Download Chapter 2 ATOMS AND ELEMENTS

Chapter 2 — Atoms, Molecules, Ions
1
Atomic Composition
• Protons
– + electrical charge
– mass = 1.672623 x 10-24 g
– mass = 1.007 atomic
mass units (u)
Chapter 2
Atoms, Molecules,
and Ions
• Electrons
–
negative electrical charge
– relative mass = 0.0005 u
• Neutrons
no electrical charge
– mass = 1.009 u
–
Jeffrey Mack
California State University,
Sacramento
Atomic Composition
The atom is mostly empty space
Atomic Number, Z
All atoms of the same element have the same
number of protons in the nucleus, Z
13
• protons and neutrons in the nucleus.
• the number of electrons is equal to the number of protons.
• electrons in space around the nucleus.
Atomic number, Z
Al
Elemental symbol
26.981
Atomic mass (u)
• extremely small. One teaspoon of water has 3 times as many
atoms as the Atlantic Ocean has teaspoons of water.
Atomic Weight
• The atomic mass of one atom of an element is relative
to one atom of another.
• Example: an Oatom has approximately 16 times more
mass than an Hatom.
• The standard is based upon the carbon-12 isotope.
• The mass of one C12 atom is 1.99265  1023 g.
• The atomic mass of C12 atom is defined as exactly 12
u.
• Therefore: 1 u = (the mass of one C12 atom 12)
= 1.66054  1024g
= 1.66054  1027 kg
Mass Number, A
•
•
•
•
C - 12 atom has 6 protons and 6 neutrons
Mass Number (A) = # protons + # neutrons
Carbon – 12 has a mass # of 12 u
Carbon – 13 has a mass # of 13 u (6 + 7) and
so on.
• Elements with different number of neutrons
are called “Isotopes”.
Chapter 2 — Atoms, Molecules, Ions
2
Isotopes
Isotopes
For a given element “X”, an isotope is written by:
Atomic number (Z) = number of protons in the
nucleus.
Mass number (A) = total number of nucleons in the
nucleus (i.e., protons and neutrons).
One nucleon has a
mass of 1 amu
(Atomic Mass Unit)
a.k.a “Dalton” or u
A
Z
X
Isotopes have the
same Z but different
total number of
nucleons (A).
For a given element “X”, an isotope is written by:
Atomic number (Z) = number of protons in the
nucleus.
Mass number (A) = total number of nucleons in the
nucleus (i.e., protons and neutrons).
A
Z
X
Hydrogen Isotopes
Shorthand notation:
Carbon – 12 = 12C
Carbon – 14 = 14C and so on...
Isotope Composition
Hydrogen has 3 isotopes
1
1
H
1 proton and 0
neutrons, protium
Isotope
2
1
H
1 proton and 1
neutron, deuterium
Sulfur-32
3
1
1 proton and 2
neutrons, tritium
H
Masses of Isotopes
determined with a mass spectrometer
74
35
Br
Electrons
Protons
Neutrons
16
16
16
35
35
39
Isotope Abundance
• The mass spectrometer gives information on the mass and
relative abundance of each element’s isotopes.
• Each isotope is represented by a Relative Abundance.
Chapter 2 — Atoms, Molecules, Ions
3
Isotope Abundance
Percent Abundance
• The percent abundance of each element’s isotopes
are found by:
Percent
=
Abundance
Atomic Weight (Mass)
Atomic
mass
æ % abundance of isotope 1 ö
=ç
÷ ? mass of isotope 1
100
è
ø
æ % abundance of isotope 2 ö
+ç
÷ ? mass of isotope 2 + ...
100
è
ø
Example:
Chlorine has two isotopes:
Cl-35
75.53 %
Isotope Problem
The average weighted atomic mass is determined by the
following mathematical expression:
 100
total number of atoms of all
isotopes of that element
Isotope Problem
35
17
The atomic masses on the periodic table are
“weighted averages” of the all of an element’s
individual isotope masses.
