Chapter Summaries
CHAPTER 2 IN REVIEW: Matter and Energy
Goal 1 Identify and explain the differences between interpreting and
describing matter at the macroscopic,
microscopic, and particulate levels.
Matter, anything that has mass, can be studied at three levels: macroscopic (seen with
the human eye), microscopic (seen with a light microscope), and particulate (cannot be
directly seen)
Goal 2 Define the term model as it is
used in chemistry to represent pieces
of matter too small to see.
Chemists use symbols and models to represent particulate matter (Fig. 2-2): chemical
formula, Lewis diagram, ball-and-stick model, space-filling model.
Goal 3 Identify and explain the differences among gases, liquids, and solids
in terms of (a) the macroscopic-level
properties shape and volume; (b) particle movement; and (c) particle spacing.
Kinetic molecular theory: Particles of matter always moving. Kinetic means motion.
Gas: variable shape and volume, independent particle movement, particles very far apart.
Liquid: variable shape, constant volume, independent particle movement beneath the
surface, particles close. Solid: constant shape and volume, particles vibrate in fixed positions, particles close.
Goal 4 Distinguish between physical
and chemical properties at both the particulate level and the macroscopic level.
Matter is described by its properties, which may be physical or chemical. Changes in matter
may also be physical or chemical. Physical property: can be observed and measured without a chemical change. Chemical property: the chemical changes possible for a
substance.
Goal 5 Distinguish between physical
and chemical changes at both the particulate level and the macroscopic level.
Physical change: alteration in physical form without change in identity. Chemical change:
identity of original substance change to something new.
Goal 6 Distinguish between a pure substance and a mixture at both the particulate level and the macroscopic level.
Pure substance: one chemical with distinct set of physical and chemical properties; cannot be separated by physical changes. Mixture: two or more pure substances; properties
vary, depending on relative amounts of pure substances; components can be separated by
physical changes.
Goal 7 Distinguish between homogeneous and heterogeneous matter.
Homogeneous: same appearance, composition, properties throughout. Solution: a homogenous mixture. Heterogeneous: different phases visible, variable properties in different parts
of sample.
Goal 8 Describe how distillation and
filtration rely on physical changes and
properties to separate components of
mixtures.
Separations usually based on different physical properties of components. Distillation
separates based on volatilities of components. Filtration separates based on particle size.
A porous medium is used to separate mixture components based on size (a physical
property).
Goal 9 Distinguish between elements
and compounds.
Element: pure substance, cannot be decomposed chemically into other pure substances.
Compound: pure substance that can be decomposed chemically into other pure
substances.
Goal 10 Distinguish between elemental
symbols and the formulas of chemical
compounds.
Elemental symbol: capital letter, sometimes followed by small letter. Formula of compound: elemental symbols of elements in compound. Subscripts show number of atoms of
each element.
Goal 11 Distinguish between atoms and
molecules.
Atom: the smallest unit particle of an element. Molecule: the smallest unit particle of a pure
substance that can exist independently and retain the identity of that substance.
Goal 12 State the Law of Definite
(or Constant) Composition, and explain
its implication for how compounds and
mixtures differ in composition.
Any compound is always made up of elements in the same proportion by mass. A compound has a definite composition; the composition of a mixture depends on the relative
quantities of the components that make up the mixture.
Goal 13 Match electrostatic forces of
attraction and repulsion with combinations of positive and negative charge.
Objects with the same charge repel. Objects with opposite charges attract.
Goal 14 Distinguish between reactants
and products in a chemical equation.
Equation: reactants -
Goal 15 Distinguish between exothermic and endothermic changes.
Exothermic: transfers energy to surroundings; endothermic: removes energy from
surroundings.
products
675
676
Chapter Summaries
Goal 16 Distinguish between kinetic
energy and potential energy.
Potential energy: energy due to the arrangement of the charged particles in a system. Tendency toward reduction of energy to the smallest amount possible is a driving force for
chemical reactions. Kinetic energy: energy of motion.
Goal 17 State the meaning of, or draw
conclusions based on, the Law of Conservation of Mass.
The Law of Conservation of Mass and Energy: Total mass + energy in the universe is
constant. The Law of Conservation of Mass: Mass is conserved in chemical change; neither created nor destroyed.
Goal 18 State the meaning of, or draw
conclusions based on, the Law of Conservation of Energy.
The Law of Conservation of Energy: Quantity of energy within an isolated system does
not change; energy is neither created nor destroyed.
CHAPTER 3 IN REVIEW: Measurement and Chemical
Calculations
Goal 1 Write in scientific notation a
number given in ordinary decimal form;
write in ordinary decimal form a number
given in scientific notation.
Any decimal number can be written in scientific notation. Scientific notation expresses a
number as a coefficient C (between 0 and 9.99 ...) multiplied by 10 raised to thee power, in
general, C x 10e. When e is larger than 0, 10e is larger than 1; when e is smaller than 0, 1oe is
smaller than 1. (Remember that 1o0 = 1.)
Goal 2 Use a calculator to add, subtract, multiply, and divide numbers
expressed in scientific notation.
To add, subtract, multiply, or divide numbers in exponential notation, following the instructions that are appropriate for your calculator.
Goal 3 Convert an equivalency into two
conversion factors.
An equivalency is two quantities that are equivalent in value. When an equivalency is
expressed as a fraction, it is called a conversion factor. Two reciprocal conversion factors
result from each equivalency.
Goal 4 Learn and apply the algorithm
for using conversion factors to solve
quantitative problems.
(1) Analyze the problem statement by writing the given quantities, their properties, the property
of the wanted quantity, and its unit. (2) Identify the equivalencies or the algebraic relationship
needed to solve the problem. (3) Construct the solution setup. (4) Check the solution at two
levels: (a) making sense and (b) what was learned.
Goal 5 Explain why the metric system
of measurement is used in the sciences.
The metric system is internationally standardized and decimal-based.
Goal 6 State and write with appropriate
metric prefixes the relationship between
any metric unit and its corresponding
kilounit, centiunit, and milliunit.
The important metric prefixes for this course are kilo- (1000), centi- (0.01), and mi/Ii- (0.001).
Goal 7 Using Table 3-1, state and write
with appropriate metric prefixes the
relationship between any metric unit and
other larger and smaller metric units.
Given the value of a metric prefix other than the three of which you have memorized their
values, you should be able to write an equivalency between a quantity with a unit with that
prefix and the basic un-prefixed unit.
Goal 8 Distinguish between mass and
weight.
The mass of an object does not change in different gravitational fields; the weight of that
object does change.
Goal 9 Identify the metric units of mass,
length, and volume.
The SI metric unit of mass is the kilogram, kg. The SI metric unit of length is the meter, m.
The SI unit of volume is the cubic meter.
Goal 10 Given a mass, length, or volume
expressed in basic metric units, kilounits, centiunits, or milliunits, express
that quantity in the other three units.
You should memorize the values needed to be able to make conversions among the unit,
kilounit, centiunit, and milliunit.
Goal 11 Given a mass, length, or volume expressed in any metric units and
Table 3-1 or the equivalent, express that
quantity in any other metric unit.
For any metric prefix other than kilo-, centi-, or mi/Ii-, given the value of the prefix, you
should be able to make conversions to another metric prefix.
Goal 12 Given a description of a measuring instrument and an associated
measurement, express the measured
quantity with the uncertain digit in the
correct location in the value.
The number of significant figures in a quantity is the number of digits that are known accurately plus the uncertain digit; the uncertain digit is the last digit written.
Goal 13 State the number of significant
figures in a given quantity.
To count significant figures, begin with the first nonzero digit and end with the uncertain
digit-the last digit shown.
Chapter Summaries
677
Goal 14 Round off given values to a
specified number of significant figures.
To round off a number to the proper number of significant figures, leave the uncertain digit
unchanged if the digit to its right is less than 5. Increase the uncertain digit by 1 if the digit to
its right is 5 or more.
Goal 15 Add or subtract given measured
quantities and express the result in the
proper number of significant figures.
In addition and subtraction, round off a sum or difference to the first column that has an
uncertain digit.
Goal 16 Multiply or divide given measured quantities and express the result in
the proper number of significant figures.
In multiplication and division, round off a product or quotient to the same number of significant figures as the smallest number of significant figures in any factor.
Goal 17 Given a metric-uses conversion
factor and a quantity expressed in any unit
in Table 3-2, express that quantity in corresponding units in the other system.
A metric-uses equivalency from Table 3-2 is converted to the appropriate conversion
factor and used in a setup to calculate a quantity in the other system of measurement.
Goal 18 Given a temperature in either
Celsius or Fahrenheit degrees, convert it
to the other scale.
TcF -32
The Celsius and Fahrenheit temperature scales are related by the equation Toe = - - 1 .8
Goal 19 Given a temperature in Celsius
degrees or kelvins, convert it to the
other scale.
The Celsius and kelvin temperature scales are related by the equation TK = Toe + 273.
Goal 20 Write a mathematical expression indicating that one quantity is
directly proportional to another quantity.
If y is directly proportional to x, it is expressed mathematically as y
Goal 21 Use a proportionality constant to
convert a proportionality to an equality.
To convert the proportionality y
k: y = kx.
Goal 22 Given the values of two quantities that are directly proportional to each
other, calculate the proportionality constant, Including its units.
If y = kx, then k = y --:-- x.
Goal 23 Write the defining equation for
a proportionality constant and identify
units in which It might be expressed.
Units are set by definition and the defining equation. For example, for density, the units are
mass units over volume units. Examples: kg/m 3 , g/cm 3 , g/ml, g/L.
Goal 24 Given two of the following for a
sample of a pure substance, calculate
the third: mass, volume, and density.
The defining equation for density is Density= mlass . Because the defining equation for
voume
density is a conversion factor, density problems are solved with that conversion factor or
its reciprocal.
ex.
x
x.
x into an equality, insert a proportionality constant,
CHAPTER 4 IN REVIEW: Introduction to Gases
Goal 1 Describe five macroscopic
characteristics unique to the gas
phase of matter.
Five macroscopic characteristics unique to the gas phase of matter are the following:
1. Gases may be compressed.
2. Gases may be expanded.
3. Gases have low densities.
4. Gases may be mixed in a fixed volume.
5. Gases exert constant pressure on the walls of their container uniformly in all
directions.
Goal 2 Use the postulates of the kinetic
molecular theory to explain the reasons
for the macroscopic characteristics
unique to the gas phase of matter.
The reasons for the macroscopic characteristics unique to the gas phase of matter are the
following:
1. A gas consists of molecules and empty space.
2. The volume occupied by the molecules in a gas is negligible when compared with
the volume of the space they occupy.
3. The attractive forces among molecules of a gas are negligible.
4. The average kinetic energy of gas molecules is proportional to the temperature,
expressed in kelvins.
5. Molecules interact with one another and with the container walls without loss of
total kinetic energy.
678
Chapter Summaries
Goal 3 Define pressure and interpret
statements in which the term pressure
is used.
Pressure is defined as force per unit area.
Goal 4 Explain the cause of the pressure of a gas.
Pressure is the effect of the force of the interactions of the huge number of rapidly moving
molecules of a gas as they collide with the surface area of an object in contact with that gas.
Goal 5 Express the relationship among
the following gas pressure units: atmospheres, torr, millimeters of mercury,
inches of mercury, pascals, kilopascals,
bars, or pounds per square inch.
Common pressure units and their relationships to one another are 1 atm = 760 torr =
760 mm Hg= 29.92 in. Hg= 1.013 x 105 Pa= 101.3 kPa = 1.013 bar= 14.69 psi.
Goal 6 Describe the operational principle of an open-end manometer.
In an open-end manometer, total pressures are equal at the lower mercury level. The pressure of the gas is exerted on the mercury surface in the closed leg; the pressure of the
atmosphere is exerted on the mercury surface in the open leg.
Goal 7 Given the pressure difference
in an open-end manometer and the
atmospheric pressure, determine the
pressure of the gas in the manometer.
p gas
Goal 8 Given the gauge pressure of a
gas and the atmospheric pressure, determine the absolute pressure of the gas.
p absolute
Goal 9 Given a pressure in atmospheres, torr, millimeters (or centimeters)
of mercury, inches of mercury, pascals,
kilopascals, bars, or pounds per square
inch, express pressure in each of the
other units.
The quantitative problem-solving methods you learned in Chapter 3 are used to convert
from one pressure unit to another.
Goal 10 When plotting the relationship
between gas volume and temperature,
explain how the straight line can be
extrapolated to determine the temperature at which the gas has zero volume,
and explain the significance of this
temperature.
Gas volume and absolute temperature are directly proportional for a fixed amount of gas at
constant pressure. When the relationship is extrapolated to zero volume, the trendline reaches
the temperature axis at OK, -273°C. This is absolute zero, the lowest temperature.
Goal 11 Describe the relationship
between the volume and temperature
of a fixed amount of a gas at constant
pressure, and express that relationship
as a proportionality, an equality, and
a graph.
Charles's Law states that the volume of a fixed quantity of gas at constant pressure is
directly proportional to absolute temperature, V rx T and V = kT. The plot of volume versus
temperature is a straight line that passes through the origin.
Goal 12 Given the initial volume (or temperature) and the initial and final temperatures (or volumes) of a fixed amount of
gas at constant pressure, calculate the
final volume (or temperature).
For a fixed amount of gas at constant pressure,
Goal 13 Describe the relationship
between the volume and pressure of
a fixed amount of a gas at constant
temperature, and express that relationship as a proportionality, an equality,
and a graph.
Boyle's Law states that for a fixed amount of gas at constant temperature, pressure is
inversely proportional to volume, P rx 1/V and P = k(1/V) or PV = k. The plot of pressure
versus the inverse of volume is a straight line that passes through the origin.
Goal 14 Given the initial volume (or
pressure) and initial and final pressures
(or volumes) of a fixed amount of gas at
constant temperature, calculate the final
volume (or pressure).
For a fixed amount of gas at constant temperature, P1V1 = P2 V2 .
Goal 15 For a fixed amount of a confined gas, given the initial volume, pressure, and temperature, and the final values of any two variables, calculate the
final value of the third variable.
=
p atmosphere
=
±
p mercury·
p gauge + p atmosphere·
V1
~
V2
T
•
2
PV
PV
1 1
The Combined Gas Law states that for a fixed amount of gas, - = ~. Given the initial
T1
T2
(or final) values of all three variables and the final (or initial) values of two, the unknown value
is calculated with the Combined Gas Law equation.
Chapter Summaries
Goal 16 State the values associated
with standard temperature and pressure
(STP) for gases.
679
Standard temperature and pressure, STP, are defined as 273 K (0°C) and 1 bar pressure.
CHAPTER 5 IN REVIEW: Atomic Theory: The Nuclear Model
of the Atom
Goal 1 State the meaning of, or draw
conclusions based on, the Law of
Multiple Proportions.
The Law of Multiple Proportions states: When two elements combine to form more than
one compound, the different weights of one element that combine with the same weight of
the other element are in a simple ratio of whole numbers.
Goal 2 Identify the main features of
Dalton's Atomic Theory.
The main features of Dalton's Atomic Theory are:
1. Each element is made up of tiny, individual particles called atoms.
2. Atoms are indivisible; they cannot be created or destroyed.
3. All atoms of any one element are identical in every respect.
4. Atoms of one element are different from atoms of any other element.
5. Atoms of one element combine with atoms of other elements to form chemical
compounds.
Goal 3 Describe the electron by
charge and approximate mass,
expressed as a comparison with the
mass of a hydrogen atom, and write
the symbol for the electron.
The electron, symbol e-, has a mass that is 1/1837 of the mass of a hydrogen atom. It has
been assigned a charge of 1-.
Goal 4 Describe and/or interpret the
Rutherford scattering experiments and
the nuclear model of the atom.
