1. CHAPTER 11 Stoichiometry 11.2 Percent Yield and Concentration

2. 2 In theory, all 100 kernels should have popped. Did you do something wrong? + 100 kernels82 popped18 unpopped

3. 3 11.2 Percent Yield and Concentration + 100 kernels82 popped18 unpopped In real life (and in the lab) things are often not perfect In theory, all 100 kernels should have popped. Did you do something wrong? No

4. 4 11.2 Percent Yield and Concentration + 100 kernels82 popped18 unpopped What you get to eat! Percent yield

5. 5 11.2 Percent Yield and Concentration + 100 kernels82 popped18 unpopped What you get to eat! Percent yield

6. 6 11.2 Percent Yield and Concentration actual yield: the amount obtained in the lab in an actual experiment. theoretical yield: the expected amount produced if everything reacted completely.

7. 7 11.2 Percent Yield and Concentration Percent yield in the lab Decomposition of baking soda: 2NaHCO 3 (s) → Na 2 CO 3 (s) + H 2 O(l) + CO 2 (g) Heating

8. 8 11.2 Percent Yield and Concentration Percent yield in the lab Decomposition of baking soda: 2NaHCO 3 (s) → Na 2 CO 3 (s) + H 2 O(l) + CO 2 (g) 10.00 g4.87 g measured experimentally Can you think of reasons why the final mass of Na 2 CO 3 may not be accurate? (What could be sources of error?)

9. 9 11.2 Percent Yield and Concentration Percent yield in the lab Decomposition of baking soda: 2NaHCO 3 (s) → Na 2 CO 3 (s) + H 2 O(l) + CO 2 (g) 10.00 g4.87 g measured experimentally - There is usually some human error, like not measuring exact amounts carefully Can you think of reasons why the final mass of Na 2 CO 3 may not be accurate? (What could be sources of error?)

10. 10 11.2 Percent Yield and Concentration Percent yield in the lab Decomposition of baking soda: 2NaHCO 3 (s) → Na 2 CO 3 (s) + H 2 O(l) + CO 2 (g) 10.00 g4.87 g measured experimentally - There is usually some human error, like not measuring exact amounts carefully - Maybe the heating time was not long enough; not all the Na 2 HCO 3 reacted Can you think of reasons why the final mass of Na 2 CO 3 may not be accurate? (What could be sources of error?)

11. 11 11.2 Percent Yield and Concentration Percent yield in the lab Decomposition of baking soda: 2NaHCO 3 (s) → Na 2 CO 3 (s) + H 2 O(l) + CO 2 (g) 10.00 g4.87 g measured experimentally - There is usually some human error, like not measuring exact amounts carefully - Maybe the heating time was not long enough; not all the Na 2 HCO 3 reacted - Maybe Na 2 CO 3 was not completely dry; some H 2 O(l) was measured too Can you think of reasons why the final mass of Na 2 CO 3 may not be accurate? (What could be sources of error?)

12. 12 11.2 Percent Yield and Concentration Percent yield in the lab Decomposition of baking soda: 2NaHCO 3 (s) → Na 2 CO 3 (s) + H 2 O(l) + CO 2 (g) 10.00 g4.87 g measured experimentally - There is usually some human error, like not measuring exact amounts carefully - Maybe the heating time was not long enough; not all the Na 2 HCO 3 reacted - Maybe Na 2 CO 3 was not completely dry; some H 2 O(l) was measured too - CO 2 is a gas and does not get measured Can you think of reasons why the final mass of Na 2 CO 3 may not be accurate? (What could be sources of error?)

13. 13 11.2 Percent Yield and Concentration Percent yield in the lab Decomposition of baking soda: 2NaHCO 3 (s) → Na 2 CO 3 (s) + H 2 O(l) + CO 2 (g) 10.00 g4.87 g measured experimentally Let’s calculate the percent yield obtained in experiment calculated

14. 14 11.2 Percent Yield and Concentration Percent yield in the lab Decomposition of baking soda: 2NaHCO 3 (s) → Na 2 CO 3 (s) + H 2 O(l) + CO 2 (g) 10.00 g4.87 g measured experimentally Let’s calculate the percent yield calculated

15. 15 11.2 Percent Yield and Concentration Percent yield in the lab Decomposition of baking soda: 2NaHCO 3 (s) → Na 2 CO 3 (s) + H 2 O(l) + CO 2 (g) 10.00 g4.87 g measured experimentally Let’s calculate the percent yield calculated

16. 16 11.2 Percent Yield and Concentration 2NaHCO 3 (s) → Na 2 CO 3 (s) + H 2 O(l) + CO 2 (g) 10.00 g Use the mass of reactant NaHCO 3 (s) to calculate the mass of the product Na 2 CO 3 (s). This is a gram-to-gram conversion:

