Download Chapter Six - La Salle University

Fundamentals of General, Organic,
and Biological Chemistry
5th Edition
Chapter Six
Chemical Reactions:
Classification and Mass
Relationships
James E. Mayhugh
Oklahoma City University
2007 Prentice Hall, Inc.
Outline
► 6.1 Chemical Equations
► 6.2 Balancing Chemical Equations
► 6.3 Avogadro’s Number and the Mole
► 6.4 Gram–Mole Conversions
► 6.5 Mole Relationships and Chemical Equations
► 6.6 Mass Relationships and Chemical Equations
► 6.7 Percent Yield
► 6.8 Classes of Chemical Reactions
► 6.9 Precipitation Reactions and Solubility Guidelines
► 6.10 Acids, Bases, and Neutralization Reactions
► 6.11 Redox Reactions
► 6.12 Recognizing Redox Reactions
► 6.13 Net Ionic Equations
Prentice Hall © 2007
Chapter Six
2
6.1 Chemical Equations
► Chemical equation: An expression in which
symbols are used to represent a chemical reaction.
► Reactant: A substance that undergoes change in a
chemical reaction and is written on the left side of
the reaction arrow in a chemical equation.
► Product: A substance that is formed in a chemical
reaction and is written on the right side of the
reaction arrow in a chemical equation.
Prentice Hall © 2007
Chapter Six
3
► The numbers and kinds of atoms must be the same on
both sides of the reaction arrow.
► Numbers in front of formulas are called coefficients;
they multiply all the atoms in a formula.
► The symbol 2 NaHCO3 indicates two units of sodium
bicarbonate, which contains 2 Na, 2 H, 2 C, and 6 O.
► Substances involved in chemical reactions may be
solids, liquids, gases, or they may be in solution.
► This information is added to an equation by placing the
appropriate symbols after the formulas:
► Solid=(s) Liquid=(l) Gas=(g) Aqueous solution=(aq)
Prentice Hall © 2007
Chapter Six
4
6.2 Balancing Chemical Equations
► Balancing chemical equations can be done using four
basic steps:
► STEP 1: Write an unbalanced equation, using the
correct formulas for all reactants and products.
► STEP 2: Add appropriate coefficients to balance the
numbers of atoms of each element.
Prentice Hall © 2007
Chapter Six
5
► A polyatomic ion appearing on both sides of an
equation can be treated as a single unit.
► STEP 3: Check the equation to make sure the
numbers and kinds of atoms on both sides of the
equation are the same.
Prentice Hall © 2007
Chapter Six
6
► STEP 4: Make sure the coefficients are reduced to
their lowest whole-number values.
► The equation:
2 H2SO4 + 4 NaOH  2 Na2SO4 + 4 H2O
is balanced, but can be simplified by dividing all
coefficients by 2:
H2SO4 + 2 NaOH  Na2SO4 + 2 H2O
► Hint: If an equation contains a pure element as a
product or reactant it helps to assign that element’s
coefficient last.
Prentice Hall © 2007
Chapter Six
7
Law of Conservation of Matter
►In a chemical reaction, matter can neither be
created nor destroyed.
►In other words… The mass of the reactants
must equal the mass of the products.
►See questions 6.28-30
Prentice Hall © 2007
Chapter Six
8
Mass of Reactants = Mass of Product
M g + O2
M gO (unbalanced)
2 M g + O2
2 M gO
2(24.31) + 1(32.00) =
80.62 amu
Prentice Hall © 2007
=
Chapter Six
2(40.31)
80.62 amu
9
Balance the Reaction
H3PO4 +
Prentice Hall © 2007
Ca(OH)2
Ca3(PO4)2 +
Chapter Six
H2O
10
First, balance the Calcium
H3PO4 +
H3PO4
+
Ca(OH)2
3 Ca(OH)2
Prentice Hall © 2007
Chapter Six
Ca3(PO4)2 +
H2O
Ca3(PO4)2
H2O
+
11
Then, Balance the Phosphate
H3PO4 +
H3PO4
+
Ca(OH)2
3 Ca(OH)2
2 H3PO4 + 3 Ca(OH)2
Prentice Hall © 2007
Ca3(PO4)2 +
H2O
Ca3(PO4)2
H2O
Ca3(PO4)2 +
Chapter Six
+
H2O
12
Finally, Balance the H and O with
H2O
H3PO4 +
H3PO4
+
Ca(OH)2
3 Ca(OH)2
2 H3PO4 + 3 Ca(OH)2
2 H3PO4
+
Prentice Hall © 2007
Ca3(PO4)2 +
H2O
Ca3(PO4)2
H2O
+
Ca3(PO4)2 +
3 Ca(OH)2
Ca3(PO4)2
Chapter Six
H2O
+ 6 H2O
13
Balance the equations
B2O3 + Mg
B + MgO
C2H2 + O2
CO 2
Pb(NO3)2 + KI
KNO3
NO2 +
Prentice Hall © 2007
H2O
HNO3
Chapter Six
+ H2O
+ PbI2
+
NO
14
6.3 Molecular Weight
Molecular weight: The sum of atomic weights of all
atoms in a molecule. (covalent compound)
Formula weight: The sum of atomic weights of all
atoms in one formula unit of any ionic compound.
