1. Clash of the Classes
Judges can be bribed.
Girl Scout Cookie purchases
Tissue box donations
Paper towel roll donations

2. Welcome to Mole Island 1 mol = molar mass 1 mole = 22.4 L @ STP
6.02 x 1023 particles

3. Stoichiometry Consider the chemical equation: 4NH3 + 5O2  6H2O + 4NO
“stochio” = Greek for element
“metry” = measurement
Stoichiometry is about measuring the amounts of elements and compounds involved in a reaction.
Consider the chemical equation:
4NH3 + 5O2  6H2O + 4NO
There are several numbers involved.
What do they all mean?

4. 4 : 5 : 6 : 4 Stoichiometry 4NH3 + 5O2  6H2O + 4NO OR
Can mean either:
4 molecules of NH3 react with 5 molecules of O2
to produce 6 molecules of H2O and 4 molecules of NO OR
4 moles of NH3 react with 5 moles of O2
to produce 6 moles of H2O and 4 moles of NO

5. Stoichiometry 4NH3 + 5O2  6H2O + 4NO
With Stoichiometry we find out that
4 : 5 : 6 : 4
do more than just multiply atoms.
4 : 5 : 6 : 4
Are what we call a mole ratio.

6. Ratios are found within a chemical equation.
Stoichiometry (more working with ratios)
Ratios are found within a chemical equation.
2HCl + Ba(OH)2  2H2O BaCl2 1 1
coefficients
give MOLAR RATIOS
2 moles of HCl react with 1 mole of Ba(OH)2 to form 2 moles of H2O and 1 mole of BaCl2

7. Visualizing a Chemical Reaction2
Na Cl NaCl 2
___ mole Na 10 10
___ mole Cl2 5 5
___ mole NaCl 10 10 ?

8. Visualizing a Chemical Reaction2
Na Cl NaCl 2
___ mole Na 10
___ mole Cl2 5
___ mole NaCl 10

9. Mass, Volume, Mole Relationship

10. Molar Volume at STP MOLES LITERS OF GAS AT STP (22.4 L/mol) MASS IN
GRAMS
MOLES
NUMBER OF
PARTICLES
Molar Mass
(g/mol)
6.02  1023
particles/mol
Molarity (mol/L) *later unit**
LITERS OF
SOLUTION
Courtesy Christy Johannesson

11. Recall that Chemical formulas have molar masses:
Stoichiometry
4NH3 + 5O2  6H2O + 4NO
Recall that Chemical formulas have molar masses:
NH3
17 g/mol O2
32 g/mol
H2O
18 g/mol NO
30 g/mol
***To find the molar mass of a chemical formula – add the atomic masses of the
elements forming the compound. Use the periodic table to determine the atomic
mass of individual elements.***

12. Use the molar mass to convert grams to moles.
Converting grams to moles. * Review!
Determine how many moles there are in 5.17 grams of Fe(C5H5)2
Given
Goal
units match
5.17 g Fe(C5H5)2
= moles Fe(C5H5)2
0.0278
Use the molar mass to convert grams to moles.
Fe(C5H5)2
2 x 5 x 1.01 =
2 x 5 x =
1 x = 55.85

13. Molar Volume at STP **Review**
1 mol of a gas=22.4 L
at STP
Standard Temperature & Pressure
0°C and 1 atm
Courtesy Christy Johannesson

14. Stoichiometry Island Diagram
Mass
Stoichiometry Island Diagram
Mass
Known Unknown
Substance A Substance B
molar mass (g)
1 mole =
1 mole =
molar mass (g)
Use coefficients
from balanced
chemical equation
Volume
Volume
1 mole =
22.4 L
@ STP
1 mole =
22.4 L
@ STP
Mole
Mole
(gases)
1 mole = x 1023 particles
(atoms or molecules)
(gases)
1 mole =
6.02 x 1023 particles
(atoms or molecules)
Particles
Particles

15. Stoichiometry Island Diagram
Known Unknown
Substance A Substance B M
Mass
Mass Mountain
Mass
Volume
Mole
Mole
Volume V
Liter Lagoon
Particles
Particles P
Particle Place
Stoichiometry Island Diagram

