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Chapter 4
Calculations and the Chemical
Equation
Denniston
Topping
Caret
5th Edition
4.1 The Mole Concept and Atoms
• Atoms are exceedingly small
– Unit of measurement for mass of an atom is
atomic mass unit (amu) – unit of measure for
the mass of atoms
• carbon-12 assigned the mass of exactly 12 amu
• 1 amu = 1.66 x 10-24 g
• Periodic table gives atomic weights in amu
4.1 The Mole Concept and
Atoms
Mass of Atoms
• What is the atomic weight of one atom of
fluorine? Answer: 19.00 amu
• What would be the mass of this one atom
in grams?
19.00amu F 1.66110-24 g 3.1561023 g F


F atom
1 amu F
F atom
• Chemists usually work with much larger
quantities
– It is more convenient to work with grams
than amu when using larger quantities
4.1 The Mole Concept and
Atoms
The Mole and Avogadro’s Number
• A practical unit for defining a collection
of atoms is the mole
1 mole of atoms = 6.022 x 1023 atoms
• This is called Avogadro’s number
– This has provided the basis for the concept
of the mole
4.1 The Mole Concept and
Atoms
The Mole
• To make this connection we must define
the mole as a counting unit
– The mole is abbreviated mol
• A mole is simply a unit that defines an
amount of something
– Dozen defines 12
– Gross defines 144
4.1 The Mole Concept and
Atoms
Atomic Mass
• The atomic mass of one atom of an element
corresponds to:
– The average mass of a single atom in amu
– The mass of a mole of atoms in grams
– 1 atom of F is 19.00 amu
19.00 amu/atom F
– 1 mole of F is 19.00 g 19.00 g/mole F
19.00 amu F 1.661024 g F 6.0221023 atomF


1 atomF
1 amu F
1 mol F
=19.00 g F/mol F or 19.00 g/mol F
4.1 The Mole Concept and
Atoms
Molar Mass
• Molar mass - The mass in grams of 1 mole of
atoms
• What is the molar mass of carbon?
12.01 g/mol C
• This means counting out a mole of Carbon
atoms (i.e., 6.022 x 1023) they would have a mass
of 12.01 g
• One mole of any element contains the same
number of atoms, 6.022 x 1023, Avogadro’s number
4.1 The Mole Concept and
Atoms
Calculating Atoms, Moles, and Mass
• We use the following conversion factors:
• Density converts grams – milliliters
• Atomic mass unit converts amu –
grams
• Avogadro’s number converts moles –
number of atoms
• Molar mass converts grams – moles
4.1 The Mole Concept and
Atoms
Strategy for Calculations
• Map out a pattern for the required
conversion
• Given a number of grams and asked for
number of atoms
• Two conversions are required
• Convert grams to moles
1 mol S/32.06 g S OR 32.06 g S/1 mol S
• Convert moles to atoms
mol S x (6.022 x 1023 atoms S) / 1 mol S
4.1 The Mole Concept and
Atoms
Practice Calculations
1. Calculate the number of atoms in 1.7
moles of boron.
2. Find the mass in grams of 2.5 mol Na
(sodium).
3. Calculate the number of atoms in 5.0 g
aluminum.
4. Calculate the mass of 5,000,000 atoms
of Au (gold)
4.1 The Mole Concept and
Atoms
Interconversion Between Moles,
Particles, and Grams
4.2 The Chemical Formula,
Formula Weight, and Molar Mass
• Chemical formula - a combination of
symbols of the various elements that make up
the compound
• Formula unit - the smallest collection of
atoms that provide two important pieces of
information
– The identity of the atoms
– The relative number of each type of atom
Formula Weight and Molar Mass
4.2 The Chemical Formula,
Chemical Formula
Consider the following formulas:
• H2 – 2 atoms of hydrogen are chemically
bonded forming diatomic hydrogen, subscript 2
• H2O – 2 atoms of hydrogen and 1 atom of
oxygen, lack of subscript means one atom
• NaCl – 1 atom each of sodium and chlorine
• Ca(OH)2 – 1 atom of calcium and 2 atoms each
of oxygen and hydrogen, subscript outside
parentheses applies to all atoms inside
Formula Weight and Molar Mass
4.2 The Chemical Formula,
Chemical Formula
Consider the following formulas:
• (NH4)3SO4 – 2 ammonium ions and 1 sulfate ion
– Ammonium ion contains 1 nitrogen and 4 hydrogen
– Sulfate ion contains 1 sulfur and 4 oxygen
– Compound contains 2 N, 8 H, 1 S, and 4 O
• CuSO4.5H2O
– This is an example of a hydrate - compounds containing
one or more water molecules as an integral part of their
structure
– 5 units of water with 1 CuSO4
Formula Weight and Molar Mass
4.2 The Chemical Formula,
Comparison of Hydrated and
Anhydrous Copper Sulfate
Hydrated copper sulfate
Anhydrous copper sulfate
Marked color difference illustrates the fact
that these are different compounds
4.2 The Chemical Formula,
Formula Weight and Molar Mass
Formula Weight and Molar Mass
• Formula weight - the sum of the atomic weights
of all atoms in the compound as represented by its
correct formula
– expressed in amu
• What is the formula weight of H2O?
