1. Chapter #8
Chemical Quantities

2. Chapter Outline 8.2 Mole relationships defined in balanced equations
8.3 Mole conversions- Stoichiometry
8.4 Mass Conversions Stoichiometry
8.5 Excess and Limiting Reactants
8.5 Theoretical Yield
8.6 Theoretical Yield From Initial Reactant Masses
8.7 Thermochemical Equations

3. STOICHIOMETRY
Stoichiometry is the use of balanced chemical equations in the conversion process.

4. 8.4 STOICHIOMETRY
Stoichiometry is the use of balanced chemical equations in the conversion process.
Examples
Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required.
2 H2 + O2 2H2O

5. 8.4 STOICHIOMETRY
Stoichiometry is the use of balanced chemical equations in the conversion process.
Examples
Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required.
2 H2 + O2 2H2O
6.33 g H2

6. 8.4 STOICHIOMETRY
Stoichiometry is the use of balanced chemical equations in the conversion process.
Examples
Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required.
2 H2 + O2 2H2O
6.33 g H2
mole H2
2.016 g H2

7. 8.4 STOICHIOMETRY
Stoichiometry is the use of balanced chemical equations in the conversion process.
Examples
Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required.
2 H2 + O2 2H2O
6.33 g H2
Mole H2
2.016 g H2

8. 8.4 STOICHIOMETRY
Stoichiometry is the use of balanced chemical equations in the conversion process.
Examples
Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required.
2 H2 + O2 2H2O
6.33 g H2
Mole H2
2 mole H2O
2.016 g H2O
2 Mole H2

9. 8.4 STOICHIOMETRY
Stoichiometry is the use of balanced chemical equations in the conversion process.
Examples
Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required.
2 H2 + O2 2 H2O
6.33 g H2
Mole H2
2 mole H2O
18.02 g H2O
2.016 g H2
2 Mole H2
Mole H2O

10. 8.4 STOICHIOMETRY
Stoichiometry is the use of balanced chemical equations in the conversion process.
Examples
Calculate the mass of water formed from 6.33 g of hydrogen. A balanced equation is required.
2 H2 + O2 2 H2O
6.33 g H2
Mole H2
2 mole H2O
18.02 g H2O
= 28.3 g H2O
2.016 g H2
2 Mole H2
Mole H2O

11. 8.5 Excess and Limiting Reactants
Reactants are substances that can be changed into something else. For example, nails and boards are reactants for carpenters, while thread and fabric are reactants for the seamstress. And for a chemist hydrogen and oxygen are reactants for making water.

12. 8.5 Building Houses
Ok, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make?

13. 8.5 Building Houses
Ok, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make? Yes, only one house!

14. 8.5 Building Houses
Ok, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make? What reactant is in excess? And how many more houses could we use if we had enough boards?

15. 8.5 Building Houses
Ok, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make? What reactant is in excess? And how many more houses could we use if we have enough boards?

16. 8.5 Building Houses
Ok, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make? What reactant is in excess? And how many more houses could we use if we have enough boards? Yes, nails are in excess!

17. 8.5 Building Houses
Ok, we want to build some houses, so we order 2 truck loads of boards and 2 truck loads of nails. If two truck loads of boards make one house and two truck loads of nails make 10 houses, then how many houses can we make? What reactant is in excess? And how many more houses could we use if we have enough boards? Yes, nails are in excess! Nine more houses if we have an adequate amount of boards.

18. 8.5 Excess and Limiting Example
If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess?

19. 8.5 Excess and Limiting Example
If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess? Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen.

20. 8.5 Excess and Limiting Example
If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess? Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen.
2 H2 + O H2O
10.0 g O2

21. 8.5 Excess and Limiting Example
If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess? Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen.
2 H2 + O H2O
10.0 g O2
mole O2
32.0 g O2

22. 8.5 Excess and Limiting Example
If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess? Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen.
2 H2 + O H2O
10.0 g O2
mole O2
2 mole H2
32.0 g O2
mole O2

23. 8.5 Excess and Limiting Example
If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess? Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen.
2 H2 + O H2O
10.0 g O2
mole O2
2 mole H2
2.02 g H2
32.0 g O2
mole O2
mole H2

24. 8.5 Excess and Limiting Example
If we react 10.0g of hydrogen with 10.0g of oxygen, which, if any, reactant will be in excess? Our conversion process can easily determine the excess reactant. We can convert 10.0 g of oxygen to grams of hydrogen to determine if there is enough hydrogen to consume the oxygen.
2 H2 + O H2O
10.0 g O2
mole O2
2 mole H2
2.02 g H2
= 1.26 g H2
32.0 g O2
mole O2
mole H2

25. 8.5 Excess and Limiting Example
Only 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water.

