Lecture 02. Matter

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Lecture PowerPoint
Chemistry
The Molecular Nature of
Matter and Change
Fifth Edition
Martin S. Silberberg

Chapter 3
© 2014 Pearson Education, Inc.
Matter
• Matter is any substance that has mass and
occupies volume.
• Matter exists in one of three physical states:
1. Solid
2. Liquid
3. Gas

Chapter 3
Solid State
• In a solid, the particles of matter are tightly packed
together.
• Solids have a definite, fixed shape.
• Solids cannot be compressed and have a definite
volume.
• Solids have the least energy of the three states of
matter.

Chapter 3
Liquid State
• In a liquid, the particles of matter are loosely
packed and are free to move past one another.
• Liquids have an indefinite shape and assume the
shape of their container.
• Liquids cannot be compressed and have a definite
volume.
• Liquids have less energy than gases, but more
energy than solids.

Chapter 3
Gaseous State
• In a gas, the particles of matter are far apart and
uniformly distributed throughout the container.
• Gases have an indefinite shape and assume the
shape of their container.
• Gases can be compressed and have an indefinite
volume.
• Gases have the most energy of the three states of
matter.

Chapter 3
Physical States of Matter

Chapter 3
Changes in Physical States
• Most substances can exist as either a solid, a
liquid, or a gas.
• Water exists as a solid below 0 °
C; as a liquid
between 0 °
C and 100 °
C; and as a gas above
100 °
C.
• A substance can change physical states as the
temperature changes.

Chapter 3
Solid ↔ Liquid Phase Changes
• When a solid changes to a liquid, the phase change
is called melting
• A substance melts as the temperature increases.
• When a liquid changes to a solid, the phase change
is called freezing
• A substance freezes as the temperature decreases.

Chapter 3
Liquid ↔ Gas Phase Changes
• When a liquid changes to a gas, the phase change
is called vaporization
• A substance vaporizes as the temperature
increases.
• When a gas changes to a liquid, the phase change
is called condensation.
• A substance condenses as the temperature
decreases.

Chapter 3
Solid ↔ Gas Phase Changes
• When a solid changes directly to a gas, the phase
change is called sublimation
• A substance sublimes as the temperature increases.
• When a gas changes directly to a
solid, the phase change is
called deposition.
• A substance undergoes
deposition as the
temperature decreases.

Chapter 3
Summary of State Changes

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Definitions for Components of Matter
Element - the simplest type of substance with unique physical and
chemical properties. An element consists of only one type of atom. It
cannot be broken down into any simpler substances by physical or
chemical means.
Molecule - a structure that consists of two or
more atoms that are chemically bound together
and thus behaves as an independent unit.
Figure 2.1

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Compound - a substance
composed of two or more elements
which are chemically combined.
Mixture - a group of two or more
elements and/o compounds that
are physically intermingled.
Definitions for Components of Matter
Figure 2.1

Chapter 3
Pure Substances
• There are two types of pure substances:
1. Compounds
2. Elements
• Compounds can be chemically separated into
individual elements.
– Water is a compound that can be separated into
hydrogen and oxygen.
• An element cannot be broken down further by
chemical reactions.

Chapter 3
Physical and Chemical Properties
• A physical property is a characteristic of a pure
substance that we can observe without changing
its composition.
• Physical properties include appearance, melting
and boiling points, density, heat and electrical
conductivity, solubility, and physical state.
• A chemical property of a pure substance describes
its chemical reactions with other substances.

Chapter 3
Chemical Properties
Sodium metal (Na)
reacts with
chlorine gas (Cl2)
to produce
sodium chloride (NaCl).

Chapter 3
Evidence for Chemical Changes
• Gas release (bubbles)
• Light or release of heat energy
• A permanent color change

Chapter 3
Classifications of Matter
• Matter can be divided into two classes:
1. Mixtures
2. Pure substances
• Mixtures are composed of more than one
substance and can be physically separated into its
component substances.
• Pure substances are composed of only one
substance and cannot be physically separated.

Chapter 3
Mixtures
• There are two types of mixtures:
1. Heterogeneous mixtures
2. Homogeneous mixtures
• Heterogeneous mixtures do not have uniform
properties throughout.
– Sand and water is a heterogeneous mixture.
• Homogeneous mixtures have uniform properties
throughout.
– Salt water is a homogeneous mixture.

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Sample Problem 2.1: Distinguishing Elements, Compounds, and
Mixtures at the Atomic Scale
PROBLEM: Theses scenes represent an atomic-scale view of three samples
of matter. Describe each sample as an element, compound, or
mixture.
SOLUTION:
PLAN: Samples that contain one type of matter are either an element or a
compound. An element contains only one type of particle and a
compound contains two or more. Mixtures contain more than one
type of matter.
(a) mixture (b) element (c) compound

Chapter 3
Names of the Elements
• Each element has a unique name.
• Names have several origins:
– Hydrogen is derived from Greek.
– Carbon is derived from Latin.
– Scandium is named for Scandinavia.
– Curium is named for Marie Curie.
– Nobelium is named for Alfred Nobel.