Molecular weight of H2O
2(1.01) + 16.00 = 18.02 amu
Formula weight of MgCl2
24.31 + 2(35.45) = 95.21 amu
Prentice Hall © 2007
Chapter Six
15
6.3 The Mole
Mole: One mole of any substance is the
amount whose mass in grams (molar
mass) is numerically equal to its molecular
or formula weight.
a mole of carbon weighs 12.01 g
a mole of H2O weighs 18.02 g
Prentice Hall © 2007
Chapter Six
16
6.3 Avogadro’s Number
Avogadro’s number: The number of molecules or
formula units in a mole. NA = 6.022 x 1023
A mole of carbon weighs 12.01 g and
has 6.02 x 1023 atoms.
A mole of H2O weighs 18.02 g and
has 6.02 x 1023 molecules.
A mole is an incredibly
incredibly tiny.
Prentice Hall © 2007
large number because atoms are so
Chapter Six
17
Analogies
Avogadro's Number compared to the Population of
the Earth.
►We will take the population of the earth to be six
billion (6 x 109 people). We can compare to
Avogadro's Number like this:
►6.022 x 1023 divided by 6 x 109 = approx. 1 x 1014
►In other words, it would take about 100 trillion Earth
populations to add up to Avogadro's number.
Prentice Hall © 2007
Chapter Six
18
Or…
►A mole of marbles would cover the United States to a
depth of 4 meters.
►If you spent 1 million dollars every second, it would
take you 1.9 x 1010 (19 billion) years to spend
$6.02 x 1023
►Or… If you started spending $4 million per second at
the time of the big bang, you would still not have
spent Avogadro’s number of dollars (although you
would be close).
Prentice Hall © 2007
Chapter Six
19
Froot Loops
If you know that the average mass of a Froot Loop is
0.20 grams, and that the average box contains 425 g
of cereal;
Approximately how many Froot Loops are in a box?
What is the fastest way to measure 1,000,000 Froot
Loops? Should we count them?? Should we weigh
them??
Prentice Hall © 2007
Chapter Six
20
1 FL = 0.20 g
1 box = 425 g of FL's
conversion factors
1 FL
425 g FL's x
0.20 g FL's = 2100 FL's in one box
1,000,000 FL's x
1 box
2100 FL's
= 480 boxes of FL's
0.20 g
1 FL
= 200,000 g FL's
OR weigh them...
1,000,000 FL's x
= 200 kg FL's
Prentice Hall © 2007
Chapter Six
21
6.4 Gram – Mole Conversions
► Molar mass
= Mass of 1 mole of a substance.
= Mass of 6.022 x 1023 molecules
of a substance.
= Molecular (formula) weight of
substance in grams.
► Molar mass serves as a conversion factor between
numbers of moles and mass. If you know how many
moles you have, you can calculate their mass; if you
know the mass of a sample, you can calculate the
number of moles.
Prentice Hall © 2007
Chapter Six
22
The molar mass of water is 18.0 g. The conversion
factor between moles of water and mass of water is
18.0 g H2O = 1 mol H2O
Prentice Hall © 2007
Chapter Six
23
Problem
►How many molecules are there in 0.25 mol of H2O?
Conversion factor is 1 mol = 6.02 x 1023
►0.25 mol H2O x 6.02 x 1023 molecules/mol H2O
known
conversion factor
= 1.51 x 1023 molecules H2O
Prentice Hall © 2007
Chapter Six
24
One Nickel…
►…weighs 4.94 g
1) How many moles if Ni are there in 4.94 g of Ni?
4.94 g Ni x 1 mol Ni / 58.69 g
= 0.0842 mol Ni
2) How many atoms of Ni in a nickel?
0.0842 mol Ni x 6.02 x 1023 atoms / mol Ni
= 5.07 x 1022 atoms Ni
Prentice Hall © 2007
Chapter Six
25
6.5 Mole Relationships and Chemical
Equations
The coefficients in a balanced chemical equation tell
how many molecules, and thus how many moles, of
each reactant are needed and how many molecules, and
thus moles, of each product are formed. See the
example below:
Prentice Hall © 2007
Chapter Six
26
► The coefficients can be put in the form of mole
ratios, which act as conversion factors when setting
up dimensional analysis calculations.