The Rutherford scattering experiments were designed so that positively charged alpha
particles were directed at thin metal foils. Most particles passed through the foils, but some
were deflected at large angles. The main features of the nuclear model of the atom are the
following:
1. Every atom contains an extremely small, extremely dense nucleus.
2. All of the positive charge and nearly all of the mass of an atom are concentrated in
the nucleus.
3. The tiny nucleus is surrounded by a much larger volume of nearly empty space that
makes up the majority of the volume of an atom.
4. The space outside the nucleus is very thinly populated by electrons, the total
charge of which exactly balances the positive charge of the nucleus.
Goal 5 Identify the three major subatomic particles by symbol, charge,
location within the nuclear atom, and
approximate atomic mass, expressed
relative to the mass of a hydrogen atom.
Subatomic
Particle
Symbol
Electron
e
Proton
p or p+
1+
Neutron
nor n°
0
-
Charge
1-
Location
Outside
nucleus
Inside
nucleus
Inside
nucleus
Mass
Relative
to H Atom
1/1837th
Same
Same
Goal 6 Explain what isotopes are and
how they differ from one another.
Atoms of the same element that have different masses are called isotopes. The mass differences between atoms of an element are caused by different numbers of neutrons.
Goal 7 For an isotope whose chemical
symbol is known, given one of the following, state the other two: (a) nuclear symbol, (b) number of protons and neutrons,
(c) atomic number and mass number.
An isotope can be represented by a nuclear symbol that has the form ~Sy, where A is the
mass number of the element, Z is the atomic number of the element, and Sy is the chemical symbol of the element. The mass number is the sum of the number of protons plus the
number of neutrons. The atomic number is the number of protons.
Goal 8 Identify the features of Dalton's
atomic theory that are no longer considered valid, and explain why.
Two features of Dalton's atomic theory are no longer considered valid:
1. Atoms are indivisible. The existence of ions, charged particles that form when
atoms gain or lose electrons, introduced in Chapter 6, invalidates this postulate.
2. Atoms of an element are identical. The existence of isotopes invalidates this
postulate.
Goal 9 Define and use the atomic mass
unit (u).
1 atomic mass unit (u) =~the mass of one carbon-12 atom
680
Chapter Summaries
Goal 10 Given the relative abundances
of the natural isotopes of an element and
the atomic mass of each isotope, calculate the atomic mass of the element.
The average mass of atoms of an element as found in nature is called the atomic mass. The
atomic mass may be calculated from the masses of the natural isotopes of that element and
the percentage abundance of each isotope.
Goal 11 Distinguish between groups
and periods in the periodic table and
identify them by number.
The periodic table arranges the elements into seven periods (horizontal rows) and 18
groups (vertical columns) in order of atomic numbers and periodic recurrence of physical
and chemical properties.
Goal 12 Given the atomic number of an
element, use a periodic table to find the
symbol and atomic mass of that element, and identify the period and group
in which it is found.
Each box in the periodic table gives the atomic number (Z), the elemental symbol, and the
atomic mass of an element.
Goal 13 Given an elemental symbol or
information from which it can be identified, classify the element as either a
main group or transition element and
either a metal, nonmetal, or metalloid.
Elements in the A groups (1, 2, and 13 to 18) of the periodic table are called main group elements. Elements in the B groups (3 to 12) are called transition elements. The stair-step
line that begins between atomic numbers 4 and 5 in Period 2 and ends between 84 and 85
in Period 6 separates the metals on the left from the nonmetals on the right.
Goal 14 Given the name or the symbol of
an element in Figure 5-19, write the other.
The names and symbols of 35 common elements are to be learned, using the periodic table
as a memory aid: H, He, Li, Be, B, C, N, 0, F, Ne, Na, Mg, Al, Si, P, S, Cl, Ar, K, Ca, Cr, Mn,
Fe, Co, Ni, Cu, Zn, Br, Kr, Ag, Sn, I, Ba, Hg, Pb.
CHAPTER 6 IN REVIEW: Chemical Nomenclature
Goal 1 Given a representation or a written
description of the particulate-level composition of an element or compound, write
the chemical formula of that substance.
A chemical formula describes the composition of the molecule in terms of the number of
each type of atom that makes up the particle. The number of each type of atom is written in
a formula as a subscript; if that number is 1, it is omitted.
Goal 2 Given a name or formula of an
element in Figure 5-19, write the other.
The names and formulas of the 35 elements in Figure 5-19 should already be in memory.
The seven diatomic elements H 2 , N2 , 0 2 , F2 , Cl 2 , Br2 , and 12 must be learned. Remember:
Horses Need Oats For Clear Brown I's.
Goal 3 Given the name or formula of
a binary molecular compound, write
the other.
Two nonmetals or a nonmetal and a metalloid form chemical bonds with each other to
form binary molecular compounds. The name of a binary molecular compound is the
name of the first element followed by the name of the second element, modified with an
-ide suffix. Prefixes are used to indicate the number of atoms of each element in the molecule. Memorize the number prefixes used in chemical names: mono- = 1, di- = 2, tri- = 3,
tetra- = 4, penta- = 5, hexa- = 6, hepta- = 7, octa- = 8, nona- = 9, deca- = 1O.
Goal 4 Given the name or the formula of
water, write the other; given the name or
the formula of ammonia, write the other.
Two common binary molecular compounds with nonsystematic names are water, H 2 0, and
ammonia, NH 3 .
Goal 5 Given the name or formula of an
ion in Figure 6-7, write the other.
Ions are charged particles. A cation has a positive charge; an anion has a negative charge. The
name of a monatomic cation is the name of the element, followed by the word ion. The name of
a monatomic anion is the name of the element, changed to end in -ide, followed by the word ion.
The formula of a monatomic ion is the symbol of the element followed by its electrical charge,
written in superscript. The charge of ions formed from many main-group elements corresponds
to the group number: 1A/1, 1+; 2A/2, 2+; 5A/15, 3-; 6A/16, 2-; 7A/17, 1-.
Goal 6 Given the name or formula of
an ion in Figure 6-9, write the other.
Some elements commonly form more than one ion. For these ions, its oxidation state is
added to the elemental name. The oxidation state is written in parentheses immediately
after the name. For example, the formula of the iron(ll) ion is Fe2 +. There are three common
ions not in Groups 1A/1 or 2A/2 with only one charge: silver ion, Ag+; zinc ion, Zn 2 +; and
aluminum ion, Al 3 +. These three formulas must be memorized.
Goal 7 Give the formula (or name) of the
ammonium ion or hydroxide ion, write
the corresponding name (or formula).
The ammonium ion is NH 4 +; the hydroxide ion is OH-.
Goal 8 Given the name of any ionic
compound made up of identifiable ions,
or other ions whose formulas are given,
write the formula of that compound.
To write the formula of an ionic compound:
1. Write the formula of the cation and the formula of the anion.
2. Mentally balance the charges. Decide what the fewest number of ions is that will
make the compound electrically neutral.
3. Write the formula, cation first, anion second, each with subscripts as needed for
charge balance. (a) If only one ion is needed, omit the subscript. (b) If a polyatomic
ion is needed more than once, enclose the formula of the ion in parentheses and
place the subscript after the closing parenthesis.
Chapter Summaries
681
Goal 9 Given the formula of an ionic
compound made up of identifiable ions,
write the name of the compound.
To write the name of an ionic compound:
Goal 10 Given the name (or formula)
of an acid in Table 6-3, write its formula
(or name).
An acid ionizes in water to give H30+ and an anion. A general equation used to describe acid
ionization is HX + H20 - H30+ + x-. Five -ic acids must be memorized: carbonic acid,
H2C03; nitric acid, HN0 3; phosphoric acid, H3P0 4 ; sulfuric acid, H2S0 4 ; chloric acid, HCI0 3.
Goal 11 Given the name (or formula) of
an anion in Table 6-5, write its formula
(or name).
A system is used to name oxoacids and oxoanions with different numbers of oxygens than
the related -ic acids and -ate anions.
1. Write the name of the cation.
2. Write the name of the anion.
Number of Oxygen Atoms
Acid Prefix
Anion Prefix
Compared with -ic Acid
and/or Suffix
and/or Suffix
One more
per- -ic
per- -ate
-ic
-ate
Same
One fewer
-ous
-ite
Two fewer
hypo- -ous
hypo- -ite
No oxygen
hydro- -ic
-ide
Remember: lck! I ate a hideous bite!
Goal 12 Given the name (or formula) of
an ion formed by the step-by-step ionization of a polyprotic acid from a Group
4A/14, 5A/15, or 6A/16 element, write its
formula (or name).
Acid anions are named in the same way as oxoanions, with the term hydrogen or dihydrogen added to indicate the number of hydrogen ions bonded to the oxoanion.
Goal 13 Given the formula of a hydrate,
state the number of water molecules
associated with each formula unit of the
anhydrous compound.
Hydrates are ionic compounds that exist with a definite number of water molecules in their
crystal structure. Waters of hydration are indicated by the "•" symbol before the number of
water molecules.
Goal 14 Given the name (or formula) of
a hydrate, write its formula (or name).
(This goal is limited to hydrates of ionic
compounds for which a name and formula can be written based on the rules
of nomenclature presented in this book.)
Prefixes are used to indicate the number of water molecules in a formula unit of a hydrate.
These are the same as the prefixes used in naming binary molecular compounds.
A BRIEF SUMMARY OF THE NOMENCLATURE SYSTEM BY EXAMPLE
Substance
Example Name
Example Formula
Element
Hydrogen
Helium
H2
He
Compound made
of two nonmetals
Diphosphorus pentoxide
Carbon tetrachloride
P205
CCl 4
Monatomic cation
Sodium ion
Aluminum ion
Na+
Al 3+
Monatomic anion
Chloride ion
Sulfide ion
c1s2-
Other polyatomic
Ions
Ammonium ion
Hydroxide ion
ow
Ionic Compound
Mercury(ll) bromite
Chromium(lll) iodate
Hg(Br0 2)2
Cr(l0 3h
Acid
Sulfuric acid
Phosphoric acid
H2S0 4
H3P04
Polyatomic anion:
Nitrate ion
N0 3-
Carbonate ion
C032-
Hydrogen carbonate ion
HC0 3-
Dihydrogen phosphate ion
H2P0 4 -
Copper(ll) sulfate pentahydrate
Sodium carbonate decahydrate
CuS0 4 • 5 H20
Na 2C0 3 · 10 H20
total ionization
Polyatomic anion:
step-by-step ionization
Hydrate
NH 4 +
682
Chapter Summaries
CHAPTER 7 IN REVIEW: Chemical Formula Relationships
Goal 1 Given the formula of a chemical
compound (or a name from which the
formula may be written), state the number of atoms of each element in the formula unit.
A chemical formula expresses the number of atoms of each element present in the formula
unit of a substance. The number of each atom is given by a subscript following the symbol
of that atom or a group of atoms. If the number is one, it is omitted in the formula.
Goal 2 Distinguish among atomic mass,
molecular mass, and formula mass.
Atomic mass is the average mass of all atoms of an element as they occur in nature. It is
measured relative to the assignment of a mass of 12 u to an atom of carbon-12. Molecular
(or formula) mass is the average mass of molecules (or formula units) compared with the
mass of an atom of carbon-12, which is 12 atomic mass units.
Goal 3 Calculate the formula (molecular)
mass of any compound whose formula is
given (or known).
The formula mass of a compound is equal to the sum of all of the atomic masses in the formula unit: Formula mass = I atomic masses in the formula unit.
Goal 4 Define the term mole. Identify the
number of objects that corresponds to 1
mole.
One mole of anything contains the same number of objects as the number of atoms in
exactly 12 grams of carbon-12. The experimentally determined value is Avogadro's
number, NM 6.02 x 1023 .
Goal 5 Given the number of moles (or
units) in any sample, calculate the number of units (or moles) in the sample.
Use conversion factors to convert between moles and number of units:
# of moles x
.
# of units x
6.02 x 10 23 units
.
= # of units
mol
1 mol
6.02 x 1 O23 units
=
# of moles
Goal 6 Define molar mass or interpret
statements in which the term molar mass
is used.
Molar mass is mass divided by amount of substance. Mass is typically expressed in grams
and amount is expressed in moles, yielding the most common units of molar mass: grams
per mole (g/mol).
Goal 7 Calculate the molar mass of any
substance whose chemical formula can
be written or is given.
The molar mass of any substance in grams per mole is numerically equal to the atomic,
molecular, or formula mass of that substance in atomic mass units.
Goal 8 Given any one of the following
for a substance with a given (or known)
formula, calculate the other two: (a)
mass; (b) number of moles; (c) number of
formula units, molecules, or atoms.
The mole is the connecting link between the macroscopic level of observation and measurement of matter, in which we measure quantities in grams, and the particulate level, in which
we count the number of units, usually grouped in moles. Using NA for Avogadro's number
and MM for molar mass,
NA
units~
mol
MM
~
g
or
MM
g ~ mol
NA
~units
Changing from units to mass or vice versa is a two-step process, requiring two conversion factors.
Goal 9 Calculate the percentage composition by mass of any compound
whose formula is given (or known).
The percentage composition by mass of a compound is the percentage by mass of each
element in the compound. Percent is the amount of one part of a mixture per 100 total parts
in the mixture. To calculate the percentage of each element,
% Element =
Total molar mass of element in compound
Molar mass of compound
x 100%
If you calculate the percentage composition of a compound correctly, the sum of all percentages must be 100%.
Goal 10 Given the mass of a sample of
any compound with a given (or known)
formula, calculate the mass of any element in the sample; or, given the mass of
any element in the sample, calculate the
mass of the sample or the mass of any
other element in the sample.
To find the amount of any element in a known amount of compound, use percentage as a
conversion factor, based on the equivalency: grams of the element per 100 grams of the
compound.
Goal 11 Distinguish between an empirical formula and a molecular formula.
An empirical formula shows the simplest whole-number ratio of atoms of the elements in a
compound. Empirical formulas are calculated from percentage composition data. They may
also be calculated from the mass of each element in a sample of a compound. Empirical
formulas may or may not be the actual molecular formulas of compounds. The molar mass
of the compound is needed to determine molecular formulas from empirical formulas.
Chapter Summaries
Goal 12 Given data from which the
mass of each element in a sample of a
compound can be determined, find the
empirical formula of the compound.
683
To determine an empirical formula:
1. Determine the percentage composition by mass or the mass of each element in a
sample of the compound.
2. Convert the masses into moles of atoms of the different elements.
3. Determine the ratio of moles of atoms.
4. Express the moles of atoms as the smallest possible ratio of integers.
5. Write the empirical formula, using the number for each atom in the integer ratio as
the subscript in the formula.
Goal 13 Given the molar mass and
empirical formula of a compound, or
information from which they can be
found, determine the molecular formula
of the compound.
To determine the molecular formula of a compound:
1. Determine the empirical formula of the compound.
2. Calculate the molar mass of the empirical formula unit.
3. Determine the molar mass of the compound (which will be given in this book}.
4. Divide the molar mass of the compound by the molar mass of the empirical formula
unit to get n, the number of empirical formula units per molecule.
5. Write the molecular formula.
CHAPTER 8 IN REVIEW: Chemical Reactions
The overarching goals for this chapter are the following:
1. Understand how a chemical equation serves as a particulate-level, symbolic representation of a macroscopic-level process.
2. Learn the mechanics of writing a chemical equation.
3. Learn how to identify four different patterns in chemical equations.
4. Learn how to predict the products of each kind of reaction and write the formulas
of those products.
5. Given potential reactants, write the equations for the probable reaction.
Goal 1 Describe five types of evidence
detectable by human senses that usually
indicate a chemical change.
