Stoichiometry
Stoichiometry
• Needs a balanced equation
• Use the balanced equation to predict
ending and / or starting amounts
• Coefficients are now mole ratios
In terms of Moles
• The coefficients tell us how many moles of
each kind.
• Mole ratio - conversion ratio that relates
the amounts in moles of any two
substances in a chemical reaction.
• Molar mass - mass, in grams, of one mole
of a substance.
Review: Molar Mass
A substance’s molar mass (molecular
weight) is the mass in grams of one mole
of the compound.
CO2 = 44.01 grams per mole
H2O = 18.02 grams per mole
Ca(OH)2 = 74.10 grams per mole
Mole Ratios
2 Al2O3(l)  4 Al(s) + 3 O2(g)
Mole Ratios
Mole Ratio (Fraction)
2 mol Al2O3
2 mol Al2O3 : 4 mol Al
2 mol Al2O3 : 3 mol O2
4 mol Al : 3 mol O2
4 mol Al
2 mol Al2O3
3 mol O2
4 mol Al
3 mol O2
III. Stoichiometric “road map” (Use the balanced chemical
equation)
aA + bB
Mass A
Mol Ratio Using the
Coefficients from
the balanced
chemical equation
Mol A
Atoms
Molecules A
Formula Units
cC + dD
Mass B
Mol B
Atoms
Molecules B
Formula Units
3A+4B2D+1F
How many moles of F are produced from 1.00 mol of A?
1 mol A
1 mol F
= 0.33 mol F
3 mol A
How many moles of D are
produced from 5.00 mol of
B?
5 mol B
Mass A
Mol Ratio
MolA
2 mol D
4 mol B
Atoms A
Molecules
= 2.50 mol D
Mass B
MolB
Atoms B
Molecules
Molecules
How many moles of lithium carbonate are
produced when 5.3 mol CO2 are reacted?
CO2(g) + 2LiOH(s)  Li2CO3(s) + H2O(l)
1. What is your starting point?
2. What is your ending point?
5.3 mol CO2
1 mol Li2CO3
5.3 mol of CO2
mol of Li2CO3
Mass A
Mol Ratio
1 mol CO2
= 5.3 mol Li2CO3
Mass B
MolA
AtomsA
Molecules
MolB
Atoms B
Molecules
3A+4B2D+1F
How many grams of F are produced from 1.00 mol of A?
If MM of F is 10.0g/mol.
1 mol A
10 g F
1 mol F
= 3.33g F
3 mol A
1 mol F
How many grams of D are
produced from 5.00 mol of
B? MM of D is 20.0g/mol
5 mol B
2 mol D
20 g D
4 mol B
1 mol D
= 50.0g D
Mass A
Mass B
Mol Ratio
MolA
AtomsA
Molecules
MolB
Atoms B
Molecules
What is the mass of glucose (C6H12O6)
produced from 3.00 mol of water (H2O)?
6CO2(g) + 6H2O(l)  C6H12O6(s) + 6O2(g)
1. What is your starting point?
2. What is your ending point?
180.81g
1 mol
3 mol H2O C6H12O6 C6H12O6
6 mol H2O 1 mol
C6H12O6
=90.1 g C6H12O6
3.00 mol of H2O
g of C6H12O6
Mass
A
Mass B
Mol Ratio
MolA
AtomsA
Molecules
MolB
AtomsB
Molecule
6CO2(g) + 6H2O(l)  C6H12O6(s) + 6O2(g)
What is the mass of oxygen (O2) produced from
2.50 mol of water (H2O)?
1. What is your starting point?
2. What is your ending point?
2.50 mol of H2O
g of O2
2.5 mol H2O 6 mol O2 32.0 g O2
6 mol H2O 1 mol O2 Mass A
Mass B
Mol Ratio
MolA
MolB
=80.0 g O2
AtomsA
Molecules
Atoms B
Molecules
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)
How many moles of NO are formed from 824 g of NH3?
1. What is your starting point?
2. What is your ending point?
824 g of NH3
mol of NO
824 g NH3 1 mol NH3 4 mol NO
17.03g
4 mol NH3 Mass A
Mass B
Mol Ratio
= 48.4 mol NO
MolA
AtomsA
Molecules
MolB
Atoms B
Molecules
3A+4B2D+1F
How many grams of F are produced from 5.00g of A? If
MM of F is 10.0g/mol and MM of A is 25.0g/mol
5gA
1 mole A
1 mole F
10 g F
25 g A
3 mole A
1 mole F
Mass A
How many grams of D are
produced from 5.00g of B?
MM of D is 20.0g and MM
of B is 10.0g/mol
5gB
1 mole B
10 g B
=5.00g D
2 mole D
= 0.67g F
Mol Ratio
MolA
20 g D
4 mole B 1 mole D
AtomsA
Molecules
Mass
B
MolB
AtomsB
Molecules
Sn(s) + 2HF(g)  SnF2(s) + H2(g)
How many grams of SnF2 are produced
from the reaction of 30.00 g HF?
1. What is your starting point?
2. What is your ending point?
30.00g HF
30.00 g of HF
g SnF2
1 mol HF 1 mol SnF2 156.71 g SnF2
20.01g
HF
2 mol HF
1 mol SnF2
Mass A
Mass B
Mol Ratio
MolA
= 117.5g SnF2
AtomsA
Molecules
MolB
AtomsB
Molecules
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?
4 Al
+ 3 O2
2 Al2O3
6.50 g of Aluminum
g of aluminum oxide
1. What is your starting point?
2. What is your ending point?
6.50 g Al
1 mol Al
2 mol Al2O3
101.96 g Al2O3
26.98 g Al
4 mol Al
1 mol Al2O3
6.50 x 2 x 101.96 ÷ 26.98 ÷ 4
= ? g Al2O3
Mass A
= 12.3 g Al2O3
Mass B
Mol Ratio
MolA
AtomsA
Molecules
MolB
Atoms B
Molecules
Sn(s) + 2HF(g)  SnF2(s) + H2(g)
How many grams of HF are produced from
the reaction of 150.5 g H2?