number of atoms of an
individual isotope
Cl- 37
Cl
37
17
&
34.96885 u
24.47 %
Cl
36.96590 u
Isotope Problem
The average weighted atomic mass is determined by the
following mathematical expression:
Average
mass of a Cl
atom
mass of a
Cl-35 atom
fraction that
are Cl-35
mass of a
Cl-37 atom
fraction that
are Cl-37
Chapter 2 — Atoms, Molecules, Ions
4
Isotope Problem
Isotope Problem
The average weighted atomic mass is determined by the
following mathematical expression:
The average weighted atomic mass is determined by the
following mathematical expression:
Average
mass of a Cl
atom
Average
mass of a Cl
atom
mass of a
Cl-35 atom
m Cl (u) = m Cl-35 
fraction that
are Cl-35
mass of a
Cl-37 atom
fraction that
are Cl-37
abundance
abundance
+ m Cl-37 
of Cl-35
of Cl-37
mass of a
Cl-35 atom
m Cl (u) = m Cl-35 
fraction that
are Cl-35
mass of a
Cl-37 atom
fraction that
are Cl-37
abundance
abundance
+ m Cl-37 
of Cl-35
of Cl-37
35.45 u = 34.96885 u  0.7553 + 36.96590 u  0.2447
(4 sig. fig)
Isotope Problem
The average weighted atomic mass is determined by the
following mathematical expression:
Average
mass of a Cl
atom
mass of a
Cl-35 atom
m Cl (u) = m Cl-35 
fraction that
are Cl-35
mass of a
Cl-37 atom
fraction that
are Cl-37
Isotope Problem:
Copper has two stable isotopes, Cu-63 and Cu-65.
If the masses are 62.929599 u and 64.927793 u respectively,
what are the relative % abundances of the two isotopes?
The weighted average atomic mass is reported as 63.546 u.
abundance
abundance
+ m Cl-37 
of Cl-35
of Cl-37
35.45 u = 34.96885 u  0.7553 + 36.96590 u  0.2447
(4 sig. fig)
This is the value that is reported on the periodic table.
Note that: 0.7553 + 0.2447 = 1.0000 (100%)
Isotope Problem:
Copper has two stable isotopes, Cu-63 and Cu-65.
If the masses are 62.929599 u and 64.927793 u respectively,
what are the relative % abundances of the two isotopes?
The weighted average atomic mass is reported as 63.546 u.
Isotope Problem:
Copper has two stable isotopes, Cu-63 and Cu-65.
If the masses are 62.929599 u and 64.927793 u respectively,
what are the relative % abundances of the two isotopes?
The weighted average atomic mass is reported as 63.546 u.
What is given:
What is given:
There are only 2 isotopes
The masses of each isotope
the average weighted mass
(from the periodic table)
What is asked:
Determine the relative
abundance of each isotope
(fraction that occurs in nature)
There are only 2 isotopes
The masses of each isotope
the average weighted mass
(from the periodic table)
Chapter 2 — Atoms, Molecules, Ions
Isotope Problem:
Copper has two stable isotopes, Cu-63 and Cu-65.
If the masses are 62.929599 u and 64.927793 u respectively,
what are the relative % abundances of the two isotopes?
The weighted average atomic mass is reported as 63.546 u.
5
Isotope Problem:
Copper has two stable isotopes, Cu-63 and Cu-65.
If the masses are 62.929599 u and 64.927793 u respectively,
what are the relative % abundances of the two isotopes?
The weighted average atomic mass is reported as 63.546 u.
Solution: Recall that the sum of the fractions that represent
the two isotopes must equal 1 exactly.
Since these quantities are unknown, let’s assign variables!
Isotope Problem:
Copper has two stable isotopes, Cu-63 and Cu-65.
If the masses are 62.929599 u and 64.927793 u respectively,
what are the relative % abundances of the two isotopes?
The weighted average atomic mass is reported as 63.546 u.
Solution: Recall that the sum of the fractions that represent
the two isotopes must equal 1 exactly.
Since these quantities are unknown, let’s assign variables!
The reported average must equal the weighted sum
of the two isotopes.
Average Mass =Mass Cu-63 × X+ Mass Cu-65 × Y
Let X equal the fraction that is Cu-63
Let Y equal the fraction that is Cu-65
Such that: X + Y = 1
The reported average must equal the weighted sum
of the two isotopes.
fraction that
is copper-63
Average Mass =Mass Cu-63 × X+ Mass Cu-65 × Y
The reported average must equal the weighted sum
of the two isotopes.
fraction that
is copper-63
fraction that
is copper-65
Average Mass =Mass Cu-63 × X+ Mass Cu-65 × Y
Chapter 2 — Atoms, Molecules, Ions
The reported average must equal the weighted sum
of the two isotopes.
fraction that
is copper-63
fraction that
is copper-65
Average Mass =Mass Cu-63 × X+ Mass Cu-65 × Y
6
Isotope Problem:
Copper has two stable isotopes, Cu-63 and Cu-65.
If the masses are 62.929599 u and 64.927793 u respectively,
what are the relative % abundances of the two isotopes?