17. 17 11.2 Percent Yield and Concentration 2NaHCO 3 (s) → Na 2 CO 3 (s) + H 2 O(l) + CO 2 (g) 10.00 g

18. 18 11.2 Percent Yield and Concentration 2NaHCO 3 (s) → Na 2 CO 3 (s) + H 2 O(l) + CO 2 (g) 10.00 g

19. 19 11.2 Percent Yield and Concentration 2NaHCO 3 (s) → Na 2 CO 3 (s) + H 2 O(l) + CO 2 (g) 10.00 g 0.1190 moles

20. 20 11.2 Percent Yield and Concentration 2NaHCO 3 (s) → Na 2 CO 3 (s) + H 2 O(l) + CO 2 (g) 10.00 g 0.1190 moles

21. 21 11.2 Percent Yield and Concentration 2NaHCO 3 (s) → Na 2 CO 3 (s) + H 2 O(l) + CO 2 (g) 10.00 g 0.1190 moles0.05950 moles

22. 22 11.2 Percent Yield and Concentration 2NaHCO 3 (s) → Na 2 CO 3 (s) + H 2 O(l) + CO 2 (g) 10.00 g 0.05950 moles

23. 23 11.2 Percent Yield and Concentration 2NaHCO 3 (s) → Na 2 CO 3 (s) + H 2 O(l) + CO 2 (g) 10.00 g 0.05950 moles6.306 g

24. 24 11.2 Percent Yield and Concentration 2NaHCO 3 (s) → Na 2 CO 3 (s) + H 2 O(l) + CO 2 (g) 10.00 g 0.05950 moles 6.306 g10.00 g 0.1190 moles 4.87 g For 10.00 g of starting material (NaHCO 3 ), the theoretical yield for Na 2 CO 3 is 6.306 g. The actual yield (measured) is 4.87 g.

25. 25 11.2 Percent Yield and Concentration 2NaHCO 3 (s) → Na 2 CO 3 (s) + H 2 O(l) + CO 2 (g) 10.00 g For 10.00 g of starting material (NaHCO 3 ), the theoretical yield for Na 2 CO 3 is 6.306 g. The actual yield (measured) is 4.87 g. 4.87 g

26. 26 11.2 Percent Yield and Concentration Decomposition of baking soda: 2NaHCO 3 (s) → Na 2 CO 3 (s) + H 2 O(l) + CO 2 (g) 10.00 g Reaction of solid zinc with hydrochloric acid: Zn(s) + 2HCl(aq) → H 2 (g) + ZnCl 2 (aq) 50.0 mL of a 3.0 M solution Convert to moles Reactions in solution Stoichiometry with solutions

27. 27 11.2 Percent Yield and Concentration A sample of zinc metal (Zn) reacts with 50.0 mL of a 3.0 M solution of hydrochloric acid (HCl) according to: Zn(s) + 2HCl(aq) → H 2 (g) + ZnCl 2 (aq). How many grams of hydrogen gas (H 2 ) will be produced? Assume zinc metal is present in excess.

28. 28 11.2 Percent Yield and Concentration A sample of zinc metal (Zn) reacts with 50.0 mL of a 3.0 M solution of hydrochloric acid (HCl) according to: Zn(s) + 2HCl(aq) → H 2 (g) + ZnCl 2 (aq). How many grams of hydrogen gas (H 2 ) will be produced? Assume zinc metal is present in excess. Asked: grams of H 2 produced Given: 50.0 mL of 3.0 M HCl reacting with excess zinc Relationships: M = mole/L Mole ratio: 2 moles HCl ~ 1 mole H 2 Molar mass of H 2 = 1.0079 x 2 = 2.02 g/mole

29. 29 11.2 Percent Yield and Concentration A sample of zinc metal (Zn) reacts with 50.0 mL of a 3.0 M solution of hydrochloric acid (HCl) according to: Zn(s) + 2HCl(aq) → H 2 (g) + ZnCl 2 (aq). How many grams of hydrogen gas (H 2 ) will be produced? Assume zinc metal is present in excess. Solve: Asked: grams of H 2 produced Given: 50.0 mL of 3.0 M HCl reacting with excess zinc Relationships: M = mole/L Mole ratio: 2 moles HCl ~ 1 mole H 2 Molar mass of H 2 = 1.0079 x 2 = 2.02 g/mole

