Introduction to Chemical Reactions
A chemical change—a chemical reaction—converts
one substance into another.
Chemical reactions involve:
• breaking bonds in the reactants (starting materials)
• forming new bonds in the products.
CH4 and O2
CO2 and H2O
1
Introduction to Chemical Reactions
A chemical equation is an expression that uses
chemical formulas and other symbols to illustrate
what reactants constitute the starting materials in a
reaction and what products are formed.
• The reactants are written on the left.
• The products are written on the right.
• Coefficients show the number of molecules of
a given element or compound that react or are
formed.
2
Introduction to Chemical Reactions
• The law of conservation of mass states that
atoms cannot be created or destroyed in a
chemical reaction.
• Coefficients are used to balance an equation.
• A balanced equation has the same number of
atoms of each element on both sides of the
equation.
3
Introduction to Chemical Reactions
4
Balancing Chemical Equations
HOW TO Balance a Chemical Equation
Example
Write a balanced chemical equation for
the reaction of propane (C3H8) with
oxygen (O2) to form carbon dioxide (CO2)
and water (H2O).
Step [1]
Write the equation with the correct formulas.
C3H8 + O2
CO2 + H2O
• The subscripts in a formula can never be changed
to balance an equation, because changing a
subscript changes the identity of a compound.
5
Balancing Chemical Equations
HOW TO Balance a Chemical Equation
Step [2]
Balance the equation with coefficients one
element at a time.
• Balance the C’s first:
• Balance the H’s next:
6
Balancing Chemical Equations
HOW TO Balance a Chemical Equation
Step [2]
Balance the equation with coefficients one
element at a time.
• Finally, balance the O’s:
7
Balancing Chemical Equations
HOW TO Balance a Chemical Equation
Step [3]
Check to make sure that the smallest set
of whole numbers is used.
8
Chemistry of an Automobile Airbag
• A severe car crash triggers the ignition of sodium
azide (NaN3) converting it to sodium (Na) and
nitrogen gas (N2).
2 NaN3
2 Na + 3 N2
• The nitrogen gas causes the bag to inflate fully in
40 milliseconds, protecting passengers from
injury.
9
The Mole and Avogadro’s Number
A mole is a quantity that contains 6.02 x 1023 items.
• 1 mole of C atoms = 6.02 x 1023 C atoms
• 1 mole of H2O molecules = 6.02 x 1023 H2O molecules
• 1 mole of vitamin C molecules = 6.02 x 1023 vitamin C
molecules
The number 6.02 x 1023 is Avogadro’s number.
It can be used as a conversion factor to relate the
number of moles of a substance to the number of
atoms or molecules:
1 mol
6.02 x 1023 atoms
or
6.02 x 1023 atoms
1 mol
10
The Mole and Avogadro’s Number
Sample Problem 5.6
How many molecules are contained in 5.0 moles
of carbon dioxide (CO2)?
Step [1]
Identify the original quantity and the
desired quantity.
5.0 mol of CO2
original quantity
? number of molecules of CO2
desired quantity
11
The Mole and Avogadro’s Number
Step [2]
Write out the conversion factors.
1 mol
or
6.02 x 1023 molecules
6.02 x 1023 molecules
1 mol
Choose this one to cancel mol.
Step [3]
5.0 mol x
Set up and solve the problem.
6.02 x 1023 molecules
1 mol
=
3.0 x 1024 molecules CO2
Unwanted unit cancels.
12
Mass to Mole Conversions
• The formula weight/ molecular mass is the sum of the
atomic weights of all the atoms in a compound, reported in
atomic mass units (amu).
Sample Problem 5.8
What is the formula weight of nicotine (C10H14N2)?
Step [1]
Determine the number of atoms of each
element from the subscripts in the chemical
formula.
• C10H14N2 contains 10 C atoms, 14 H atoms,
and 2 N atoms.
13
Mass to Mole Conversions
Step [2]
Multiply the number of atoms of each
element by the atomic weight and add
the results.
10 C atoms x 12.01 amu = 120.1 amu
14 H atoms x 1.08 amu = 14.11 amu
2 N atoms x 14.01 amu = 28.02 amu
Formula weight of C10H14N2 = 162.23 amu
14
Mass to Mole Conversions
Molar Mass
• The molar mass is the mass of one mole of any
substance, reported in grams.
• The value of the molar mass of a compound in
grams equals the value of its formula weight in
amu.
15
Mass to Mole Conversions
Relating Grams to Moles
• The molar mass relates the number of moles to
the number of grams of a substance.
• In this way, molar mass can be used as a
conversion factor.
Sample Problem 5.10
How many moles are present in 100. g of aspirin
(C9H8O4, molar mass 180.2 g/mol)?
Step [1]
Identify the original quantity and the
desired quantity.