Chapter 3
Element Symbols
• Each element is abbreviated using a chemical
symbol.
• The symbols are one or two letters long.
• Most of the time, the symbol is derived from the
name of the element.
– C is the symbol for carbon.
– Cd is the symbol for cadmium.
• When a symbol has two letters, the first is
capitalized and the second is lowercase.

Chapter 3
Other Element Symbols
• For some elements, the chemical symbol is
derived from the original Latin name.
Gold – Au Sodium – Na
Silver – Ag Antimony – Sb
Copper – Cu Tin – Sn
Mercury – Hg Iron – Fe
Potassium – K Zinc– Zn

Chapter 3
Critical Thinking: Aluminum or Aluminium?
• Most metals have names that end in –ium.
• However, element #13 is called aluminum in the
USA and Canada, and aluminium in the rest of the
world.
• The different spelling is believed to be from a
spelling error which caught on in the USA and
Canada.
• The official IUPAC name is “aluminium”;
however, in 1993, IUPAC recognized the alternate
spelling “aluminum.”

Chapter 3
Chemical Formulas
• A unit of matter composed of two or more
nonmetal atoms is a molecule.
• A chemical formula is an expression of the
number of atoms of each element in a compound.
• The chemical formula of
sulfuric acid is H2SO4.

Chapter 3
Writing Chemical Formulas
• The number of each type of atom in a molecule is
indicated with a subscript in a chemical formula.
• If there is only one atom of a certain type, no “1”
is used.
• A molecule of vitamin B3 has 6 carbon atoms,
6 hydrogen atoms, 2 nitrogen atoms, and 1 oxygen
atom. What is the chemical formula?
C6H6N2O

Chapter 3
Interpreting Chemical Formulas
• Some chemical formulas use parentheses to clarify
atomic composition.
• Ethylene glycol, a component of some antifreezes,
has a chemical formula of C2H4(OH)2. It contains
2 carbon atoms, 4 hydrogen atoms, and 2 OH
units, for a total of 6 hydrogen atoms and 2 oxygen
atoms. How many total atoms are in ethylene
glycol?
• Ethylene glycol has a total of 10 atoms.

Chapter 3
Matter Summary

Chapter 3
Conservation of Mass
• Antoine Lavoisier found that the mass of reactants
before a chemical change was always equal to the
mass of products after a chemical change.
• This is the law of conservation of mass.
• Matter is neither created nor destroyed in a
chemical reaction.

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Figure 2.2 The law of mass conservation:
mass remains constant during a chemical reaction.

Chapter 3
Conservation of Mass Example
• If 1.0 gram of hydrogen combines with 8.0 grams
of oxygen, 9.0 grams of water is produced.
• Consequently, 3.0 grams of hydrogen combine
with 24.0 grams of oxygen to produce 27.0 grams
of water.
• If 50.0 grams of water decompose to produce
45.0 grams of oxygen, how many grams of
hydrogen are produced?
50.0 g water – 45.0 g oxygen = 5.0 g hydrogen

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The total mass of substances does not change during a chemical
reaction.
reactant 1 + reactant 2 product
total mass total mass
=
calcium oxide + carbon dioxide calcium carbonate
CaO + CO2 CaCO
3
56.08 g + 44.00 g 100.08 g
Law of Mass Conservation

Chapter 3
Law of Conservation of Energy
• Just like matter, energy cannot be created or destroyed,
but it can be converted from one form to another.
• This is the law of conservation of energy.
• There are six forms of energy:
1. Heat
2. Light
3. Chemical
4. Electrical
5. Mechanical
6. Nuclear

Chapter 3
Energy and Chemical Changes
• In a chemical change, energy is transformed from
one form to another. For example:

Chapter 3
Critical Thinking: Lower Gasoline Bills
• In terms of expense, is it better to fill a gas tank in
the cool morning, or in the warm afternoon?
• No matter the temperature, the number of gallons
delivered is always the same.
• When the temperature is lower, a greater mass of
gasoline is delivered for the same volume.
• However, the difference in mass between 40°
F and
70°
F is only about 1%.

Chapter 3
Law of Conservation of Mass and Energy
• Mass and energy are related by Einstein’s theory
of relativity, E = mc2.
• Mass and energy can be interchanged.
• The law of conservation of
mass and energy states that
the total mass and energy of
the universe is constant.