Five types of evidence indicate the possibility of a chemical change:
1. A color change
2. The formation of a solid
3. The formation of a gas
4. The absorption or release of heat energy
5. The emission of light energy
Goal 2 Distinguish between an unbalanced and a balanced chemical equation, and explain why a chemical equation needs to be balanced.
Goal 3 Given an unbalanced chemical
equation, balance it by inspection.
A chemical equation is a shorthand description of a chemical reaction. A balanced chemical
equation reflects the Law of Conservation of Mass. The subscripts in a chemical equation may
never be changed simply to balance the equation. Changing a subscript in the formula of a
substance changes the chemical identity of that substance (Law of Definite Composition).
A formal approach to balancing a chemical equation follows:
1. Place a "1" in front of the formula with the largest number of atoms. If two formulas
have the same number of atoms, select the one with the greater number of elements.
2. Insert coefficients that balance the elements that appear in compounds. Use fractional coefficients, if necessary. Choosing elements in the following order is usually
easiest: (a) elements in the starting formula that are in only one other compound,
(b} all other elements from the starting formula, (c} all other elements in compounds.
3. Place coefficients in front of formulas of uncombined elements that balance those
elements. Use fractional coefficients, if necessary.
4. Clear fractions, if any, by multiplying all coefficients by the lowest common denominator. Remove any "1" coefficients that remain.
5. Check to be sure the final equation is balanced.
Goal 4 Given a balanced chemical
equation or information from which it
can be written, describe its meaning on
the particulate, molar, and macroscopic
levels.
The coefficients in a balanced chemical equation have two common interpretations:
1. Particulate-level interpretation: The coefficients represent the number of atoms,
molecules, or formula units of each species.
2. Molar interpretation: The coefficients represent the number of moles of each species.
To interpret coefficients at the macroscopic level, you need to combine the molar interpretation with the molar mass of each species.
684
Chapter Summaries
Goal 5 Write the equation for the reaction in which a single product compound
is formed by the combination of two or
more substances.
A reaction in which two or more substances combine to form a single product is a combination reaction or synthesis reaction.
Reactants: Any combination of elements and/or compounds
Product: One compound
Equation type: A
Goal 6 Given a single compound that is
decomposed into two or more substances, either compounds or elements,
write the equation for the reaction.
+ X - AX
A decomposition reaction occurs when a single compound breaks down into simpler
substances.
Reactant: One compound
Products: Any combination of elements and compounds
Equation type: AX - A
Goal 7 Given the reactants of a singlereplacement reaction, write the equation
for the reaction.
+X
In a single-replacement reaction, it looks as if one element is replacing another in a
compound. If the reactant element is a metal, it replaces the metal or hydrogen in the
compound. If the element is a nonmetal, it replaces the nonmetal in a compound.
Reactants: Element (A) plus a solution of either an acid or an ionic compound (BX)
Products: An ionic compound (usually in solution) (AX) plus an element (B)
Equation type: A + BX - AX + B
Goal 8 Given the reactants in a doublereplacement precipitation or neutralization reaction, write the equation.
In a double-replacement reaction, it looks as if the ions of the two reactants change partners. When one or more products of a double-replacement reaction is a solid that is formed
from reactants in solution, the reaction is a precipitation reaction. When one reactant is an
acid and the other is a base, the double-replacement reaction yields water as one product,
and the reaction is a neutralization reaction.
Reactants: Solutions of two compounds, each with positive and negative ions (AX + BY)
Products: Two new compounds (AY + BX), which may be a solid, water, an acid, or an aqueous
ionic compound
Equation type: AX + BY -AY + BX
CHAPTER 9 IN REVIEW: Chemical Change
Goal 1 Distinguish between strong
electrolytes, weak electrolytes, and
nonelectrolytes.
A solution is a homogeneous mixture. A solute may be classified as a strong electrolyte, a
weak electrolyte, or a nonelectrolyte according to the ability of its water solution to conduct
electricity strongly, weakly, or not at all. If a solution conducts electricity, ions must be present as solute particles.
Goal 2 Given the formula of an ionic
compound (or its name), write the formulas of the ions present when it is dissolved in water.
When an ionic compound dissolves in water, its solution consists of water molecules and
ions surrounded by water molecules. The species in solution are aqueous ions, occurring in
the same ratio as in the formula of the ionic compound.
Goal 3 Explain why the solution of an
acid may be a good conductor or a poor
conductor of electricity.
An acid is a substance that has a proton that can be removed by a water molecule when in
a water solution. Acids are classified as strong or weak, depending on the extent to which
the original compound ionizes when dissolved in water. When a strong acid dissolves, it
dissociates into ions. The major species present in the solution are ions, and the minor
species present are un-ionized molecules. Strong acids are strong electrolytes because
the major species in solution are ions, and the presence of ions in a solution makes the solution a good conductor. When a weak acid dissolves, it does not dissociate into ions to a
large extent. The solution inventory is mainly un-ionized molecules, and the solution is a
poor conductor.
Goal 4 Given the formula or the name
of a soluble acid, write the major and
minor species present when it is dissolved in water.
There are seven common strong acids. Their names and formulas must be memorized:
nitric acid, HN0 3 ; sulfuric acid, H2 S0 4 ; hydrochloric acid, HCI; hydrobromic acid, HBr;
hydroiodic acid, HI; chloric acid, HCl0 3 ; perchloric acid, HCl0 4 . If an acid is not one of the
seven strong acids, it is a weak acid. Ions are the major species in the solutions of two
kinds of substances: (1) all soluble ionic compounds and (2) the seven strong acids. Neutral
molecules are the major species in solutions of everything else, primarily (1) compounds
with all nonmetal atoms (except strong acids), (2) weak bases, and (3) water.
Chapter Summaries
Goal 5 Distinguish among conventional,
total ionic, and net ionic equations.
685
A conventional equation shows the formulas of the reactants written on the left side of
an arrow and the formulas of the products on the right side of the arrow. Strong acids
and ionic compounds designated (aq) in a conventional equation are rewritten in a total
ionic equation with the formulas of the major species in solution. A total ionic equation
is made into a net ionic equation by removing the spectators, those species that are on
both sides of the total ionic equation. The procedure for writing a net ionic equation
is this:
1. Write the conventional equation, including state symbols-(g), (f), (s), and (aq). Balance the equation.
2. Write the total ionic equation by replacing each aqueous (aq) substance that is a
strong acid or a soluble ionic compound with its major species. Do not separate a
weak acid into ions, even though its state is aqueous (aq). Also, never change solids (s), liquids (f), or gases (g) into ions. Be sure the equation is balanced in both
atoms and charge.
3. Write the net ionic equation by removing the spectators from the total ionic equation. Reduce coefficients to lowest terms, if necessary. Be sure the equation is balanced in both atoms and charge.
Goal 6 Given two substances that may
engage in a single-replacement redox
reaction and an activity series by which
the reaction may be predicted, write the
conventional, total ionic, and net ionic
equations for the reaction that will occur,
if any.
A single-replacement redox reaction has the form A + BX___.,. AX + B. The compounds
BX and AX are usually aqueous and the elements A and B are usually solid metals or
H2 {g). Prediction of a single-replacement redox reaction is made by referring to an activity series. A more active element will replace the dissolved ions of any less active
element.
Goal 7 Write the equation for the
complete oxidation or burning of any
compound containing only carbon and
hydrogen or only carbon, hydrogen,
and oxygen.
The general equation for a complete oxidation (burning) of a compound that consists of
only carbon and hydrogen or only carbon, hydrogen, and oxygen is CxHPz + 0 2 (g) ___.,.
C0 2(g) + H2 0(f). As a rule, these equations are most easily balanced if you take the elements carbon, hydrogen, and oxygen in that order.
Goal 8 Predict whether a precipitate
will form when known solutions are
combined; if a precipitate forms, write
the net ionic equation. (Reference to a
solubility table or a solubility guidelines list may or may not be allowed,
depending upon the preference of your
instructor.)
An ion-combination reaction occurs when the cation (positively charged ion) from one
reactant combines with the anion (negatively charged ion) from another to form a product
compound. The conventional equation is a double-replacement type in which the ions
appear to change partners: AX + BY___.,. AY + BX. When a product in this type of reaction is
an insoluble ionic compound, the solid is called a precipitate and the reaction is a precipitation reaction.
Goal 9 Given the product of a precipitation reaction, write the net ionic
equation.
Your instructor will have you memorize solubility guidelines or allow you to have access to
the guidelines or a solubility table. Use the method suggested by your instructor to predict
precipitation reactions.
Goal 10 Given reactants for a doublereplacement reaction that yield a molecular product, write the conventional,
total ionic, and net ionic equation.
The reaction of an acid often leads to an ion combination that yields a molecular product instead of a precipitate. Except for the difference in the product, the equations are
written in exactly the same way. Just as you had to recognize an insoluble product and
not break it up in total ionic equations, you must recognize a molecular product and not
break it into ions. Water and weak acids are the two kinds of molecular products you will
find. Neutralization reactions are the most common molecular-product reactions, HX +
MOH___.,. HOH + MX. There are two points by which you can identify a molecular-product
reaction: (1) one reactant is an acid, usually strong, and (2) one product is water or a
weak acid.
Goal 11 Given reactants that form
H2 C0 3 , H2 S0 3 , or "NH 4 0H" by ion combination, write the net ionic equation for
the reaction.
Three ion combinations yield molecular products that are not the products you would
expect:
2 W(aq) + COl-(aq) ___.,. H2 C0 3 (aq) ___.,. C02 (g) + H2 0(f)
2 W(aq) + 80 3 2 -(aq) ___.,. H2 S0 3 (aq) ___.,. S0 2 (aq) + H2 0(f)
NH 4 +(aq) + OH-(aq) ___.,. "NH 4 0H" ___.,. NH 3 (aq) + Hp(f)
When ion combinations yield unstable substances, the right side of the net ionic equation
has the formulas of the stable decomposition products.
686
Chapter Summaries
CHAPTER 10 IN REVIEW: Quantity Relationships
in Chemical Reactions
Goal 1 Given a chemical equation, or
a reaction for which the equation is
known, and the number of moles of one
species in the reaction, calculate the
number of moles of any other species.
The coefficients in a chemical equation express the mole relationships between the different substances in the reaction. The coefficients may be used in a conversion factor from
moles of one substance to moles of another.
Goal 2 Given a chemical equation, or a
reaction for which the equation can be
written, and the mass in grams or moles
of one species in the reaction, find the
mass in grams or moles of any other
species.
The mass-to-mass stoichiometry path is the following:
Goal 3 Given two of the following,
or information from which two of the
following may be determined, calculate
the third: ideal yield, actual yield, percentage yield.
1. Change the mass of the given species to moles.
2. Change the moles of the given species to moles of the wanted species.
3. Change the moles of the wanted species to mass.
This three-step method is at the heart of almost all stoichiometry problems.
The efficiency of a reaction is stated in percentage yield, in which the actual yield is
expressed as a percent of the ideal yield calculated by stoichiometry:
% yield=
actual yield
ideal yield
X 100%
You need to be able to solve three types of percentage yield problems:
1. Given: Actual and theoretical yields
actual yield
Solve by: % yield = 'd
. Id X 100%
1 ea 1 y1e
Wanted: Percentage yield
2. Given: Reactant quantity and percentage yield
Wanted: Product quantity
Solve by: Use percentage yield as a conversion factor
3) Given: Product quantity and percentage yield
Wanted: Reactant quantity
Solve by: Use percentage yield as a conversion factor
Goal 4 Identify and describe or explain
limiting reactants and excess reactants.
A limiting reactant is the reactant totally consumed in a reaction, thereby determining the
maximum yield possible. An excess reactant is a reactant that has a quantity that is in
excess of the amount needed to completely react with the limiting reactant.
Goal 5 Given a chemical equation, or
information from which it may be determined, and initial quantities of two or
more reactants, (a) identify the limiting
reactant, (b) calculate the ideal yield of a
specified product, assuming complete
use of the limiting reactant, and (c) calculate the quantity of the reactant initially in excess that remains unreacted.
The comparison-of-moles method for solving limiting reactant problems is the following:
1. Convert the number of grams of each reactant to moles.
2. Identify the limiting reactant.
3. Calculate the number of moles of each species that reacts or is produced.
4. Calculate the number of moles of each species that remains after the reaction.
5. Change the number of moles of each species to grams.
The smaller-amount method for solving limiting reactant problems is the following:
1. Calculate the amount of product that can be formed by the initial amount of each
reactant. (a) The reactant that yields the smaller amount of product is the limiting
reactant. (b) The smaller amount of product is the amount that will be formed when
all of the limiting reactant is used up.
2. Calculate the amount of excess reactant that is used by the total amount of limiting
reactant.
3. Subtract from the amount of excess reactant present initially the amount that is used by
all of the limiting reactant; the difference is the amount of excess reactant that is left.
Goal 6 Given energy in one of the following units, plus variations created by adding metric prefixes, calculate the other
two: joules, calories, and food Calories.
The SI unit of energy is the joule, a force of one newton applied for a distance of one meter.
Another energy unit used by chemists is the calorie, which is equal to 4.184 joules. The joule (J)
and the calorie (cal) are small energy units, so units 1000 times larger, the kilojoule, kJ, and the
kilocalorie, kcal, are often used. The food energy Calorie is the thermochemical kilocalorie.
Goal 7 Given a chemical equation, or
infonnation from which it may be written,
and the heat (enthalpy) of reaction, write
the thennochemical equation either (a)
with ~H to the right of the conventional
equation or (b) as a reactant or product.
The amount of heat given off or absorbed in a chemical reaction is called the enthalpy of
reaction, symbolized by ~rH. The t'.1rH of a reaction may be included in the chemical equation as a reactant or a product, or it may be written next to the equation. The equation is
then called a thermochemical equation. For an endothermic reaction, t'.1rH is positive; heat
is a reactant in the thermochemical equation. For an exothermic reaction, L1rH is negative;
heat is a product in the thermochemical equation.
Chapter Summaries
Goal 8 Given a thermochemical equation, or information from which it may be
written, calculate the amount of energy
released or added for a given amount of
reactant or product; alternately, calculate the mass of reactant required to
produce a given amount of energy.
687
Using the ilrH of a thermochemical equation and the coefficient of any substance in that
equation, you can convert in either direction from moles of that substance to amount of heat
transferred.
Summary of stoichiometry pattern:
General Stoichiometry Pattern
Particulate
quantity
Product
Specific Applications
Mole ratio
from balanced
Quantity in moles
equation
Quantity in moles
Reactant
If------~
Product
Quantity in
number of particles
Quantity in
number of particles
Reactant
Product
CHAPTER 11 IN REVIEW: Atomic Theory: The Quantum Model
of the Atom
Goal 1 Define and describe electromagnetic radiation.
Electromagnetic radiation is a form of energy that has wavelike properties that includes
gamma rays, x-rays, ultraviolet radiation, visible light, infrared radiation, microwaves, and
radio waves. It travels at the speed of light, c: c = X. (wavelength} x v (frequency).
Goal 2 Distinguish between continuous
and line spectra.
A continuous spectrum is a spectrum that is distributed over a continuous range of wavelengths. A line spectrum has discrete lines.
Goal 3 Describe the Bohr model of the
hydrogen atom.
The Bohr model of the hydrogen atom features a relatively small volume, extremely dense
nucleus that contains all of the atom's positive charge and nearly all of its mass. The negatively charged electron of the hydrogen atom has a very small mass, and it travels in one of
a series of circular orbits around the nucleus.
Goal 4 Explain the meaning of quantized energy levels in an atom and show
how these levels relate to the discrete
lines in the spectrum of that atom.
The Bohr model of the hydrogen atom restricts the electron to certain quantized energy
levels. The electron can have certain definite energies, but never may it have an energy
between the quantized values. The spectrum of the atoms of an element is the result of
energy released as electrons in an excited state drop to a lower energy level via a quantum
jump or quantum leap.