The weighted average atomic mass is reported as 63.546 u.
63.546 u = 62.929599 u × X +
64.927793 u × Y
Plugging in the numbers:
63.546 u = 62.929599 u × X +
64.927793 u × Y
Isotope Problem:
Copper has two stable isotopes, Cu-63 and Cu-65.
If the masses are 62.929599 u and 64.927793 u respectively,
what are the relative % abundances of the two isotopes?
The weighted average atomic mass is reported as 63.546 u.
63.546 u = 62.929599 u × X +
64.927793 u × Y
Isotope Problem:
Copper has two stable isotopes, Cu-63 and Cu-65.
If the masses are 62.929599 u and 64.927793 u respectively,
what are the relative % abundances of the two isotopes?
The weighted average atomic mass is reported as 63.546 u.
63.546 u = 62.929599 u × X +
Since X + Y = 1
64.927793 u × Y
Since X + Y = 1
We can write the equation in terms of one variable,
then solve.
Isotope Problem:
Copper has two stable isotopes, Cu-63 and Cu-65.
If the masses are 62.929599 u and 64.927793 u respectively,
what are the relative % abundances of the two isotopes?
The weighted average atomic mass is reported as 63.546 u.
63.546 u = 62.929599 u × X +
64.927793 u × Y
Since X + Y = 1
We can write the equation in terms of one variable,
then solve.
63.546 u = 62.929599 u × X+ 64.927793 u × (1−X)
Isotope Problem:
Copper has two stable isotopes, Cu-63 and Cu-65.
If the masses are 62.929599 u and 64.927793 u respectively,
what are the relative % abundances of the two isotopes?
The weighted average atomic mass is reported as 63.546 u.
Solving for X:
X = 0.69500
Chapter 2 — Atoms, Molecules, Ions
Isotope Problem:
Copper has two stable isotopes, Cu-63 and Cu-65.
If the masses are 62.929599 u and 64.927793 u respectively,
what are the relative % abundances of the two isotopes?
The weighted average atomic mass is reported as 63.546 u.
Solving for X:
7
Isotope Problem:
Copper has two stable isotopes, Cu-63 and Cu-65.
If the masses are 62.929599 u and 64.927793 u respectively,
what are the relative % abundances of the two isotopes?
The weighted average atomic mass is reported as 63.546 u.
Solving for X:
X = 0.69500
Y must therefore be: Y = 0.30500
X = 0.69500
recall,
Y=1X
Y must therefore be: Y = 0.30500
Isotope Problem:
Copper has two stable isotopes, Cu-63 and Cu-65.
If the masses are 62.929599 u and 64.927793 u respectively,
what are the relative % abundances of the two isotopes?
The weighted average atomic mass is reported as 63.546 u.
Isotopes
• For more information on Elements, Isotopes
and Isotope abundance go to:
http://www.webelements.com
Solving for X:
X = 0.69500
recall,
Y=1X
Y must therefore be: Y = 0.30500
This means that the % abundances for the two isotopes are:
69.500 % for Cu-63 and
30.500 % for Cu-65
The Periodic Table
• Dmitri Mendeleev (18341907) developed the modern
periodic table.
• Argued that elemental
properties are periodic
functions of atomic weights.
• We now know that element
properties are periodic
functions of the ATOMIC
NUMBERS.
• Here you can explore more about the
periodic table by looking at interesting
elements such as boron, silicon, chromium
and antimony.
• The web site will provide you with practical
applications and usage of elements.
Features of the Periodic Table
Periods & Groups
Chapter 2 — Atoms, Molecules, Ions
Regions in the Periodic Table
8
Group 1A: Alkali Metals
Li, Na, K, Rb, Cs
sodium metal
Group 2A: Alkaline Earth Metals
Be, Mg, Ca, Sr, Ba, Ra
• The second group in the periodic table,
Group 2A, is also composed entirely of
metals that occur naturally only in
compounds.
• Except for beryllium (Be), these elements
react with water to produce basic solutions.
• Most of these oxides (such as lime, CaO)
form alkaline solutions; hence, they are
known as the alkaline earth metals.
• Magnesium (Mg) and calcium (Ca) are the
seventh and fifth most abundant elements
in the earth’s crust, respectively
Magnesium
Magnesium
oxide
Gems & Minerals
• Sapphire: Al2O3
with Fe3+ or Ti3+
impurity gives
blue whereas V3+
gives violet.
• Ruby: Al2O3 with
Cr3+ impurity
• Elements in the leftmost
column, Group 1A, are known
as the alkali metals (except
H).
