1. mass reactants= mass products
Stoichiometry
This word sounds scary, but it’s really just a big word for the relationships between reactants and products in a chemical reaction.
-based on law of conservation of mass
mass reactants= mass products

2. Proportional Relationships
2 1/4 c. flour
1 tsp. baking soda
1 tsp. salt
1 c. butter
3/4 c. sugar
3/4 c. brown sugar
1 tsp vanilla extract
2 eggs
2 c. chocolate chips
Makes 5 dozen cookies.
I have 5 eggs. Assuming I have plenty of the other ingredients, how many cookies can I make?
Ratio of eggs to cookies
5 dozen cookies
2 eggs
5 eggs
= 12.5 dozen cookies

3. Proportional Relationships
Stoichiometry
mass relationships between substances in a chemical reaction
based on the mole ratio
Mole Ratio
indicated by coefficients in a balanced equation
2 Mg + O2  2 MgO

4. Mole Ratios
2 Mg + O2  2 MgO
Relationships between coeffecients are used as conversion factors in stoichiometry!
A mole ratio is a ratio between the numbers of moles of any 2 substances.

5. 2 Mg + O2  2 MgO Mole Ratios Total of 6 mole ratios!
What are the mole ratios in this equation?
2 molMg
2 molMg
2 mol MgO
1 mol O2
1 mol O2
1 mol O2
2 mol Mg
2 mol MgO
2 mol MgO
2 mol MgO
2 mol Mg
1 mol O2
Total of 6 mole ratios!

6. 2KClO3→2KCl + 3O2 Mole Ratios Total of 6 mole ratios!
What are the mole ratios in this equation?
2 molKClO3
2 molKClO3
3 mol O2
2 mol KCl
2 mol KCl
2 mol KCl
2 mol KClO3
3 mol O2
3 mol O2
3 mol O2
2 mol KClO3
2 mol KCl
Total of 6 mole ratios!

7. Stoichiometry Steps Core step in all stoichiometry problems!!
1. Write a balanced equation.
2. Identify known & unknown.
3. Line up conversion factors.
Mole ratio - moles  moles
Molar mass - moles  grams
Molarity - moles  liters soln
Molar volume - moles  liters gas
Mole ratio - moles  moles
Core step in all stoichiometry problems!!
4. Check answer.

8. Mole ratios CO2(g) + 2LiOH(s) → Li2CO3(s) + H2O What is the ratio of
CO2 molecules to LiOH molecules?
CO2 molecules to Li2CO3 molecules
Li2CO3 molecules to LiOH molecules

9. Molar Conversions MASS IN GRAMS MOLES NUMBER OF PARTICLES Molar Mass
(g/mol)
6.02  1023
particles/mol

10. Chemical Equations and the Mole
P4O H2O  4 H3PO4
1 molecule of P4O10 reacts with 6 molecules of H2O to give 4 molecules of H3PO4
1 mole of P4O10 reacts with 6 moles of H2O to give 4 moles of H3PO4

11. Mole to Mole conversions
-use mole ratios from last notes
Formula used:
Moles of known
Moles of unknown
Moles of known
= moles of unknown

12. Mole to Mole conversions
Steps in Solving
Determine moles given in problem
Find mole ratio showing unknown on top and known on bottom
Fill into formula above and calculate moles of unknown

13. Stoichiometry Problems
How many moles of KClO3 must decompose in order to produce 9 moles of oxygen gas?
2KClO3  2KCl O2
? mol
9 mol
9 mol O2
2 mol KClO3
3 mol O2
= 6 mol KClO3

14. Stoichiometry and Chemical Equations
The coefficients in a chemical equation not only describe the number of atoms/molecules being reacted/produced, they also describe the number of moles of atoms/molecules being reacted/produced.
N2(g) + 3H2(g)
2NH3(g)
Suppose 5.46 mol N2 completely react in this fashion:
5.46 mol N2
3 mol H2
1 mol N2
= 16.4 mol H2 are needed to react
5.46 mol N2
2 mol NH3
1 mol N2
= 10.9 mol NH3 must be produced