Goal 5 Distinguish between ground
state and excited state.
The electron is normally found in its ground state, the condition when the electron is in n = 1
(for a hydrogen atom) or when all electrons in an atom occupy the lowest possible energy
levels (for atoms with multiple electrons). If energy is transferred to an atom, an electron can
be raised to an excited state, the condition at which one or more electrons in an atom has
an energy level above ground state.
Goal 6 Identify the principal energy levels in an atom and state the energy trend
among them.
The quantum mechanical model of the atom identifies principal energy levels, and for atoms
with two or more electrons, sublevels within each principal energy level. In general, energies
increase as the principal quantum numbers increase: n = 1 < n = 2 < n = 3 ... < n = 7.
Goal 7 For each principal energy level,
state the number of sublevels, identify
them, and state the energy trend
among them.
The total number of sublevels within a given principal energy level is equal ton, the principal
quantum number. For any given value of n, energy increases through the sublevels in the
order s < p < d < f.
Goal 8 Sketch the shapes of s and p
orbitals.
Figure 11-16 shows the shapes of the sand p (and d) orbitals. Ans orbital is spherical. A p
orbital is dumbbell-shaped.
Goal 9 State the number of orbitals in
each sublevel.
There is one orbital for every s sublevel. All p sublevels have three orbitals, all d sublevels
have five, and all f sublevels have seven. This 1-3-5-7 sequence of odd numbers continues through higher sublevels.
Goal 10 State the restrictions on the
electron population of an orbital.
An orbital may be unoccupied, occupied by one electron, or occupied by two electrons.
688
Chapter Summaries
Goal 11 Use a periodic table to list
electron sublevels in order of increasing
energy.
Two rules guide the assignments of electrons to orbitals:
1. At ground state the electrons fill the lowest-energy orbitals available.
2. No orbital can have more than two electrons.
The periodic table is a guide to the order of increasing energy of sublevels: 1s < 2s < 2p <
3s < 3p < 4s < 3d < 4p and so on.
Goal 12 Referring only to a periodic
table, write the ground state electron
configuration of an atom of any element
up to atomic number 36.
The procedure for writing an electron configuration follows:
1. Locate the element in the periodic table. From its position in the table, identify and
write the electron configuration of its highest occupied energy sublevel.
2. To the left of what has already been written, list all lower-energy sublevels in order
of increasing energy.
3. For each filled lower-energy sublevel, write as a superscript the number of electrons that fill that sublevel.
4. Confirm that the total number of electrons is the same as the atomic number.
Goal 13 Using n for the highest occupied energy level, write the configuration
of the valence electrons of any main
group element.
Atoms in the same group of the periodic table (same column) have the same highest occupied sublevel electron configurations, or the same nsxnpr valence electron configurations.
These range from ns 1 for Group 1A/1 to ns 2 np 6 for Group 8A/18. The highest occupied principal energy level value (n) increases as you go down a group.
Goal 14 Write the Lewis (electron dot)
symbol for an atom of any main group
element.
Valence electrons are depicted by Lewis symbols, which are also called electron dot symbols. The symbol of the element is surrounded by the number of dots that matches the
number of valence electrons.
Goal 15 Predict how and explain why
atomic size varies with position in the
periodic table.
The sizes of atoms in the periodic table increase as you move down a group, but decrease
as you move left to right across a period. This is explained by the highest occupied energy
level and nuclear charge.
Goal 16 Predict how and explain why
first ionization energy varies with position in the periodic table.
First ionization energy, the energy required to remove an electron from a neutral atom,
decreases as you move down a group (highest occupied energy level is farther from the
nucleus) and increases as you move across a period (more positive charge in the nucleus).
Goal 17 Explain, from the standpoint of
electron configuration, why certain
groups of elements make up chemical
families.
Groups of elements in the periodic table exhibit similar behavior and are called chemical
families. The chemical properties of the elements in a family are similar because they have
the same valence electron configuration.
Goal 18 Identify in the periodic table the
following chemical families: alkali metals,
alkaline earths, halogens, noble gases.
Alkali metals, ns 1 Group 1A/1; alkaline earths, ns 2 , Group 2A/2; halogens, ns 2 np 5 , Group
?A/17; noble gases, ns 2np 6 , Group 8A/18.
Goal 19 Identify metals and nonmetals
in the periodic table.
The elements in the periodic table are classified as metals or nonmetals, based on their
chemical behavior. Metals are to the left of the stair-step line, nonmetals are to the right, and
six metalloids hug the line.
Goal 20 Predict how and explain why
metallic character varies with position in
the periodic table.
Metallic character increases from right to left across any row of the periodic table and
from top to bottom in any group. A metal can lose one or more electrons and become a
positively charged ion. The lower the energy required to remove an electron, the greater the
metallic character of an element.
CHAPTER 12 IN REVIEW: Chemical Bonding
Goal 1 Define and distinguish between
cations and anions.
A cation is a positively charged ion; an anion is a negatively charged ion.
Goal 2 Identify the monatomic ions that
are isoelectronic with a given noble gas
atom and write the electron configuration of those ions.
The formation of monatomic ions that are isoelectronic with neon atoms illustrates the pattern that is duplicated for other noble gases. A neon atoms has 10 electrons, including a full
octet of valence electrons. Its electron configuration is 1s 2 2s 2 2p 6 . Nitrogen, oxygen, and
fluorine atoms form anions by gaining enough electrons to reach the same configuration.
Sodium, magnesium, and aluminum atoms form cations by losing valence electrons to
reach the same configuration.
Goal 3 Use Lewis symbols to illustrate
how an ionic bond can form between monatomic cations from Groups 1A, 2A, and
3A (1, 2, 13) and anions from Groups SA,
6A, and 7A (15-17) of the periodic table.
Ionic bonds between atoms form when atoms of a metal lose one, two, or three electrons to
form a cation that is isoelectronic with a noble gas and atoms of a nonmetal gain one, two,
or three electrons to form an anion that is also isoelectronic with a noble gas. An ionic
bond, also called an electron-transfer bond, is formed because of the electrostatic attraction between oppositely charged ions.
Chapter Summaries
689
Goal 4 Describe, use, or explain each
of the following with respect to forming a
covalent bond: electron cloud, charge
cloud, or charge density; valence electrons; half-filled electron orbital; filled
electron orbital; electron sharing; orbital
overlap; octet rule or rule of eight.
Two atoms in a molecule are held together by a covalent bond when they share one or
more pairs of electrons. The electron cloud or charge cloud formed by the two bonding
electrons is concentrated in the region between the two nuclei. The bonding electrons count
as valence electrons for each bonded atom. Covalent bonds form by the overlap of halffilled electron orbitals. The stability of a noble-gas electron configuration-the octet rule or
rule of eight-is a result of the minimization of energy associated with that configuration.
Goal 5 Use Lewis symbols to show how
covalent bonds are formed between two
nonmetal atoms.
A covalent bond is formed between two nonmetal atoms, both of which have atoms that
are one, two, or even three electrons short of a noble gas electron configuration. This covalent bonding process is characterized by valence electron pairs that are shared.
Goal 6 Distinguish between bonding
electron pairs and lone pairs.
Bonding electron pairs are represented in a Lewis diagram as two dots or a straight line
drawn between atoms. Both formats represent the covalent bond that holds the atoms
together. Unshared pairs are also shown in Lewis diagrams. These are also called lone pairs.
Goal 7 Distinguish between polar and
nonpolar covalent bonds.
In a nonpolar covalent bond, the bonding electrons are shared equally by the bonded
atoms. In a polar covalent bond, the nucleus of one atom attracts the shared electrons
more strongly than the other.
Goal 8 Predict which end of a polar
bond between identified atoms is positive and which end is negative.
The relative ability of atoms of an element to attract electron pairs in covalent bonds is expressed
by the electronegativity of the element. The polarity of a bond is estimated by the difference in
electronegativities of the bonded atoms. The atom with the higher electronegativity is the negative end of the bond. The atom with the lower electronegativity is the positive end of the bond.
Goal 9 Rank bonds in order of increasing or decreasing polarity based on periodic trends in electronegativity values or
actual values, if given.
There is a periodic trend in the electronegativity of elements. In general, electronegativity
increases across a period and decreases down a group. The greater the difference in electronegativity between two bonded elements, the more polar the bond.
Goal 10 Distinguish among single, double, and triple bonds, and identify these
bonds in a Lewis diagram.
The sharing of one pair of electrons by two bonded atoms is called a single bond. When
two atoms are bonded by two pairs of electrons, it is a double bond. When two atoms are
bonded by three pairs of electrons, the bond is called a triple bond. All four electrons in a
double bond and all six electrons in a triple bond are counted as valence electrons for the
bonded atoms. Multiple bonds is a general term that includes double and triple bonds.
Goal 11 Describe how metallic bonding
differs from ionic and covalent bonding.
Metallic bonding occurs because of attractive forces between negatively charged valence
electrons moving among positively charged metal ions. In covalent bonds, the bonding
electrons are localized between two specific atoms. Electrons in a metallic bond are delocalized because the bonding electrons do not stay near any single atom or pair of atoms.
The nature of the metallic bond explains many of the properties of metals, such as electrical
conductivity and the ability to bend and be stretched into thin wires.
Goal 12 Sketch a particulate-level illustration of the electron-sea model of
metallic bonding.
The electron-sea model of a metallic crystal is characterized by monatomic ions in a crystal pattern with the highest-energy valence electrons free to move among the ions. The positively charged metal ions are held in fixed positions in the crystal because of their attraction
to the negatively charged valence electrons that move among the ions.
CHAPTER 13 IN REVIEW: Structure and Shape
Goal 1 Draw the Lewis diagram for any
molecule or polyatomic ion made up of
main group elements.
The procedure for drawing a Lewis diagram follows:
1. Count the total number of valence electrons. Adjust for charge on ions.
2. Place the least electronegative atom(s) in the center of the molecule.
3. Draw a tentative diagram. Join atoms by single bonds. Add unshared pairs to complete the octet around all atoms except hydrogen.
4. Calculate the number of valence electrons in your tentative diagram and compare
it with the actual number of valence electrons. If the tentative diagram has too
many electrons, remove a lone pair from the central atom and from a terminal
atom, and replace them with an additional bonding pair between those atoms.
If the tentative diagram still has too many electrons, repeat the process.
5. Check the Lewis diagram. Hydrogen atoms must have only one bond, and all other
atoms should have a total of four electron pairs.
Goal 2 Describe the electron-pair
geometry when a central atom is surrounded by two, three, or four regions
of electron density.
Electron-pair geometry describes the arrangement of two, three, or four regions of electron density around a central atom.
Electron Pairs
2
3
4
Geometry
Linear
Trigonal planar
Tetrahedral
Electron-Pair Angles
180°
120°
109.5°
690
Chapter Summaries
Goal 3 Given or having derived the
Lewis diagram of a molecule or polyatomic ion in which a central atom is
surrounded by two, three, or four
regions of electron density, predict
and sketch the molecular geometry
around that atom.
Goal 4 Draw a wedge-and-dash diagram of any molecule for which a Lewis
diagram can be drawn.
Molecular geometry describes the arrangement of two, three, or four atoms around a
central atom to which they are all bonded.
Electron Pairs
Bonded Atoms
2
2
3
3
4
4
4
3
2
4
3
2
Molecular Geometry
Bond Angle
Linear
Trigonal planar
Angular
Tetrahedral
Trigonal pyramidal
Bent
180°
120°
120°
109.5°
109.5°
109.5°
The following procedure is for drawing a wedge-and-dash diagram:
1. When two atoms are in the same plane as the page, they are connected with a
solid line of uniform width.
2. When an atom is behind the plane of the page, it is connected to the central atom
by a line that is dashed. The width of the dashed line increases as it moves away
from the central atom.
3. When an atom is in front of the plane of the page, it is connected to the central
atom by a line that is wedge-shaped. The width of the wedge-shaped line
increases as it moves away from the central atom.
Goal 5 For a molecule with more than
one central atom and/or multiple bonds,
draw the Lewis diagram and predict and
sketch the molecular geometry around
each central atom, and draw a wedgeand-dash diagram of the molecule.
It is the number of regions of electron density that surround a central atom that determines the electron-pair geometry around that atom. A region of electron density can be a
single, double, or triple bond, or a lone pair. No matter the number of pairs of bonding electrons between two atoms, each region of electron density is distributed as far away from
other regions of electron density as possible, as predicted by VSEPR theory.
Goal 6 Given or having determined the
Lewis diagram of a molecule, predict
whether the molecule is polar or
non polar.
Molecular polarity depends on both bond polarity and molecular geometry. A polar molecule is one in which there is an asymmetrical distribution of charge, resulting in positive
and negative poles. A nonpolar molecule either has nonpolar bonds or has polar bonds
that cancel, resulting in no overall regions of positive and negative charge. If the central
atom of a molecule has no lone pairs and all atoms bonded to it are identical, the molecule is
nonpolar. If these conditions are not met, the molecule is polar.
Goal 7 Distinguish between organic
compounds and inorganic compounds.
The majority of all chemical compounds that have been characterized are classified as
organic compounds-compounds based upon the carbon atom. Inorganic compounds
are those without carbon atoms, plus the following carbon-containing compounds: carbonates, cyanides, and oxides of carbon.
Goal 8 Distinguish between hydrocarbons and other organic compounds.
Hydrocarbons are made of only carbon and hydrogen. The alkanes are a hydrocarbon
family with all single bonds.
Goal 9 On the basis of structure and the
geometry of the identifying group, distinguish among alcohols, ethers, and carboxylic acids.
If a hydrogen atom in a hydrocarbon, CH 4 , for example, is replaced by a hydroxyl group,
-OH, the resulting molecule is an alcohol, CH 3 -0H. An alcohol may be thought of as a
water molecule (H-OH) in which one hydrogen atom is replaced by a hydrocarbon group
(CH 3 -0H). An ether may be thought of as a water molecule (H-0-H) in which both
hydrogen atoms are replaced by hydrocarbon groups (CH 3 -0-CH 3). Alcohols and ethers
with the same number of carbon atoms are isomers. If a hydrogen atom in a hydrocarbon is
replaced by a carboxyl group, -COOH, the resulting molecule is a carboxylic acid. The
most common carboxylic acid is acetic acid, written as HC 2 H3 0 2 or CH 3COOH.
CHAPTER 14 IN REVIEW: The Ideal Gas Law
and Its Applications
Important ideas to review from Chapter 4:
Charles's Law states that at constant pressure, the volume of a fixed amount of a gas is
directly proportional to the absolute temperature, V x T.
Boyle's Law states that at constant temperature, the volume of a fixed amount of a gas is
inversely proportional to its pressure, V x 1/P.
P1 V1
Charles's and Boyle's Laws can be coupled as the Combined Gas Law: - T1
Goal 1 If pressure and temperature are
constant, state how volume and amount
of gas are related and explain phenomena or make predictions based on that
relationship.
P2 V2
= --.
T2
Avogadro's Law states that equal volumes of two gases at the same temperature and pressure contain the same number of molecules, V x n.
Chapter Summaries
691
Goal 2 Explain how the ideal gas equation can be constructed by combining
Charles's Law (Section 4-4), Boyle's
Law (Section 4-5), and Avogadro's Law
(Section 14-2), and explain how the ideal
gas equation can be used to derive each
of the three two-variable laws.
Since V -x T, V -x 1/P, and V -x n, it follows that V -x T x (1/P) x n. Inserting a proportionality
constant R, V = RT(1/P)n, or, rearranging, PV = nRT. When pressure and amount are constant, V = kT, which is Charles's Law. When temperature and amount are constant, PV = k,
which is Boyle's Law. When pressure and temperature are constant, V = kn, which is
Avogadro's Law.