– 16.00 amu + 2(1.008 amu) = 18.02 amu
• Molar mass – mass of a mole of compound in
grams / mole
– Numerically equal to the formula weight in amu
• What is the molar mass of H2O?
– 18.02 g/mol H2O
4.2 The Chemical Formula, Formula
Weight and Molar Mass
Formula Unit
• Formula unit – smallest
collection of atoms from which
the formula of a compound can
be established
• When calculating the formula
weight (or molar mass) of an
ionic compound, the smallest
unit of the crystal is used
What is the molar mass of (NH4)3PO4?
3(N amu) + 12(H amu) + P amu + 4(O amu)=
3(14.01) + 12(1.008) + 30.97 + 4(16.00)=
149.10 g/mol (NH4)3PO4
4.3 The Chemical Equation and the
Information It Conveys
A Recipe For Chemical Change
• Chemical equation - shorthand notation of a
chemical reaction
– Describes all of the substances that react and all
the products that form, physical states, and
experimental conditions
– Reactants – (starting materials) – the substances
that undergo change in the reaction
– Products – substances produced by the reaction
and the Information It Conveys
4.3 The Chemical Equation
Features of a Chemical Equation
1. Identity of products and reactants must
be specified using chemical symbols
2. Reactants are written to the left of the
reaction arrow and products are written
to the right
3. Physical states of reactants and products
may be shown in parentheses
4. Symbol  over the reaction arrow
means that energy is necessary for the
reaction to occur
5. Equation must be balanced
and the Information It Conveys
4.3 The Chemical Equation
Features of a Chemical Equation
2HgO(s)

 2Hg(l ) 
O2 (g)
Products and reactants must be
specified using chemical symbols
Reactants – written on the left of arrow
Products – written on the right
 – energy is needed
Physical states are shown in parentheses
and the Information It Conveys
4.3 The Chemical Equation
The Experimental Basis of a
Chemical Equation
We know that a chemical equation
represents a chemical change
• One or more substances changed into
new substances
• Different chemical and physical
properties
and the Information It Conveys
4.3 The Chemical Equation
Evidence of a Reaction Occurring
The following can be visual evidence of a reaction:
•Release of a gas
– CO2 is released when acid is placed in a solution
containing CO32- ions
•Formation of a solid (precipitate)
– A solution containing Ag+ ions mixed with a solution
containing Cl- ions
•Heat is produced or absorbed
– Acid and base are mixed together
•Color changes
and the Information It Conveys
4.3 The Chemical Equation
Subtle Indications of a Reaction
• Heat or light is absorbed or emitted
• Changes in the way the substances
behave in an electrical or magnetic
field before and after a reaction
• Changes in electrical properties
and the Information It Conveys
4.3 The Chemical Equation
Writing Chemical Reactions
• We will learn to identify the following
patterns of chemical reactions:
–
–
–
–
combination
decomposition
single-replacement
double-replacement
• Recognizing the pattern will help you
write and understand reactions
and the Information It Conveys
4.3 The Chemical Equation
Combination Reactions
• The joining of two or more elements or
compounds, producing a product of
different composition
A + B  AB
• Examples:
2Na(s) + Cl2(g)  2NaCl(s)
MgO(s) + CO2(g)  MgCO3(s)
and the Information It Conveys
4.3 The Chemical Equation
Types of Combination
Reactions
1. Combination of a metal and a nonmetal
to form a salt
2. Combination of hydrogen and chlorine
molecules to produce hydrogen chloride
3. Formation of water from hydrogen and
oxygen molecules
4. Reaction of magnesium oxide and
carbon dioxide to produce magnesium
carbonate
and the Information It Conveys
4.3 The Chemical Equation
Decomposition Reactions
• Produce two or more products from a
single reactant
• Reverse of a combination reaction
AB  A + B
• Examples:
2HgO(s)  2Hg(l) + O2(g)
CaCO3(s)  CaO(s) + CO2(g)
and the Information It Conveys
4.3 The Chemical Equation
Types of Decomposition
Reactions
1. Heating calcium carbonate to produce
calcium oxide and carbon dioxide