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No matter the source, a particular compound is
composed of the same elements in the same parts
(fractions) by mass.
Calcium carbonate
Analysis by Mass
(grams/20.0 g)
Mass Fraction
(parts/1.00 part)
Percent by Mass
(parts/100 parts)
8.0 g calcium
2.4 g carbon
9.6 g oxygen
20.0 g
40% calcium
12% carbon
48% oxygen
100% by mass
0.40 calcium
0.12 carbon
0.48 oxygen
1.00 part by mass
Law of Definite (or Constant) Composition
Figure 2.3

Chapter 3
Law of Definite Composition
• The law of definite composition states that
“Compounds always contain the same elements in
a constant proportion by mass.”
• Water is always 11.2% hydrogen and 88.8%
oxygen by mass, no matter what its source.

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Sample Problem 2.2 Calculating the Mass of an Element in a Compound
PROBLEM: Analysis of 84.2 g of the uranium containing compound
pitchblende shows it is composed of 71.4 g of uranium, with
oxygen as the only other element. How many grams of uranium
can be obtained from 102 kg of pitchblende?
PLAN: The mass ratio of uranium/pitchblende
is the same no matter the source. We
can use the ratio to find the answer.
SOLUTION:
mass (kg) of pitchblende
mass (kg) of uranium
mass (g) of uranium
= 86.5 kg
uranium
= 102 kg pitchblende x
mass (kg) pitchblende x
mass (kg) uranium in pitchblende
mass (kg) pitchblende
71.4 kg uranium
84.2 kg pitchblende
mass (kg) of uranium =
86.5 kg uranium x
1000 g
kg
= 8.65 x 104 g uranium

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If elements A and B react to form two compounds, the different
masses of B that combine with a fixed mass of A can be expressed
as a ratio of small whole numbers.
Example: Carbon Oxides A & B
Carbon Oxide I : 57.1% oxygen and 42.9% carbon
Carbon Oxide II : 72.7% oxygen and 27.3% carbon
Assume that you have 100 g of each compound.
In 100 g of each compound: g O = 57.1 g for oxide I & 72.7 g for oxide II
g C = 42.9 g for oxide I & 27.3 g for oxide II
g O
g C
=
57.1
42.9
= 1.33
=
g O
g C
72.7
27.3
= 2.66
2.66 g O/g C in II
1.33 g O/g C in I
2
1
=
Law of Multiple Proportions

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Dalton’s Atomic Theory
1. All matter consists of atoms.
2. Atoms of one element cannot be converted into
atoms of another element.
3. Atoms of an element are identical in mass and other
properties and are different from atoms of any other
element.
4. Compounds result from the chemical combination of
a specific ratio of atoms of different elements.
The Postulates
https://www.britannica.com/biography/John-Dalton/media/1/150287/164616

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explains the mass laws
Mass conservation
Atoms cannot be created or destroyed
or converted into other types of atoms.
postulate 1
postulate 2
Since every atom has a fixed mass,
during a chemical reaction atoms are combined
differently, and therefore, there is no mass change
overall.
postulate 3

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Definite composition
Atoms are combined in compounds in
specific ratios
and each atom has a specific mass.
Each element has a fixed fraction of the total mass in a
compound.
postulate 3
postulate 4

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Multiple proportions
Atoms of an element have the same mass
and atoms are indivisible.
When different numbers of atoms of elements
combine, they must do so in ratios of small, whole
numbers.
postulate 3
postulate 1

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The Modern Reassessment of the Atomic Theory
1. All matter is composed of atoms. The atom is the smallest body that
retains the unique identity of the element.
2. Atoms of one element cannot be converted into atoms of another
element in a chemical reaction. Elements can only be converted
into other elements in nuclear reactions.
3. All atoms of an element have the same number of protons and
electrons, which determines the chemical behavior of the element.
Isotopes of an element differ in the number of neutrons, and thus
in mass number. A sample of the element is treated as though its
atoms have an average mass.
4. Compounds are formed by the chemical combination of two or more
elements in specific ratios.

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Sample Problem 2.3: Visualizing the Mass Laws
PROBLEM: Theses scenes represent an atomic-scale view of a chemical
reaction. Which of the mass laws: mass conservation, definite
composition, or multiple proportions is (are) illustrated?
SOLUTION:
PLAN: Mass conservation illustrated if number of each atom before and after
reaction remains constant. Definite composition illustrated by formation
of compounds that always have the same atom ratio. Different
compounds made of same elements have small whole number ratios of
those elements illustrates multiple proportions.
Seven purple and nine green atoms in each circle, mass conserved.
One compound formed has one purple and two green, definite
composition. Law of multiple proportions does not apply.