150.5 g H2
1 mol H2
2 mol HF
2.02g H2 1 mol H2
= 2982 g HF
20.01g HF
1 mol HF
Mass A
Mass B
Mol Ratio
MolA
AtomsA
Molecules
MolB
Atoms B
Molecules
II. Limiting Reagent
A. Stoichiometric amounts: The proportions indicated in
balanced rxn.
the _________
B. Most reactions do not have stoichiometric amounts.
Generally, one reactant will be _________
depleted before the
other. The reactant that is depleted first is known as the
limiting reagent (LR) The reactant that is left at the
___________________.
excess reagent (ER)
end of the reaction is called the ___________________.
C. Analogy: How to make a cheese sandwich.
2 slices of bread + 1 slice of cheese → 1 cheese sandwich
If you have 8 slices of bread and 6 slices of cheese, how
many sandwiches can you make? __
4 (theoretical yield)
bread
What is the limiting reagent? ______
cheese
What is the excess reagent? _______
How much of the excess reagent is left at the end of the
rxn? 2______________
slices cheese
D. Theoretical yield: The amount of product in
limiting reagent
grams that forms if all of the ________________
has reacted. (This number is CALCULATED on
paper! Units are in grams only!)
E. Actual yield: The amount of product (in
grams) that is actually made (This number is
from the EXPERIMENT).
Actual yield
Percentyield 
x 100
T heoretica
l yield
F. Percent yield: The comparison of the actual
yield to the theoretical yield.
Zn + 2 HCl  ZnCl2 + H2
If you have 1 mol of Zn, how much H2 would you
make?
1 mol Zn
1 mol H2
1 mol Zn
= 1 mol H2
If you have 1 mol of HCl, how much H2 would you
make?
1 mol HCl
1 mol H2
2 mol HCl
= 0.5 mol H2
What is the limiting reagent?
HCl
How much H2 is produced? 0.5 mol – theoretical yield
Zn + 2 HCl  ZnCl2 + H2
If you have 0.25 mol of Zn, how much H2 would you
make?
0.25 mol Zn
1 mol H2
1 mol Zn
= 0.25 mol H2
If you have 1.00 mol of HCl, how much H2 would
you make?
1.00 mol HCl
1 mol H2
2 mol HCl
= 0.5 mol H2
What is the limiting reagent?
Zn
How much H2 is produced? 0.25 mol - theoretical yield
PCl3 + 3 H2O  H3PO3 + 3 HCl
3.00 mol PCl3 and 3.00 mol H2O react. Determine the
limiting reactant and theoretical yield of HCl.
1. Determine the limiting reactant
3.00 mol PCl3 3 mol HCl
= 9.00 mol HCl
 EXCESS
= 3.00 mol HCl
 LIMITING
1 mol PCl3
3.00 mol H2O 3 mol HCl
3 mol H2O
2. Determine the theoretical yield of HCl
3.00 mol
PCl3 + 3 H2O  H3PO3 + 3 HCl
2. Determine the theoretical yield of HCl
3.00 mol
3. Determine the theoretical yield of HCl in grams
3.00 mol HCl 36.46g HCl
= 109 g HCl
1 mol HCl
PCl3 + 3 H2O  H3PO3 + 3 HCl
Determine the limiting reactant and theoretical yield (g)
of H3PO3 if 225 g of PCl3 and 123 g of H2O are
reacted.
1. Determine the limiting reactant
LIMITING
225 g PCl3 1 mol PCl3 1 mol H3PO3 82.00g H3PO3
137.32g
PCl3
1 mol
PCl3
1 mol H3PO3
123 g H2O 1 mol H2O 1 mol H3PO3 82.00g H3PO3
18.02g
H2O
3 mol H2O
1 mol H3PO3
= 134g H3PO3
EXCESS
= 187g H3PO3
PCl3 + 3 H2O  H3PO3 + 3 HCl
The theoretical yield of this reaction is 134g H3PO3.
However, the actual yield from the experiment is
120g. Calculate the percent yield.
Actual yield
Percentyield 
x 100
T heoretica
l yield
120 g H3PO4
134 g H3PO4
X 100%
= 89.6 %
N2 + 3 H2  2 NH3
Determine the limiting reagent, the theoretical yield
and the percentage yield if 14.0g N2 are mixed with
9.0g H2 and the 16.1g NH3 actually formed.
LIMITING
14 g N2
1 mol N2
2 mol NH3
28.01g N2 1 mol N2
9 g H2
17.04g NH3
1 mol NH3
1 mol H2
2 mol NH3
17.04g NH3
2.02g H2
3 mol H2
1 mol NH3
16.1 g NH3
17.0 g NH3
X 100%
= 94.2 %
= 17.0g NH3
EXCESS
= 50.6g NH3
16.1g of bromine are mixed with 8.42g of chlorine to
give an actual yield of 21.1g of bromine
monochloride. Determine the limiting reactant and
the percentage yield.
Br2 + Cl2  2 BrCl
LIMITING
16.1 g Br2 1 mol Br2 2 mol BrCl
159.8g Br2 1 mol Br2
115.35g BrCl
= 23.2g BrCl
1 mol BrCl
EXCESS
8.42 g Cl2 1 mol Cl2 2 mol BrCl
115.35g BrCl
70.9g Cl2
1 mol BrCl
21.1 g BrCl
23.2 g BrCl
1 mol Cl2
X 100%
= 27.4g BrCl
= 90.9 %