► In the ammonia synthesis, the mole ratio of H2 to N2
is 3:1, the mole ratio of H2 to NH3 is 3:2, and the
mole ratio of N2 to NH3 is 1:2 leading to the
following conversion factors:
(3 mol H2)/(1 mol N2)
(3 mol H2)/(2 mol NH3)
(1 mol N2)/(2 mol NH3)
Prentice Hall © 2007
Chapter Six
27
6.6 Mass Relationships and
Chemical Equations
Mole to mole conversions are carried out using mole
ratios as conversion factors.
Prentice Hall © 2007
Chapter Six
28
Balanced Reaction:
Mole to Mole Conversions
►In the reaction below, how many moles of Ag will
form from 6.0 moles of Cu?
►How many moles of Cu are needed to react with 3.5
mol of AgNO3?
AgNO3(aq) + Cu(s)
Prentice Hall © 2007
Ag(s) + Cu(NO3)2(aq)
Chapter Six
29
► Mole to mass and mass to mole conversions are
carried out using molar mass as a conversion factor.
► Mass to mass conversions are frequently needed,
but cannot be carried out directly.
► Overall, there are four steps for determining mass
relationships among reactants and products.
Prentice Hall © 2007
Chapter Six
30
Mass to mass conversions:
►STEP 1: Write the
balanced chemical equation.
►STEP 2: Choose molar
masses and mole ratios to
convert known information
into needed information.
►STEP 3: Set up the factorlabel expression, and
calculate the answer.
►STEP 4: Estimate or check
the answer using a ballpark
solution.
Prentice Hall © 2007
Chapter Six
31
Turning Rust into Iron
Fe2O3 +
Prentice Hall © 2007
C
Fe +
Chapter Six
CO 2
32
Turning Rust into Iron
Fe2O3 +
2 Fe2O3 +
Prentice Hall © 2007
C
Fe +
3C
CO 2
4 Fe +
Chapter Six
3 CO2
33
Mole to Mole
Mole to Gram
►How many moles of Fe will form from 6.0 moles of
Fe2O3?
2 Fe2O3 +
3C
4 Fe +
3 CO2
4.0 mol Fe
6.0 mol Fe2O3 x
2.0 mol Fe2O3 = 12 mol Fe
►How many grams of Fe will form from 6.0 moles
Fe2O3?
4.0 mol Fe
6.0 mol Fe 2O3 x
x
2.0 mol Fe 2O3
Prentice Hall © 2007
Chapter Six
55.85 g
1 mol Fe
= 670 g Fe
34
Gram to Gram conversions
►How many grams of Carbon are needed to react with
26.3 grams of Fe2O3?
2 Fe2O3 +
Prentice Hall © 2007
3C
4 Fe +
Chapter Six
3 CO2
35
Gram to Gram conversions
►How many grams of Carbon are needed to react with
26.3 grams of Fe2O3?
2 Fe2O3 +
3C
4 Fe +
3 CO2
conversion factors
MW
g Fe2O3
mol Fe2O3
bal.
rxn
given
Prentice Hall © 2007
At. Wt.
mol C
gC
need
Chapter Six
36
Gram to Gram conversions
26.3 g Fe 2O3 x
1 mol Fe2O3
159.7 g Fe 2O3
mol Fe2O3
Prentice Hall © 2007
x
3 mol C
2 mol Fe 2O3
mol C
Chapter Six
x
12.0 g C
1 mol C
= 2.96 g C
gC
37
6.7 Percent Yield
► The amount of product actually formed in a
chemical reaction is often somewhat less than the
amount predicted by theory.
► Unwanted side reactions and loss of product during
handling prevent one from obtaining a perfect
conversion of all the reactants to desired products.
► The amount of product actually obtained in a
chemical reaction is usually expressed as a percent
yield.
Prentice Hall © 2007
Chapter Six
38
► Percent yield is defined as:
(Actual yield ÷ Theoretical yield) x 100%
► The actual yield is found by weighing the product
obtained.
► The theoretical yield is found by a mass-to-mass
calculation.
Prentice Hall © 2007
Chapter Six
39
►How many grams of H2O will be produced in the
combustion of 225 g of methane (CH4)?
Prentice Hall © 2007
Chapter Six
40
►How many grams of H2O will be produced in the
combustion of 225 g of methane (CH4)?
