CHAPTER 11
Stoichiometry
11.2 Percent Yield
and Concentration
In theory, all 100 kernels should have popped.
Did you do something wrong?
+
100 kernels
‹#›
82 popped
18 unpopped
11.2 Percent Yield and Concentration
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
11.2 Percent Yield and Concentration
Percent yield
What you get to eat!
percent yield 
amount of corn popped
 100
amount of kernels in the bag
82
percent yield 
 100  82%
100
+
100 kernels
‹#›
82 popped
18 unpopped
11.2 Percent Yield and Concentration
Percent yield
What you get to eat!
percent yield 
amount of corn popped
 100
amount of kernels in the bag
82
percent yield 
 100  82%
100
+
100 kernels
‹#›
82 popped
18 unpopped
11.2 Percent Yield and Concentration
actual yield
percent yield 
 100
theoretical yield
percent yield 
actual yield
 100
theoretical yield
actual yield: the amount obtained in the lab in an actual
experiment.
theoretical yield: the expected amount produced if
everything reacted completely.
‹#›
11.2 Percent Yield and Concentration
Percent yield in the lab
Decomposition of baking soda:
2NaHCO3(s)
→
Na2CO3(s) + H2O(l) + CO2(g)
Heating
‹#›
11.2 Percent Yield and Concentration
Percent yield in the lab
Decomposition of baking soda:
2NaHCO3(s)
10.00 g
→
Na2CO3(s) + H2O(l) + CO2(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?)
‹#›
11.2 Percent Yield and Concentration
Percent yield in the lab
Decomposition of baking soda:
2NaHCO3(s)
10.00 g
→
Na2CO3(s) + H2O(l) + CO2(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
‹#›
11.2 Percent Yield and Concentration
Percent yield in the lab
Decomposition of baking soda:
2NaHCO3(s)
10.00 g
→
Na2CO3(s) + H2O(l) + CO2(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
‹#›
11.2 Percent Yield and Concentration
Percent yield in the lab
Decomposition of baking soda:
2NaHCO3(s)
10.00 g
→
Na2CO3(s) + H2O(l) + CO2(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
‹#›
11.2 Percent Yield and Concentration
Percent yield in the lab
Decomposition of baking soda:
2NaHCO3(s)
10.00 g
→
Na2CO3(s) + H2O(l) + CO2(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
‹#›
11.2 Percent Yield and Concentration
Percent yield in the lab
Decomposition of baking soda:
2NaHCO3(s)
10.00 g
→
Na2CO3(s) + H2O(l) + CO2(g)
4.87 g
measured experimentally
Let’s calculate the percent yield
obtained in experiment
actual yield
percent yield 
 100
theoretical yield
calculated
‹#›
11.2 Percent Yield and Concentration
Percent yield in the lab
Decomposition of baking soda:
2NaHCO3(s)
10.00 g
→
Na2CO3(s) + H2O(l) + CO2(g)
4.87 g
measured experimentally
Let’s calculate the percent yield
4.87 g
percent yield 
 100
theoretical yield
calculated
‹#›
11.2 Percent Yield and Concentration
Percent yield in the lab
Decomposition of baking soda:
2NaHCO3(s)
10.00 g
→
Na2CO3(s) + H2O(l) + CO2(g)
4.87 g
measured experimentally
Let’s calculate the percent yield
4.87 g
percent yield 
 100
theoretical yield
calculated
‹#›
11.2 Percent Yield and Concentration
2NaHCO3(s)
→
Na2CO3(s) + H2O(l) + CO2(g)
10.00 g
Use the mass of reactant NaHCO3(s) to calculate the mass of the product
Na2CO3(s).
This is a gram-to-gram conversion:
‹#›
11.2 Percent Yield and Concentration
2NaHCO3(s)
→
Na2CO3(s) + H2O(l) + CO2(g)
10.00 g
10.00 g
‹#›
11.2 Percent Yield and Concentration
2NaHCO3(s)
→
Na2CO3(s) + H2O(l) + CO2(g)
10.00 g
10.00 g
molar mass of NaHCO3  22.99  1.0079  12.011  3(15.999)  84.01 g / mole
10.00 g NaHCO3 
‹#›
1 mole NaHCO3
 0.1190 moles NaHCO3
84.01 g NaHCO3
11.2 Percent Yield and Concentration
2NaHCO3(s)
→
Na2CO3(s) + H2O(l) + CO2(g)
10.00 g
10.00 g
0.1190 moles
molar mass of NaHCO3  22.99  1.0079  12.011  3(15.999)  84.01 g / mole
10.00 g NaHCO3 
‹#›
1 mole NaHCO3
 0.1190 moles NaHCO3
84.01 g NaHCO3
11.2 Percent Yield and Concentration
2NaHCO3(s)
→
Na2CO3(s) + H2O(l) + CO2(g)
10.00 g
0.1190 moles
1 mole Na2CO3
0.1190 moles NaHCO3 
 0.05950 moles NaHCO3
2 moles NaHCO3
‹#›
11.2 Percent Yield and Concentration
2NaHCO3(s)
→
Na2CO3(s) + H2O(l) + CO2(g)
10.00 g
0.1190 moles
0.05950 moles
1 mole Na2CO3
0.1190 moles NaHCO3 
 0.05950 moles NaHCO3
2 moles NaHCO3
‹#›
11.2 Percent Yield and Concentration
2NaHCO3(s)
→
Na2CO3(s) + H2O(l) + CO2(g)
10.00 g
0.05950 moles
molar mass of Na2CO3   22.99  2   12.011  15.999  3   105.99 g / mole
105.99 g Na2CO3
0.05950 moles Na2CO3 
 6.306 g Na2CO3
1 mole Na2CO3
‹#›
11.2 Percent Yield and Concentration
2NaHCO3(s)
→
Na2CO3(s) + H2O(l) + CO2(g)
10.00 g
0.05950 moles
6.306 g
molar mass of Na2CO3   22.99  2   12.011  15.999  3   105.99 g / mole
105.99 g Na2CO3
0.05950 moles Na2CO3 
