1. Unit 4 – Conservation of Mass and Stoichiometry
Cartoon courtesy of NearingZero.net
Unit 4 – Conservation of Mass and Stoichiometry

2. Stoichiometry
“In solving a problem of this sort, the grand thing is to be able to reason backward. This is a very useful accomplishment, and a very easy one, but people do not practice it much.”
Sherlock Holmes, in Sir Arthur Conan Doyle’s A Study in Scarlet
Reaction Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions.

3. Review: Atomic Masses Carbon = 98.89% 12C 1.11% 13C <0.01% 14C
Elements occur in nature as mixtures of isotopes
Carbon = 98.89% 12C
1.11% 13C
<0.01% 14C
Carbon’s atomic mass = amu

4. Review: The Mole
The number equal to the number of carbon atoms in exactly 12 grams of pure 12C.
1 mole of anything = ´ 1023 units of that thing

5. Review: Molar Mass
A substance’s molar mass (molecular weight) is the mass in grams of one mole of the compound.
CO2 = grams per mole
H2O = grams per mole
Ca(OH)2 = grams per mole

6. Standard Molar Volume 1 mole of a gas occupies 22.4 liters of volume
Equal volumes of all gases at the same temperature and pressure contain the same number of molecules.
- Amedeo Avogadro
At STP (Standard Temperature and Pressure):
1 mole of a gas occupies 22.4 liters of volume

7. The Mole

8. Review: Chemical Equations
Chemical change involves a reorganization of
the atoms in one or more substances.
C2H5OH + 3O2 ® 2CO H2O
reactants
products
When the equation is balanced it has quantitative significance:
1 mole of ethanol reacts with 3 moles of oxygen
to produce 2 moles of carbon dioxide and 3 moles of water

9. Using Compound Masses Using Chemical Equations:
Compound A to Compound B conversions
Using Compound Masses
Compound A
Compound B

10. Mole Relations

11. Calculating Masses of Reactants and Products
Balance the equation.
Convert to moles.
Set up mole ratios.
Use mole ratios to calculate moles of desired substituent.
Convert moles to desired unit.

12. Working a Stoichiometry Problem
6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed.
1. Identify reactants and products and write the balanced equation. 4 Al + 3 O2 2
Al2O3
a. Every reaction needs a yield sign!
b. What are the reactants?
c. What are the products?
d. What are the balanced coefficients?

13. Working a Stoichiometry Problem
6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed?
4 Al O2  2Al2O3
6.50 g Al
1 mol Al
2 mol Al2O3
g Al2O3 =
? g Al2O3
26.98 g Al
4 mol Al
1 mol Al2O3
6.50 x 2 x ÷ ÷ 4 =
12.3 g Al2O3

14. Gas Stoichiometry #1
How many liters of ammonia can be produced when 12 liters of hydrogen react with an excess of nitrogen?
3 H2(g) N2(g)  NH3(g)
1 mole H2
22.4
L NH3
12 L H2
2 mol NH3
= L NH3
8.0
3 mole H2
1 mol NH3
22.4 L H2

15. Gas Stoichiometry #2
How many liters of oxygen gas, at STP, can be collected from the complete decomposition of 50.0 grams of potassium chlorate?
2 KClO3(s)  2 KCl(s) + 3 O2(g)
50.0 g KClO3
1 mol KClO3
3 mol O2
22.4 L O2
g KClO3
2 mol KClO3
1 mol O2
= L O2
13.7

16. Gas Stoichiometry #3
How many liters of oxygen gas, at 37.0C and atmospheres, can be collected from the complete decomposition of 50.0 grams of potassium chlorate?
2 KClO3(s)  2 KCl(s) + 3 O2(g)
50.0 g KClO3
1 mol KClO3
3 mol O2
= “n” mol O2
0.612
mol O2
g KClO3
2 mol KClO3
= 16.7 L

17. Limiting Reactant The limiting reactant is the reactant
that is consumed first, limiting the amounts of products formed.