15. Stoichiometry Practice
Bismuth(III) chloride will react with hydrogen sulfide to form bismuth(III) sulfide and hydrochloric acid. Write the balanced equation for this reaction, then calculate how many moles of acid would be formed if 15.0 mol of hydrogen sulfide react.
2 BiCl3 + 3 H2S
Bi2S3 + 6 HCl
15.0 mol H2S
6 mol HCl
3 mol H2S
= 30.0 mol HCl

16. Mole to Mass conversions
-use if you know # moles of one thing and you want to know mass of another thing
Steps in solving
Convert moles of known to moles of unknown.
Convert moles of unknown to grams of unknown

17. Stoichiometry Practice
Sodium nitride can be formed by reacting sodium metal with nitrogen gas. Write the balanced equation for this reaction, then calculate how many moles of sodium nitride can be produced from 25.0 g of sodium.
6 Na + N2
2 Na3N
25.0 g Na
1 mol Na
2 mol Na3N
22.99 g Na
6 mol Na
= mol Na3N

18. Stoichiometry Practice
When magnesium burns in air, it combines with oxygen to form magnesium oxide according to the following equation:
What mass (in grams) of magnesium oxide is produced from 2.00 mol of magnesium?
2Mg(s) + O2(g)
2MgO(s)
2.0 mol Mg
2 mol MgO
40.3g MgO
2 mol Mg
1 mol MgO
= g MgO

19. Mass to Mass conversions
Use if you know the mass of 1 thing and want to find mass of another
Steps in solving
Convert mass of known to moles of known
Do mole-mole conversion
Convert moles of unknown to mass of unknown

20. 2KClO3 → 2KCl + 3O2 Calculate the mass of O2 produced if 2
2KClO3 → 2KCl + 3O2 Calculate the mass of O2 produced if 2.5 g of KClO3 is completely decomposed when burned
2.5 g KClO3
1 mol KClO3
3 mol O2
32 g O2
122.6g KClO3
2 mol KClO3
1 mol O2
= .98g O2
Mole ratio

21. NH4NO3 → N2O + 2H2O Determine the mass of water produced from the decomposition of 25.0g of solid ammonium nitrate
25.0 g NH4NO3
1 mol NH4NO3
2 mol H2O
18.02 g H2O
80.04 g NH4NO3
1 mol NH4NO3
1 mol H2O
= 11.26g H2O
Mole ratio

22. Summary of Conversions
Mole to Mole: 1 step
Mass to Mole: 2 steps
Mass to Mass: 3 steps
What setup do all conversions require?
Mole:mole ratio

23. Limiting Reactants

24. How many bicycles can be assembled from the parts shown?
From eight wheels four bikes can be constructed.
From three pedal assemblies three bikes can be constructed.
From four frames four bikes can be constructed.
The limiting part is the number of pedal assemblies.

25. Likewise, in a chemical reaction, it is possible for one reactant to be used up before the other(s).
A limiting reactant is a substance that is completely used up in a chemical reaction and causes the reaction to stop.
Therefore, it is the limiting reactant that also determines how much product can actually be formed.

26. The bicycle wheels and frames were in excess in our example
An excess reactant are the left-over reactants that are in abundance in a reaction.
The bicycle wheels and frames were in excess in our example

27. To determine which reactant is limiting in a given reaction:
1. Convert grams of each reactant to moles
Divide both #’s by the smallest # to determine actual mole ratio
Determine balanced mole ratio using moles in balanced equation
Which reactant isn’t as large as it’s supposed to be? THAT is your limiting reactant.

28. Limiting Reactant Example
S8 + 4Cl2 → 4S2Cl2
If we are given 100 grams of Cl2 and 200 grams of S8, what is the limiting reactant?
Step 1: Convert grams into moles
100 g Cl2
1 mol Cl2
= 1.4 mol Cl2
70 g Cl2
200 g S8
1 mol S8
= 0.8 mol S8
256 g S8

29. For Cl2: 1.4 mol Cl2 For S8: 0.8 mol S8 = 1.75 =1 0.8 mol S8
Step 2: Divide both #’s by the smallest # to determine actual mole ratio
For Cl2:
1.4 mol Cl2
For S8:
0.8 mol S8
= 1.75 =1
0.8 mol S8
0.8 mol S8
Divide by smallest #

30. Step 3: Determine balanced mole ratio using moles in balanced equation
S8 + 4Cl2 → 4S2Cl2
Balanced mole ratio
Actual mole ratio
1 mol S8
1 mol S8
4 mol Cl2
1.75 mol Cl2

31. Step 4: Determine limiting reactant based on which reactant isn’t as large as it’s supposed to be
Balanced mole ratio
Actual mole ratio
1 mol S8
1 mol S8
4 mol Cl2
1.75 mol Cl2
Cl2 is limiting b/c it’s actual mole ratio is smaller that the balanced mole ratio!!!