26. 8.5 Excess and Limiting Example
Only 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water.
10.0 g O2
mole O2
32.0 g O2

27. 8.5 Excess and Limiting Example
Only 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water.
10.0 g O2
mole O2
2 mole H2O
32.0 g O2
mole O2

28. 8.5 Excess and Limiting Example
Only 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water.
10.0 g O2
mole O2
2 mole H2O
18.0 g H2O
32.0 g O2
mole O2
mole H2O

29. 8.5 Excess and Limiting Example
Only 1.26 g of hydrogen are required to react with 10.0 g of oxygen. Since there are 10.0 g of hydrogen available, then hydrogen must be the excess reactant and oxygen is the limiting reactant. The remainder of hydrogen = 8.7 g is called the amount in excess. The amount of water produced is determined by using the limiting reactant and converting it into water.
10.0 g O2
mole O2
2 mole H2O
18.0 g H2O
= 11.3 g H2O
32.0 g O2
mole O2
mole H2O

30. 8.5 Percentage Yield
The percent yield is a comparison of the laboratory answer to the correct answer which is determined by the conversion process. Suppose a student combined 10.0 g of oxygen and 10.0 g of hydrogen in the lab and recovered 8.66 g of water. What would be the percent yield?

31. 8.5 Percentage Yield
The percent yield is a comparison of the laboratory answer to the correct answer which is determined by the conversion process. Suppose a student combined 10.0 g of oxygen and 10.0 g of hydrogen in the lab and recovered 8.66 g of water. What would be the percent yield?
Yield (the lab amount)
percent yield =
X 100
Theoretical Yield (by conversions)
8.66
percent yield =
X 100 = 76.6%
11.3

32. 8.7 Thermochemistry Exothermic Reaction Endothermic Reaction Eact + ΔH

33. 8.7 Thermochemistry
When a chemical or physical change takes place energy is either lost of gained. A Thermochemical equation describes this change. Equations gaining energy are called endothermic and equations losing energy are called exothermic.

34. 8.7 Thermochemical Equations
When a chemical or physical change takes place energy is either lost of gained. A Thermochemical equation describes this change. Equations gaining energy are called endothermic and equations losing energy are called exothermic.
Examples:
C3H6O (l ) O2 (g) CO2(g) H2O (g)
ΔH = kj
Exothermic
H2O (l) H2O (g)
ΔH = kj
Endothermic

35. 8.7 Thermochemical Equations
How many kj of heat are released when 709 g of C3H6O are burned?

36. 8.7 Thermochemical Equations
How many kj of heat are released when 709 g of C3H6O are burned?
C3H6O (l ) O2 (g) CO2(g) H2O (g)
ΔH = kj
709 g C3H6O

37. 8.7 Thermochemical Equations
How many kj of heat are released when 709 g of C3H6O are burned?
C3H6O (l ) + 4O2 (g) CO2(g) H2O (g)
ΔH = kj
709 g C3H6O
mole C3H6O
58.1 g C3H6O

38. 8.7 Thermochemical Equations
How many kj of heat are released when 709 g of C3H6O are burned?
C3H6O (l ) O2 (g) CO2(g) H2O (g)
ΔH = kj
709 g C3H6O
mole C3H6O
58.1 g C3H6O

39. 8.7 Thermochemical Equations
How many kj of heat are released when 709 g of C3H6O are burned?
C3H6O (l ) O2 (g) CO2(g) H2O (g)
ΔH = kj
709 g C3H6O
mole C3H6O
1790 kj
= kj
mole C3H6O
58.1 g C3H6O

40. The End