16. Stoichiometry Island Diagram
Known Unknown
Substance A Substance B M
Mass
Mass Mountain
Mass
Mole Island
Volume
Mole
Mole
Volume V
Liter Lagoon
Particles
Particles P
Particle Place
Stoichiometry Island Diagram

17. Formation of Ammonia N2 (g) + 3 H2 (g) 2 NH3 (g) + + + + + + + +
2 atoms N and
2 atoms N +
6 atoms H
6 atoms H
1 molecule N2 +
3 molecules H2
2 molecules NH3
10 molecule N2 +
30 molecules H2
20 molecules NH3
6.02 x 1023
molecules N2
6.02 x 1023
molecules H2
6.02 x 1023
molecules NH3
1 x +
3 x
2 x
1 mol N2 +
3 mol H2
2 mol NH3
28 g N2 +
3 x 2 g H2
2 x 17 g NH3
34 g reactants
34 g products
Assume
STP
22.4 L
22.4 L
22.4 L
22.4 L
22.4 L
22.4 L +
22.4 L N2
67.2 L H2
44.8 L NH3

18. 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. How many cookies can I make?
Ratio of eggs to cookies
5 eggs
5 doz.
2 eggs
= 12.5 dozen cookies
150 cookies
Courtesy Christy Johannesson

19. 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
A balanced chemical equation gives the identity of the reactants and products and the accurate number of molecules or moles of each that are consumed or produced.
Stoichiometry is a collective term for the quantitative relationships between the masses, numbers of moles, and numbers of particles (atoms, molecules, and ions) of the reactants and products in a balanced reaction.
A stoichiometric quantity is the amount of product or reactant specified by the coefficients in a balanced chemical equation.
2 Mg + O2  2 MgO
Courtesy Christy Johannesson

20. 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 **later unit
Molar volume - moles  liters gas
Mole ratio - moles  moles
Core step in all stoichiometry problems!!
4. Check answer.
Courtesy Christy Johannesson

21. Mol – Mol Conversions “1 step” Mol is in the middle!
When N2O5 is heated, it decomposes:
2N2O5(g)  4NO2(g) + O2(g)
a. How many moles of NO2 can be produced from 4.3 moles of N2O5?
2N2O5(g)  4NO2(g) + O2(g)
4.3 mol
? mol
Units match
4.3 mol N2O5
= moles NO2
8.6
b. How many moles of O2 can be produced from 4.3 moles of N2O5?
2N2O5(g)  4NO2(g) + O2(g)
4.3 mol
? mol
4.3 mol N2O5
= mole O2
2.2

22. Stoichiometry Question (1)
4NH3 + 5O2  6H2O + 4NO
How many moles of H2O are produced if 2.00 moles of O2 are used?
6 mol H2O
5 mol O2
2.00 mol O2 =
2.40 mol H2O
Notice that a correctly balanced equation is essential to get the right answer

23. Stoichiometry Question (2)
4 NH O2  6 H2O NO
How many moles of NO are produced in the reaction if 15 mol of H2O are also produced?
4 mol NO
6 mol H2O
15 mol H2O =
10. mol NO

24. 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
Courtesy Christy Johannesson

25. 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
2 mol KClO3
6 mol
x mol KClO3 = 9 mol O2
= 6 mol KClO3
3 mol O2 O2
KClO3
Courtesy Christy Johannesson

26. First write a balanced equation.
Gram to Gram Conversions
“3 step” Grams-mol in the middle
Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid?
Al(s) HCl(aq)  AlCl3(aq) H2(g)
First write a balanced equation.

27. Now let’s get organized. Write the information below the substances.
Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid?
Gram to Gram Conversions
Al(s) HCl(aq)  AlCl3(aq) H2(g)
3.45 g
? grams
Now let’s get organized. Write the information below the substances.

28. 17.0 Al(s) + HCl(aq)  AlCl3(aq) + H2(g) 2 6 2 3 3.45 g ? grams
gram to gram conversions
Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid?
Al(s) HCl(aq)  AlCl3(aq) H2(g)
3.45 g
? grams
Units match
3.45 g Al
= g AlCl3
17.0
Now use the molar mass to convert to grams.
Now use the molar ratio.
We must always convert to moles.
Let’s work the problem.