100. g of aspirin
? mol of aspirin
original quantity
desired quantity
16
Mass to Mole Conversions
Relating Grams to Moles
Step [2]
Write out the conversion factors.
• The conversion factor is the molar mass, and it
can be written in two ways.
• Choose the one that places the unwanted unit,
grams, in the denominator so that the units cancel:
180.2 g aspirin
1 mol
or
1 mol
180.2 g aspirin
Choose this one to
cancel g.
17
Mass to Mole Conversions
Relating Grams to Moles
Step [3]
Set up and solve the problem.
100. g aspirin
x
1 mol
= 0.555 mol aspirin
180.2 g aspirin
Unwanted unit cancels.
18
Mole Calculations in Chemical Equations
A balanced chemical equation also tells us:
• the number of moles of each reactant that combine
• the number of moles of each product formed.
1 N2(g)
+
1 O2(g)
1 mole of N2
1 mole of O2
1 molecule N2 1 molecule O2
2 NO(g)
2 moles of NO
2 molecules NO
(The coefficient “1” has been written for emphasis.)
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Mole Calculations in Chemical Equations
Coefficients are used to form mole ratios, which can
serve as conversion factors.
N2(g)
+
Mole ratios:
1 mol N2
1 mol O2
O2(g)
1 mol N2
2 mol NO
2 NO(g)
1 mol O2
2 mol NO
Use the mole ratios from the coefficients in the
balanced equation to convert moles of one
compound (A) into moles of another compound (B).
20
Mole Calculations in Chemical Equations
Sample Problem 5.11
Using the balanced chemical equation, how
many moles of CO are produced from 3.5 moles
of C2H6?
2 C2H6(g) + 5 O2(g)
Step [1]
4 CO(g) + 6 H2O(g)
Identify the original and desired quantities.
3.5 mol C2H6
original quantity
? mol CO
desired quantity
21
Mole Calculations in Chemical Equations
Step [2]
Write out the conversion factors.
2 mol C2H6
4 mol CO
4 mol CO
2 mol C2H6
or
Choose this one to
cancel mol C2H6.
Step [3]
Set up and solve the problem.
3.5 mol C2H6 x
4 mol CO =
2 mol C2H6
7.0 mol CO
Unwanted unit cancels.
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Mass Calculations in Chemical Equations
HOW TO Convert Moles of Reactant to Grams of Product
Example
Using the balanced equation, how many
grams of O3 are formed from 9.0 mol of O2?
3 O2(g)
Moles of
reactant
[1]
mole–mole
conversion
factor
sunlight
Moles of
product
2 O3(g)
[2]
Grams of
product
molar mass
conversion
factor
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Mass Calculations in Chemical Equations
HOW TO Convert Moles of Reactant to Grams of Product
Step [1]
Convert the number of moles of reactant
to the number of moles of product using
a mole–mole conversion factor.
Step [2]
Convert the number of moles of product
to the number of grams of product using
the product’s molar mass.
3 mol O2 or
2 mol O3
2 mol O3
3 mol O2
Cancel mol O2
in step [1].
1 mol O3 or
48.00 g O3
48.00 g O3
1 mol O3
Cancel mol O3
in step [2].
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Mass Calculations in Chemical Equations
HOW TO Convert Moles of Reactant to Grams of Product
• Set up and solve the conversion.
Moles of
reactant
9.0 mol O2
mole–mole
conversion
factor
2 mol O3
x
3 mol O2
Mol O2 cancel.
x
molar mass
conversion
factor
Grams of
product
48.00 g O3
1 mol O3 =
290 g O3
Mol O3 cancel.
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Mass Calculations in Chemical Equations
HOW TO Convert Grams of Reactant to Grams of Product
Example
Ethanol (C2H6O, molar mass 46.07 g/mol) is
synthesized by reacting ethylene (C2H4,
molar mass 28.05 g/mol) with water.
How many grams of ethanol are formed
from 14 g of ethylene?
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Mass Calculations in Chemical Equations
HOW TO Convert Grams of Reactant to Grams of Product
mole–mole
conversion
factor
Moles of
reactant
molar mass
conversion
factor
[1]
Grams of
reactant
[2]
Moles of
product
[3]
molar mass
conversion
factor
Grams of
product
27
Mass Calculations in Chemical Equations
HOW TO Convert Grams of Reactant to Grams of Product
Grams of
reactant
molar mass
conversion
factor
mole–mole
conversion
factor
molar mass
conversion
factor
1 mol C2H4
1 mol C2H6O
46.07 g C2H6O
14 g C2H4 x 28.05 g C H x 1 mol C H x 1 mol C H O
2 4
2 4
2 6
Grams C2H4
cancel.
=
Moles C2H4
cancel.
23 g C2H6O
Moles C2H6O
cancel.
Grams of
product
28