1. Stoichiometry and the Mole
Chapter 9
Stoichiometry and the Mole

2. 9.2 The Mole The mole Relates number of atoms/molecules to grams
Conversion unit that is necessary due to small size of individual atoms and molecules
Equal to a set number of particles

3. 9.2 The Mole (Continued)
Consider the chemical equation :
H2 + I2  2HI
This equation expresses amounts in molecules and atoms:
In terms of moles:

4. 9.2 The Mole (Continued) Molar mass
Allows conversion between mass and moles
Simply replace atomic mass amu with grams to find the mass of one mole.

5. 9.2 The Mole (Continued) Suppose you obtain 393.934 g of gold.
(a) What is the molar mass of gold?
(b) How many moles of gold do you have?
(c) How many atoms of gold do you have?

6. Molar mass of molecules
9.2 The Mole (Continued)
Molar mass of molecules
Number of atoms = number of moles
Consider CO2…

7. Conversions with molar mass
9.2 The Mole (Continued)
Conversions with molar mass
1 mole of water has a mass of g/mol.
Moles=
How many moles of water are in g of H2O?
How many grams of water are in moles of water?

8. 9.2 The Mole (Continued)
Stoichiometry Conversions

9. Part 2 9.3 Reaction Stoichiometry
According to the equation below…
H2 + I2  2HI
“One mole of H2 reacts with one mole of I2 to form 2 moles of HI.”
What do we say if there are 2 moles of each reactant?

10. 9.3: Reaction Stoichiometry
Theoretical yield
Maximum amount of product that can be produced from a certain reactant amount

11. 9.3 Reaction Stoichiometry
Actual yield
Amount actually produced in a chemical reaction
Often is less than theoretical yield due to side reactions going on simultaneously
Is directly measured rather than derived (the mass of the ppt from lab is the actual yield!)

12. 9.3 Reaction Stoichiometry
Percent yield
Percentage of theoretical yield that was measured
What would be the percent yield if g of HI was actually produced?

13. 9.3 Reaction Stoichiometry Practice
Calculating Actual Yield from a Reaction- STEP BY STEP!
4 Al + 3 O2  2 Al2O3
Step 1: Determine the molar ratio
“__moles of Al react with__ moles of oxygen to produce ___moles of aluminum oxide”

14. 9.3 Reaction Stoichiometry
Calculating Yield from a reaction:
4 Al + 3 O2  2 Al2O3
Step 2: Convert moles to masses
Al: 4 moles x ______ g/mol =
O2 : 3 mol x ______ g/mol=
Aluminum Oxide 3mol x _______g/mol=

15. 9.3 Reaction Stoichiometry
Calculating Yield from a reaction:
4 Al + 3 O2  2 Al2O3
Therefore…
4 mol of Al require 3mol O2 to completely react
ANOTHER INTERPRETATION…
4 mol of Al react with 3mol O2 to produce 2mol Al2O3

16. 9.3 Reaction Stoichiometry (Continued)
Calculating Yield from a reaction:
4 Al + 3 O2  2 Al2O3
Step 3: Calculate % Yield
The theoretical yield = g Al2O3
After the reaction is complete, the actual yield =200.3g
Calculate the % yield

17. 9.4 Dealing with a Limiting Reactant
Reactant that is in short supply
“Limits” the amount of product that can be made
Will be completely used up in a reaction
The reactants that is consumed (runs out) first
Excess reactant
Present in excess in a reaction
Will be left over after the reaction stops
Is done sometimes to make sure a reaction goes to completion (to make sure there’s enough)

18. 9.4 Dealing with a Limiting Reactant
Finding the limiting reactant
Write out the balanced equation
Calculate the number of moles of each reactant
Divide moles by the coefficient in the balanced equation.
The smallest mole-to-coefficient ratio is the limiting reactant! SIMPLE!!

19. 9.4 Dealing with a Limiting Reactant
6 moles of propane are available to react with 29 moles of oxygen…What is the limiting reactant?
Balanced Equation:
C3H O2  CO2 + H2O

20. 9.4 Limiting Reactant Look Familiar? 4 Al + 3 O2  2 Al2O3
Find the limiting reactant if 13.3mol of Al react with 11.8mol O2

21. 9.4 Limiting Reactant
Look Familiar?
4 Al + 3 O2  2 Al2O3
Find the limiting reactant if 13.3mol of Al react with 11.8mol O2
Al:
O2:

22. 9.4 Limiting Reactant
What happens with yield when there’s a limiting reactant?
Yield is calculated based on a revised molar ratio…
4 Al + 3 O2  2 Al2O3
Molar ratio = 4: 3: 2
IF 13.3mol of Al reacts (which makes it limiting) (see previous page)…
Theoretical to actual molar ratio of Al is proportional to theoretical to actual molar ratio of Al2O3…

23. 9.4 Limiting Reactant
What happens with yield when there’s a limiting reactant?
Yield is calculated based on a revised molar ratio…
4 Al + 3 O2  2 Al2O3
Molar ratio = 4: 3: 2
IF 13.3mol of Al reacts, 6.65 moles of Al2O3 are produced

24. 9.4 Limiting Reactant
What happens with yield when there’s a limiting reactant?
Yield is calculated based on a revised molar ratio…
4 Al + 3 O2  2 Al2O3
The actual yield must be measured in the lab!
If the actual yield is only 612g of Al2O3 what is the % yield of Al2O3?