Chemistry 101 : Chap. 3
Stoichiometry
(1) Chemical Equations
(2) Some Simple Patterns of Chemical Reactivity
(3) Formula Weight
(4) Avogadro’s Number and the Mole
(5) Empirical Formulas from Analyses
(6) Quantitative Information from Balanced Equations
(7) Limiting Reactants
Stoichiometry
A study of mass relationships that exist between substances
consumed (reactants) and produced (products) in
chemical reactions.
Stoichiometry is build upon an understanding of atomic
masses, chemical formulas and law of conservation of mass.
“… nothing is created: an equal quantity of matter
exists both before and after the experiment.”
Antoine Lavoisier (1743 ~ 1794)
With the Dalton’s atomic theory, law of conservation
of mass is known as law of conservation of atom
Balancing Chemical Equations
 Chemical Equation:
2H2
+
O2
=
2H2O
coefficient
reactants
products
+
 Balanced Chemical Equation: Equations that represent the
correct amount of reactants and products in chemical reaction.
Balanced chemical equations satisfy
the law of conservation of atoms
Balancing Chemical Equations
 Balancing Chemical Equations: Determining the coefficients
that provide equal numbers of each type of atom on
each side of the equation
(1) You only determine the coefficients, not the subscripts.
(2) Subscripts should never be changed in balancing equations.
 It changes the identities of reactants or products
(3) Balanced equation should contain the smallest possible
whole-number coefficients
(4) It is usually best to balance first those elements that occur
in the fewest chemical formulas on each side of equation
Balancing Chemical Equations
→
+
2 CO
+
→
+
CH4
+
→
O2
4 Cl2
2 CO2
+
→
CCl4
+
4 HCl
Balancing Chemical Equations
C 2H 4
2
3
Al
NH4NO3
+
6 HCl
→
2 NH4NO3
H 2O
N2
→
→
+
2 N2
2CO2
2
3
→
2 HCl
+
2 Al
→
3 O2
+
AlCl3
+
2 AlCl3
1
O
2 2
+
+
O2
+
2 H 2O
+
4
+
2 H 2O
H2
x3
3 H2
x2
Balancing Chemical Equations
Example: Law of conservation of atom (or mass)
How many NH3 molecules should be shown in the
container on the right?
Three Basic Reactions Types
 Combination Reactions: A + B  C
2Mg (s) + O2 (g)  2MgO (s)
Two reactants combine to form a single product
NOTE: The symbols (s), (l), (g), (aq) specify the physical state of
each substance
Examples of combination reactions [ Not balanced!]
(1)
N2
+
H2
NH3
(2)
Fe
+
O2
Fe2O3 Iron (III) oxide
(3)
S
+
O2
SO3
Three Basic Reactions Types
 Decomposition Reactions: A  B + C
2 HgO (s)  2 Hg (l) + O2 (g)
A single reactant decomposes to form two or more products
Examples of decomposition reactions [not balanced!]
(1)
NaN3 (s)
Na (s)
+
N2 (g)
(2)
CaCO3 (s)
CaO + CO2 (g)
NOTE: Over 22 million tons of limestone (CaCO3) are converted to
lime (CaO) each year for use in making glass, mortar, and for
extracting iron from its ore
Three Basic Reactions Types
 Combustion Reactions : Organic compounds (or hydrocarbons)
is burned (combusted) in oxygen to produce CO2 and H2O
C3H8 (g) + 5 O2 (g)
3 CO2 (g) + 4 H2O (g)
NOTE: When balancing combustion reaction, it is best
to balance the oxygen atoms last.
Examples of combustion reactions [not balanced]
(1) CH4(g) +
(2) C4H10 (g) +
O2 (g)
O2 (g)
CO2 (g) +
H2O (g)
CO2 (g) + H2O (g)
Formula Weight
Chemical equations tell us the exact numbers of molecules and atoms
involved in chemical reactions. But, in laboratory, what we usually
measure is the weight of each compound. How can we relate them?
 Formula Weight (FW) : The sum of the atomic weight of
each atom in its chemical formula
FW of H2SO4 = 2 x (AW of H) + (AW of S) + 4 x (AW of O)
= 2 x 1.0 + 32.1 + 4 x 16.0 = 98.1 (amu)
NOTE: If the chemical formula is that of a molecule, the formula
weight is also called molecular weight (MW).