• All the alkali metals are solids
at room temperature and all
are reactive with water to
produce hydrogen and basic
solutions.
• These metals are only found
in nature combined in
compounds (such as NaCl),
never as free elements.
Group 3A: B, Al, Ga, In, Tl
• Group 3A (aluminum,
gallium, indium, and
thallium) are metals,
whereas boron (B) is a
metalloid.
• Aluminum (Al) is the most
abundant metal in the
earth’s crust at 8.2% by
mass. It is exceeded in
abundance only by the
nonmetal oxygen and
metalloid silicon.
Gallium is one of the few
metals that can be liquid
close to room temp.
Group 4A: C, Si, Ge, Sn, Pb
• Beginning with Group 4A, however, the
groups contain more and more
nonmetals.
• In Group 4A there is a nonmetal,
carbon, two metalloids, silicon and
germanium, and two metals, tin and
lead
• Because of the change from nonmetallic
to metallic behavior, more variation
occurs in the properties of the elements
of this group than in most others.
• Nonetheless, there are similarities. For
example, these elements form
compounds with analogous formulas
such as CO2, SiO2, GeO2, and PbO2.
Diamond
Quartz, SiO2
Chapter 2 — Atoms, Molecules, Ions
Carbon
9
Group 5A: N, P, As, Sb, Bi
• One interesting aspect nonmetals (like carbon) is that an element of this
type can often exist in several different and distinct forms, called
allotropes.
• Each allotrope has its own physical and chemical properties.
Polyatomic Elements
• Nitrogen in Group 5A occurs
naturally in the form of the
diatomic molecule (N2) It
makes up about three-fourths
of earth’s atmosphere.
• Phosphorus is essential to life.
It is an important constituent in
bones, teeth, and DNA. The
element glows in the dark if it is
in the air. Its name is based on
Greek words meaning “lightbearing”.
Phosphorous also has
several allotropes, the
most important being
white & red phosphorus.
Group 6A: O, S, Se, Te, Po
• In addition to nitrogen there are several elements
that exist in nature as diatomic or polyatomic
molecules.
• When one speaks of oxygen in its natural state, one
means O2.
• A single oxygen atom is referred to as “atomic
oxygen” and so on.
• Oxygen, which constitutes about 20%
of earth’s atmosphere and which
combines readily with most other
elements, is at the top of Group 6A.
• Most of the energy that powers life on
earth is derived from reactions in
which oxygen combines with other
substances.
• Sulfur, selenium, and tellurium are
often referred to collectively as
chalcogens (from the Greek word,
khalkos, for copper) because most
copper ores contain these elements.
One elemental
form of S has a
ring structure.
Group 7A: Halogens
Group 8A: Noble Gases
F, Cl, Br, I, At
He, Ne, Ar, Kr, Xe, Rn
• The Group 7A elements, fluorine,
chlorine, bromine, iodine, and
radioactive astatine are nonmetals.
• All exist as diatomic molecules.
• These elements are among the most
reactive of all elements. All combine
violently with alkali metals to form
salts such as table salt, NaCl.
• Collectively they are know as the
halogens, from the Greek “hals”,
meaning “salt,” & genes, “forming.”
Br2 gas
Chapter 2 — Atoms, Molecules, Ions
Group 8A: Noble Gases
He, Ne, Ar, Kr, Xe, Rn
• The Group 8A elements, helium, neon, argon,
krypton, xenon, and radioactive radon are
collectively know as the “Noble” or “Inert” gases.
• The are the least reactive elements of the period.
• All are gases, and none is abundant on earth or in
earth’s atmosphere.
• It was first thought that the Noble gases would not
combine chemically to form stable compounds.
However in 1962, a xenon & fluorine compound
was first prepared, (XeF4) opening the way to the
preparation of a number of other such compounds.
Transition Elements
• Stretching between Groups 2A and 3A is a series of
elements called the transition elements.
• These fill the B-groups (1B through 8B) in the fourth through
the seventh periods in the center of the periodic table.
• All are metals and 13 of them are in the top 30 elements in
terms of abundance in the earth’s crust
• Most occur naturally in combination with other elements, but
a few, Cu, Ag, Au & Pt are much less reactive where they
can be found in nature as pure elements.
• Two rows at the bottom of the table accommodate the
lanthanides [the series of elements between the elements
lanthanum (Z = 57) and hafnium (Z = 72) and the actinides
the series of elements between actinium
(Z = 89) and rutherfordium (Z = 104).
Compounds & Molecules
• COMPOUNDS are a combination of 2 or
more elements in definite ratios by mass.