29. Moving along the stoichiometry path
We always use the same type of information to make the jumps between steps:
grams (x)  moles (x)  moles (y)  grams (y)
Molar mass of y
Molar mass of x
Mole ratio from balanced equation

30. How many grams of silver will be formed from 12.0 g copper?
Cu AgNO3  2 Ag Cu(NO3)2
12.0 g
? g
12.0
g Cu
1 mol Cu
63.55
g Cu
2 mol Ag
1 mol Cu
107.87
g Ag
1 mol Ag
= 40.7 g Ag
1 mol Cu
2 mol Ag
g Ag
40.7 g
x g Ag = 12.0 g Cu
= 40.7 g Ag
63.55 g Cu
1 mol Cu
1 mol Ag Cu Ag
Courtesy Christy Johannesson

31. Stoichiometry Question (5)
4 NH O2  6 H2O NO
How many grams of NO is produced if 12 g of O2 is combined with excess ammonia?
1 mol O2
32 g O2 x
4 mol NO
5 mol O2 x
30.01 g NO
1 mol NO x
12 g O2
9.0 g NO =

32. gram ↔ mole and mole↔ gram conversions
“2 step”
When N2O5 is heated, it decomposes:
2N2O5(g)  4NO2(g) + O2(g)
a. How many moles of N2O5 were used if 210g of NO2 were produced?
2N2O5(g)  4NO2(g) + O2(g)
? moles
210g
Units match
210 g NO2
= moles N2O5
2.28
b. How many grams of N2O5 are needed to produce 5.0 molesof O2?
2N2O5(g)  4NO2(g) + O2(g)
? grams
75.0 g
5.0 mol O2
= grams N2O5
1080
Moles are lazy and lean on

33. Stoichiometry Question (3)
4 NH O2  6 H2O NO
How many grams of H2O are produced if 2.2 mol of NH3 are combined with excess oxygen?
6 mol H2O
4 mol NH3
18.02 g H2O
1 mol H2O
59 g H2O =
2.2 mol NH3

34. Stoichiometry Question (4)
4 NH O2  6 H2O NO
How many grams of O2 are required to produce 0.3 mol of H2O?
5 mol O2
6 mol H2O
32 g O2
1 mol O2
0.3 mol H2O
8 g O2 =

35. Have we learned it yet?
Try these on your own NH3 + 5 O2  6 H2O + 4 NO
a) How many moles of H2O can be made using 1.6 mol NH3?
b) what mass of NH3 is needed to make 0.75 mol NO?
c) how many grams of NO can be made from 47 g of NH3?

36. Answers 2.4 mol H2O 13 g NH3 83 g NO 4 NH3 + 5 O2  6 H2O + 4 NO a) b)c)
6 mol H2O
4 mol NH3 x
2.4 mol H2O =
1.6 mol NH3
4 mol NH3
4 mol NO x
17.04 g NH3
1 mol NH3 x
13 g NH3 =
0.75 mol NO
1 mol NH3
17.04 g NH3 x
4 mol NO
4 mol NH3 x
30.01 g NO
1 mol NO x
47 g NH3
83 g NO =

37. Caution: this stuff is difficult to follow at first.
Limiting Reactants
Caution: this stuff is difficult to follow at first.
Be patient.

38. Limiting Reactants Limiting Reactant Excess Reactant
used up in a reaction
determines the amount of product
Excess Reactant
added to ensure that the other reactant is completely used up
cheaper & easier to recycle
If one or more of the reactants is not used up completely but is left over when the reaction is completed, then the amount of product that can be obtained is limited by the amount of only one of the reactants
A limiting reactant is the reactant that restricts the amount of product obtained. The reactant that remains after a reaction has gone to completion is present in excess.
Courtesy Christy Johannesson

39. Grilled Cheese Sandwich
Bread Cheese  ‘Cheese Melt’
2 B C  B2C
100 bread
30 slices
30 sandwiches
? sandwiches
Multiple Guess:
130 sandwiches
100 sandwiches
90 sandwiches
60 sandwiches
30 sandwiches
Not enough information given

40. Limiting Reactants Available Ingredients Limiting Reactant
4 slices of bread
1 jar of peanut butter
1/2 jar of jelly
Limiting Reactant
bread
Excess Reactants
peanut butter and jelly
Courtesy Christy Johannesson