Goal 3 Given values for all except one of
the variables in the ideal gas equation,
calculate the value of the remaining
variable.
The ideal gas equation is PV = nRT. Two values of the gas constant, R, are 0.0821 L ·
atm/mol · K and 62.4 L · torr/mol · K. Given all the values in the ideal gas equation except
one, the remaining value may be calculated. Substituting m/MM for its equivalent, n, in the
Goal 4 Calculate the density of a known
gas at any specified temperature and
pressure.
Solving the PV = (m/MM)RT form of the ideal gas equation for m/V, which is density, yields
D = m = (MM)P
RT .
V
Goal 5 Given the density of a pure gas
at specified temperature and pressure,
or information from which the density
may be found, calculate the molar mass
of that gas.
The density of a gas at constant temperature and pressure is directly proportional to its
molar mass, D o: MM. Either molar mass or density can be calculated from the other using
the ideal gas equation.
Goal 6 Calculate the molar volume of
any gas at any given temperature and
pressure.
Molar volume is the volume occupied by one mole of a gas, Vin. The molar volume of
any ideal gas at STP (0°C and 1 bar) is 22.7 L/mol. This quantity is useful in calculations
involving moles, mass, volume, density, and molar mass of a gas measured at STP.
Goal 7 Given the molar volume of a gas at
any specified temperature or pressure, or
information from which the molar volume
may be determined, and either the number
of moles in or the volume of a sample of
that gas, calculate the other quantity.
Molar volume
ideal gas equation gives PV
.
IS
MV =
nV
=
=
MmM RT.
RT
p· Once molar volume is determined, it can be used to convert
between the macroscopic volume of a gas and the particulate-level number of particles,
grouped in moles.
Goal 8 Given a chemical equation, or a
reaction for which the equation can be
written, and the mass or number of
moles of one species in the reaction, or
the STP volume of a gaseous species,
find the mass or number of moles of
another species, or the STP volume of
another gaseous species.
22.7 liters = 1 mole is an equivalency that can be used to convert between the volume of a
gas at STP and the number of particles of that gas, counted in moles. 22.7 L/mol can be
used only for ideal gases at STP. If your stoichiometry skills are rusty, review Section 10-1.
Goal 9 Given a chemical equation, or
a reaction for which the equation can
be written, and the mass or number of
moles of one species in the reaction, or
the volume of any gaseous species at a
given temperature and pressure, find
the mass or number of moles of any
other species, or the volume of any other
gaseous species at a given temperature
and pressure.
A gas stoichiometry problem at non-STP conditions can be solved by finding the molar
volume of the gas and then following the stoichiometry path (molar volume method presented in Section 14-7) or by applying the ideal gas equation and then following the stoichiometry path or by following the stoichiometry path and then applying the ideal gas equation
(ideal gas equation method presented in Section 14-8).
Goal 10 Given a chemical equation, or a
reaction for which the equation can be
written, and the volume of any gaseous
species at a given temperature and
pressure, find the volume of any other
gaseous species at a given temperature
and pressure.
The ratio of volumes of gases in a reaction is the same as the ratio of moles, provided that
the gas volumes are measured at the same temperature and pressure. Thus the coefficients
in a balanced chemical equation can be used to convert between volumes, as long as the
volumes are at the same temperature and pressure.
CHAPTER 15 IN REVIEW: Gases, Liquids, and Solids
Goal 1 Given the partial pressure of
each component in a mixture of gases,
find the total pressure.
The partial pressure of a gas in a gaseous mixture is the pressure that gas alone would
exert in the same volume at the same temperature.
692
Chapter Summaries
Goal 2 Given the total pressure of a
gaseous mixture and the partial pressures of all components except one, or
information from which those partial
pressures can be obtained, find the partial pressure of the remaining
component.
The total pressure of a gas mixture is the sum of the partial pressures of all gases in that
mixture, P = p1 + p2 + p 3 + .... This is Dalton's Law of Partial Pressures.
Goal 3 Explain the differences between
the physical behavior of liquids and
gases in terms of the relative distances
among particles and the effect of those
distances on intermolecular attractions.
Important properties of liquids (and comparisons with gases) include the following:
1. Gases may be compressed; liquids cannot. Liquid particles are "touchingly close"
to one another.
2. Gases expand to fill their containers; liquids do not. The strong attractions between
liquid particles hold them together at the bottom of a container.
3. Gases have low densities; liquids have relatively high densities. If the particles of a
liquid are close together compared with the particles of a gas, a given number of
liquid particles will occupy a much smaller volume than the same number of particles will occupy as a gas.
4. Gases may be mixed in a fixed volume; liquids cannot. There is no space between
particles of a liquid, so combining liquids must increase volume.
5. Gases exert constant pressure on the walls of their container uniformly in all
directions; the pressure in a liquid container increases with increasing depth.
Liquid pressures depend on the depth of the liquid due to variation in weight at
varying depth.
Goal 4 For two liquids, given comparative values of physical properties that
depend on intermolecular attractions,
predict the relative strengths of those
attractions; or, given a comparison of
the strengths of the intermolecular
attractions, predict the relative values of
physical properties that the attractions
cause.
Properties of liquids are related to intermolecular attractions:
Vapor pressure is the partial pressure of a vapor in equilibrium with its liquid state at a
given temperature. Liquids with strong intermolecular attractions have lower vapor pressures than liquids with weak intermolecular attractions.
Heat of vaporization is the quantity of energy required to change 1 mole of a liquid to a
gas, while at constant temperature and pressure. Liquids with strong intermolecular attractions have higher heats of vaporization than liquids with weak intermolecular attractions.
Boiling point is the temperature at which vapor pressure becomes equal to the pressure
above a liquid. Liquids with strong intermolecular attractions have higher boiling points than
liquids with weak intermolecular attractions.
Viscosity is the resistance of a liquid to flow. Liquids with strong intermolecular attractions
have higher viscosities than liquids with weak intermolecular attractions.
Surface tension is the force exerted on the molecules at the surface of a liquid by the molecules below the surface. Liquids with strong intermolecular attractions have greater surface tension than liquids with weak intermolecular attractions.
Goal 5 Identify and describe or explain
induced dipole forces, dipole forces, and
hydrogen bonds.
Induced dipole forces are comparatively weak intermolecular attractions between all molecules. They are the result of temporary dipoles caused by shifting electron density in molecules. Induced dipole forces vary directly with surface area and may be large if the molecules
are large.
Dipole forces are electrostatic attractions between polar molecules (dipoles).
Exceptionally strong dipole-like forces called hydrogen bonds arise between molecules
that have hydrogen atoms covalently bonded to a highly electronegative atom. This atom,
usually nitrogen, oxygen, or fluorine, must have at least one unshared electron pair.
Goal 6 Given the structure of a molecule, or information from which it may be
determined, identify the significant intermolecular forces present.
In general, a liquid with nonpolar molecules will have only induced dipole forces acting
among the molecules. The larger the molecules, the greater the strength of the attractive
forces. Liquids with polar molecules have dipole forces acting among the molecules.
The more polar the molecules, the stronger the forces. Liquids with molecules with a
hydrogen atom bonded to an atom that is small, highly electronegative, and that has at
least one unshared pair of electrons have hydrogen bonds acting among the
molecules.
Goal 7 Given the molecular structures
of two substances, or information from
which they may be obtained, compare or
predict relative values of physical properties that are related to them.
All other things being equal, intermolecular attractive forces increase in the order of induced
dipoles < dipole forces < hydrogen bonds. In large molecules, however, induced dipole
forces can be the most important forces acting to determine values of physical properties.
Chapter Summaries
693
Goal 8 Describe or explain the equilibrium between a liquid and Its own vapor
and the process by which It is reached.
Equilibrium is defined as the condition in which the rates of opposing changes are equal. ln
a liquid-vapor equilibrium, the rate of evaporation is equal to the rate of condensation. Such
an equilibrium is achieved in a closed flask by starting with the movement of molecules is
one direction, from liquid to vapor. The fraction of molecules with an energy greater than the
escape energy, as illustrated on a kinetic energy distribution curve, will move into the
vapor phase when they reach the liquid surface. The condensation rate is initially zero. As
the number of molecules in the vapor state increases, the condensation rate increases.
Simultaneously, the evaporation rate stays constant. As long as the rate of evaporation is
greater than the rate of condensation, the vapor concentration will rise. Eventually, the evaporation and condensation rates become equal, and equilibrium is achieved.
Goal 9 Describe the relationship
between vapor pressure and temperature for a liquid-vapor system In equilibrium; explain this relationship In terms of
the kinetic molecular theory.
The partial pressure exerted by a vapor in equilibrium with a liquid is the equilibrium vapor
pressure at the existing temperature. Equilibrium vapor pressure increases with increasing
temperature because the shape of the kinetic energy distribution curve changes. At higher
temperatures, a larger fraction of the liquid sample has enough energy to evaporate.
Goal 10 Describe the process of bolling
and the relationships among bolling
point, vapor pressure, and surrounding
pressure.
The boiling point of a liquid is the temperature at which the vapor pressure of that liquid is
equal to or slightly greater than the surrounding pressure. Normal boiling point is the boiling
point at one atmosphere of pressure.
Goal 11 Describe the typical relative
density relationship between the solid
and liquid phase of a substance, and
explain why water is an exception to
this trend.
Water, a molecule necessary for life, breaks almost all the rules for predicting physical
properties of liquids due to its extremely strong hydrogen bonding, which leads to exceptionally strong intermolecular attractions among its molecules. Some of the unusual properties of water include the following: an anomalously high boiling point, high surface tension, high heat of vaporization, low vapor pressure, high viscosity, an exceptional ability as a
solvent, and its unusual state (liquid) at common temperatures and pressures. The fact that
solid water (ice) floats on liquid water is also unusual; for most other substances, the solid
phase is denser than the liquid phase.
Goal 12 Distinguish between amorphous and crystalline solids.
Solids can be classified based upon particle arrangement. A crystalline solid has its particles arranged in a repeating pattern. An amorphous solid has no long-range order among
its particles.
Goal 13 Distinguish among the following types of crystalline solids: Ionic,
molecular, covalent, and metallic.
Crystalline solids can be classified based upon the forces that hold the particles together:
Ionic crystals are composed of oppositely charged ions that are held together by strong
ionic bonds. Ionic crystals typically have a high melting temperature, they are frequently
water soluble, and they have very low electrical conductivities.
Molecular crystals are made of small, discrete molecules held together by relatively weak
intermolecular forces. Molecular crystals are typically soft, have a low melting temperature,
and are generally insoluble in water. They are usually nonconductors.
Covalent network solids are composed of atoms that are covalently bonded to each other to
form a single, indefinite-sized network. Covalent network solids are almost always insoluble in
any common solvent, are poor conductors of electricity, and they have high melting points.
Metallic crystals are made of a crystal lattice of positive ions through which valence electrons move freely. The freely moving electrons make metals excellent conductors of electricity. Metallic crystals are insoluble in common solvents, malleable, and ductile (can be
pressed into thin sheets and drawn into wire), and they have a wide range of melting points.
Goal 14 Given two of the following, calculate the third: (a) mass of a pure substance changing between the liquid and
vapor (gaseous) states; (b) heat of vaporization; (c) energy change.
In the change between a liquid and a gas, vaporization is endothermic and condensation
is exothermic. The energy required to vaporize a substance, q, is proportional to the amount
of substance: q = ~Hvap x m. The proportionality constant, ~Hvap• is called the heat of
vaporization. It is the energy transferred when one gram of a substance changes between
the liquid and gaseous states.
Goal 15 Given two of the following, calculate the third: (a) mass of a pure substance changing between the solid and
liquid states; (b) heat of fusion; (c)
energy change.
It takes energy to melt a solid, an endothermic change. Energy is transferred out of a liquid
when it freezes, an exothermic change. The energy required to melt a substance, q, is proportional to the amount of substance: q = ~Htus x m. The proportionality constant, ~Htus• is
called the heat of fusion. It is the energy transferred when one gram of a substance
changes between the liquid and solid states.
Goal 16 Given three of the following
quantities, calculate the fourth: (a)
energy change; (b) mass of a pure substance; (c) specific heat of the substance; (d) temperature change, or Initial
and final temperatures.
The heat flow, q, in heating or cooling a substance is proportional to both the mass of the
sample, m, and its temperature change, Lff: q = m x c x ~T. The proportionality constant,
c, is a property of a pure substance called its specific heat. It is the amount of energy
needed to change the temperature of one gram of a substance one degree Celsius.
694
Chapter Summaries
Goal 17 Sketch, interpret, or identify
regions in a graph of temperature versus
energy for a pure substance over a temperature range from below the melting
point to above the boiling point.
Goal 18 Given (a) the mass of a pure
substance, (b) l1Hvap and/or l1H 1us of
the substance, and (c) the average specific heat of the substance in the solid,
liquid, and/or vapor state, calculate the
total heat flow in going from one state
and temperature to another state and
temperature.
A graph of temperature versus heat energy for a pure substance has five sections: (1) a positively sloped section representing warming or cooling the solid, (2) a zero-slope section
representing the solid melting or the liquid freezing, (3) a positively sloped section representing warming or cooling the liquid, (4) a zero-slope section representing the liquid vaporizing or the gas condensing, and (5) a positively sloped section representing warming or
cooling the gas.
To calculate the total heat flow for a change in temperature plus a change in state:
1. Sketch a graph of temperature versus heat energy for the substance. Mark the
beginning and ending point for each phase of matter and each change of state.
2. Calculate the heat flow, q, for each sloped and horizontal portion of the graph
between the starting and ending points.
3. Add the heat flows calculated in Step 2.
CHAPTER 16 IN REVIEW: Solutions
Goal 1 Define the term solution, and,
given a description of a substance,
determine if it is a solution.
A solution is a homogeneous mixture. Solutions have variable physical properties that
depend on the composition of the mixture.
Goal 2 Distinguish among terms in the
following groups: solute and solvent;
concentrated and dilute; solubility, saturated, unsaturated, and supersaturated;
miscible and immiscible.
The major component of a solution is called the solvent; the minor components are called
the solutes. A concentrated solution has a relative large quantity of solute per quantity of
solvent; a dilute solution has a relatively small quantity of the same solute per quantity of
solvent. A solution whose concentration is at the solubility limit for a given temperature is a
saturated solution. If the concentration is less than the solubility limit, the solution is unsaturated. A supersaturated solution has a concentration greater than the normal solubility
limit. Liquids are miscible if they dissolve in each other in all proportions. Liquids that are
insoluble in each other are immiscible.
Goal 3 Describe the formation of a saturated solution from the time excess solid
solute is first placed into a liquid solvent.
When a soluble ionic solid solute is placed in water, polar water molecules surround the
ions, helping them move away from their positions in the crystal. In solution, the ions are
hydrated, or surrounded by water molecules. The dissolving rate is constant throughout
the process. When dissolving has just begun, the crystallization rate is zero, and it
increases until it equals the dissolving rate and a dynamic equilibrium is established.
Goal 4 Identify and explain the factors
that determine the time required to dissolve a given amount of solute or to
reach equilibrium.
A finely divided solid dissolves more rapidly because of greater surface area. Stirring or
agitating a solution makes it dissolve more rapidly because of prevention of buildup of concentration at the solute surface. Higher temperature causes a solution to dissolve more
rapidly because of faster particle movement.
Goal 5 Given the structural formulas
of two molecular substances, or other
information from which the strength of
their intermolecular forces may be
estimated, predict if they will dissolve
appreciably in each other, and state
the criteria on which your prediction
is based.
Intermolecular forces fall into three categories: induced dipole forces, dipole forces, and
hydrogen bonds. Substances with similar intermolecular forces will usually dissolve in
one another.
Goal 6 Predict how and explain why the
solubility of a gas in a liquid is affected
by a change in the partial pressure of
that gas over the liquid.