►Set the reaction up and balance it.
CH4
Prentice Hall © 2007
+
O2
CO 2
Chapter Six
+
H2O
41
►How many grams of H2O will be produced in the
combustion of 225 g of methane (CH4)?
►Set the reaction up and balance it.
CH4
Prentice Hall © 2007
+
2 O2
CO2
Chapter Six
+
2 H2O
42
Cannot convert grams to grams
directly
►How many grams of H2O will be produced in the
combustion of 225 g of methane (CH4)?
►Set the reaction and balance it.
CH4 + 2 O2
CO2 + 2 H2O
grams
X
moles
Prentice Hall © 2007
grams
moles
Chapter Six
43
CH4 +
2 O2
1 mol CH 4
225 g CH 4 x
16.0 g CH 4
mol CH 4
Prentice Hall © 2007
CO2 +
x
2 mol H 2O
1 mol CH 4
mol H 2O
Chapter Six
x
2 H2O
18.0 g H 2O
= 506 g H 2O
1 mol H 2O
g H 2O
44
6.8 Classes of Chemical Reactions
► When learning about chemical reactions it is helpful
to group the reactions of ionic compounds into three
general classes: precipitation reactions, acid–base
neutralization reactions, and oxidation–reduction
reactions.
► Precipitation reactions are processes in which an
insoluble solid (s) called a precipitate forms when
reactants are combined in aqueous solution.
Prentice Hall © 2007
Chapter Six
45
► Acid–base neutralization reactions are processes
in which H+ ions from an acid react with OH- ions
from a base to yield water. An ionic compound
called a salt is also produced. The “salt” produced
need not be common table salt. Any ionic compound
produced in an acid–base reaction is called a salt.
► Oxidation–reduction reactions, or redox
reactions, are processes in which one or more
electrons are transferred between reaction partners
(atoms, molecules, or ions). As a result of this
transfer, the charges on atoms in the various
reactants change.
Prentice Hall © 2007
Chapter Six
46
6.9 Precipitation Reactions and
Solubility Guidelines
► Reaction of aqueous Pb(NO3)2
with aqueous KI gives a
yellow precipitate of PbI2.
► To predict whether a
precipitation reaction will
occur on mixing aqueous
solutions of two ionic
compounds, you must know
the solubility of the potential
products.
Prentice Hall © 2007
Chapter Six
47
If a potential product does not contain at least one of
the ions listed below, it is probably not soluble and
will precipitate from solution when formed.
Prentice Hall © 2007
Chapter Six
48
Precipitation Reactions
Pb(NO3)2(aq) + 2 HBr(aq)
PbBr2(s)
ppt
AgNO3(aq) + Na2CO3(aq)
Prentice Hall © 2007
+
2 HNO3(aq)
Ag2CO3(s) + NaNO3(aq)
ppt
Chapter Six
49
6.10 Acids, Bases, and Neutralization
Reactions
► When acids and bases are mixed together in correct
proportion, acidic and basic properties disappear.
► A neutralization reaction produces water and a salt.
HA(aq) + MOH(aq)  H2O(l) + MA(aq)
acid + base
 water + salt
► The reaction of hydrochloric acid with potassium
hydroxide to produce potassium chloride is an
example:
► HCl(aq) + KOH(aq)  H2O(l) + KCl(aq)
Prentice Hall © 2007
Chapter Six
50
Acid + Base = Salt + H20
Ca(OH)2 + H2SO4
base
acid
CaSO 4 + 2 H2O
salt
water
3 KOH + H3PO4
base
Prentice Hall © 2007
K3PO4 + 3 H2O
acid
salt
Chapter Six
water
51
6.11 Redox Reactions
► Oxidation–reduction (redox) reaction: A reaction in
which electrons transfer from one atom to another.
► Oxidation: Loss of one or more electrons by an atom.
► Reduction: Gain of one or more electrons by an atom.
Prentice Hall © 2007
Chapter Six
52
► Oxidation and reduction always occur together.
► A substance that is oxidized gives up an electron,
causes reduction, and is called a reducing agent.
► A substance that is reduced gains an electron, causes
oxidation, and is called an oxidizing agent.
► The charge on the reducing agent increases during
the reaction, and the charge on the oxidizing agent
decreases.
Prentice Hall © 2007
Chapter Six
53
Reducing agent:
► Loses one or more electrons
► Causes reduction
► Undergoes oxidation
► Becomes more positive (or less negative)
Oxidizing agent:
► Gains one or more electrons
► Causes oxidation
► Undergoes reduction
► Becomes more negative (or less positive)
Prentice Hall © 2007
Chapter Six
54
6.12 Recognizing Redox Reactions
► One can determine whether atoms are oxidized or
reduced in a reaction by keeping track of changes
in electron sharing by the atoms. Each atom in a
substance is assigned a value called an oxidation
number or oxidation state.