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PROBING THE ATOM

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Figure 2.4
Experiments to determine the properties of cathode rays.
Observation Conclusion
Ray bends in magnetic field Consists of charged particles
Ray bends toward positive plate in
electric field
Consists of negative particles
Ray is identical for any cathode Particles found in ALL matter

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Figure 2.5
Millikan’s oil-drop experiment
for measuring an electron’s charge.
https://www.britannica.com/science/Millikan-oil-drop-experiment/media/1/382908/111169

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Millikan used his findings to also calculate the mass of an
electron.
mass of electron =
mass
charge
x charge
= (-5.686x10-12 kg/C) x (-1.602x10-19 C)
determined by J.J. Thomson and
others
= 9.109x10-31 kg = 9.109x10-28 g

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Figure 2.6
Rutherford’s a-scattering experiment
and discovery of the atomic nucleus.

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Figure 2.7 General features of the atom today.
•The atom is an electrically neutral, spherical entity composed of a positively
charged central nucleus surrounded by one or more negatively charged
electrons.
•The atomic nucleus consists of protons and neutrons.

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Charge Mass
Relative
1+
0
1-
Absolute(C)*
+1.60218x10-19
0
-1.60218x10-19
Relative(amu)†
1.00727
1.00866
0.00054858
Absolute(g)
1.67262x10-
24
1.67493x10-24
9.10939x10-28
Location
in the Atom
Nucleus
Outside
Nucleus
Nucleus
Name(Symbol)
Electron (e-)
Neutron (n0)
Proton (p+)
Table 2.2 Properties of the Three Key Subatomic Particles
* The coulomb (C) is the SI unit of charge.
† The atomic mass unit (amu) equals 1.66054x10-24 g.

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ATOM
ELECTRON NUCLEUS
NEUTRON
PROTON
consists of
consists of
contributes
very little to
Mass of the atom
together account for
most of

Chapter 3
The Periodic Table

Chapter 3
Periodic Table of the Elements
• Each element is assigned a number to identify it.
It is called the atomic number.
• Hydrogen’s atomic number is 1; helium is 2; up to
uranium, which is 92.
• The elements are arranged by atomic number on
the periodic table.

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Figure 2.8
Atomic Symbols, Isotopes, Numbers
X = Atomic symbol of the element
A = mass number; A = Z + N
Isotope = atoms of an element with the same
number of protons, but a different number
of neutrons
A
Z
Z = atomic number
(the number of protons in the nucleus)
N = number of neutrons in the nucleus
X = The symbol of the atom or isotope
See Laboratory Tools

Chapter 4
Isotopes
• All atoms of the same element have the same
number of protons.
• Most elements occur naturally with varying
numbers of neutrons.
• Atoms of the same element that have a different
number of neutrons in the nucleus are called
isotopes.
• Isotopes have the same atomic number, but
different mass numbers.

Chapter 4
Isotopes, Continued
• We often refer to an isotope by stating the name of
the element followed by the mass number.
– Protium is
– Deuterium is
• How many protons and neutrons does an atom of
lead-206 have?
– The atomic number of Pb is 82, so it has 82 protons.
– Pb-206 has 206 – 82 = 124 neutrons.
H
1
1
H
2
1

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Sample Problem 2.4 Determining the Number of Subatomic
Particles in the Isotopes of an Element
PROBLEM: Silicon (Si) has three naturally occurring isotopes: 28Si, 29Si,
and 30Si. Determine the number of protons, neutrons, and
electrons in each silicon isotope.
PLAN: Mass number (A), protons + neutrons, is given for the listed
isotopes. Atomic number (Z), number of protons, for each element
is given in the periodic table and equal to the number of electrons.
Number of neutrons is determined using equation 2.2.
SOLUTION: The atomic number of silicon is 14. Therefore
28Si has 14p+, 14e- and 14n0 (28-14)
29Si has 14p+, 14e- and 15n0 (29-14)
30Si has 14p+, 14e- and 16n0 (30-14)

Chapter 4
Simple and Weighted Averages
• A simple average assumes the same number of
each object.
• A weighted average takes into account the fact that
we do not have equal numbers of all the objects.
• A weighted average is calculated by multiplying
the percentage of the object (as a decimal number)
by its mass for each object and adding the numbers
together.

Chapter 4
Average Atomic Mass
• Since not all isotopes of an atom are present in
equal proportions, we must use the weighted
average.
• Gallium has two isotopes:
1. 69Ga, with a mass of 68.926 amu and 60.11% abundance.
2. 71Ga, with a mass of 70.925 amu and 39.89% abundance.
• The average atomic mass of gallium is:
(68.926 amu)(0.6011) + (70.925 amu)(0.3989)
= 69.72 amu