• The character of each element is lost when
forming a compound.
• MOLECULES are the smallest unit of a
compound that retains the characteristics of
the compound. (non-metal combined with a
non-metal)
10
Transition Elements
Lanthanides and actinides
Iron in air gives
iron(III) oxide
Molecules, Ions, and Their
Compounds
Elements combine to form compounds. These new
species have different chemical and physical
properties than the original elements.
Compounds
• A compound is a distinct substance that contains
two or more elements combined in a definite
proportion by weight.
• Atoms of the elements that constitute a
compound are always present in simple whole
number ratios.
• They are never present as fractional parts.
Examples:
Never:
AB
A2B
A½B
AB2
Chapter 2 — Atoms, Molecules, Ions
Molecular Formulas
11
Molecular Modeling
Structural formula of
glycine:
•
•
•
•
Molecular Formula: #’s & types of each atom.
Condensed Formula: indicates structure
Structural Formula: Shows the connections (bonds)
Molecular Model: Give 3-D perspective
Molecular Modeling
Ball & stick
H H O
H N C C O H
H
Space-filling
Ions & Ionic Compounds
Various ways to represent a molecule.
Ionic Compounds
• Ionic compounds (metals & nonmetals) constitute another major
class of compounds.
• They consist of ions, atoms or
groups of atoms that bear a positive
or negative electric charge.
• Many familiar compounds are
composed of ions. Table salt, or
sodium chloride (NaCl) is one
example.
• These are generically referred to as
salts.
Ions & Ionic Compounds
• IONS are atoms or groups of atoms with a
formal positive or negative charge.
• Removing electrons from an atom produces a
CATION with a positive charge.
• Adding a electrons to an atom gives an
ANION with a negative charge.
NaCl
Chapter 2 — Atoms, Molecules, Ions
Ions & Ionic Compounds
Predicting Ion Charges
12
Forming Cations & Anions
A CATION forms
when an atom
loses one or
more electrons.
An ANION forms
when an atom
gains one or
more electrons
Mg  Mg2+ + 2 e-
F + e-  F-
Charges on Common Ions
-4 -3 -2 -1
+1
+2
+3
In general
• metals (Mg) lose electrons forming cations
• nonmetals (F) gain electrons forming anions
Transition Metal Cations
By losing or gaining electrons, an atom has same
number of electrons as the nearest Noble gas atom.
Nonmetal Ions Names
NONMETAL + n e-  Xnwhere n = 8 - Group no.
• These elements may have more than one charge
states.
• Oxidation Number (charge) found from the formula.
Group 4A
C4-,carbide
Example:
FeCl2
chlorine forms a -1 charge so Fe = +2
Group 5A
N3-,
nitride
Group 6A
O2-,
FeCl3
chlorine forms a -1 charge so Fe = +3
+3 + 3  (-1) = 0
oxide F-, fluoride
S2-, sulfide Cl-, chloride
+2 + 2  (-1) = 0
Example:
Group 7A
Br-, bromide
Name derived by
adding “ide” suffix.
I-, iodide
Chapter 2 — Atoms, Molecules, Ions
13
Metal Ions Names
Metal Ions Names
Main group elements
All metal cations are named after there representative
element followed by the word “ion”.
sodium
barium
Na
Na+
sodium ion
Ba
Ba2+
Barium ion
Al3+
aluminum ion
aluminum Al
Transition elements
Since the transition elements have multiple charge
states, each must be identified uniquely.
Cu
Cu+
copper (I) ion
Cu
Cu2+
copper (II) ion
Polyatomic Ions
• Polyatomic anions are groups of atoms
(molecules) with a net charge.
• You must MEMORIZE the names and
formulas in Table 2.4
• This requires practice in the form of time!
• Flash cards often help. Have your friends
quiz you!
Remember:
chem – is – try
Note: many O
containing anions
have names ending
in –ate (or -ite).
Polyatomic Anions Names
Two polyatomic anions are named like simple ions
with the suffix (ending) -ide.
Example:
OH-
hydroxide ion
CN-
cyanide ion
Polyatomic Ions
NH4+
ammonium ion
Most other complex ions end in -ate or -ite.
SO 24-
sulfate ion
SO32-
sulfite ion
One of the few common polyatomic cations
Chapter 2 — Atoms, Molecules, Ions
Compounds Formed From Ions
CATION + ANION  COMPOUND
Na+ + Cl  NaCl
All ionic compounds are neutral in
charge. The individual charges of
the cations and anions must sum to
zero.