 6.306 g Na2CO3
1 mole Na2CO3
‹#›
11.2 Percent Yield and Concentration
2NaHCO3(s)
→
10.00 g
10.00 g
Na2CO3(s) + H2O(l) + CO2(g)
4.87 g
0.1190 moles
0.05950 moles
6.306 g
For 10.00 g of starting material (NaHCO3),
the theoretical yield for Na2CO3 is 6.306 g.
The actual yield (measured) is 4.87 g.
‹#›
11.2 Percent Yield and Concentration
2NaHCO3(s)
10.00 g
→
Na2CO3(s) + H2O(l) + CO2(g)
4.87 g
actual yield
percent yield 
 100
theoretical yield
4.87 g
percent yield 
 100  77.2%
6.306 g
For 10.00 g of starting material (NaHCO3),
the theoretical yield for Na2CO3 is 6.306 g.
The actual yield (measured) is 4.87 g.
‹#›
11.2 Percent Yield and Concentration
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)
50.0 mL of a
3.0 M solution
Reactions in
solution
Convert to moles
‹#›
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) → H2(g) + ZnCl2(aq). How many grams of
hydrogen gas (H2) will be produced? Assume zinc metal is present in excess.
‹#›
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) → 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: 50.0 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 = 1.0079 x 2
= 2.02 g/mole
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) → 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: 50.0 mL of 3.0 M HCl reacting
with excess zinc
Solve:
mole
HCl  0.0500 mL  0.150 moles HCl
L
1 mole H2
0.150 moles HCl 
 0.0750 moles H 2
2 moles HCl
moles of HCl  3.0
0.0750 moles H2 
‹#›
Relationships:
M = mole/L
Mole ratio: 2 moles HCl ~ 1 mole H2
Molar mass of H2 = 1.0079 x 2
= 2.02 g/mole
2.02 g H2
 0.15 g H2
1 mole H2
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) → 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: 50.0 mL of 3.0 M HCl reacting
with excess zinc
Solve:
mole
HCl  0.0500 mL  0.150 moles HCl
L
1 mole H2
0.150 moles HCl 
 0.0750 moles H 2
2 moles HCl
moles of HCl  3.0
0.0750 moles H2 
‹#›
Relationships:
M = mole/L
Mole ratio: 2 moles HCl ~ 1 mole H2
Molar mass of H2 = 1.0079 x 2
= 2.02 g/mole
2.02 g H2
 0.15 g H2
1 mole H2
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) → 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: 50.0 mL of 3.0 M HCl reacting
with excess zinc
Solve:
mole
HCl  0.0500 mL  0.150 moles HCl
L
1 mole H2
0.150 moles HCl 
 0.0750 moles H 2
2 moles HCl
moles of HCl  3.0
0.0750 moles H2 
‹#›
Relationships:
M = mole/L
Mole ratio: 2 moles HCl ~ 1 mole H2
Molar mass of H2 = 1.0079 x 2
= 2.02 g/mole
2.02 g H2
 0.15 g H2
1 mole H2
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) → 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: 50.0 mL of 3.0 M HCl reacting
with excess zinc
Solve:
Relationships:
M = mole/L
Mole ratio: 2 moles HCl ~ 1 mole H2
Molar mass of H2 = 1.0079 x 2
= 2.02 g/mole
mole
HCl  0.0500 mL  0.150 moles HCl
L
1 mole H2
0.150 moles HCl 
 0.0750 moles H 2
2 moles HCl
moles of HCl  3.0
0.0750 moles H2 
2.02 g H2
 0.15 g H2
1 mole H2
Answer: 0.15 grams of H2 are produced
‹#›
11.2 Percent Yield and Concentration
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
mass % of compound 
‹#›
mass of compound
 100
total mass of solution
11.2 Percent Yield and Concentration
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.)
‹#›
11.2 Percent Yield and Concentration
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
‹#›
mass % 
mass of acetic acid
 100
mass of solution
11.2 Percent Yield and Concentration
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
mass % 
mass of acetic acid
 100
mass of solution
Solve: mass % mass of acetic acid

100
mass of solution
5 % mass of acetic acid

100
120 g
‹#›
11.2 Percent Yield and Concentration
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: mass % mass of acetic acid

100
mass of solution
5 % mass of acetic acid

100
120 g
‹#›
mass % 
mass of acetic acid
 100
mass of solution
0.05 
mass of acetic acid
120 g
mass of acetic acid  0.05  120 g
mass of acetic acid  6.0 g
11.2 Percent Yield and Concentration
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: mass % mass of acetic acid

100
mass of solution
5 % mass of acetic acid

100
120 g
mass % 
mass of acetic acid
 100
mass of solution
0.05 
mass of acetic acid
120 g
mass of acetic acid  0.05  120 g
mass of acetic acid  6.0 g
Answer: 6.0 g of acetic acid.
‹#›
11.2 Percent Yield and Concentration
actual yield
percent yield 
 100
theoretical yield
Obtained from the experiment
percent yield 
actual yield
 100
theoretical yield
Calculate using molar masses
and mole ratios
mass % of compound 
‹#›
mass of compound
 100
total mass of solution
11.2 Percent Yield and Concentration