30. 30 11.2 Percent Yield and Concentration A sample of zinc metal (Zn) reacts with 50.0 mL of a 3.0 M solution of hydrochloric acid (HCl) according to: Zn(s) + 2HCl(aq) → H 2 (g) + ZnCl 2 (aq). How many grams of hydrogen gas (H 2 ) will be produced? Assume zinc metal is present in excess. Solve: Asked: grams of H 2 produced Given: 50.0 mL of 3.0 M HCl reacting with excess zinc Relationships: M = mole/L Mole ratio: 2 moles HCl ~ 1 mole H 2 Molar mass of H 2 = 1.0079 x 2 = 2.02 g/mole

31. 31 11.2 Percent Yield and Concentration A sample of zinc metal (Zn) reacts with 50.0 mL of a 3.0 M solution of hydrochloric acid (HCl) according to: Zn(s) + 2HCl(aq) → H 2 (g) + ZnCl 2 (aq). How many grams of hydrogen gas (H 2 ) will be produced? Assume zinc metal is present in excess. Solve: Asked: grams of H 2 produced Given: 50.0 mL of 3.0 M HCl reacting with excess zinc Relationships: M = mole/L Mole ratio: 2 moles HCl ~ 1 mole H 2 Molar mass of H 2 = 1.0079 x 2 = 2.02 g/mole

32. 32 11.2 Percent Yield and Concentration A sample of zinc metal (Zn) reacts with 50.0 mL of a 3.0 M solution of hydrochloric acid (HCl) according to: Zn(s) + 2HCl(aq) → H 2 (g) + ZnCl 2 (aq). How many grams of hydrogen gas (H 2 ) will be produced? Assume zinc metal is present in excess. Solve: Asked: grams of H 2 produced Given: 50.0 mL of 3.0 M HCl reacting with excess zinc Relationships: M = mole/L Mole ratio: 2 moles HCl ~ 1 mole H 2 Molar mass of H 2 = 1.0079 x 2 = 2.02 g/mole Answer: 0.15 grams of H 2 are produced

33. 33 11.2 Percent Yield and Concentration Reaction of solid zinc with hydrochloric acid: Zn(s) + 2HCl(aq) → H 2 (g) + ZnCl 2 (aq) 50.0 mL of a 3.0 M solution Convert molarity to moles Vinegar is 5% acetic acid by mass Sometimes the concentration is written in mass percent

34. 34 11.2 Percent Yield and Concentration Commercial vinegar is reported to be 5% acetic acid (C 2 H 4 O 2 ) by mass. How many grams of acetic acid are in 120 mL of commercial vinegar? (Assume the density of vinegar is the same as pure water, 1.0 g/mL.)

35. 35 11.2 Percent Yield and Concentration Commercial vinegar is reported to be 5% acetic acid (C 2 H 4 O 2 ) by mass. How many grams of acetic acid are in 120 mL of commercial vinegar? (Assume the density of vinegar is the same as pure water, 1.0 g/mL.) Asked: grams of acetic acid in 120 mL of vinegar Given: 120 mL of vinegar and 5% acetic acid by mass Relationships: 120 mL = 120 g, given a density of 1.0 g/mL

36. 36 11.2 Percent Yield and Concentration Commercial vinegar is reported to be 5% acetic acid (C 2 H 4 O 2 ) by mass. How many grams of acetic acid are in 120 mL of commercial vinegar? (Assume the density of vinegar is the same as pure water, 1.0 g/mL.) Asked: grams of acetic acid in 120 mL of vinegar Given: 120 mL of vinegar and 5% acetic acid by mass Relationships: 120 mL = 120 g, given a density of 1.0 g/mL Solve:

37. 37 11.2 Percent Yield and Concentration Commercial vinegar is reported to be 5% acetic acid (C 2 H 4 O 2 ) by mass. How many grams of acetic acid are in 120 mL of commercial vinegar? (Assume the density of vinegar is the same as pure water, 1.0 g/mL.) Asked: grams of acetic acid in 120 mL of vinegar Given: 120 mL of vinegar and 5% acetic acid by mass Relationships: 120 mL = 120 g, given a density of 1.0 g/mL Solve:

38. 38 11.2 Percent Yield and Concentration Commercial vinegar is reported to be 5% acetic acid (C 2 H 4 O 2 ) by mass. How many grams of acetic acid are in 120 mL of commercial vinegar? (Assume the density of vinegar is the same as pure water, 1.0 g/mL.) Asked: grams of acetic acid in 120 mL of vinegar Given: 120 mL of vinegar and 5% acetic acid by mass Relationships: 120 mL = 120 g, given a density of 1.0 g/mL Solve: Answer: 6.0 g of acetic acid.

39. 39 11.2 Percent Yield and Concentration Calculate using molar masses and mole ratios Obtained from the experiment