14
Ionic Compounds
NH4+
Cl-
ammonium chloride, NH4Cl
Naming Ionic Compounds
Metal of fixed charge with a complex ion
Cation
Anion
Formula
Naming Ionic Compounds
Al2(SO4)3
Name
spoken as: Al - two, sulfate, taken three times
K+
OH
Ca2+
OH
Ca(OH)2
Calcium hydroxide
Na+
SO24-
Na2SO4
Sodium sulfate
Al3+
24
Al2(SO4)3
Aluminum sulfate
SO
KOH
Potassium hydroxide
Naming Ionic Compounds
Metal of variable charge with a complex ion
Cation
Anion
Formula
Name
Pb2+
SO24-
PbSO4
lead (II) sulfate
Pb4+
SO24-
Pb(SO4)2
lead (IV) sulfate
Fe3+
NO3-
Fe(NO3)3
Iron (III) nitrate
Fe2+
NO2-
Fe(NO2)2
Iron (II) nitrite
or Al -two, parenthesis, sulfate, three
Properties of Ionic Compounds
Forming NaCl from Na(s) and Cl2(g)
• A metal atom can transfer an electron to a
nonmetal atom.
• The resulting cation and anion are attracted
to each other by electrostatic forces.
Chapter 2 — Atoms, Molecules, Ions
Electrostatic Forces
COULOMB’S LAW
15
Electrostatic Forces
• The oppositely charged ions in ionic
compounds are attracted to one another by
ELECTROSTATIC FORCES.
• These forces are governed by COULOMB’S
LAW.
• As ion charges increase, the attractive forces
between oppositely charged ions increases.
• As the distance between ions increase, the
attractive forces decreases.
Electrostatic Forces
Affect of Coulomb’s Law
COULOMB’S LAW
NaCl, Na+ and Cl-,
m.p. 804 oC
The bond distance in an ionic compound is measures as a
function of the sum of the individual ionic radii.
Molecular Compounds
Compounds without Ions
MgO, Mg2+ and O2m.p. 2800 oC
MgO with the greater charge and smaller bond
distance has the higher melting point.
Naming Molecular Compounds
When non-metals combine, they form molecules.
They may do so in multiple forms:
CO “carbon monoxide”
CO2 Carbon dioxide
CH4 methane
BCl3 boron trichloride
CO2 “carbon dioxide”
Because of this we need to specify the number of
each atom by way of a prefix.
1
mono
6
hexa
2
di
7
hepta
3
tri
8
octa
4
tetra
9
nona
5
penta
10
deca
Chapter 2 — Atoms, Molecules, Ions
Examples:
Formula
Naming Molecular Compounds
Name:
BCl3
boron trichloride
SO3
sulfur trioxide
NO
nitrogen monoxide
we don’t write:
16
CO2 Carbon dioxide
nitrogen monooxide
or mononitrogen monoxide
CH4 methane
N2O4
BCl3
boron trichloride
dinitrogen tetraoxide
Counting Atoms
Mg burns in air (O2) to produce
white magnesium oxide, MgO.
All are formed from
two or more
nonmetals.
Ionic compounds
generally involve a
metal and
nonmetal
(NaCl)
Counting Atoms: The Mole
Chemistry is a quantitative
science—we need a
“counting unit” the:
MOLE
How can we figure out how
much oxide is produced from a
given mass of Mg?
Counting Atoms: The Mole
Avogadro’s Number and the Mole
• The concept of a mole is defined so that we may equate the
amount of matter (mass) to the number of particles (mole).
• The Standard is based upon the C -12 isotope.
• The mass of one C - 12 atom is 1.99265  10-23 g.
• The atomic mass of C - 12 is defined as exactly 12 u.
• Therefore: 1u = (the mass of one 12C atom 12)
= 1.66054  10-24g
= 1.66054  10-27 kg
1 mole is the amount of
substance that contains as
many particles (atoms,
molecules) as there are in
12.0 g of 12C.
518 g of Pb, 2.50 mol
Counting Atoms: The Mole
Avogadro’s Number and the Mole
• Since one mole of C -12 has a mass of 12g (exactly), 12g of
12C contains 6.022142  1023 C -12 atoms.
• But carbon exists as 3 isotopes: C -12, C -13 & C -14
• The average atomic mass of carbon is 12.011 u.
• From this we conclude that 12.011g of carbon contains
6.022142  1023 C-atoms
Is this a valid assumption?
Yes, since NA is so large, the statistics hold.