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Sample Problem 2.5 Calculating the Atomic Mass of an Element
PLAN: Find the weighted average of the
isotopic masses.
SOLUTION:
PROBLEM: Silver’s(Ag: Z = 47) naturally occurring isotopes, 107Ag and 109Ag,
give this mass spectrometric data, calculate the atomic mass of Ag:
Isotope Mass(amu) Abundance(%)
107Ag
109Ag
106.90509
108.90476
51.84
48.16
mass portion from 107Ag =
106.90509 amu x 0.5184 = 55.42 amu
mass portion from 109Ag =
108.90476amu x 0.4816 = 52.45amu
atomic mass of Ag = 55.42amu + 52.45amu = 107.87amu
mass(g) of each isotope
portion of atomic mass
from each isotope
atomic mass
multiply by fractional
abundance of each isotope
add isotopic portions

Chapter 4
Chemistry Connection: Heavy Water
• Heavy water still has the formula H2O, but the
hydrogen atoms are the isotope hydrogen-2.
• Hydrogen-2 is often referred to as deuterium, and
is given the symbol D.
• Heavy water (D2O) is slightly more dense than
light water (H2O), and has slightly higher melting
and boiling points.
• Heavy water is used in nuclear reactors to slow
down neutrons released during the fission process.

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Figure 2.9 The modern periodic table.

Chapter 3
Metals, Nonmetals, and Semimetals
• Metals are on the left side of the periodic table,
nonmetals are on the right side, and the semimetals
are in between.

Chapter 3
Physical States of the Elements
• Shown are the physical states of the elements at
25 °
C on the periodic table.

Chapter 3
Chemistry Connection:
Elements 104 and Beyond
• Scientists continue to discover new, heavier
elements beyond the current periodic table.
• Sometimes disagreements arise over naming of the
new elements.
• IUPAC assigns names to new elements.
• Until IUPAC assigns a name, the elements are
named using Latin prefixes for the numbers
followed by the suffix –ium.
– Hence, element 104 is unnilquadium.

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Figure 2.10 Metals, metalloids, and nonmetals.
Chromium
Copper Cadmium Lead
Bismuth
Boron
Silicon
Arsenic
Antimony
Tellurium
Carbon
(graphite)
Sulfur
Chlorine Bromine
Iodine

Chapter 3
Types of Elements
• Elements can be divided into three classes:
1. Metals
2. Nonmetals
3. Semimetals or metalloids
• Semimetals have properties midway between
those of metals and nonmetals.

Chapter 3
Metal Properties
• Metals are typically solids with high melting
points and high densities and have a bright,
metallic luster.
• Metals are good conductors of heat and electricity.
• Metals can be hammered into thin sheets and are
said to be _ _ _ _ _ _ _ _ _ (9)
• Metals can be drawn into fine wires and are said to
be _ _ _ _ _ _ _ _ (7)

Chapter 3
Nonmetal Properties
• Nonmetals typically have low melting points and
low densities and have a dull appearance.
• Nonmetals are poor conductors of heat and
electricity.
• Nonmetals are not malleable or ductile and crush
into a powder when hammered.
• Eleven nonmetals occur naturally in the
gaseous state.

Chapter 3
Summary of Properties

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Figure 2.11
The formation of an ionic compound.
Transferring electrons from the atoms of one
element to those of another results in an ionic
compound.

2-76
Figure 2.12 Factors that influence the strength of ionic bonding.

2-77

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• Ionic radii follow similar trends to atomic radii with one
critical difference. Cations have very different radii than
anions. Cations are all smaller than their neutral analogs
while anions are all larger. This is easy to understand since
cations have lost electrons. As a result they have both
fewer electrons in the highest energy atomic orbitals
farthest from the nuclei and the remaining electrons feel a
stronger pull from the nucleus. Look at Na+ which is
isoelectronic with Ne. It goes from being one of the
"largest" atoms on the left-hand side of the periodic table
to effectively one of the smallest with an electron
configuration that is the same a neon (all the way on the
right hand side). Moreover, Ne has a nuclear charge
(atomic number) of Z=10 and Na+has a Z=11. Thus the
Na+ should be smaller than Ne.

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• Conversely adding an electron to F to make F- also
generates an ion that isoelectronic with Ne. However,
now you have added electrons and kept the number of
protons constant. Thus F- will be larger than Ne (and
larger than Na+).
• The general trends will continue to hold. From top to
bottom of the periodic table ions will increase in
radii. However, now left to right the radius is more of
a function of the number of electrons. Mg2+ is smaller
than Na+. Both have 10 electrons but Z=12 for Mg
and Z=11 for Na. Similarly, O2- will be larger than F-
as both have 10 electrons but Z=8 for oxygen and
Z=9 for fluorine.

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Ca2+ Calcium is a metal in Group 2A(2). It loses two
electrons to have the same number of electrons as 18Ar.
Sample Problem 2.6 Predicting the Ion and Element Forms
PROBLEM: What monatomic ions do the following elements form?
PLAN: Use Z to find the element. Find its relationship to the nearest
noble gas. Elements occurring before the noble gas gain
electrons and elements following lose electrons.
SOLUTION:
(a) Iodine (Z = 53) (b) Calcium (Z = 20) (c) Aluminum (Z = 13)
I- Iodine is a nonmetal in Group 7A(17). It gains one
electron to have the same number of electrons as 54Xe.
Al3+ Aluminum is a metal in Group 3A(13). It loses three
electrons to have the same number of electrons as 10Ne.