NOTE: The masses used in the formula weight are from the
periodic table (weighted average of all isotope’s masses)
Formula Weight
 Examples
: Determined the formula weights of following substances
(1) FW of Mg(OH)2 =
(2) FW of Ca(NO3)2 =
Formula Weight
 Percentage
Composition : The percentage by mass contributed
by each element in the substance
% element 
total mass of element
100
formular w eight of compound
Example : Calculate the % weight of C, H and O in C12H22O11
The Mole
The mole is the SI unit for amount (number)
of substance. It was originally proposed by
Wilhelm Ostward in 1896
 Mole (mol): A quantity of objects equal to
the number of atoms of carbon in exactly
12 g of 12C.
Wilhelm Ostward (1853 ~ 1932)
He was awarded the novel prize for
chemistry for his work on catalysis
in 1909.
A mole of particles is an extremely large number of
objects and it is approximately equal to 6.022  1023
particles. This number is referred to as Avogadro’s
Number (NA) in honor of Amadeo Avogadro.
Amadeo Avogadro
(1776 ~ 1856)
1 mol = 6.022  1023 particles
The Mole
The mole is simply a unit of counting
1 dozen eggs = 12 individual eggs
½ dozen eggs = 6 individual eggs
1 mol of eggs = 6.022  1023 individual eggs
½ mol of eggs = 3.011  1023 individual eggs
1 mol of oxygen atoms = 6.022 1023 oxygen atoms (O)
1 mol of oxygen molecules = 6.022 1023 oxygen molecules (O2)
= 2 mol of oxygen atoms
The Mole
 Example : How many moles of eggs are in an egg carton holding
12 eggs?
 Example : How many moles of oxygen atoms are in 1.5 moles
of Ca(NO3)2 ?
Molar Mass
Molar mass = mass of 1 mole of a substance in gram
(1) Molar mass is nothing to do with Molecule.
(2) 1 mole of He atoms includes the same number of particles
as 1 mole of Ne atoms. However, 1 mole of He atoms and
1 mole of Ne atoms have different weights because
He and Ne have different molar mass.
(3) 12 amu = mass of a single 12C atom (exact)
12 gram = mass of 1 mole of 12C atoms (exact)
molar mass of 12C = 12 g/mol
FW (or MW) in amu = molar mass in g/mol
Molar Mass
 1 mol of Cu = 6.02  1023 copper atoms
 AW of Cu = 63.5 amu = average mass of naturally occurring
(FW or MW)
single Cu atom
(This number is from the periodic table)
 molar mass of Cu = 63.5 g/mol = mass of 6.02  1023 copper atoms
amount of a substance in gram
 molar mass
 molar mass
amount of a substance in mol
Molar Mass
 Example: Calculate the mass of 0.433 mole of calcium nitrate.
 Example: How many grams of oxygen are in 0.433 mole of Ca(NO3)2
Molar Mass
 Problem: How many hydrogen atoms are in 4.5g of CH3OH ?
1. Compute the molecular weight or molar mass of CH3OH:
2. Determine the amount of CH3OH in mol.
3. Determine the amount of H
Molar Mass
 Problem: How many grams of carbon are in 42g of CH3CH2OH ?
1. Compute the molecular weight or molar mass of CH3CH2OH:
2. Determine the amount of CH3CH2OH in mol.
3. Determine the amount of C
Molar Mass and
Avogadro’s Number
1.08  1024
Copper atoms
1.8 mol of
Copper
114.3 g of
Copper
molar mass
Avogadro’s
number
Empirical Formulas from Analyses
 Empirical Formula : Indicate the relative number of atoms
of each type in a molecule
If we know the relative number of atoms in mol, we can
determine the compound’s empirical formula
 Example : A compound made of Hg and Cl has 73.9% mercury
(Hg) and 26.1% chlorine (Cl) by mass. What is the
empirical formula for this compound?
Empirical Formulas from Analyses
 Example
: A compound is found to contain by mass 47.7% C,
10.5% H and 42.1% O. What is the empirical formula
of the compound?
Empirical Formulas from Analyses
 Example: A compound is found to have by mass 53.5% C,
11.1% H and 35.6% O. The experimentally determined
molecular weight is 90 amu. What is the molecular
formula of this compound?
