1. Unit 10
Stoichiometry

2. Stoichiometry
Looking at quantitative relationships of the reactants and products of a chemical equation
MUST use a balanced equation

3. Using Balanced Equations
You can determine the quantities of reactants and products in a chemical reaction from a balanced equation (like adjusting a recipe for more/less)

4. Relationships Derived from a Balanced Equation
Iron +
Oxygen
Iron (III) oxide
4Fe (s) +
3O2 (g)
2Fe2O3 (s)
4 atoms Fe
3 molecules O2
2 formula units Fe2O3
4 moles Fe
3 moles O2
2 moles Fe2O3
223.4 g Fe
96.0 g O2
319.4 g Fe2O3
319.4 g reactants
319.4 g products

5. Mole Ratios
Ratio of the coefficients of any two parts of a balanced chemical reaction
Example: 2 Al (s) + 3 Br2 (l) → 2 AlBr3 (s)
What is the ratio of…..
Al : Br2?
Al : AlBr3?
Br2 : Al?
AlBr3 : Br2?

6. Mole to Mole Conversions
If you start with a given amount of moles of one reactant, how many moles of the product will be created?
Just like finding out how many cookies you can make from 5 eggs…
Mole ratios can be used as conversion factors
First, write a balanced equation.
Multiply moles given by the correct mole ratio to find what you’re looking for

7. Mole to Mole conversions
How many moles of O2 are produced when 3.34 moles of Al2O3 decompose?
2 Al2O3  4Al + 3O2
3.34 moles Al2O3
3 mole O2
= 5.01 moles O2
2 moles Al2O3

8. Practice
2C2H2 + 5 O2 ® 4CO2 + 2 H2O
If 3.84 moles of C2H2 are burned, how many moles of O2 are needed?
How many moles of C2H2 are needed to produce 8.95 moles of H2O?
If 2.47 moles of C2H2 are burned, how many moles of CO2 are formed?

9. You can now…
Calculate the number of moles of a reactant needed to create a given product.
Calculate the number of moles of a product that will be created if you start with a given amount of a reactant

10. One step further…
You can also calculate the mass of reactant needed or product created if given a certain number of moles
These are called MOLE-MASS CONVERSIONS

11. Mole-Mass Conversions
You will be given an amount in moles (either reactant or product)
Multiply it by the correct mole ratio
Then convert to mass by multiplying by the molar mass of the compound or element you’re looking for

12. How many grams of NaCl will be produced when 1.25 mol of Cl2 reacts?
2 Na + Cl2 → 2 NaCl
1.25mol Cl2 2mol NaCl g NaCl = 146 g
1mol Cl mol NaCl
(mole ratio) (molar mass)

13. Let’s add one more step…
You can do the exact same thing if your given is in GRAMS instead of MOLES
This is called a MASS-MASS CONVERSION
You will be given an amount in grams

14. Mass-mass conversions
Convert mass to moles of given by using the molar mass
Next multiply that number by the correct mole ratio
Then multiply by the molar mass of the compound or element that the problem is asking for.

15. For example...
If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how much solid copper would form?
Fe + CuSO4 ® Fe2(SO4)3 + Cu
2Fe + 3CuSO4 ® Fe2(SO4)3 + 3Cu
1 mol Fe
63.55 g Cu
10.1 g Fe
3 mol Cu
55.85 g Fe
2 mol Fe
1 mol Cu =
17.2 g Cu

16. How many grams of water will be produced when 25
How many grams of water will be produced when 25.0 g of NH4NO3 decomposes?
NH4NO3 → N2O + 2H2O
25.0gNH4NO3 x 1molNH4NO3 x 2molH2O x gH2O = 11.2g
80.0gNH4NO3 1mol NH4NO3 1molH2O
(given) (molar mass) (mole ratio) (molar mass)

17. Mass-Mass Conversions
These problems are also known as “gram-mole-mole-gram” problems for obvious reasons…

18. In a perfect world…
We would have the exact amount of each reactant needed for a chemical equation
2H2 + O2  2H2O
This would be a COMPLETE reaction & uses ALL of the given reactants

19. But in reality…
Our measurements may not be so precise or we may not have that perfect amount on hand
So the reaction will continue until one of the reactants runs out
There will be “leftovers” of the other reactant(s)

20. The reactant that runs out first is called the LIMITING REAGENT or LIMITING REACTANT
This is what determines how much product you can make

21. An example…
A recipe calls for 2 TB peanut butter & 2 pieces of bread to make 1 sandwich
You have full jar of PB, but only 4 pieces of bread
The bread is the limiting reactant & ultimately determines how much you can make

22. Which is the limiting reactant?
Make sure you have a balanced equation.
For EACH reactant, find out how much product would be produced (grams or moles).
The reactant that produces the SMALLER amount is the limiting reactant.

23. Determine the limiting reactant if you have 25. 0 g of P4 and 50
Determine the limiting reactant if you have 25.0 g of P4 and 50.0 g of O2 in the following equation.
P O2 → P4O10
25.0 g P4 1
124
mole P4
g P4 1
mole P4O10
mole P4
= mole P4O10 produced

24. Determine the limiting reactant if you have 25. 0 g of P4 and 50
Determine the limiting reactant if you have 25.0 g of P4 and 50.0 g of O2 in the following equation.
P4 + O2 → P4O10
50.0 g O2 1 32
mole O2
g O2 1 5
mole P4O10
mole O2
= mole P4O10 produced

25. So… 25.0 g of P4 produces 0.202 moles of P4O10
50.0 g of O2 produces moles of P4O10
0.202 < 0.313
P4 is the limiting reactant

26. Another variation…
How much of the product will you make if given certain reactants
2 slices bread + 2 T PB  1 Sandwich
How many sandwiches can I make with 4 slices of bread and 20 T of peanut butter?

27. P O2 → P4O10
How much P4O10 could you make if you have 10.0 g of P4 and 16.0 g of O2?

28. Percent Yield Percent Yield = Actual Yield x 100% Theoretical Yield
Actual Yield = what you experimentally determine in lab
Theoretical Yield = how much you should produce in a perfect world

29. How to determine percent yield
Determine theoretical yield (using given in a gram-mole-mole-gram problem)
Calculate percent yield using formula

30. 2AgNO3 + K2CrO4 → Ag2CrO4 + 2KNO3
In a lab experiment, you complete the reaction above and produce g of Ag2CrO4. You started with g of AgNO3. What is the percent yield of Ag2CrO4?

31. 2AgNO3 + K2CrO4 → Ag2CrO4 + 2KNO3 = 0.491 g Ag2CrO4
STEP 1 - Determine Theoretical Yield
0.50 g AgNO3 1
169
mole AgNO3
g AgNO3
1 mole Ag2CrO4
2 moles AgNO3
332 g Ag2CrO4
1 mole Ag2CrO4
= g Ag2CrO4

32. STEP 2 – Calculate Percent Yield
2AgNO3 + K2CrO4 → Ag2CrO4 + 2KNO3
STEP 2 – Calculate Percent Yield
0.455 g
x 100%
= 92.7%
0.491 g