41. Limiting Reactants + CB + 4 T CT4 plus 16 tires excess 8 car bodies
8 car bodies
48 tires
8 cars
CB T CT4

42. Limiting Reactants + C + 2 H2 CH4 Methane, CH4 plus 8 hydrogen
molecules
excess
plus 16
hydrogen
atoms
excess
8 carbon
atoms
24 hydrogen
molecules
8 methane
molecules
C H CH4

43. Limiting Reactants – Method 2
Begin by writing a correctly balanced chemical equation
Write down all quantitative values under equation (include units)
Convert ALL reactants to units of moles
Divide by the coefficient in front of each reactant
The smallest value is the limiting reactant!

44. Container 1
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 269

45. Before and After Reaction 1
N H NH3 3 2
Before the reaction
After the reaction
All the hydrogen and nitrogen atoms combine.
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 269

46. Container 2
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 270

47. Before and After Reaction 2
N H NH3 3 2
excess
limiting
Before the reaction
After the reaction
LIMITING REACTANT DETERMINES AMOUNT OF PRODUCT
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 270

48. Real-World Stoichiometry: Limiting Reactants
Fe S FeS
Ideal
Stoichiometry
Limiting
Reactants
S =
Fe =
excess
LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 366

49. The Limiting Reactant 3 B + 2 M + EE B3M2EE With… …and… …one can make…
A balanced equation for making a Big Mac® might be:
3 B M + EE B3M2EE
With…
…and…
…one can make…
excess B and
excess EE
30 M
15 B3M2EE
excess M and
excess EE
30 B
10 B3M2EE
30 B and
excess EE
30 M
10 B3M2EE

50. The Limiting Reactant 3 W + 2 P + S + H + F W3P2SHF With… …and…
A balanced equation for making a tricycle might be:
3 W P + S + H + F W3P2SHF
With…
…and…
…one can make…
excess of all
other reactants
50 P
25 W3P2SHF
excess of all
other reactants
50 S
50 W3P2SHF
50 S and
excess of all
other reactants
50 P
25 W3P2SHF

51. Limiting Reactants aluminum + chlorine gas  aluminum chloride
Al(s) Cl2(g)  AlCl3
2 Al(s) Cl2(g)  AlCl3
100 g g ? g
A g B g C g D. 494 g

52. Limiting Reactants 2 Al(s) + 3 Cl2(g)  2 AlCl3 100 g 100 g x g
aluminum chlorine gas  aluminum chloride
2 Al(s) Cl2(g)  AlCl3
100 g g x g
How much product would be made if we begin with 100 g of aluminum?
1 mol Al
2 mol AlCl3
133.5 g AlCl3
x g AlCl3 = 100 g Al
= 494 g AlCl3
27 g Al
2 mol Al
1 mol AlCl3 Al
AlCl3
How much product would be made if we begin with 100 g of chlorine gas?
1 mol Cl2
2 mol AlCl3
133.5 g AlCl3
x g AlCl3 = 100 g Cl2
= 125 g AlCl3
71 g Cl2
3 mol Cl2
1 mol AlCl3
Cl2
AlCl3

53. Percent Composition
What is the percent carbon in C5H8NO4 (the glutamic acid used to make MSG monosodium glutamate), a compound used to flavor foods and tenderize meats?

54. Percent Composition What is the percent carbon in C5H8NO4?
Step 1: First calculate the Formula weight for the compound (FW for C5H8NO4 is ________amu)
Step 2: % composition = (mass of part ÷mass of whole) X 100 or
% element = (# atoms of that element)(atomic weight of the element) x 100
Formula weight of the compound

55. What is the % carbon in MSG?
Step 1: FW for C5H8NO4 is ________amu C H N O

56. % Composition of Carbon in C5H8NO4
Step 2 % C = (mass of part ÷mass of whole) X 100 or
% element = (# atoms of element)(atomic weight of the element) x 100
Formula weight of the compound
Answer?

57. Percent Composition
What is the percent carbon in C5H8NO4 (the glutamic acid used to make MSG monosodium glutamate), a compound used to flavor foods and tenderize meats?
a) %C
b) %C
c) %C