2. Removal of water from a hydrated
material
and the Information It Conveys
4.3 The Chemical Equation
Replacement Reactions
1. Single-replacement
•
One atom replaces another in the
compound producing a new compound
A + BC  B + AC
•
Examples:
Cu(s)+2AgNO3(aq)  2Ag(s)+Cu(NO3)2(aq)
2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g)
and the Information It Conveys
4.3 The Chemical Equation
Types of Replacement
Reactions
1. Replacement of copper by zinc in
copper sulfate
2. Replacement of aluminum by
sodium in aluminum nitrate
and the Information It Conveys
4.3 The Chemical Equation
Replacement Reactions
2. Double-replacement
• Two compounds undergo a “change
of partners”
• Two compounds react by
exchanging atoms to produce two
new compounds
AB + CD  AD + CB
and the Information It Conveys
4.3 The Chemical Equation
Types of Double-Replacement
• Reaction of an acid with a base to
produce water and salt
HCl(aq)+NaOH(aq) NaCl(aq)+H2O(l)
• Formation of solid lead chloride from
lead nitrate and sodium chloride
Pb(NO3)2(aq) + 2NaCl(aq) 
PbCl2(s) + 2NaNO3(aq)
AB + CD  AD + CB
and the Information It Conveys
4.3 The Chemical Equation
Types of Chemical Reactions
Precipitation Reactions
• Chemical change in a solution that
results in one or more insoluble products
• To predict if a precipitation reaction can
occur it is helpful to know the
solubilities of ionic compounds
and the Information It Conveys
4.3 The Chemical Equation
Solubilities of Some Common
Ionic Compounds
and the Information It Conveys
4.3 The Chemical Equation
Predicting Whether Precipitation
Will Occur
• Recombine the ionic compounds to have
them exchange partners
• Examine the new compounds formed
and determine if any are insoluble
according to the rules in Table 4.1
• Any insoluble salt will be the precipitate
Pb(NO3)2(aq) + NaCl(aq) 
PbCl2 (s)
(?) + NaNO3 ((aq)
?)
and the Information It Conveys
4.3 The Chemical Equation
Predict Whether These Reactions
Form Precipitates
• Potassium chloride and silver nitrate
• Potassium acetate and silver nitrate
and the Information It Conveys
4.3 The Chemical Equation
Reactions with Oxygen
• Reactions with oxygen generally release
energy
• Combustion of natural gas
– Organic compounds CO2 and H2O are
usually the products
CH4+2O2CO2+2H2O
• Rusting or corrosion of iron
4Fe + 3O2  2Fe2O3
and the Information It Conveys
4.3 The Chemical Equation
Acid-Base Reactions
• These reactions involve the transfer of a
hydrogen ion (H+) from one reactant
(acid) to another (base)
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
The H+ on HCl was
transferred to the oxygen
in OH-, giving H2O
and the Information It Conveys
4.3 The Chemical Equation
Oxidation-Reduction Reactions
• Reaction involves the transfer of one or
more electrons from one reactant to
another
Zn(s) + Cu2+(aq) Cu(s) + Zn2+(aq)
Two electrons are
transferred from Zn to
Cu2+
and the Information It Conveys
4.3 The Chemical Equation
Writing Chemical Reactions
Consider the following reaction:
hydrogen reacts with oxygen to produce water
• Write the above reaction as a chemical
equation
H2 + O2  H2O
• Don’t forget the diatomic elements
and the Information It Conveys
4.3 The Chemical Equation
Law of Conservation of Mass
• Law of conservation of mass - matter
cannot be either gained or lost in the
process of a chemical reaction
– The total mass of the products must equal
the total mass of the reactants
4.4 Balancing Chemical
Equations
A Visual Example of the Law of
Conservation of Mass
4.4 Balancing Chemical Equations
• A chemical equation shows the molar
quantity of reactants needed to produce a
particular molar quantity of products
• The relative number of moles of each
product and reactant is indicated by
placing a whole-number coefficient
before the formula of each substance in
the chemical equation
4.4 Balancing Chemical
Equations
Balancing
Coefficient - how many of that substance
are in the reaction
2HgO(s)


 2Hg(l )  O2 ( g )
• The equation must be balanced
– All the atoms of every reactant must also
appear in the products
• Number of Hg on left?
