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Chapter 3 Stoichiometry:
Calculations with Chemical Formulas
and Equations
Stoichiometry
Stoichiometry
Law of Conservation of Mass
“We may lay it down as an
incontestable axiom that, in all
the operations of art and nature,
nothing is created; an equal
amount of matter exists both
before and after the experiment.
Upon this principle, the whole art
of performing chemical
experiments depends.”
--Antoine Lavoisier, 1789
Stoichiometry
Poor Antoine
Stoichiometry
Dalton’s Postulate #4
• Atoms of elements are neither created nor
destroyed, simply rearranged during a
chemical reaction.
Stoichiometry
Chemical Equations
Chemical equations are concise representations
of chemical reactions.
Write a balanced equation for this reaction.
CH4 (g) +
2 O2 (g)
CO2 (g) + 2 H2O (g)
Stoichiometry
Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (g)
Stoichiometry
Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g)
Reactants appear on the left
side of the equation.
CO2 (g) + 2 H2O (g)
Stoichiometry
Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g)
Products appear on the
right side of the equation.
CO2 (g) + 2 H2O (g)
Stoichiometry
Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (g)
The states of the reactants and products
are written in parentheses to the right of
each compound.
Stoichiometry
Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g)
Coefficients are inserted
to balance the equation.
CO2 (g) + 2 H2O (g)
Stoichiometry
Subscripts and Coefficients Give
Different Information
• Subscripts tell the number of atoms of
each element in a molecule.
Stoichiometry
Subscripts and Coefficients Give
Different Information
• Subscripts tell the number of atoms of
each element in a molecule
• Coefficients tell the number of
molecules.
Stoichiometry
Reaction Types
Synthesis
Decomposition
Single Replacement
Combustion
DoubleStoichiometry
Replacement
Combination or Synthesis Reactions
• In this type of
reaction two or
more substances
react to form one
product.
Examples:
2 Mg (s) + O2 (g)  2 MgO (s)
N2 (g) + 3 H2 (g)  2 NH3 (g)
C3H6 (g) + Br2 (l)  C3H6Br2 (l)
Stoichiometry
Decomposition Reactions
• In a decomposition
one substance breaks
down into two or more
substances.
Examples:
CaCO3 (s)  CaO (s) + CO2 (g)
2 KClO3 (s)  2 KCl (s) + O2 (g)
2 NaN3 (s)  2 Na (s) + 3 N2 (g)
Stoichiometry
NaN3 = Sodium Azide
•
•
- colorless salt is the gas-forming
component in many car airbag
systems.
- an ionic substance, is highly soluble
in water, and is very acutely toxic.
Azide is the anion with the
formula N-3.
Stoichiometry
Combustion Reactions
• These are generally
rapid reactions that
produce a flame.
• Most often involve
hydrocarbons reacting
with oxygen in the air.
• Examples:
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g)
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g)
Stoichiometry
Formula Weights
Molecular Weights
Percent Composition
Molar Mass
Stoichiometry
Formula Weight (FW)
• A formula weight is the sum of the atomic
weights for the atoms in a formula unit.
• So, the formula weight of calcium chloride,
CaCl2, would be
Ca: 1(40.1 amu)
+ Cl: 2(35.5 amu)
111.1 amu
• Formula weights are generally reported for
ionic compounds.
Stoichiometry
Molecular Weight (MW)
• A molecular weight is the sum of the atomic
weights of the atoms in a molecule.
• For the molecule ethane, C2H6, the molecular
weight would be
C: 2(12.0 amu)
+ H: 6(1.0 amu)
30.0 amu
• Molecular weights are generally reported for
covalent compounds.
Stoichiometry
Percent Composition
One can find the percentage of the mass of a
compound that comes from each of the
elements in the compound by using this
equation:
(number of atoms)(atomic weight)
% element =
(FW of the compound)
x 100
Stoichiometry
Percent Composition
So … the percentage of carbon in
ethane (C2H6) is…
(2)(12.0 amu)
%C =
(30.0 amu)
24.0 amu
x 100
=
30.0 amu
= 80.0%
Stoichiometry
Moles
Stoichiometry
Stoichiometry
Avogadro’s Number
• 6.02 x 1023
• 1 mole of 12C has a
mass of 12 g.
Stoichiometry
Molar Mass
• By definition, a molar mass is the mass of 1
mol of a substance (i.e., g/mol).
– The molar mass of an element is the mass
number for the element that we find on the
periodic table.
– The formula weight (in amu’s) will be the same
number as the molar mass (in g/mol).
Stoichiometry
Stoichiometry
Stoichiometry
Using Moles
Moles provide a bridge from the molecular scale to
the real-world scale.
Stoichiometry
Mole Relationships
• One mole of atoms, ions, or molecules contains
Avogadro’s number of those particles.
• One mole of molecules or formula units contains
Avogadro’s number times the number of atoms or
ions of each element in the compound.
Stoichiometry
Calculating
Empirical +
Molecular
Formulas
Stoichiometry
Empirical vs. Molecular
Stoichiometry
Calculating Empirical Formulas
÷ by small
% to mass
mass to mole
× ‘til whole
One can calculate the empirical formula from the
percent composition.
Stoichiometry
Stoichiometry
Calculating Empirical Formulas
The compound para-aminobenzoic acid (you may have seen
it listed as PABA on your bottle of sunscreen) is composed of
carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and
oxygen (23.33%). Find the empirical formula of PABA.