Chapter 2 — Atoms, Molecules, Ions
17
Particles and the Mole
“Avogadro’s Number”
Named for Amedeo Avogadro
1776-1856
Molar Mass
Since we can equate mass (how much matter) with moles
(how many particles) we now have a conversion factor that
relates the two.
mols 
6.0221415  1023
molar mass (g/mol) = grams
The Molar Mass of a substance is the amount of matter
that contains one-mole or 6.022  1023 particles.
aka: Avogadro's number (NA)
There is Avogadro’s number of particles in a
mole of any substance.
One-Mole Amounts
The atomic masses on the Periodic Table also represent
the molar masses of each element in grams per mole
(g/mol)
Grams, Moles, & Molar Mass
The molar mass of an atom is a conversion factor that
relates mass (grams) to moles and vice versa.
(how much matter)
(number of atoms)
g
mol
Avogadro's number (NA) relates moles numbers of individual
particles:
1 mol 
Mass/Moles Conversions
6.022  1023 particles
 6.022  1023 particles
mol
Problem: How many magnesium atoms are there in
150.0 g of magnesium?
Chapter 2 — Atoms, Molecules, Ions
18
Problem: How many magnesium atoms are there in
150.0 g of magnesium?
Problem: How many magnesium atoms are there in
150.0 g of magnesium?
Solution: Use the molar mass of Mg from the periodic table
and Avogadro's number.
Solution: Use the molar mass of Mg from the periodic table
and Avogadro's number.
150.0g Mg
Problem: How many magnesium atoms are there in
150.0 g of magnesium?
Problem: How many magnesium atoms are there in
150.0 g of magnesium?
Solution: Use the molar mass of Mg from the periodic table
and Avogadro's number.
Solution: Use the molar mass of Mg from the periodic table
and Avogadro's number.
150.0g Mg ´
1 mol Mg
24.305g Mg
150.0g Mg ´
1 mol Mg
6.022 x 1023 Mg atoms
´
24.305g Mg
1 mol Mg
Problem: How many magnesium atoms are there in
150.0 g of magnesium?
Problem: How many magnesium atoms are there in
150.0 g of magnesium?
Solution: Use the molar mass of Mg from the periodic table
and Avogadro's number.
Solution: Use the molar mass of Mg from the periodic table
and Avogadro's number.
150.0g Mg ´
1 mol Mg
6.022 x 1023 Mg atoms
´
24.305g Mg
1 mol Mg
150.0g Mg ´
1 mol Mg
6.022 x 1023 Mg atoms
´
24.305g Mg
1 mol Mg
= 3.717 ´ 1024 Mg atoms
Chapter 2 — Atoms, Molecules, Ions
Problem: How many magnesium atoms are there in
150.0 g of magnesium?
19
How many grams of chromium are there in
25.1g of chromium (III) acetate?
Solution: Use the molar mass of Mg from the periodic table
and Avogadro's number.
Cr(C2H3O2)3 = 229.13 g/mol
1 mol Mg
6.022 x 1023 Mg atoms
150.0g Mg ´
´
24.305g Mg
1 mol Mg
25.1g Cr(C2 H3O2 )3 ´
1 mol Cr(C2 H3O 2 )3
229.13g Cr(C2 H3O 2 )3
´
= 3.717 ´ 1024 Mg atoms
5.70 g Cr 3 sf
(4 sig figs) big number as expected!
Molecular Weight & Molar Mass
1 mol Cr
52.00g Cr
´
=
1 mol Cr(C2 H3O2 )3
1mol Cr
What is the molar mass of ethanol,
C2H6O?
Molecular weight = sum of the atomic
weights of all atoms in the molecule.
1 mol contains:
Molar mass = molecular weight in grams
2 mol  C (12.01g C/1 mol) = 24.02 g C
6 mol  H (1.01g H/1 mol)
= 6.06 g H
1 mol  O (16.00g O/1 mol) = 16.00 g O
TOTAL = molar mass
Molar Masses
Note that the mass of water is
included in the molar mass of a
compound.
= 46.08 g/mol
Percent Composition
• Each element in a compound my be
represented by its weight (mass) percent.
• This is calculated by the following:
Wt.% =
mass (g) of indiv. element in one formula unit
´ 100
mass of one formula unit (g)
• The sum of the wt. % of each element in the
compound must sum to be 100%.
Chapter 2 — Atoms, Molecules, Ions
20
Percent Composition
Percent Composition
• A pure compound always consists of the
same elements combined in the same
proportions by weight.