In most dilute solutions, the solubility of a gas is directly proportional to the partial
pressure of the gas over the surface of the liquid.
Goal 7 Given the mass of solute and of
solvent or solution, calculate percentage
concentration by mass.
.
% concentration by mass
Goal 8 Given mass of solution and percentage concentration by mass, calculate mass of solute and solvent.
.
mass solute
.
mass solution x
.
= mass solute; mass solvent = mass solution - mass solute
mass so 1ut1on
Goal 9 Given two of the following, calculate the third: moles of solute (or data
from which it may be found), volume of
solution, molarity.
Molarity is the number of moles of solute in one liter of solution. Units of molarity are moles
per liter, or mol/L. The symbol for molarity is M.
M =moles solute
liter solution
mol
L
=
mass solute
.
x 100%
mass solution
=
mass solute
x 100%
mass solute+ mass solvent
Chapter Summaries
Goal 10 Given two of the following, calculate the third: moles of solute (or data
from which it may be found), mass of
solvent, molality.
695
Molality is moles of solute in one kilogram of solvent. The symbol for molality is m (note that
molality is lower case m and molarity is uppercase M).
mol solute
m=---kg solvent
Molality is used in situations where temperature independence is important. Neither the number
of solute particles nor the mass of the solvent varies with temperature. In contrast, molarity is
temperature dependent because the volume of a solution varies with temperature.
Goal 11 Given an equation for a neutralization reaction, state the number of
equivalents of acid or base per mole and
calculate the equivalent mass of the acid
or base.
One equivalent of an acid is defined as that amount of acid that yields one mole of hydrogen ion in a specific reaction. One equivalent of a base is the amount of base that reacts
with one equivalent of an acid. The equivalent mass of a substance is the number of grams
of the substance per equivalent. To calculate equivalent mass, divide molar mass by equivalents per mole:
g/mol
g
eq/mol
mol
--- = --
Goal 12 Given two of the following, calculate the third: equivalents of acid or
base (or data from which they may be
found), volume of solution, normality.
mol
x -eq
g
.
= - , equivalent mass
eq
Normality is the number of equivalents of solute in one liter of solution. Units of normality
are equivalents per liter, or eq/L. The symbol for normality is N.
equivalents solute
N=-------liter solution
eq
L
Goal 13 Given any three of the following, calculate the fourth: (a) volume of
concentrated solution, (b) molarity
of concentrated solution, (c) volume
of dilute solution, (d) molarity of dilute
solution.
The key relationship for calculations involving dilution of concentrated solutions is
Goal 14 Given the quantity of any species participating in a chemical reaction
for which the equation can be written,
find the quantity of any other species,
either quantity being measured in (a)
grams, (b) volume of solution at specified molarity, or (c) (if gases have been
studied) volume of gas at given temperature and pressure.
For all stoichiometry problems, a macroscopic measurable quantity (mass, energy, volume of a gas at known pressure and temperature, or volume of a solution of known concentration) is changed to the number of particles, grouped in moles. The mole ratio from
the balanced chemical equation is then used to change to the number of particles of
another species involved in the chemical change. Finally, the number of moles is changed
to a macroscopic measurable quantity. For solution stoichiometry, volume of solution and
concentration, typically molarity, can be used to convert to number of moles.
Goal 15 Given the volume of a solution
that reacts with a known mass of a
primary standard and the equation for
the reaction, calculate the molarity of
the solution.
Titration is the controlled and measured addition of one solution into another. Titration
problems are solution stoichiometry problems. A buret is a device that measures delivered
volumes precisely. An indicator is a substance that exhibits different color in solution at different solution acidities. A solution is standardized when its concentration is determined by
reaction with a substance that can be weighed accurately, which is called a primary
standard.
Goal 16 Given the volumes of two solutions that react with each other in a titration, the molarity of one solution, and the
equation for the reaction or information
from which it can be written, calculate
the molarity of the second solution.
Once a solution is standardized, we may use it to find the concentration of another solution.
Volume times molarity for the solution of known concentration yields amount in moles for
that solute. A balanced chemical equation is used to convert to moles of the other species.
The definition of molarity, moles per liter, is used to calculate the molarity of the second
solution.
Goal 17 Given the volume of a solution
that reacts with a known mass of a
primary standard and the equation for
the reaction, calculate the normality of
the solution.
In a reaction, the number of equivalents of acid and base that react with each other
are equal. This idea of equal numbers of equivalents is the basis of the normality system.
There are two ways to calculate the number of equivalents in a sample of a substance:
Mc
x Ve=
Md
x Vd, in which Mis molarity, Vis volume, c is concentrated and dis dilute.
1. If you know the mass of the substance and its equivalent mass, use the equivalent mass
as a conversion factor to get equivalents.
2. If the sample is a solution and you know its volume and normality, multiply one by the
other: V x N = eq.
Goal 18 Given the volumes of two solutions that react with each other in a titration and the normality of one solution,
calculate the normality of the second
solution.
For an acid-base titration, Vacid x Nacid = Vbase x Nbase·
696
Chapter Summaries
Goal 19 Given (a) the molality of a solution, or data from which it may be found,
(b) the normal freezing or boiling point of
the solvent, and (c) the freezing or boiling point constant, find the freezing or
boiling point of the solution.
The properties of a solution that depend only on the number of solute particles present,
without regard to their identity, are called colligative properties. Freezing point depression and boiling point elevation are colligative properties that are directly proportional to
the molal concentration of any solute. These proportionality constants are called the molal
freezing point constant and the molal boiling point constant, respectively. The values of
these constants depend only on the chemical identity of the solvent in the solution. For
freezing point depression, Lff1 = K1 m. For boiling point elevation, /1T b = Kbm.
Goal 20 Given the freezing point
depression or boiling point elevation and
the molality of a solution, or data from
which they may be found, calculate the
molal freezing point constant or molal
boiling point constant.
K1 and Kb, are, respectively, the molal freezing point depression constant and the molal
.1T1
.:iTb
boiling point elevation constant: K1 = - and Kb = - .
Goal 21 Given (a) the mass of solute and
solvent in a solution, (b) the freezing
point depression or boiling point elevation, or data from which they may be
found, and (c) the molal freezing/boiling
point constant of the solvent, find the
approximate molar mass of the solute.
Freezing point depression experiments may be used to determine molar mass. To calculate
the molar mass of a solute from freezing point depression or boiling point elevation data,
m
m
1. Calculate molality from m = D.T/K1 or m = D.T b/Kb. Express as mol solute/kg solvent.
2. Using molality as a conversion factor between moles of solute and kilograms of solvent,
find the number of moles of solute.
3. Use the defining equation for molar mass, MM
the solute.
=
g/mol, to calculate the molar mass of
CHAPTER 17 IN REVIEW: Acid-Base (Proton Transfer) Reactions
Goal 1 Distinguish between an acid and
a base according to the Arrhenius theory
of acids and bases.
An Arrhenius acid is a substance that increases the concentration of hydrogen ions in solution when dissolved in water, and an Arrhenius base is a substance that increases the concentration of hydroxide ions when dissolved in water.
Goal 2 Given the equation for a Br0nstedLowry acid-base reaction, explain how
or why it can be so classified.
According to the Brensted-Lowry theory, an acid-base reaction involves a transfer of a
proton from one substance, the acid, to another, the base. An acid is a proton source; a
base is a proton remover. When writing an equation, there must be a Bmnsted-Lowry base
whenever there is a Bmnsted-Lowry acid.
Goal 3 Given the formula of a BrenstedLowry acid and the formula of a BrenstedLowry base, write the net ionic equation
for the reaction between them.
Acid-base reactions are reversible; they reach a state of equilibrium according to the general equation HA+ B ~A-+ HB+.
Goal 4 Distinguish between a Lewis
acid and a Lewis base. Given the structural formula of a molecule or ion, state if
it can be a Lewis acid, a Lewis base, or
both, and explain why.
According to the Lewis theory of acids and bases, an acid is an electron-pair acceptor and
a base is an electron-pair donor.
Goal 5 Given the structural equation for
a Lewis acid-base reaction, explain how
or why it can be so classified.
Structurally, the most common feature of the Lewis acids introduced in this course is that
they tend to be positive ions or species with a central atom with less that a full octet of
valence electrons. Lewis bases tend to be negative ions or species with central atoms with
one or more unshared electron pairs.
Goal 6 Define and identify conjugate
acid-base pairs.
Two substances whose formulas differ only by a proton (a hydrogen ion) are a conjugate
acid-base pair.
Goal 7 Given the formula of an acid or a
base, write the formula of its conjugate
base or acid.
If an acid has the general form HA, its conjugate base has the general form A-. If a base has
the general form B, its conjugate acid has the general form HB+.
Goal 8 Given a table of the relative
strengths of acids and bases, arrange a
group of acids or a group of bases in
order of increasing or decreasing
strength.
The more readily protons are surrendered, the stronger the acid. The stronger the ability to
remove protons is, the stronger the base will be. The relative strengths of Bremsted-Lowry
acids and bases may be decided from their positions in a table of relative strengths of acids
and bases.
Goal 9 Given the formulas of a potential
Brensted-Lowry acid and a BrenstedLowry base, write the equation for the
possible proton transfer reaction
between them.
Proton transfer reactions involve two conjugate acid-base pairs. Strong acids release
protons readily; weak acids do not. Strong bases remove protons readily; weak bases do
not. When a proton transfer reaction reaches equilibrium, the proton transfer yields the
weaker conjugate acid and the weaker conjugate base.
Chapter Summaries
Goal 10 Given a table of the relative
strengths of acids and bases and information from which a proton transfer
reaction equation between two species
in the table may be written, write the
equation and predict the direction in
which the reaction will be favored.
Goal 11 (If Chapter 19 has been studied)
Compare and contrast acid-base reactions with redox reactions.
697
To predict the favored direction of an acid-base reaction, do the following:
1. For a given pair of reactants, write the equation for the transfer of one proton from
one species to the other.
2. Label the acid and base on each side of the equation.
3. Determine which side of the equation has both the weaker acid and the weaker
base. That side identifies the products in the favored direction.
Acid-base reactions resemble redox reactions in the following ways:
1. Transfer of a subatomic particle: An acid-base reaction is a transfer of protons; a
redox reaction is a transfer of electrons.
2. Special names for the species involved in the transfer: An acid is a proton source;
a base is a proton remover. A reducing agent is an electron source; an oxidizing
agent is an electron remover.
3. Species that can both remove and release subatomic particles, depending on relative strength.
4. Varying strengths of abilities to remove and release subatomic particles.
5. Most systems are equilibrium reactions; the favored direction of reaction can be
predicted by comparison of relative strengths.
Goal 12 Given the hydrogen or hydroxide ion concentration of water or a water
solution, calculate the other value.
Water itself is both a weak acid and a weak base. Kw = [H+][QH-] = 1.0 x 10-14 at 25°C. A
neutral water solution has pH = 7.00 at 25°C. An acidic water solution has pH less than 7.00
at 25°C; a basic water solution has pH greater than 7.00 when measured at 25°C.
Goal 13 Given any one of the following,
calculate the remaining three: hydrogen
or hydroxide ion concentration
expressed as 10 raised to an integral
power or its decimal equivalent, pH, and
pOH expressed as an integer.
Because concentrations of H+(aq) or OH-(aq) are usually small, but can vary over wide
ranges, they are usually expressed in scientific notation and as logarithms. pH = -log [H+]
and pOH = -log [OH-]. From these equations, you can obtain [H+] = 10-pH and [OH-] =
10-poH. In water solutions at 25°C, pH + pOH = 14.00.
Goal 14 Given any one of the following,
calculate the remaining three: hydrogen
ion concentration, hydroxide ion concentration, pH, and pOH.
The new skills needed to complete this optional section are not chemical, but mathematical.
The ideas concerning pH, pOH, [W], and [OH-] are the same as in Section 17-9; only the
numbers have been changed. Note that in a logarithm, the digits to the left of the decimal
are not counted as significant figures. Counting significant figures in a logarithm begins at
the decimal point.
CHAPTER 18 IN REVIEW: Chemical Equilibrium
Goal 1 Identify a chemical equilibrium
by the conditions it satisfies.
Four conditions characterize every equilibrium:
1. The change is reversible and can be represented by an equation with a double
arrow.
2. The equilibrium system is "closed"-closed in the sense that no substance can
enter or leave the immediate vicinity of the equilibrium.
3. The equilibrium is dynamic.
4. The things that are equal in an equilibrium are the forward rate of change (from left
to right in the equation) and the reverse rate of change (from right to left).
Goal 2 Distinguish between reactionproducing molecular collisions and
molecular collisions that do not yield
reactions.
According to the collision theory of chemical reactions, all chemical reactions start with
molecular collisions, but not all molecular collisions give a chemical reaction. The particles
must have (1) enough kinetic energy and (2) the proper orientation.
Goal 3 Sketch and/or interpret an
energy versus reaction progress graph.
Identify the (a) transition state region,
(b) activation energy, and (c) f1E for
the reaction.
An energy versus reaction progress graph shows potential energies of reactants, the
transition state, and products in a reaction, and the activation energy as the reaction proceeds in either direction, as well as f1E for the reaction. The graph illustrates the activation
energy, Ea, for the forward and reverse reactions: the minimum kinetic energy needed to
produce an effective collision.
Goal 4 State and explain the relationship between reaction rate and
temperature.
Reaction rates are higher at higher temperatures because a larger fraction of the sample has
enough kinetic energy to participate in reaction-producing collisions. This may be illustrated
with a kinetic energy distribution curve. The energy of collision must be enough to overcome the mutual repulsion of the valence electrons of the reacting particles.
698
Chapter Summaries
Goal 5 Using an energy versus reaction
progress graph, explain how a catalyst
affects reaction rate.
A catalyst increases reaction rate by providing an alternative reaction path with a lower
activation energy. A larger fraction of the molecules in a sample are able to enter into reactionproducing collisions, so the reaction rate increases.
Goal 6 Identify and explain the relationship between reactant concentration
and reaction rate.
Collision rates are higher at higher concentrations, so reaction rates are higher at higher
reactant concentrations.
Goal 7 Trace and explain the changes in
concentrations of reactants and products that lead to a chemical equilibrium.
For any reversible reaction in a closed system, whenever the opposing reactions are occurring
at different rates, the faster reaction will gradually become slower, and the slower reaction will
become faster. Finally, the reaction rates become equal, and equilibrium is established.
Goal 8 Given the equation for a chemical equilibrium, predict the direction in
which the equilibrium will shift because
of a change in the concentration of one
species.
Le Chatelier's Principle states that if an equilibrium system is subjected to change, processes occur that tend to partially counteract the initial change, thereby bringing the system
to a new position of equilibrium. When the concentration of a species in an equilibrium reaction is changed, a shift will occur in the direction that tries to return the substance disturbed
to its original condition.
Goal 9 Given the equation for a chemical equilibrium involving one or more
gases, predict the direction in which the
equilibrium will shift because of a
change in the volume of the system.
If a gaseous equilibrium is compressed, the increased pressure will be partially relieved by a
shift in the direction of fewer gaseous molecules; if the system is expanded, the reduced
pressure will be partially restored by a shift in the direction of more gaseous molecules.
Goal 10 Given a thermochemical equation for a chemical equilibrium, or information from which it can be written,
predict the direction in which the
equilibrium will shift because of a
change in temperature.
A thermochemical equation is one that includes a change in energy. Including the energy
term in a thermochemical equation, rather than showing b.rH separately, makes it easier to
predict the Le Chatelier effect of a change in temperature. In the equation, think of energy as
you would a substance being added or removed. An increase in temperature is interpreted
as the "addition of heat," and a lowering of temperature is the "removal of heat."