► The oxidation number indicates whether the atom
is neutral, electron-rich, or electron-poor.
► By comparing the oxidation state of an atom before
and after reaction, we can tell whether the atom has
gained or lost electrons.
Prentice Hall © 2007
Chapter Six
55
► Rules for assigning oxidation numbers:
► An atom in its elemental state has an oxidation
number of zero.
► A monatomic ion has an oxidation number equal to
its charge.
Prentice Hall © 2007
Chapter Six
56
► In a molecular compound, an atom usually has the
same oxidation number it would have if it were a
monatomic ion.
► Examples: H often has an oxidation number of +1,
oxygen often has an oxidation number of -2,
halogens often have an oxidation number of -1.
Prentice Hall © 2007
Chapter Six
57
► For compounds with more than one nonmetal
element, such as SO2, NO, and CO2, the more
electronegative element—oxygen in these
examples—has its preferred negative oxidation
number.
► The less electronegative element is assigned a
positive oxidation number so that the sum of the
oxidation numbers in a neutral compound is 0.
Prentice Hall © 2007
Chapter Six
58
What is oxidized; what is reduced?
Si + 2Cl 2
Cl2 +
SiCl 4
2 NaBr
CuCl2 +
Prentice Hall © 2007
Br2
Zn
+
ZnCl2 +
Chapter Six
2 NaCl
Cu
59
Oxidized species loses eReduced species gains eSi + 2Cl 2
ox.# 0
0
oxidized reduced
SiCl 4
+4 -1
Cl2 + 2 NaBr
ox. # 0
-1
reduced oxidized
Br2
0
CuCl2 + Zn
ox. # +2
0
reduced oxidized
Prentice Hall © 2007
+
2 NaCl
-1
ZnCl2 + Cu
+2
0
Chapter Six
60
6.13 Net Ionic Equations
► Ionic equation: An equation in which ions are
explicitly shown.
► Spectator ion: An ion that appears unchanged on
both sides of a reaction arrow.
► Net ionic equation: An equation that does not
include spectator ions.
Prentice Hall © 2007
Chapter Six
61
Ionic Equation
Pb(NO3)2(aq)
+
2 HBr(aq)
PbBr2(s)
ppt
+
2 HNO3(aq)
write all aqueous compounds as ions
+2
ionic equation: Pb
+ 2NO3
-
net ionic equation: Pb
(remove spectator ions)
Prentice Hall © 2007
+
+ 2H
+2
-
+ 2 Br
-
+ 2 Br
Chapter Six
+
PbBr2 + 2H + 2NO3
PbBr 2
62
-
Write net ionic equations for…
Ba(ClO 4)2(aq) + Na2SO 4(aq)
BaSO 4(s) + 2 NaClO4(aq)
AgNO 3(aq) + KI(aq)
Prentice Hall © 2007
AgI(s) + KNO3(aq)
Chapter Six
63
Chapter Summary
► Chemical equations must be balanced; the numbers
and kinds of atoms must be the same in both the
reactants and the products.
► To balance an equation, coefficients are placed
before formulas but the formulas themselves cannot
be changed.
► A mole refers to Avogadro’s number of formula
units of a substance. One mole of any substance has
a mass equal to its formula weight in grams.
► Molar masses act as conversion factors between
numbers of molecules and masses in grams.
Prentice Hall © 2007
Chapter Six
64
Chapter Summary Cont.
► The coefficients in a balanced chemical equation
represent the numbers of moles of reactants and
products in a reaction.
► Mole ratios relate amounts of reactants and/or
products. Using molar masses and mole ratios in
factor-label calculations relates unknown masses to
known masses or molar amounts.
► The yield is the amount of product obtained.
► The percent yield is the amount of product obtained
divided by the amount theoretically possible and
multiplied by 100%.
Prentice Hall © 2007
Chapter Six
65
Chapter Summary Cont.
► Precipitation reactions are processes in which an
insoluble solid called a precipitate is formed.
► In acid–base neutralization reactions an acid reacts
with a base to yield water plus a salt.
► Oxidation–reduction (redox) reactions are processes
in which one or more electrons are transferred
between reaction partners.
► Oxidation is the loss of electrons by an atom, and
reduction is the gain of electrons by an atom.
► Oxidation numbers are assigned to provide a
measure of whether an atom is neutral, electron-rich,
or electron-poor.
Prentice Hall © 2007
Chapter Six
66