32. Calculating the amount of product from a limiting reactant:
1. Determine the limiting reactant
Determine the amount of limiting reactant (given in problem)
Multiply amount of limiting reactant by the mole ratio of product/limiting rgnt.
Convert moles of product into grams (multiply by molar mass)

33. S8 + 4Cl2 → 4S2Cl2 Step 1: Determine the limiting reactant
We know it’s Cl2 already!
Step 2: Determine the amount of limiting reactant- given in problem
1.4 mol Cl2

34. S8 + 4Cl2 → 4S2Cl2
Step 3: Multiply amount of limiting reactant by the mole ratio of product/limiting reactant from balanced equation
Step 4: Multiply by molar mass
1.4 mol Cl2
4 mol S2Cl2
135 g S2Cl2
=190.4g S2Cl2
4mol Cl2
1 mol S2Cl2

35. Calculating amount of excess reactant:
1. Determine the limiting reactant
Determine the amount of limiting reactant (given in problem)
Multiply amount of limiting reactant by the mole ratio of excess reactant/limiting rgnt.
Convert moles of excess reactant into grams (multiply by molar mass)
Subtract mass in #4 from original mass

36. 1.4 mol Cl2 S8 + 4Cl2 → 4S2Cl2 Step 1: Determine limiting reactant
Step 2: Determine amount (in moles of limiting reactant)
1.4 mol Cl2

37. S8 + 4Cl2 → 4S2Cl2
Step 3: Multiply amount of limiting reactant by the mole ratio of
Excess reactant
Limiting reactant
Step 4: Multiply by molar mass of excess reactant
1.4 mol Cl2
1 mol S8
256.5g S8
= g S8
4mol Cl2
1 mol S8

38. Step 5: Subtract mass in #4 from original mass given at the beginning of the problem. That is the amount left over
200g S g S8 =
109.6g S8 in excess

39. Limiting Reactant Practice
Sodium metal will react with gaseous ammonia to produce solid sodium amide, NaNH2. The unbalanced equation for this reaction is…
Na + NH3
NaNH2 + H2
If 60.0 g of sodium are mixed with 48.0 g of ammonia, how many grams of sodium amide can be produced? Which substance is the limiting reactant?

40. Limiting Reactant Practice2
Na NH3 2 2
NaNH2 + H2
smallest
60.0 g Na
1 mol Na
2 mol NaNH2
= mol NaNH2
22.99 g Na
2 mol Na
Na is limiting
48.0 g NH3
1 mol NH3
2 mol NaNH2
= mol NaNH2
17.04 g NH3
3 mol NH3

41. Theoretical, Actual & Percent Yields
Theoretical yield: the amount of product you would get if the reaction occurs with complete efficiency.
Actual yield: what you “actually” get when you perform the reaction.
Percent yield: the actual yield divided by the theoretical yield multiplied by 100.
% Yield = (Actual/Theoretical) x 100

42. Percent Yield
measured in lab
calculated on paper

43. When 45. 8 g of K2CO3 react with excess HCl, 46. 3 g of KCl are formed
When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl.
K2CO3 + 2HCl  2KCl + H2O + CO2
45.8 g
? g
actual: 46.3 g

44. K2CO3 + 2HCl  2KCl + H2O + CO2
45.8 g
? g
actual: 46.3 g
Theoretical Yield:
45.8 g
K2CO3
2 mol
KCl
1 mol
K2CO3
74.55
g KCl
1 mol
KCl
1 mol
K2CO3 g
= 49.4
g KCl

45. K2CO3 + 2HCl  2KCl + H2O + CO2
45.8 g
49.4 g
actual: 46.3 g
Theoretical Yield = 49.4 g KCl
46.3 g
49.4 g
 100 =
93.7%
% Yield =