Quantitative Information from
Balanced Chemical Equations
Conservation of atom
FOUR H atoms
2H2
TWO O atoms
+
TWO hydrogen
molecules
FOUR H atoms + TWO O atoms

O2
ONE oxygen
molecule
2H2O
TWO water
molecules
 6.021023
2 mol H2 + 1 mol O2  2 mol H2O
 molar mass
4g H2
+ 32 g O2  36g H2O
Conservation
of mass
Quantitative Information from
Balanced Chemical Equations
 Example: How many moles of carbon dioxide would be produced
by burning 3 moles of carbon monoxide?
(1) Write down the reactants and products:
(2) Balance the chemical equation :
(3) Compute the amount of CO2 :
Quantitative Information from
Balanced Chemical Equations
 Example: How many grams of H2O are formed from the complete
combustion of 3.0 g of C2H6?
(1) Write down the reactants and products :
(2) Balance the chemical equation :
(3) Convert the mass (gram) to mol
(4) Compute the amount of water in mol :
(5) Convert the mol to mass (gram) :
Quantitative Information from
Balanced Chemical Equations
Stoichiometric relations between compounds A and B
Quantitative Information from
Balanced Chemical Equations
 Example: Octane (C8H18), which is liquid in room temperature,
has a density of 0.692 g/ml at 20oC. How many grams
of O2 are required to completely burn 1.00 gal of octane?
(1) Balance the combustion reaction:
(2) Compute the mass of octane :
(3) Convert the mass of octane to mol
(4) Compute the amount of oxygen to react (in mol)
(5) Compute the amount of oxygen to react (in gram)
Limiting Reactants
 Limiting reactant : The reactant that is completely consumed
in a reaction.
2H2
+
O2

2H2O
Suppose we have a mixture of 10 mol H2 and 7 mol O2.
Then, how many moles of O2 will be used?
Since 2 moles of H2 consume 1 mole of O2, 10 moles of H2
will consume 5 moles of O2 to produce 10 moles of H2O
 We have 7 mol – 5 mol = 2 mol of excess oxygens.
Hydrogen is the limiting reactant !
Limiting Reactants
2H2
+
O2

2H2O
Initial quantities :
Changes :
10 mol
10 mol
7 mol
5 mol
0 mol
10 mol
Final quantities :
0 mol
2 mol
10 mol
Limiting Reactants
What if we had 4 moles of oxygens to start the same reaction?
2H2
+
O2

2H2O
Initial quantities :
Changes :
10 mol
8 mol
4 mol
4 mol
0 mol
8 mol
Final quantities :
2 mol
0 mol
8 mol
Oxygens are completely used up.
Therefore, O2 is the limiting reactant
Limiting Reactants
How many compete cars can be built from these parts?
Limiting Reactants
 Example : How many grams of P4O10 can be produced by the
reaction of 1.0g of phosphorous and 3.0g of oxygens?
P4
+
MW=124
5O2 
MW=32
P4O10
MW=284
(1) Convert grams to mol :
(2) Find the limiting reactant :
(3) Set up the stoichiometic table (if you want)
(4) Convert mol to grams
Yields
 Theoretical Yield : The quantity of product that is calculated to
form when all of the limiting reactants react.
 This is from balanced chemical equations
 Actual Yield : The amount of product actually obtained in a
reaction.
 This is from actual experiments
Actual yield ≤ theoretical yield
 Percent Yield : Ratio between the theoretical yield and the
actual yield
actual yield
Percent Yield 
100 %
theoretica l yield
Yields
 Example : 2.50 g of copper is reacted with excess sulfur and
3.00 g of copper(I) sulfide is produced. What is the
%yield of the reaction?
16Cu(s) + S8(s)  8Cu2S (s)
AW = 63.5
FW=159
(1)What is the limiting reactant?
(2) Calculate the theoretical yield of copper sulfide
(3) Calculate % yield :
Yields
 Example : Lithium and nitrogen react to produce lithium nitride.
If 5.00 g of each reactant undergoes a reaction with a 88.5%
yield, how many grams of Li3N are obtained?
6 Li (s) +
N2 (g)  2 Li3N (s)
(1) Convert grams to mol:
(2) Determine the limiting reactant
(3) Compute the theoretical yield
(4) Compute the actual yield