Download Lecture 14

Chemistry 103
Lecture 14
Outline
I. Empirical/Molecular Formulas
II. Chemical Reactions
- basic symbols
- balancing
- classification
III. Stoichiometry
Types of Formulas
The molecular formula
 Is the true or actual number of the atoms in a
molecule
The empirical formula

Is the simplest whole number ratio of the atoms
(this is the formula for ionic compounds)
H2O2
molecular formula
HO
empirical formula
Empirical & Molecular Formulas

Ionic Compounds - Only need Empirical
formula.

Molecules - Need information on both to
determine exact make-up of your system
Molecular Formula
C 6H 6
Empirical Formula
CH
Relating Molecular and Empirical
Formulas
A molecular formula
 Is a multiple (or equal) of its empirical
formula
 Has a molar mass that is the empirical
mass multiplied by a whole number
molar mass
= a whole number
empirical mass
 Is obtained by multiplying the empirical
formula by a whole number
Finding the Molecular Formula
Determine the molecular formula of
compound that has a molar mass of
78.11 g/mole and an empirical formula
of CH.
Molecular Formula
A compound is 24.27% C, 4.07% H, and
71.65% Cl. The molar mass is known to
be 99.0 g. What are the empirical and
molecular formulas?
Chemical Reactions
Physical Change
In a physical change,
• The identity and composition
of the substance do not
change
• The state can change or the
material can be torn into
smaller pieces
Copyright © 2005 by Pearson Education, Inc.
Publishing as Benjamin Cummings
Chemical Change
In a chemical change,
• Reacting substances form
new substances with
different compositions and
properties
• A chemical reaction takes
place
Copyright © 2005 by Pearson Education, Inc.
Publishing as Benjamin Cummings
Chemical Reaction
In a chemical reaction,
 Old bonds are broken and
new bonds are formed
 Atoms in the reactants are
rearranged to form one or
more different substances

Fe and O2 form rust (Fe2O3)
Chemical Reaction
In a chemical reaction,
• A chemical change
produces one or more new
substances
• There is a change in the
composition of one or more
substances
Copyright © 2005 by Pearson Education, Inc.
Publishing as Benjamin Cummings
Evidence of a Chemical Reaction

Changes that can be
seen are evidence of a
chemical reaction.
Writing a Chemical Reaction

Chemists use a shorthand approach when
writing the specifics of a chemical reaction.
This approach is called the chemical
equation.
Reactants -----> Products
Chemical Equations
A chemical equation,
• Gives the chemical formulas of the reactants on the left of an
arrow and the products on the right
Reactants
Product
O2 (g)
C(s)
CO2 (g)
Symbols Used in Equations
Symbols used in chemical
equations show:

The states of the reactants

The states of the products

The reaction conditions
Chemical Equations Are Balanced
In a balanced chemical reaction,
• Atoms are not gained or lost
Chemical Equations Are Balanced
In a balanced chemical reaction,
• The number of reactant atoms are equal to the
number of product atoms
Chemical Equations
•Chemical equations:
symbolic descriptions of
chemical reactions.
•Two parts to an equation:
•reactants and products
H2 + O2  H2O
A Chemical Equation must
also be “balanced”.
2H2 + O2 --> 2H2O
Balanced Chemical Equations

Chemical Equations must be balanced

There must be equal numbers of atoms of each
element on both sides of the equation (both sides
of the arrow)