– on right
2
• Number of O on left?
– on right
2
2
2
4.4 Balancing Chemical
Equations
Examine the Equation
H2 + O2  H2O
• Is the law of conservation of mass obeyed
as written? NO
• Balancing chemical equations uses coefficients
to ensure that the law of conservation of mass is
obeyed
• You may never change subscripts!
• WRONG: H2 + O2  H2O2
4.4 Balancing Chemical
Equations
Steps in Equation Balancing
H2 + O2  H2O
The steps to balancing:
Step 1. Count the number of moles of
atoms of each element on both
product and reactant sides
Reactants
Products
2 mol H
2 mol O
2 mol H
1 mol O
4.4 Balancing Chemical
Equations
Steps in Equation Balancing
H2 + O2  H2O
Step 2. Determine which elements are not
balanced – do not have same number on
both sides of the equation
– Oxygen is not balanced
Step 3. Balance one element at a time by
changing the coefficients
H2 + O2  2H2O
This balances oxygen, but is hydrogen
still balanced?
4.4 Balancing Chemical
Equations
Steps in Equation Balancing
H2 + O2  2H2O
How will we balance hydrogen?
2H2 + O2  2H2O
Step 4. Check! Make sure the law of
conservation of mass is obeyed
Reactants
Products
4 mol H
2 mol O
4 mol H
2 mol O
4.4 Balancing Chemical
Equations
Balancing an Equation
4.4 Balancing Chemical
Equations
Practice Equation Balancing
Balance the following equations:
1. C2H2 + O2  CO2 + H2O
2. AgNO3 + FeCl3  Fe(NO3)3 + AgCl
3. C2H6 + O2  CO2 + H2O
4. N2 + H2  NH3
4.5 Calculations Using the
Chemical Equation
• Calculation quantities of reactants and
products in a chemical reaction has many
applications
• Need a balanced chemical equation for the
reaction of interest
• The coefficients represent the number of
moles of each substance in the equation
4.5 Calculations Using the
Chemical Equation
General Principles
1. Chemical formulas of all reactants and
products must be known
2. Equation must be balanced to obey the
law of conservation of mass
•
Calculations of an unbalanced equation
are meaningless
3. Calculations are performed in terms of
moles
•
Coefficients in the balanced equation
represent the relative number of moles of
products and reactants
4.5 Calculations Using the
Chemical Equation
Using the Chemical Equation
• Examine the reaction:
2H2 + O2  2H2O
• Coefficients tell us?
– 2 mol H2 reacts with 1 mol O2 to produce 2
mol H2O
• What if 4 moles of H2 reacts with 2 moles of
O2?
– It yields 4 moles of H2O
4.5 Calculations Using the
Chemical Equation
Using the Chemical Equation
2H2 + O2  2H2O
• The coefficients of the balanced equation
are used to convert between moles of
substances
• How many moles of O2 are needed to
react with 4.26 moles of H2?