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Formation of a covalent bond between two H atoms.
Figure 2.14
Covalent bonds form when elements share electrons, which usually
occurs between nonmetals.

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Figure 2.15 Elements that occur as molecules.
1A 2A 3A 4A 5A 6A 7A 8A
(1) (2) (13) (14) (15) (16) (17) (18)
H2
N2 O2 F2
P4 S8 Cl2
Se8 Br2
I2
diatomic molecules tetratomic molecules octatomic molecules

2-83
A polyatomic ion
Figure 2.16
Elements that are polyatomic.

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Figure 2.17 A biological periodic table.

Chapter 3
Elements in the Human Body
• Oxygen is the most common element in Earth’s
crust and in the human body.
• While silicon is the second most abundant element
in Earth’s crust, carbon is the second most
abundant in the body.

Chapter 3
Occurrence of the Elements
• There are over 100 elements that occur in nature;
81 of those elements are stable.
• Only 10 elements account for more than 95% of
the mass of Earth’s crust, water, and atmosphere:

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Types of Chemical Formulas
❖An empirical formula indicates the relative number of atoms of
each element in the compound. It is the simplest type of formula.
❖A molecular formula shows the actual number of atoms of
each element in a molecule of the compound.
❖A structural formula shows the number of atoms and the
bonds between them, that is, the relative placement and
connections of atoms in the molecule.
A chemical formula is comprised of element symbols and numerical
subscripts that show the type and number of each atom present in the
smallest unit of the substance.
The empirical formula for hydrogen peroxide is HO.
The molecular formula for hydrogen peroxide is H2O2.
The structural formula for hydrogen peroxide is H-O-O-H.

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Figure 2.18 Some common monatomic ions of the elements.
Can you see any patterns?

2-89
Table 2.3 Common Monoatomic Ions
H- hydride
Na+ sodium
H+ hydrogen
Li+ lithium fluoride
F-
Cs+ cesium
K+ potassium
Ag+ silver
chloride
Cl-
bromide
Br-
iodide
I-
Mg2+ magnesium
Sr2+ strontium
Ca2+ calcium
Zn2+ zinc
Ba2+ barium
Cd2+ cadmium
Al3+ aluminum
+1
+2
+3
Cations
Charge Formula Name
Anions
Charge Formula Name
-1
-2
-3
oxide
O2-
sulfide
S2-
nitride
N3-
Common ions are in blue.

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Naming Binary Ionic Compounds
The name of the cation is the same as the name of the metal.
Many metal names end in -ium.
The name of the anion takes the root of the nonmetal name
and adds the suffix -ide.
Calcium and bromine form calcium bromide.
The name of the cation is written first, followed by that of the anion.

2-91
Sample Problem 2.7 Naming Binary Ionic Compounds
PROBLEM: Name the ionic compound formed from the following pairs of
elements:
PLAN:
(a) magnesium and nitrogen
SOLUTION:
Use the periodic table to decide which element is the metal and
which the nonmetal. The metal (cation) is named first and we
use the -ide suffix on the nonmetal name root.
(b) iodine and cadmium
(c) strontium and fluorine (d) sulfur and cesium
(a) magnesium nitride
(b) cadmium iodide
(c) strontium fluoride
(d) cesium sulfide

2-92
Sample Problem 2.8 Determining Formulas of Binary Ionic Compounds
PROBLEM: Write empirical formulas for the compounds named in Sample
Problem 2.7:
PLAN:
SOLUTION:
Compounds are neutral. We find the smallest number of each
ion which will produce a neutral formula. Use subscripts to the
right of the element symbol.
(a) Mg2+ and N3-; three Mg2+(6+) and two N3-(6-); Mg3N2
(b) Cd2+ and I-; one Cd2+(2+) and two I-(2-); CdI2
(c) Sr2+ and F-; one Sr2+(2+) and two F-(2-); SrF2
(d) Cs+ and S2-; two Cs+(2+) and one S2- (2-); Cs2S
(a) magnesium nitride (b) cadmium iodide
(c) strontium fluoride (d) cesium sulfide

2-93
Element
Table 2.4 Some Metals That Form More Than One Monatomic Ion
Ion Formula Systematic Name Common Name
Copper Cu+1
Cu+2
copper(I)
copper(II)
cuprous
cupric
Cobalt
Co+2
Co+3
cobalt(II)
cobalt (III)
ferrous
Iron
Fe+2 iron(II)
Fe+3 iron(III) ferric
Lead
Pb+2 lead(II)
Pb+4 lead(IV)
Tin
Sn+2 tin(II)
Sn+4 tin(IV)
stannous
stannic
(partial table)