Stoichiometry
Calculating Empirical Formulas
% to mass
Assuming 100.00 g of para-aminobenzoic acid,
C:
H:
N:
O:
1 mol
12.01 g
1 mol
5.14 g x
1.01 g
1 mol
10.21 g x
14.01 g
1 mol
23.33 g x
16.00 g
61.31 g x
= 5.105 mol C
= 5.09 mol H
= 0.7288 mol N
= 1.456 mol O
mass to mole
Stoichiometry
Calculating Empirical Formulas
Calculate the mole ratio by dividing by the smallest number
of moles:
÷ by small
C:
5.105 mol
0.7288 mol
= 7.005  7
H:
5.09 mol
0.7288 mol
= 6.984  7
N:
0.7288 mol
0.7288 mol
= 1.000
O:
1.458 mol
0.7288 mol
= 2.001  2
Stoichiometry
Calculating Empirical Formulas
These are the subscripts for the empirical formula:
C7H7NO2
Coefficients become subscripts
Stoichiometry
Calculating Molecular from Empirical
Mesitylene, a hydrocarbon that occurs in small amounts in crude
oil, has an empirical formula of C3H4. The experimentally
determined molecular weight of this substance is 121 amu. What
is the molecular formula of mesitylene?
1. Calculate the empirical formula weight of C3H4
3 (12 amu) + 4 (1 amu) = 40.0 amu = FW
2. Divide the molecular weight by the formula weight to obtain
the factor used to multiply the subscripts in C3H4:
MW = 121 = 3.02 ~ 3
FW
40
3. Multiply each subscript in the empirical formula by 3 to
give
Stoichiometry
the molecular formula: C9H12
Let’s Try and Example!
Practice Exercise 3.14 pg. 94
Stoichiometry
Combustion Analysis Technique
Lab technique used to determine Empirical Formula contiaining
hydrocarbons mainly. Go to p. 95 Figure 3.14!
Stoichiometry
Combustion Analysis
• Compounds containing C, H and O are routinely
analyzed through combustion in a chamber like this.
– C is determined from the mass of CO2 produced.
– H is determined from the mass of H2O produced.
– O is determined by difference after the C and H have been
determined.
Stoichiometry
Elemental Analyses
Compounds
containing other
elements are
analyzed using
methods analogous to
those used for C, H
and O.
Complete Sample Exercise 3.15 pg. 95
Stoichiometry
Combustion Analysis Problems
1. Use Stoichiometry to find mass of C, H, O
2. Mass (g) to moles of C, H, O
3. Divide by small, multiply ‘til whole
Stoichiometry
What is the mole ratio of smores?
2 Gc + 1 M + 4 Cp  1 Smore
Where do mole ratios come from?
Stoichiometry
Stoichiometric Calculations
The coefficients in the balanced equation give the
ratio of moles of reactants and products.
Stoichiometry
Stoichiometric Calculations
Mole Ratio
Stoichiometry
Ex. How many grams of water are produced
in the ocidation of 1.00 g of glucose?
C6H12O6 + 6 O2  6 CO2 + 6 H2O
Circle the known.
1g C6H12O6
GIVEN
1 mol C6H12O6
180 g C6H12O6
C: 6 x 12 = 72
H: 12 x 1 = 12
O: 6 x 16 = 96 +
180 g/ mol
Box what you are trying to find.
6 mol H2O
1 mol C6H12O6
18.0 g H O
2
= 0.600 g H2O
1 mol H O
2
H: 2 x 1 = 2
O: 1 x 16 = 16 +
Stoichiometry
18 g/mol
Limiting
Reactants
Stoichiometry
What’s the max amount of product
that I can yield?
2 Gc + 1 M + 4 Cp  1 Smore
What is limiting?
What is excess?
7 Gc + 1 M + 8 Cp  ? Smore
Stoichiometry
Which is limiting the amount of
product?
Stoichiometry
How Many Cookies Can I Make?
• You can make cookies
until you run out of one
of the ingredients.
• Once this family runs
out of sugar, they will
stop making cookies
(at least any cookies
you would want to eat).
Stoichiometry
How Many Cookies Can I Make?
• In this example the
sugar would be the
limiting reactant,
because it will limit the
amount of cookies you
can make.
Stoichiometry
Limiting Reactants
• The limiting reactant is the reactant present in the
smallest stoichiometric amount.
– In other words, it’s the reactant you’ll run out of first
(in this case, the H2).
Stoichiometry
Limiting Reactants
In the example below, the O2 would be the excess
reagent.
Stoichiometry
Limiting Reactants Practice
p. 101 Practice Exercise
Stoichiometry
Theoretical Yield
• The theoretical yield is the maximum amount
of product that can be made.
– In other words it’s the amount of product
possible as calculated through the stoichiometry
problem.
• This is different from the actual yield, which
is the amount one actually produces and
measures.
Stoichiometry
Theoretically vs. Actually
Stoichiometry
Theoretical Yield Example
• What is the theoretical yield of CaO if 24.8 g of
CaCO3 is heated?
1
1
1
_____
CaCO3  _____
CaO + ______
CO2
mol CaCO3 x _________
1 mol CaO x __________
56 g CaO
24.8 g CaCO3 x 1________
100.1 g CaCO3
1 mol CaCO3
= 13.9 g CaO
1 mol CaO
• Theoretical Yield = 13.9 g CaO
Stoichiometry
Percent Yield
One finds the percent yield by comparing the
amount actually obtained (actual yield) to the
amount it was possible to make (theoretical
yield).
Actual Yield
Percent Yield =
Theoretical Yield
x 100
Stoichiometry
Percent Yield Example
• What is the percent yield if 13.1 g of CaO was
actually produced when 24.8 g of CaCO3 was
heated?
% Yield
=
13.1 g x 100 = 94.2 %
13.9 g
Stoichiometry