of Fe 2 O 3 is available in this reaction. 50.0 grams of Fe are made. What is the % yield? 3. Convert moles of product to grams of product. 1.062077689 mol Fe x 55.845 g Fe 1 1 mol Fe = 59.31172857 g Fe Rounded to 59.3 g Fe (this is your theoretical yield).

23. Fe 2 O 3 + 3CO  2Fe + 3CO 2 84.8 g of Fe 2 O 3 is available in this reaction. 50.0 grams of Fe are made. What is the % yield? 4. Calculate the % yield.

24. Fe 2 O 3 + 3CO  2Fe + 3CO 2 84.8 g of Fe 2 O 3 is available in this reaction. 50.0 grams of Fe are made. What is the % yield? 4. Calculate the % yield. 50.0g x 100 = 84.317... Rounded to 84.3% 59.3g

25. Last one: CaCO 3  CaO + CO 2 What is the percent yield if 13.1 g CaO is actually produced when 24.8g CaCO 3 is available? 1. Change grams of reactant to moles of reactant. 2. Convert moles of reactant to moles of product. 3. Convert moles of product to grams of product. 4. Calculate the % yield.

26. CaCO 3  CaO + CO 2 What is the percent yield if 13.1 g CaO is actually produced when 24.8g CaCO 3 is available? 1. Change grams of reactant to moles of reactant.

27. CaCO 3  CaO + CO 2 What is the percent yield if 13.1 g CaO is actually produced when 24.8g CaCO 3 is available? 1. Change grams of reactant to moles of reactant. 24.8 g CaCO 3 x 1 mol CaCO 3 1 100.086 g CaCO 3 =0.247786903 mol CaCO 3

28. CaCO 3  CaO + CO 2 What is the percent yield if 13.1 g CaO is actually produced when 24.8g CaCO 3 is available? 2. Convert moles of reactant to moles of product.

29. CaCO 3  CaO + CO 2 What is the percent yield if 13.1 g CaO is actually produced when 24.8g CaCO 3 is available? 2. Convert moles of reactant to moles of product. 0.247786903 mol CaCO 3 x 1 mol CaO 1 1 mol CaCO 3 = 0.247786903 mol CaO

30. CaCO 3  CaO + CO 2 What is the percent yield if 13.1 g CaO is actually produced when 24.8g CaCO 3 is available? 3. Convert moles of product to grams of product.

31. CaCO 3  CaO + CO 2 What is the percent yield if 13.1 g CaO is actually produced when 24.8g CaCO 3 is available? 3. Convert moles of product to grams of product. 0.247786903 mol CaO x 56.077 g CaO 1 1 mole CaO = 13.89514617 g CaO rounded to 13.9 g CaO This is your theoretical yield

32. CaCO 3  CaO + CO 2 What is the percent yield if 13.1 g CaO is actually produced when 24.8g CaCO 3 is available? 4. Calculate the % yield.

33. CaCO 3  CaO + CO 2 What is the percent yield if 13.1 g CaO is actually produced when 24.8g CaCO 3 is available? 4. Calculate the % yield. 13.1 g CaO x 100 = 94.2446... = 94.2% 13.9 g CaO

34. Practice Complete the % yield practice sheet. If you do not finish, this will become homework, due at the beginning of our next class.