Goal 11 Given any chemical equilibrium
equation, or information from which it
can be written, write the equilibrium constant expression.
An equilibrium system can be described by an equilibrium constant, K. The constant K is a
concentration ratio; the form of the ratio depends on how the equilibrium equation is written.
For the general reaction a A+ b B
~ c C + d D, K =
[Cl [DJ:. When writing an equilibrium
[AJa [BJ
constant expression, use only the concentrations of gases, (g), or dissolved substances,
(aq). Do not include solids, (s), or liquids, (f).
Goal 12 Given an equilibrium equation
and the value of the equilibrium constant, identify the direction in which the
equilibrium is favored.
If an equilibrium constant is very large (> 100), the forward reaction is favored; if the constant is very small(< 100), the reverse reaction is favored. If the constant is neither large nor
small, appreciable quantities of all species are present at equilibrium.
Goal 13 Given the solubility product
constant or the solubility of a slightly
soluble compound (or data from which
the solubility can be found), calculate
the other value.
No ionic compound is completely insoluble. The equilibrium constant for a low-solubility
compound is the solubility product constant, Ksp- The solubility product constant for any
ionic compound has the form
The first two steps in solving any solubility product constant problem are the same:
1. Write the equilibrium equation for the reaction.
2. Write the solubility product constant expression.
To find Ksp from solubility:
Determine the molar concentrations of the ions in solution.
Substitute into the K5 P expression and solve.
To find solubility from K 5 P:
3. Assign a variable to represent one of the ionic species in the equilibrium;
4. Determine the concentration of the other ionic species in terms of the same variable;
5. Substitute the concentrations from Steps 3 and 4 into the Ksp expression, equate to
the Ksp value, and solve.
Goal 14 Given the solubility product
constant of a slightly soluble compound
and the concentration of a solution having a common ion, calculate the solubility of the slightly soluble compound in
the solution.
The solubility of a low-solubility substance is reduced when a common ion-one already
present in the solution-is introduced from another source. This is called the common ion
effect.
Chapter Summaries
Goal 15 Given the formula of a weak
acid, HA, write the equilibrium equation
for its ionization and the expression for
its acid constant, Ka.
699
If HA is the formula of a weak acid, its ionization equation and equilibrium constant expression are
HA(aq)
~
W(aq) + A-(aq)
The equilibrium constant is the acid constant, Ka. The undissociated molecule is the major
species in the solution and the W ion and the conjugate base of the acid, A-, are the minor
species. Weak acids ionize only slightly when dissolved in water. The ionization of a weak
acid is usually so small that it is negligible compared with the initial concentration of the
acid.
Goal 16 Given any two of the following
three values for a weak acid, HA, calculate the third: (a) the initial concentration of the acid; (b) the pH of the solution, the percentage dissociation of
the acid, or [W] or [A-] at equilibrium;
(c) Ka for the acid.
Goal 17 For a weak acid, HA, given Ka,
[W], and [A-], or information from which
they may be obtained, calculate the pH
of the buffer produced.
Percentage ionization, as all other percentage concepts, is the ratio of the part to the
whole, multiplied by 100:
)1'. •
•
t'
amount of solute ionized x
)/'.
070
100070.
1oniza ion =
tota 1 so 1ute present
To determine Ka from pH or percent ionization of an acid with a given concentration,
determine the values of [H+] and [A-] with [H+] = 10-pH, substitute the values into the Ka
expression, and solve. To determine percent ionization and pH from known molarity and
Ka, if the ionization of the acid is the only source of H+ and A- ions, start with the acid
constant equation, multiply both sides by [HA], and substitute [H+] for its equal [A-]:
Ka [HA] = [W][A-] = [W]2;
[W] = YKa [HA].
A buffer is a solution that resists changes in pH because it contains relatively high concentrations of both a weak acid and a weak base. The acid is able to consume any oH- that
may be added, and the base can react with H+, both without significant change in either
[HA] or [A-]. To find the pH of a buffer solution, solve the acid constant expression for [W]:
[HA]
[W] =Ka x [AT
Goal 18 Given Ka for a weak acid,
HA, determine the ratio between [HA]
and [A-] that will produce a buffer of
specified pH.
Goal 19 Given equilibrium concentrations of species in a gas phase equilibrium, or information from which they
can be found, and the equation for the
equilibrium, calculate the equilibrium
constant.
A buffer can be tailor-made for any pH simply by adjusting the [HA]/[A-] ratio to the proper
.
h
'd
. f
h
. .
[HA] [W]
va Iue. So Iv1ng t e ac1 constant expression or t at ratio gives [A-]
=Ka.
When an equilibrium involves only gases, the changes in the starting concentrations are not
negligible. It often helps to trace these changes by assembling them into a table. The columns
are headed by the species in the equilibrium just as they appear in the reaction equation. The
three lines give the initial concentration of each substance, the change in concentration as
the system reaches equilibrium, and the equilibrium concentration:
aA
+
bB
cc
+
dD
mol/L at start
mol/L change, + or mol/L at equilibrium
CHAPTER 19 IN REVIEW: Oxidation-Reduction (Electron
Transfer) Reactions
Goal 1 Describe and explain oxidation
and reduction in terms of electron
transfer.
Oxidation is a loss of electrons or an increase in oxidation number. Reduction is a gain of
electrons or a reduction in oxidation number.
Goal 2 Given an oxidation half-reaction
equation and a reduction half-reaction
equation, combine them to form a
net ionic equation for an oxidationreduction reaction.
Oxidation-reduction (redox) reactions can be divided into an oxidation half-reaction equation and a reduction half-reaction equation. Addition of these equations gives a balanced
equation for a redox reaction.
Goal 3 Distinguish among electrolytic
cells, voltaic cells, and galvanic cells.
A voltaic cell is a cell in which an electrical potential is developed by a spontaneous chemical change. It is also called a galvanic cell. An electrolytic cell is a cell in which electrolysis occurs as a result of an externally applied electrical potential.
700
Chapter Summaries
Goal 4 Describe and identify the parts
of an electrolytic or voltaic (galvanic) cell
and explain how it operates.
An electrolytic cell is made up of a container holding an ionic solution called an electrolyte
and two electrodes. Ions in the electrolyte move to oppositely charged electrodes, where
chemical reactions occur. The anode is the electrode at which oxidation occurs; reduction
occurs at the cathode. In a voltaic cell, chemical changes occur at the electrodes, causing
electricity to flow in an outside circuit. The solutions in each half-cell are connected by a
salt bridge.
Goal 5 Given the formula of an element,
molecule, or ion, assign an oxidation
number to each element in the formula.
Oxidation numbers are assigned as follows: an element is O; a monatomic ion is the charge
on the ion; combined oxygen is -2 (peroxides, -1; superoxides, -1/2; OF2 , +2); combined
hydrogen is -1 (H-, -1); the sum of oxidation numbers of all atoms in a polyatomic species
is equal to its charge.
Goal 6 Describe and explain oxidation
and reduction in terms of change in oxidation numbers.
Oxidation is an increase in oxidation number. Reduction is a decrease in oxidation number.
Goal 7 Given a redox equation, identify
the oxidizing agent, the reducing agent,
the element oxidized, and the element
reduced.
The species that removes electrons in a redox reaction, that is, the species that is itself
reduced, is referred to as an oxidizing agent. The species from which the electrons are
removed so reduction of another element can occur is called a reducing agent. Thus, the
reducing agent is itself oxidized.
Goal 8 Distinguish between strong and
weak oxidizing agents.
A strong oxidizing agent has a strong attraction for electrons. A weak oxidizing agent
attracts electrons only slightly. A strong reducing agent releases electrons readily. A weak
reducing agent holds on to its electrons.
Goal 9 Given a table of the relative
strengths of oxidizing and reducing
agents, arrange a group of oxidizing
agents or a group of reducing agents in
order of increasing or decreasing
strength.
Table 19-2 lists oxidizing agents in order of decreasing strength on the left side of the halfreaction equation and lists reducing agents in order of increasing strength on the right side.
Goal 10 Given a table of the relative
strengths of oxidizing agents, and
information from which an electron
transfer reaction equation between two
species in the table may be written,
write the equation and predict the
direction in which the reaction will be
favored.
When a reversible reaction equation is read from left to right, the forward reaction, or the
reaction in the forward direction, is described; from right to left, the change is the reverse
reaction, or in the reverse direction. A strong oxidizing agent takes electrons from a
strong reducing agent to produce weaker oxidizing and reducing agents. The reaction is
said to be favored in the direction pointing to the weaker oxidizing and reducing agents.
Use Table 19-2 to identify the stronger and weaker oxidizing and reducing agents.
Goal 11 Compare and contrast redox
reactions with acid-base reactions.
Acid-base reactions resemble redox reactions in the following ways:
1. Transfer of a subatomic particle: An acid-base reaction is a transfer of protons; a
redox reaction is a transfer of electrons.
2. Special names for the species involved in the transfer: An acid is a proton source; a
base is a proton remover. A reducing agent is an electron source; an oxidizing
agent is an electron remover.
3. Species that can both remove and release subatomic particles, depending on
relative strength.
4. Varying strengths of abilities to remove and release subatomic particles.
5. Most systems are equilibrium reactions; the favored direction of reaction can be
predicted by comparison of relative strengths.
Goal 12 Given the before and after formulas of species containing elements
that are oxidized and reduced in an
acidic solution, write the oxidation and
reduction half-reaction equations and
the net ionic equation for the reaction.
To write a redox equation for a half-reaction in acidic solution:
1. After identifying the element oxidized or reduced, write a skeleton half-reaction
equation with the element in its original form on the left and in its final form on
the right.
2. Balance the element oxidized or reduced.
3. Balance elements other than hydrogen or oxygen, if any.
4. Balance oxygen by adding water molecules when necessary.
5. Balance hydrogen by adding H+ when necessary.
6. Balance charge by adding electrons to the more positive side.
7. Recheck the equation to be sure it is balanced in both atoms and charge.
Chapter Summaries
701
CHAPTER 20 IN REVIEW: Nuclear Chemistry
Goal 1 Define and describe
radioactivity.
Radioactivity is the spontaneous emission of particles or electromagnetic radiation resulting from the decay, or breaking up, of an atomic nucleus.
Goal 2 Name, identify from a description, or describe three types of radioactive emissions.
Three types of natural radioactivity, alpha (a), beta {fl}, and gamma ('y) rays, can be formed
when a nucleus decays. Alpha particles are much more massive than beta particles, and
gamma rays have no mass. Alpha particles have a positive charge twice as large as the
magnitude of the negative charge on beta particles; gamma rays have no charge. Penetrating power increases in the sequence alpha < beta < gamma.
Goal 3 Identify the function of a Geiger
counter and describe how it operates.
The hand piece of a Geiger counter has a thin window through which ionizing radiation
passes, entering a tube filled with argon gas. The gas is momentarily ionized, and the ions
complete a circuit between two electrodes. The signal is amplified to produce clicking from a
speaker and register on a meter. The frequency of the clicks indicates the radiation intensity.
Goal 4 Explain how exposure to radiation may harm or help living systems.
Radioisotopes are atoms with an unstable nucleus. The harmful effects of radiation on living
systems come from its ability to break chemical bonds and thereby destroy healthy tissue.
The key to radiation therapy is selective radiation of unhealthy tissue, which can eliminate it or
reduce it.
Goal 5 Describe or illustrate what is
meant by the half-life of a radioactive
substance.
The rate at which a radioactive substance decays is measured by its half-life, the time it
takes for one half of the radioactive atoms in a sample to decay.
Goal 6 Given the starting quantity of a
radioactive substance, Figure 20-11, and
two of the following, calculate the third:
half-life, elapsed time, quantity of isotope remaining.
If S is the starting quantity of a radioactive substance and R is the amount that remains after
n half-lives, then R = 5 x (1/2)". Figure 20-11 is a half-life decay curve for a radioactive substance, a plot of fraction of substance remaining versus half-lives elapsed.
Goal 7 Describe a natural radioactive
decay series.
Products of radioactive decay may be other nuclei that undergo further decay, forming a
natural radioactive decay series. When a radioisotope emits an alpha or beta particle,
there is a transmutation of an element, that is, a change from one element to another. A
decay series ends when a stable nucleus is formed.
Goal 8 Given the identity of a radioactive
isotope and the particle it emits, write a
nuclear equation for the emission.
An equation for radioactive decay is balanced for nuclear charge (number of protons) and
nuclear mass (number of protons and neutrons). Emitted particles may be alpha particles,
~He, or beta particles, _01e.
Goal 9 List or identify four ways in
which nuclear reactions differ from ordinary chemical reactions.
1. Chemical properties are the same for all isotopes of an element. Nuclear properties of the isotopes of an element are quite different.
There are four primary ways in which nuclear reactions differ from ordinary chemical reactions:
2. Radioactivity is independent of the state of chemical combination of the radioactive isotope.
3. Nuclear reactions can result in the formation of different elements. In chemical
reactions, atoms keep their elemental identities.
4. The amount of energy per quantity of reactant for nuclear changes is much larger
than for chemical changes.
Goal 10 Define or identify nuclear bombardment reactions.
A nuclear bombardment reaction occurs when one nuclide is bombarded, or continuously
exposed to a stream of particles, by another.
Goal 11 Distinguish natural radioactivity
from induced radioactivity produced by
bombardment reactions.
Induced or artificial radioactivity is generated when a stable nuclide is made radioactive
by combination with another nuclide.
Goal 12 Define or identify transuranium
elements.
The elements having atomic numbers greater than 92 are called the transuranium elements. All of these elements are radioactive.
Goal 13 Identify and describe uses for
synthetic radioisotopes.
Synthetic radionuclides have many uses, including medical applications; ionizing air in a
smoke detector; killing bacteria, molds, and yeasts to preserve food; and industrial and scientific applications.
Goal 14 Define or identify a nuclear
fission reaction.
A nucleus that splits into lighter nuclei undergoes nuclear fission.
Goal 15 Define or identify a chain
reaction.
A chain reaction occurs when a product of one reaction is a reactant in the next step of the
reaction pathway, thereby continuing the process. The minimum quantity of fissionable isotope required for this purpose is called the critical mass.
702
Chapter Summaries
1S Describe how a nuclear power
plant differs from a fossil-fueled power
plant.
The turbine, generator, and condenser in a nuclear power plant are similar to those
found in any fuel-burning power plant. The primary difference is the source of energy
used to turn the turbines: nuclear fuel versus fossil fuel (or water from a dam flowing).
The nuclear fission reaction has three main components: the fuel elements, control rods,
and moderator.
Define or identify a nuclear
fusion reaction.
Two light nuclei that are joined to form a heavier nucleus undergo nuclear fusion.
CHAPTER 21 IN REVIEW: Organic Chemistry
The overarching goals for this chapter are the following:
A. Distinguish between organic and inorganic chemistry.
8. Define the term hydrocarbon.
C. Distinguish between saturated and unsaturated hydrocarbons.
D. Write, recognize, or otherwise identify (a) the structural unit, or functional group,
(b) the general formula, and (c) the molecular or structural formulas and/or names
of specific examples of the following classes of organic compounds: alkanes,
alkenes, alkynes, cycloalkanes, aromatic hydrocarbons, alcohols, ethers, aldehydes, ketones, carboxylic acids, esters, amines, and amides.
E. Define and give examples of isomerism.
F. Define and give examples of monomers and polymers.
l Distinguish between organic and
inorganic compounds.
Organic compounds are those that contain carbon atoms. Inorganic compounds
have no carbon atoms; they are made up of elements other than carbon. Carbonates,
cyanides, and oxides of carbon contain carbon, but they are traditionally classified as
inorganic.
·.;;:.:ai 2. Given Lewis diagrams, balland-stick models, or space-filling
models of two or more organic molecules with the same molecular formula,
distinguish between isomers and different orientations of the same
molecule.
