1. CHAPTER 11
Stoichiometry
11.2 Percent Yield and Concentration

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

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

4. Percent yield
What you get to eat! +
100 kernels
82 popped
18 unpopped

5. Percent yield
What you get to eat! +
100 kernels
82 popped
18 unpopped

6. actual yield: the amount obtained in the lab in an actual experiment.
theoretical yield: the expected amount produced if everything reacted completely.

7. Percent yield in the lab
Decomposition of baking soda:
2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)
Heating

8. Percent yield in the lab
Decomposition of baking soda:
2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)
10.00 g
4.87 g
measured experimentally
Can you think of reasons why the final mass of Na2CO3 may not be accurate? (What could be sources of error?)

9. Percent yield in the lab
Decomposition of baking soda:
2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)
10.00 g
4.87 g
measured experimentally
Can you think of reasons why the final mass of Na2CO3 may not be accurate? (What could be sources of error?)
- There is usually some human error, like not measuring exact amounts carefully

10. Percent yield in the lab
Decomposition of baking soda:
2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)
10.00 g
4.87 g
measured experimentally
Can you think of reasons why the final mass of Na2CO3 may not be accurate? (What could be sources of error?)
- There is usually some human error, like not measuring exact amounts carefully
- Maybe the heating time was not long enough; not all the Na2HCO3 reacted

11. Percent yield in the lab
Decomposition of baking soda:
2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)
10.00 g
4.87 g
measured experimentally
Can you think of reasons why the final mass of Na2CO3 may not be accurate? (What could be sources of error?)
- There is usually some human error, like not measuring exact amounts carefully
- Maybe the heating time was not long enough; not all the Na2HCO3 reacted
- Maybe Na2CO3 was not completely dry; some H2O(l) was measured too

12. Percent yield in the lab
Decomposition of baking soda:
2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)
10.00 g
4.87 g
measured experimentally
Can you think of reasons why the final mass of Na2CO3 may not be accurate? (What could be sources of error?)
- There is usually some human error, like not measuring exact amounts carefully
- Maybe the heating time was not long enough; not all the Na2HCO3 reacted
- Maybe Na2CO3 was not completely dry; some H2O(l) was measured too
- CO2 is a gas and does not get measured

13. Percent yield in the lab
Decomposition of baking soda:
2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)
10.00 g
4.87 g
measured experimentally
Let’s calculate the percent yield
obtained in experiment
calculated

14. Percent yield in the lab
Decomposition of baking soda:
2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)
10.00 g
4.87 g
measured experimentally
Let’s calculate the percent yield
calculated

15. Percent yield in the lab
Decomposition of baking soda:
2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)
10.00 g
4.87 g
measured experimentally
Let’s calculate the percent yield
calculated

16. 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)
Use the mass of reactant NaHCO3(s) to calculate the mass of the product Na2CO3(s).
This is a gram-to-gram conversion:

17. 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)

18. 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)

19. 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)
moles

20. 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)
moles

21. 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)
moles
moles

22. 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)
moles

23. 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)
moles
6.306 g

24. 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)
moles
moles
6.306 g
For g of starting material (NaHCO3), the theoretical yield for Na2CO3 is g.
The actual yield (measured) is 4.87 g.

25. 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)
For g of starting material (NaHCO3), the theoretical yield for Na2CO3 is g.
The actual yield (measured) is 4.87 g.

26. Stoichiometry with solutions
Decomposition of baking soda:
2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)
10.00 g
Convert to moles
Reaction of solid zinc with hydrochloric acid:
Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq)
Reactions in solution
50.0 mL of a 3.0 M solution
Convert to moles

27. A sample of zinc metal (Zn) reacts with 50. 0 mL of a 3
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) → H2(g) + ZnCl2(aq). How many grams of hydrogen gas (H2) will be produced? Assume zinc metal is present in excess.

28. A sample of zinc metal (Zn) reacts with 50. 0 mL of a 3
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) → H2(g) + ZnCl2(aq). How many grams of hydrogen gas (H2) will be produced? Assume zinc metal is present in excess.
Asked: grams of H2 produced
Given: mL of 3.0 M HCl reacting
with excess zinc
Relationships: M = mole/L
Mole ratio: 2 moles HCl ~ 1 mole H2
Molar mass of H2 = x 2
= 2.02 g/mole

29. A sample of zinc metal (Zn) reacts with 50. 0 mL of a 3
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) → H2(g) + ZnCl2(aq). How many grams of hydrogen gas (H2) will be produced? Assume zinc metal is present in excess.
Asked: grams of H2 produced
Given: mL of 3.0 M HCl reacting
with excess zinc
Relationships: M = mole/L
Mole ratio: 2 moles HCl ~ 1 mole H2
Molar mass of H2 = x 2
= 2.02 g/mole
Solve:

30. A sample of zinc metal (Zn) reacts with 50. 0 mL of a 3
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) → H2(g) + ZnCl2(aq). How many grams of hydrogen gas (H2) will be produced? Assume zinc metal is present in excess.
Asked: grams of H2 produced
Given: mL of 3.0 M HCl reacting
with excess zinc
Relationships: M = mole/L
Mole ratio: 2 moles HCl ~ 1 mole H2
Molar mass of H2 = x 2
= 2.02 g/mole
Solve:

31. A sample of zinc metal (Zn) reacts with 50. 0 mL of a 3
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) → H2(g) + ZnCl2(aq). How many grams of hydrogen gas (H2) will be produced? Assume zinc metal is present in excess.
Asked: grams of H2 produced
Given: mL of 3.0 M HCl reacting
with excess zinc
Relationships: M = mole/L
Mole ratio: 2 moles HCl ~ 1 mole H2
Molar mass of H2 = x 2
= 2.02 g/mole
Solve:

32. A sample of zinc metal (Zn) reacts with 50. 0 mL of a 3
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) → H2(g) + ZnCl2(aq). How many grams of hydrogen gas (H2) will be produced? Assume zinc metal is present in excess.
Asked: grams of H2 produced
Given: mL of 3.0 M HCl reacting
with excess zinc
Relationships: M = mole/L
Mole ratio: 2 moles HCl ~ 1 mole H2
Molar mass of H2 = x 2
= 2.02 g/mole
Solve:
Answer: 0.15 grams of H2 are produced

33. Reaction of solid zinc with hydrochloric acid:
Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq)
50.0 mL of a 3.0 M solution
Convert molarity to moles
Sometimes the concentration is written in mass percent
Vinegar is 5% acetic acid by mass

34. Commercial vinegar is reported to be 5% acetic acid (C2H4O2) by mass
Commercial vinegar is reported to be 5% acetic acid (C2H4O2) 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. Commercial vinegar is reported to be 5% acetic acid (C2H4O2) by mass
Commercial vinegar is reported to be 5% acetic acid (C2H4O2) 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. Commercial vinegar is reported to be 5% acetic acid (C2H4O2) by mass
Commercial vinegar is reported to be 5% acetic acid (C2H4O2) 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. Commercial vinegar is reported to be 5% acetic acid (C2H4O2) by mass
Commercial vinegar is reported to be 5% acetic acid (C2H4O2) 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. Commercial vinegar is reported to be 5% acetic acid (C2H4O2) by mass
Commercial vinegar is reported to be 5% acetic acid (C2H4O2) 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. Obtained from the experiment
Calculate using molar masses and mole ratios