2-94
Sample Problem 2.9 Determining Names and Formulas of Ionic
Compounds of Elements That Form More
Than One Ion
PLAN:
SOLUTION:
Compounds are neutral. Find the smallest number of each ion
which will produce a neutral formula.
PROBLEM: Give the systematic names for the formulas or the formulas for
the names of the following compounds:
(a) tin(II) fluoride (b) CrI3
(c) ferric oxide (d) CoS
(a) Tin (II) is Sn2+; fluoride is F-; so the formula is SnF2.
(b) The anion I- is iodide(I-); 3I- means that Cr(chromium) is +3.
CrI3 is chromium(III) iodide.
(c) Ferric is a common name for Fe3+; oxide is O2-, therefore the
formula is Fe2O3.
(d) Co is cobalt; the anion S2- is sulfide(S2-); the compound is
cobalt (II) sulfide.

2-95
(partial table)
Formula
Cations
NH4
+
Common Anions
H3O+
Formula
ammonium hydronium
Name Name
CH3COO- acetate
CN- cyanide
OH- hydroxide
ClO3
- chlorate
NO2
- nitrite
NO3
- nitrate
MnO4
- permanganate
CO3
-2 carbonate
CrO4
-2 chromate
Cr2O7
-2 dichromate
O2
-2 oxide
SO4
-2
sulfate
PO4
-3 phosphate
Table 2.5 Some Common Polyatomic Ions

2-96
Naming oxoanions
Prefixes Root Suffixes Examples
root
per ate ClO4
- perchlorate
ate
root ClO3
- chlorate
ite
root ClO2
- chlorite
ite
hypo root ClO- hypochlorite
No.
of
O
atoms Figure 2.19
Number Prefix Number Prefix Number Prefix
1 mono-
2 di-
3 tri-
4 tetra-
5 penta-
6 hexa-
7 hepta-
8 octa-
9 nona-
10 deca-
Table 2.6 Numerical Prefixes for Hydrates and Binary Covalent Compounds

2-97
Sample Problem 2.10 Determining Names and Formulas of Ionic
Compounds Containing Polyatomic Ions
PLAN:
SOLUTION:
Note that polyatomic ions have an overall charge so when
writing a formula with more than one polyatomic unit, place the
ion in a set of parentheses.
PROBLEM: Give the systematic names or the formula or the formulas for the
names of the following compounds:
(a) Fe(ClO4)2 (b) sodium sulfite
(a) ClO4
- is perchlorate; iron must have a 2+ charge. This is
iron(II) perchlorate.
(b) The anion sulfite is SO3
2-; therefore you need 2 sodiums per
sulfite. The formula is Na2SO3.
(c) Hydroxide is OH- and barium is a 2+ ion. When water is
included in the formula, we use the term “hydrate” and a prefix
which indicates the number of waters. So it is barium hydroxide
octahydrate.
(c) Ba(OH)2 8H2O

2-98
Sample Problem 2.9 Recognizing Incorrect Names and Formulas
of Ionic Compounds
SOLUTION:
PROBLEM: Something is wrong with the second part of each statement.
Provide the correct name or formula.
(a) Ba(C2H3O2)2 is called barium diacetate.
(b) Sodium sulfide has the formula (Na)2SO3.
(a) Barium is always a +2 ion and acetate is -1. The “di-” is
unnecessary.
(b) An ion of a single element does not need parentheses.
Sulfide is S2-, not SO3
2-. The correct formula is Na2S.
(c) Since sulfate has a 2- charge, only 1 Fe2+ is needed. The
formula should be FeSO4.
(d) The parentheses are unnecessary. The correct formula is
Cs2CO3.
(c) Iron(II) sulfate has the formula Fe2(SO4)3.
(d) Cesium carbonate has the formula Cs2(CO3).

2-99
Naming Acids
1) Binary acids solutions form when certain gaseous compounds
dissolve in water.
For example, when gaseous hydrogen chloride (HCl) dissolves in
water, it forms a solution called hydrochloric acid.
Prefix hydro- + anion nonmetal root + suffix -ic + the word acid -
hydro + chlor + ic + acid
hydrochloric acid
2) Oxoacid names are similar to those of the oxoanions, except for
two suffix changes:
-ate in the anion becomes –ic in the acid
-ite in the anion becomes –ous in the acid
The oxoanion prefixes hypo- and per- are retained. Thus,
BrO4
- is perbromate, and HBrO4 is perbromic acid;
IO2
- is iodite, and HIO2 is iodous acid.