Compounds with the same molecular formula but different molecular structures are called
isomers.
J.oa.! 3 Distinguish between saturated
and unsaturated hydrocarbons {or other
compounds).
Hydrocarbons are made of carbon and hydrogen. A saturated hydrocarbon has only single bonds. An unsaturated hydrocarbon has one or more double or triple bonds between
carbon atoms.
·l ;,a:\ 4 Distinguish between alkanes,
alkenes, and alkynes {Sections 21-3
and 21-4).
The alkanes are a hydrocarbon family where each carbon atom forms single bonds to four
other atoms and there are no multiple bonds. The alkenes are hydrocarbons with at least
one double bond; the alkynes are hydrocarbons with at least one triple bond.
· xii s Given a formula of a hydrocarbon or information from which it can be
written, determine whether the compound can be a normal alkane or a
cycloalkane.
A hydrocarbon with all carbon atoms in the molecule in a continuous chain is a normal
alkane. The normal and branched alkanes have the general formula CnH 2 nt 2 , where n is the
number of carbon atoms in the compound. Cycloalkanes have all carbon-carbon single
bonds, with at least some of the carbon atoms forming a ring.
6 Given the name {or structural
diagram) of a normal, branched, or halogen-substituted alkane, write the structural diagram (or name).
To name an alkane:
1. Identify as the parent alkane the longest continuous chain. An alkane is named
by using a prefix to indicate the number of carbon atoms in the longest chain:
meth- = 1, eth- = 2, prop- = 3, but- = 4, pent- = 5, hex- = 6, hept- = 7, act- = 8,
non- = 9, dee- = 10. The suffix -ane denotes an alkane.
2. Removing an H atom from an alkane gives an alkyl functional group. Identify by
number the carbon atom to which the alkyl group (or other species) is bonded to
the chain.
3. Identify the branched group (or other species).
4. If the same alkyl group, or other species, appears more than once, indicate the
number of appearances by di-, tri-, tetra-, etc., and show the location of each
branch by number.
5. If two or more different alkyl groups, or other species, are attached to the parent
chain, they are named in alphabetical order.
Chapter Summaries
703
Goal 7 Given the name (or structural
diagram) of a cycloalkane, write the
structural diagram (or name).
Cycloalkanes are named according to the number of carbon atoms in the ring with the
prefix cyclo-.
Goal 8 Given a formula of a hydrocarbon or information from which it can
be written, determine whether the
compound can be an alkene or an
alkyne.
The alkenes are hydrocarbons with at least one double bond; the alkynes are hydrocarbons with at least one triple bond.
Goal 9 Given the name (or structural
diagram) of an alkene or an alkyne, write
the structural diagram (or name).
The IUPAC nomenclature system for the alkenes and alkynes matches that of the
alkanes. The suffix -ene denotes an alkene. The suffix -yne denotes an alkyne. In naming unsaturated compounds, the position of the double or triple bond is specified by
number.
Goal 10 Identify and distinguish
between cis and trans geometric
isomers.
Double bonds can give geometric isomers, also called cis-trans isomers. Two alkyl groups
can be on the same side (cis) or on opposite sides (trans) of the double bond.
Goal 11 Distinguish between aliphatic
and aromatic hydrocarbons.
Any hydrocarbon that does not contain a benzene ring is an aliphatic hydrocarbon. A
hydrocarbon with one or more benzene rings, which may be substituted, is an aromatic
hydrocarbon.
Goal 12 Given the name (or structural
diagram) of an alkyl- or halogensubstituted benzene compound, write
the structural diagram (or name).
The carbon on a benzene ring to which a functional group is bonded is identified by number.
Ortho-, meta-, and para- prefixes are also used for disubstituted rings.
Goal 13 Given the reactants in an
addition or substitution reaction
between (a) an alkane, alkene, alkyne,
or benzene and (b) a hydrogen or halogen molecule, predict the products of
the reaction.
An alkene or alkyne is capable of reacting via an addition reaction, where atoms of an element (or compound) are added to the unsaturated hydrocarbon. A reaction in which a
hydrogen atom in an alkane is replaced by an atom of another element is a substitution
reaction.
Goal 14 Identify the structural formulas
of the functional groups that distinguish
alcohols and ethers.
The general formula of an alcohol is R-OH. The general formula for an ether is R-0-R'.
alcohol
ether
Goal 15 Given the name (or structural
diagram) of an alcohol or ether, write the
structural diagram (or name).
The suffix -o/ denotes an alcohol. The word ether preceded by the names of the attached
alkyl groups is the name of the ether.
Goal 16 Given the reactants (or products) of a dehydration reaction between
two alcohols, predict the products (or
reactants) of the reaction.
Ethers can be prepared by dehydrating alcohols. The -H from the hydroxyl group of one
molecule reacts with the -OH hydroxyl group of another molecule, producing an R-0-R'
ether and a water molecule.
Goal 17 Given the molecular structures
of alcohols or ethers, or information from
which they may be obtained, predict relative values of boiling points or solubility
in water.
In general, the boiling points of alcohols increase as the number of carbon atoms increases.
The boiling point of an alcohol is always much higher than that of the alkane with the same
number of carbon atoms. The 1-to-3-carbon alcohols are completely soluble in water, and
the solubility drops off as the alkyl chain lengthens. The boiling points of ethers are lower
than the corresponding alcohols. The solubility of ethers in water is about the same as the
solubility of the isomeric alcohols.
Goal 18 Identify the structural formulas
of the functional groups that distinguish
aldehydes and ketones.
Aldehydes and ketones have a carbon atom double bonded to an oxygen atom. In an aldehyde, the carbon is at the end of the chain; in a ketone, the carbon is inside the chain.
0
II
Goal 19 Given the name (or structural
diagram) of an aldehyde or ketone,
write the structural diagram (or name).
0
II
c
R/C"'H
R/ "'R'
aldehyde
ketone
The IUPAC system uses the suffix -a/ for aldehydes and the suffix -one for ketones. Ketones
may also be named like ethers, but using the word ketone.
704
Chapter Summaries
Goal 20 Write structural diagrams to
show how a specified aldehyde or
ketone Is prepared from an alcohol.
Aldehydes and ketones are prepared by oxidation of alcohols or hydrations of alkynes.
Aldehydes are themselves easily oxidized; ketones resist oxidation. Aldehydes and ketones
may be reduced to alcohols.
Goal 21 Identify the structural formulae
of the functional groups that distinguish
carboxyllc acids and eeter1.
The general formula of a carboxylic acid is RCOOH. The functional group, -COOH, is a
combination of a carbonyl group and a hydroxyl group called a carboxyl group. In an ester,
the carboxyl hydrogen is replaced by another alkyl group, RCOOR'.
0
0
//
R-C
\
//
R-C
\
0-H
0-R'
ester
carboxylic acid
Goal 22 Given the name (or structural
diagram) of a carboxyllc acid or ester,
write the structural diagram (or name).
Goal 23 Given the reactants (or prod·
uct1) of an oatorlflcatlon reaction, predict
tho product• (or reactant•) of the
reaction.
Goal 24 Identify tho structural formulae
of tho functional groups that dl1tlngul1h
camlnoa and omldot.
Goal 25 Given tho nomo (or 1tructural
dlAgrAm) of An omlno or amide, wrlto tho
1tructural diagram (or namo).
IUPAC uses the suffix -oic and the word acid to denote carboxylic acids. Esters have two
word names. The first word is the alkyl group from the alcohol and the second is the anion
derived from the acid. (Remember -ic - -ate in anions of acids.)
The reaction between an acid and an alcohol is called esterification. The products of the
reaction are an ester and water. The acid contributes the entire hydroxyl group, while the
alcohol furnishes only the hydrogen.
Amines are organic derivatives of ammonia, NH 3 . An amine is formed by replacing one, two,
or three hydrogens in an ammonia molecule with an alkyl group. An amide is a derivative
of a carboxylic acid in which the hydroxyl part of the carboxyl group is replaced by an
-NH 2 group.
The IUPAC system names amines like ethers, but using the word amine. Name an amide by
replacing the -oic acid suffix with the word amide.
Goal 26 Qlvon tho 1tructural diagram of
An amino, or Information from which It
cAn bo written, cla11lfy tho amino H prl·
miry, 11cond1ry, or tertiary.
The number of ammonia hydrogens replaced determines whether an amine is primary (one
hydrogen replaced), secondary (two), or tertiary (three).
Goal 27 Qlvon tho rooctAnt1 of a roaction botwoon A c1rboxyllc 1cld 1nd
im Amino, prodlct tho product• of tho
rHctlon.
A carboxylic acid and an amine react by removing a water molecule:
0
H-N-R'
II
I
R-C-OH
+
0
~
H
I
R-C-N-R'
+
H20
I
H
Goal 28 Qivon tho 1tructural dl1gram or
nAmo of can othylono·llko monomer, pro·
diet tho 1tructurAI dlAQrAm of tho prod·
uet chcaln=growth polymon ;Ivon tho
1truotur1I diagrnm of 1 ohAln·growth
polymer, prodlet tho 1tructur1I dlAgr1m
And/or tho nAmo of tho othylono·llko
monomor from which it e1n bo formed.
Small molecules called monomers join together to form polymers. Chain-growth
polymers are formed by repeated addition reaction of an alkene monomer to give an
alkane-like polymer chain. For example, polyethylene is formed from ethylene
monomers:
H H
H H
I I
C=C
I I
of A dicArboxylic Acid and A dl1lcohol,
prodict tho 1truoturAI diAgram of tho
product 1ttm=;rowth polymer; given tho
1tnrnturAI diagrAm of ft 1topagrowth
polymor, prodiot tho 1trueturAI di1grAm1
of tho dicarboxylic Acid And dinleohol
from which it can bo formed.
I
I
I
H H
I
I
I
I
+ C=C + C=C
H H
Goal 29 Qivon tho 1tructur1I dlagr1m1
I
H H
H
H
I
-
I
H
I
H
I
H
H
I
I
-c-c-c-c-c-c1 I
H H
H
H
I
H
I
H
I
I
H
H
One type of step-growth polymer is formed by repeated condensation reactions to give a
polymer chain with repeated ester or amide functional groups. For example, nylon 66 is
formed from a 6-carbon dicarboxylic acid and a 6-carbon diamine:
H
+
0
II
0
I
H
I
H
I
H-N-(CH
I
2 ) 6-N-H
H
I
-C-(CH 2 ) 4 -C-N-(CH2)6-N-
+ H 20
-
Chapter Summaries
705
CHAPTER 22 IN REVIEW: Biochemistry
The overarching goals for this chapter are the following:
A. Identify, describe the distinguishing features of, and give an example of a molecule
in each of the four major classes of biological molecules: proteins, carbohydrates,
lipids, and nucleic acids.
B. Identify the monomers and describe how the polymeric molecules are assembled
in the following: proteins, carbohydrates, nucleic acids.
C. Describe how an enzyme functions as a biological catalyst.
D. Describe the process by which protein molecules are constructed from the information encoded in DNA molecules.
Goal 1 Given a Lewis diagram of a
polypeptide or information from
which it may be written, identify the
C-terminal amino acid and the
N-terminal amino acid.
An amino acid is a molecule that contains both an amine group and a carboxylic acid group.
The amino acid with the free carboxyl group at the end of a polypeptide chain is called the
C-terminal acid; the amino acid with the free amine group is called the N-terminal acid.
Goal 2 Given a table of Lewis diagrams
of amino acids and their corresponding
three-letter and one-letter abbreviations,
draw the Lewis diagram of a polypeptide
from its abbreviation.
The bond between amino acids is called a peptide linkage, which is formed when the
hydroxyl part of the carboxyl group of an amino acid molecule reacts with a hydrogen of the
-NH 2 group of another amino acid molecule to form a molecule of water.
Goal 3 Explain the meaning of the terms
primary, secondary, tertiary, and quaternary structure as they apply to proteins.
The primary structure of a protein is its sequence of amino acids. The secondary structure of a protein is the local spatial layout of the amino acid backbones. The tertiary structure of a protein describes the spatial arrangement of the entire polypeptide chain. The
quaternary structure of a protein describes how its polypeptide chains are arranged to
form the protein molecule.
Goal 4 Describe how hydrogen bonding
results in (a) ex-helix and (b) fl-pleated
sheet secondary protein structures.
Secondary structures allow for maximum hydrogen bonding and the greatest stability. The
ex-helix structure is a continuous series of loops. The fl-pleated-sheet structure is like a
curtain, but with sharp angles, like pleats.
Goal 5 Define the following terms as
they apply to enzymes: substrate, active
site, inhibitor.
An enzyme's substrate binds at the enzyme's active site. Enzyme inhibitors compete with
the substrate for the active site.
Goal 6 Use the induced fit model to
explain enzyme activity.
In the induced fit model, the shape of the substrate is a close, but not exact, match to the
shape of the active site of the enzyme. As the substrate binds to the enzyme, either or both
molecules change shape slightly.
Goal 7 Given a Lewis diagram of a
monosaccharide in its open chain form,
determine whether the sugar is an
aldose or a ketose.
If the carbonyl group of a monosaccharide is at the end of the carbon chain, the sugar is an
aldose; if the carbonyl group is within the chain, the sugar is a ketose.
Goal 8 Distinguish among monosaccharides, disaccharides, and
polysaccharides.
Monosaccharides are simple sugars that cannot be converted to smaller carbohydrates.
Disaccharides are formed from two simple sugars. Polysaccharides are formed from
many sugars.
Goal 9 Given the Lewis diagrams of two
monosaccharides and a description of
the bond linking the molecules, draw
the Lewis diagram of the resulting
disaccharide.
Two monosaccharides are combined by a reaction between two -OH groups, resulting in
an - 0 - bond between the monomers and a water molecule.
Goal 10 State (a) the defining characteristics and (b) the three major subclassifications of lipids.
Lipids are found in living organisms and are insoluble in water but soluble in nonpolar solvents. The three major classes are (1) fats, oils, and phospholipids, (2) waxes, and (3) lipids
(usually) without ester groups such as steroids.
Goal 11 Identify the physical property
that distinguishes fats from oils.
Fats are triacylglycerols that are solids at room temperature; oils are triacylglycerols that
are liquids at room temperature.
Goal 12 Identify the structural feature
common to all steroid molecules.
Steroids are lipids that all contain the same four-fused-ring system, three six-carbon rings
and one five-carbon ring.
706
Chapter Summaries
Goal 13 Describe the biological roles of
DNAand RNA.
Deoxyribonucleic acid (DNA) stores genetic information and transmits that information to
the next generation during cell division. Ribonucleic acid (RNA) assists in this process by
serving as a "messenger" and as a "switching engine" to transfer the correct amino acid
during protein synthesis.
Goal 14 Describe the components of a
nucleotide.
Nucleic acid monomers are called nucleotides. Each nucleotide has three parts: (1) a nitrogencontaining cyclic molecule called a base; (2) a sugar, either ribose in RNA or deoxyribose in
DNA; and (3) one or more phosphate groups, attached to the hydroxyl groups of the sugar.
Goal 15 Draw Lewis diagrams of
adenine, cytosine, guanine, thymine,
and uracil.
There are five nucleic acid nitrogen bases: thymine (T), cytosine (C), adenine (A), guanine
(G), and uracil (U). They are illustrated in Figure 22-22.
Goal 16 Determine whether any two
DNA or RNA nitrogen bases are
complementary.
In DNA, adenine (A) and thymine (T) form a complementary pair; cytosine (C) and guanine
(G) form the other complementary pair. In RNA, adenine (A) always pairs with uracil (U), a
thymine without a methyl group.
Goal 17 Describe the process by which
a protein molecule is formed.
In translation, messenger RNA is decoded and individual amino acids are brought by different transfer RNA molecules to be assembled into proteins.