1. Write the correct symbols and formulas for all of the
reactants and products.
2. Count the number of each type of atom on BOTH
sides of the equation.
3. Insert coefficients until there are the equal numbers
of each kind of atom on both sides of the equation.
Example - making ammonia
Hydrogen gas and Nitrogen gas combine to
make ammonia in the gaseous state.
Example - making ammonia
Hydrogen gas and Nitrogen gas combine to
make ammonia in the gaseous state.
H2(g)
+
N2(g)
NH3 (g)
Example - making ammonia
Hydrogen gas and Nitrogen gas combine to
make ammonia in the gaseous state.
H2(g)
+
N2(g)
NH3 (g)
Example - making ammonia
Hydrogen gas and Nitrogen gas combine to
make ammonia in the gaseous state.
H2(g)
+
N2(g)
NH3 (g)
Example - making ammonia
Hydrogen gas and Nitrogen gas combine to
make ammonia in the gaseous state.
H2(g)
+
N2(g)
2 NH3 (g)
Example - making ammonia
Hydrogen gas and Nitrogen gas combine to
make ammonia in the gaseous state.
3 H2(g)
+
N2(g)
2 NH3 (g)
Example - making ammonia
Hydrogen gas and Nitrogen gas combine to
make ammonia in the gaseous state.
BEFORE
3 H2(g) + N2(g)
AFTER
2 NH3 (g)
Balancing Equations

Methane reacts with oxygen (combustion
reaction) to form carbon dioxide and water.
Write a properly balanced chemical equation
The Numbers
in Chemical Equations
More Practice:
Balancing Reactions

C 2H 6 +
O2 
CO2 +
H 2O

C 3H 6 +
O2 
CO2 +
H 2O

NH3 +
O2 
NO +
H 2O
Stoichiometry

Chemical Stoichiometry: using mass and quantity
relationships among reactants and products in a
chemical reaction to make predictions about how much
product will be made.
Quantities in a Chemical Reaction
4NH3(g) + 5O2(g)
4NO(g) + 6H2O(g)
Four molecules NH3 react with five molecules O2
to produce
four molecules NO and six molecules H2O
Quantities in a Chemical Reaction
4NH3(g) + 5O2(g)
4NO(g) + 6H2O(g)
Four molecules NH3 react with five molecules O2
to produce
four molecules NO and six molecules H2O
and
Four mol NH3 react with five mol O2
to produce
four mol NO and six mol H2O
Moles in Equations
 We can read the equation in “moles” by placing the
word “mole” or “mol” between each coefficient and
formula.
4Fe(s)
+
4 mol Fe +
3O2(g)
2Fe2O3(s)
3 mol O2
2 mol Fe2O3
Conservation of Mass
The Law of Conservation of Mass indicates:
• No change in total mass occurs in a reaction
• Mass of products is equal to mass of
reactants
Law of Conservation of Mass
In an ordinary chemical reaction,
• Matter cannot be created or destroyed
• The number of atoms of each element are equal
• The mass of reactants equals the mass of products
H2(g)
+
Cl2(g)
2HCl(g)
2 mol H,
2 mol Cl
=
2(1.008) + 2(35.45)
72.92 g
2 mol H, 2 mol Cl
=
=
2(36.46)
72.92 g
Conservation of Mass
2 mol Ag
+
2 (107.9 g) +
1 mol S
=
1 mol Ag2S
1(32.07 g)
=
1 (247.9 g)
=
247.9 g product
247.9 g reactants
Writing Mole-Mole Factors
A mole-mole factor is a ratio of the moles for two
substances in an equation.
4Fe(s)
+
Fe and O2
Fe and Fe2O3
O2 and Fe2O3
3O2(g)
2Fe2O3(s)
4 mol Fe
and
3 mol O2
4 mol Fe
and
2 mol Fe2O3
3 mol O2
and
2 mol Fe2O3
3 mol O2
4 mol Fe
2 mol Fe2O3
4 mol Fe
2 mol Fe2O3
3 mol O2
Calculations with Mole Factors
How many moles of Fe2O3 can form from 6.0 mol O2?
4Fe(s)
+
3O2(g)
Relationship:
2Fe2O3(s)
3 mol O2 = 2 mol Fe2O3
Write a mole-mole factor to determine the moles of Fe2O3.
6.0 mol O2 x
2 mol Fe2O3 = 4.0 mol Fe2O3
3 mol O2
Mole relations
How many moles of Fe are needed to react with 12.0 mol O2?
4Fe(s)
+
3O2(g)
A) 3.00 mol Fe
B) 9.00 mol Fe
C) 16.0 mol Fe
2 Fe2O3(s)