2-
Sample Problem 2.10 Determining Names and Formulas of Anions
and Acids
SOLUTION:
PROBLEM: Name the following anions and give the names and formulas of
the acids derived from them:
(a) Br - (b) IO3
- (c) CN - (d) SO4
2- (e) NO2
-
(a) The anion is bromide; the acid is hydrobromic acid, HBr.
(b) The anion is iodate; the acid is iodic acid, HIO3.
(c) The anion is cyanide; the acid is hydrocyanic acid, HCN.
(d) The anion is sulfate; the acid is sulfuric acid, H2SO4.
(e) The anion is nitrite; the acid is nitrous acid, HNO2.

2-
Sample Problem 2.13 Determining Names and Formulas of Binary
Covalent Compounds
SOLUTION:
PROBLEM: (a) What is the formula of carbon disulfide?
(c) Give the name and formula of the compound whose
molecules each consist of two N atoms and four O
atoms.
(b) What is the name of PCl5?
(a) Carbon is C, sulfide is sulfur S and di-means two - CS2.
(b) P is phosphorous, Cl is chloride, the prefix for 5 is penta-.
Phosphorous pentachloride.
(c) N is nitrogen and is in a lower group number than O (oxygen).
Therefore the compound formula is N2O4 and name dinitrogen
tetraoxide.

2-
Sample Problem 2.14 Recognizing Incorrect Names and Formulas
of Binary Covalent Compounds
SOLUTION:
(a) SF4 is monosulfur pentafluoride.
(c) N2O3 is dinitrotrioxide.
(b) Dichlorine heptaoxide is Cl2O6.
(a) The prefix mono- is not needed for one atom; the prefix for
four is tetra-. So the name is sulfur tetrafluoride.
(b) Hepta- means 7; the formula should be Cl2O7.
(c) The first element is given its elemental name so this is
dinitrogen trioxide.
PROBLEM: Explain what is wrong with the name of formula in the second
part of each statement and correct it:

2-
Sample Problem 2.15 Calculating the Molecular Mass of a Compound
SOLUTION:
(a) tetraphosphorous trisulfide (b) ammonium nitrate
PROBLEM: Using the data in the periodic table, calculate the molecular (or
formula) mass of:
PLAN: Write the formula and then multiply the number of atoms by the
respective atomic masses. Add the masses for the compound.
(a) P4S3
molecular mass = (4 x atomic mass of P) + (3 x atomic mass of S)
= (4 x 30.97 amu) + (3 x 32.07 amu) = 220.09 amu
(b) NH4NO3
molecular mass = (2 x atomic mass of N) + (4 x atomic mass of H) +
(3 x atomic mass of O)
= (2 x 14.01 amu) + (4 x 1.008 amu) + (3 x 16.00 amu)
= 80.05 amu

2-
Sample Problem 2.16 Naming Compounds from Their Depictions
PROBLEM: Each circle contains a representation of a binary compound.
Determine its formula, name, and molecular (formula) mass.
SOLUTION:
(a) There is 1 sodium (brown) for every fluorine (green), so the formula is NaF.
(b) There are 3 fluorines (green) for every nitrogen (blue), so the formula is NF3.
molecular mass = (3x atomic mass of F) (1x atomic mass of N)
+
formula mass = (1x atomic mass of Na) (1x atomic mass of F)
+
= 22.99 amu + 19.00 amu = 41.99 amu
= (3x 19.00 amu) + 14.01 amu = 71.01 amu
PLAN: Each compound contains only two elements. Find simplest whole number
ratio of one atom to the other to determine formula, name, and mass.

2-
Figure 2.20 The distinction between mixtures and compounds.
Allowed to react chemically
therefore cannot be separated by
physical means.
S2-
Fe2+
Physically mixed therefore can be
separated by physical means; in
this case by a magnet.

2-
Mixtures
Heterogeneous mixture: has one or more visible boundaries
between the components.
Homogeneous mixture: has no visible boundaries because the
components are mixed as individual atoms, ions, and molecules.
Solutions: A homogeneous mixture is also called a solution.
Solutions in water are called aqueous solutions, and are very
important in chemistry. Although we normally think of solutions as
liquids, they can exist in all three physical states.

2-
Filtration: Separates components of a mixture based upon
differences in particle size. Normally separating a precipitate
from a solution, or particles from an air stream.
Crystallization: Separation is based upon differences in solubility of
components in a mixture.
Distillation: separation is based upon differences in volatility.
Extraction: Separation is based upon differences in solubility in
different solvents (major material).
Chromatography: Separation is based upon differences in solubility
in a solvent versus a stationary phase.
Basic Separation Techniques
Tools of the Laboratory

2-
Basic Separation Techniques
Figure B2.3 Filtration Figure B2.4 Crystallization
return to previous slide

2-
Figure B2.5

2-
Figure B2.6

2-
Procedure for Column Chromatography
Figure B2.7

2-
Figure B2.8 Separation by Gas - Liquid Chromatography

Lecture 02. Matter

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