• Use the factor-label method to perform
this calculation
4.5 Calculations Using the
Chemical Equation
Use of Conversion Factors
2H2 + O2  2H2O
__mol
O2
1
4.26 molH 2 
 2.13 mol O2
__
2 mol H 2
• Digits in the conversion factor come
from the balanced equation
4.5 Calculations Using the
Chemical Equation
Conversion Between Moles
and Grams
• Requires only the formula weight
• Convert 1.00 mol O2 to grams
of
grams of
– Plan the path moles
Oxygen
Oxygen
– Find the molar mass of oxygen
• 32.0 g O2 = 1 mol O2
– Set up the equation
– Cancel units 1.00 mol O2 x 32.0 g O2
1 mol O2
– Solve equation 1.00 x 32.0 g O2 = 32.0 g O2
4.5 Calculations Using the
Chemical Equation
Conversion of Mole Reactants to
Mole Products
• Use a balanced equation
• C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)
• 1 mol C3H8 results in:
– 5 mol O2 consumed
1 mol C3H8 /5 mol O2
– 3 mol CO2 formed
1 mol C3H8 /3 mol CO2
– 4 mol H2O formed
1 mol C3H8 /4 mol H2O
• This can be rewritten as conversion
factors
4.5 Calculations Using the
Chemical Equation
Calculating Reacting Quantities
• Calculate grams O2 reacting with 1.00 mol C3H8
• Use 2 conversion factors
– Moles C3H8 to moles O2
– Moles of O2 to grams O2
moles
C 3 H8
moles
Oxygen
grams
Oxygen
– Set up the equation and cancel units
– 1.00 mol C3H8 x 5 mol O2 x 32.0 g O2 =
1 mol C3H8 1 mol O2
– 1.00 x 5 x 32.0 g O2 = 1.60 x 102 g O2
4.5 Calculations Using the
Chemical Equation
Calculating Grams of Product
from Moles of Reactant
• Calculate grams CO2 from combustion of 1.00
mol C3H8
• Use 2 conversion factors
– Moles C3H8 to moles CO2
– Moles of CO2 to grams CO2
moles
C 3 H8
moles
CO2
grams
CO2
– Set up the equation and cancel units
– 1.00 mol C3H8 x 3 mol CO2 x 44.0 g CO2 =
1 mol C3H8 1 mol CO2
– 1.00 x 3 x 44.0 g CO2 = 1.32 x 102 g CO2
4.5 Calculations Using the
Chemical Equation
Relating Masses of Reactants
and Products
• Calculate grams C3H8 required to produce
36.0 grams of H2O
• Use 3 conversion factors
– Grams H2O to moles H2O
– Moles H2O to moles C3H8
– Moles of C3H8 to grams C3H8
grams
H2 O
moles
H2 O
moles
C 3 H8
grams
C 3 H8
– Set up the equation and cancel units
36.0 g H2O x 1 mol H2O x 1 mol C3H8 x 44.0 g C3H8
18.0 g H2O 4 mol H2O 1 mol C3H8
– 36.0 x [1/18.0] x [1/4] x 44.0 g C3H8 = 22.0 g C3H8
4.5 Calculations Using the
Chemical Equation
Calculating a Quantity of Reactant
• Ca(OH)2 neutralizes HCl
• Calculate grams HCl neutralized by 0.500 mol
Ca(OH)2
– Write chemical equation and balance
• Ca(OH)2(s) + 2HCl(aq)
CaCl2(s) + 2H2O(l)
– Plan the path
moles
Ca(OH)2
moles
HCl
grams
HCl
– Set up the equation and cancel units
0.500 mol Ca(OH)2 x 2 mol HCl x 36.5 g HCl
1 mol Ca(OH)2 1 mol HCl
Solve equation 0.500 x [2/1] x 36.5 g HCl = 36.5 g HCl
4.5 Calculations Using the
Chemical Equation
General Problem-solving Strategy
4.5 Calculations Using the
Chemical Equation
Sample Calculation
Na + Cl2  NaCl
1. Balance the equation 2Na + Cl2  2NaCl
2. Calculate the moles Cl2 reacting with
5.00 mol Na
3. Calculate the grams NaCl produced when
5.00 mol Na reacts with an excess of Cl2
4. Calculate the grams Na reacting with
5.00 g Cl2
4.5 Calculations Using the
Chemical Equation
Theoretical and Percent Yield
• Theoretical yield - the maximum amount of
product that can be produced
– Pencil and paper yield
• Actual yield - the amount produced when
the reaction is performed
– Laboratory yield
• Percent yield:
actual yield
% yield 
100%
theoretical yield
= 125 g CO2 actual x 100% = 97.4%
132 g CO2 theoretical
4.5 Calculations Using the
Chemical Equation
Sample Calculation
If the theoretical yield of iron was 30.0 g
and actual yield was 25.0 g, calculate the
percent yield:
2 Al(s) + Fe2O3(s)  Al2O3(aq) + 2Fe(aq)
• [25.0 g / 30.0 g] x 100% = 83.3%
• Calculate the % yield if 26.8 grams iron
was collected in the same reaction