18. Limiting Reagents - Combustion

19. Limiting Reagent Limiting reactant
Limits or determines the amount of product that can be formed in a reaction.
is the reagent (reactant) that you run out of first.
When you run out then the reaction stops.

20. Excess Reagent More than enough to react with limiting reagent.
You have extra
Left overs

21. Cake recipe 3 cups flour 2 cups sugar 1 T baking soda 1.5 cups milk
.5 cup butter
1.5 tsp vanilla
3 eggs

22. Recipe  I have  can make
6 cups flour
6 cups sugar
5 T baking soda
8 cups milk
4 cups butter
6 tsp vanilla
3 eggs 2 3 5 5+ 8 4 1
3 cups flour
2 cups sugar
1 T baking soda
1.5 cups milk
.5 cup butter
1.5 tsp vanilla
3 eggs

23. Which do I run out of first?
Eggs are my limiting reagent because I only have enough eggs to make one cake. I run out of eggs first.
What is the maximum amount I can produce? ONE

24. Limiting Reagent Calculation s
Convert each reagent to the SAME product. (It does not matter which product you use. However, if the question asks about a specific product, convert to that product.)
The reagent/reactant that produces the LEAST amount of the product is the limiting reagent.

25. Limiting Reagent Example
N H2  2 NH3
How many grams of NH3 can be produced from the reaction of 28g of N2 and 25g of H2?

26. Limiting Reagent K: 28 g N2 ; 25 g H2 P: gN2 mol N2  mol NH3  g NH3
N H2  2 NH3
U: limiting reagent  will either be N2 or H2
K: 28 g N2 ; 25 g H2
P: gN2 mol N2  mol NH3  g NH3
gH2 mol H2  mol NH3  g NH3
reagent that produced the least NH3 is limiting reagent
LEAST amount of NH3 is the amount produced

27. Limiting Reagent = 34 g NH3 = 140 g NH3 28 g N2 2 g H2
N H2  2 NH3
S: 28g N2
1 mole N2
2 mole NH3
17 g NH3
= 34 g NH3
1 mole N2
28 g N2
1 mole NH3
25 g H2
1 mole H2
2 mole NH3
17 g NH3
= 140 g NH3
3 mole H2
2 g H2
1 mole NH3

28. Nitrogen is the limiting reagent because it
N H2  2 NH3
S: 28g N2
1 mole N2
2 mole NH3
17 g NH3
= 34 g NH3
1 mole N2
28 g N2
1 mole NH3
25 g H2
1 mole H2
2 mole NH3
17 g NH3
= 140 g NH3
3 mole H2
2 g H2
1 mole NH3
Nitrogen is the limiting reagent because it
produces the LEAST NH3

29. Excess Reagent K: 28 g N2 limiting; 25 g H2
N H2  2 NH3
U: excess reagent  how much H2 is left over
K: 28 g N2 limiting; 25 g H2
P: gN2mol N2  mol H2  g H2
subtract the amount used based on limiting reagent from the amount of H2 that you have.

30. Excess Reagent - = = 6.0 g H2 used 28 g N2 N2 + 3 H2  2 NH3 S: 28g N2
1 mole N2
3 mole H2
2 g H2
= 6.0 g H2
used
1 mole N2
28 g N2
1 mole H2
19 g H2
left over
25 g H2
Have -
6.0 g H2
used =

31. Limiting Reagent
Stop and work practice problems.

32. Solving a Limiting Reagent Stoichiometry Problem
Balance the equation.
Convert each reagent to the SAME product. (It does not matter which product you use. However, if the question asks about a specific product, convert to that product.)
The reagent/reactant that produces the LEAST amount of the product is the limiting reagent

33. % yield = actual yield * 100%
Percent yield
Amount actually recovered/produced compared to what should have been recovered/produced.
Tells what percent of the predicted amount you are able to produce.
% yield = actual yield * 100%
Theoretical yield

34. Percent yield U: % yield K: theoretical yield = 21.8g
actual yield = 13.9 g
P: % yield = actual yield * 100%
Theoretical yield
S: % yield = g * 100%
21.8g
% yield = 63.8%

35. Percent Yield
Stop and work practice problems.