• Therefore, we can express molecular
composition as PERCENT BY WEIGHT
• Consider some of the family of nitrogenoxygen compounds:
• NO2, nitrogen dioxide and closely related,
NO, nitrogen monoxide (or nitric oxide)
Ethanol, C2H6O
52.13% C
13.15% H
34.72% O
Structure of NO2
Percent Composition
Percent Composition
Consider NO2: What is the weight percent of each
element in the compound?
Consider NO2: What is the weight percent of each
element in the compound?
Solution: From the formula of the compound, find the
moles and mass of each element, then use the
masses and molar mass to find the mass
percentages.
Wt% N =
1mol N
14.01g 1mol NO2
´
´
´ 100 = 30.45% N
1mol NO2 1mol N 46.01g NO2
Wt% O =
2mol O 16.00g O 1mol NO2
´
´
´ 100 = 69.55%O
1mol NO2
1mol O
46.01g NO2
1 mol N ?
14.01g N
16.00g O
+ 1 mol O ?
= 46.01 g/mol
1mol N
1mol O
Describing Compound Formulas:
Percent Composition
Given the relative percentages of each element in
a compound,
10 % X,
20 % Y,
30 % Z
etc…
one can find the empirical formula of the compound.
The empirical formula of a compound or molecule
represents the simplest ratio of each element in 1
mol of the compound or molecule.
Or: 100  30.45% = 69.55% O
Example: A compound is found to be 64.82 %
carbon, 21.59 % oxygen and 13.59 % hydrogen.
What is the empirical formula for this compound?
Chapter 2 — Atoms, Molecules, Ions
21
Problem Strategy
Map
Example: A compound is found to be 64.82 %
The solution to
the problem
involves the
following series
of steps.
Solution: determine X, Y & Z in (CXHYOZ)
Example: A compound is found to be 64.82 %
Example: A compound is found to be 64.82 %
carbon, 21.59 % oxygen and 13.59 % hydrogen.
What is the empirical formula for this compound?
carbon, 21.59 % oxygen and 13.59 % hydrogen.
What is the empirical formula for this compound?
Solution: determine X, Y & Z in (CXHYOZ)
Solution: determine X, Y & Z in (CXHYOZ)
1. Since the percentages for each element sum to
1. Since the percentages for each element sum to
100%, if one equates % to grams (g), the sum of
100%, if one equates % to grams (g), the sum of
the masses must be 100g.
the masses must be 100g.
(i.e. one can assume 100g of the compound)
carbon, 21.59 % oxygen and 13.59 % hydrogen.
What is the empirical formula for this compound?
(i.e. one can assume 100g of the compound)
64.82 g C
21.59 g O
13.59 g H
Example: A compound is found to be 64.82 %
Example: A compound is found to be 64.82 %
carbon, 21.59 % oxygen and 13.59 % hydrogen.
What is the empirical formula for this compound?
carbon, 21.59 % oxygen and 13.59 % hydrogen.
What is the empirical formula for this compound?
2. Convert the grams of each element to moles.
2. Convert the grams of each element to moles.
(g element X  mole X etc…)
(g element X  mole X etc…)
1 mol C
64.82g C ´
= 5.397 mol C
12.011 g C
Chapter 2 — Atoms, Molecules, Ions
22
Example: A compound is found to be 64.82 %
Example: A compound is found to be 64.82 %
carbon, 21.59 % oxygen and 13.59 % hydrogen.
What is the empirical formula for this compound?
carbon, 21.59 % oxygen and 13.59 % hydrogen.
What is the empirical formula for this compound?
2. Convert the grams of each element to moles.
2. Convert the grams of each element to moles.
Oxygen in a
compound is
always
monatomic…
(g element X  mole X etc…)
1 mol C
64.82g C ´
= 5.397 mol C
12.011 g C
21.59g O ´
1 mol O
16.00 g O
= 1.349 mol O
Oxygen in a
compound is
always
monatomic…
The same
applies to
hydrogen!
(g element X  mole X etc…)
1 mol C
64.82g C ´
= 5.397 mol C
12.011 g C
21.59g O ´
1 mol O
16.00 g O
13.59g H ´
1 mol H
= 13.48 mol H
1.0079 g H
= 1.349 mol O
Example: A compound is found to be 64.82 %
Example: A compound is found to be 64.82 %
carbon, 21.59 % oxygen and 13.59 % hydrogen.
What is the empirical formula for this compound?
carbon, 21.59 % oxygen and 13.59 % hydrogen.
What is the empirical formula for this compound?
3. Divide each of the individual moles by the smallest
number of moles to gain the molar ratios for each
element in the compound.
5.397 mol C 1.349 mol O 13.48 mol H
These are the formula subscripts. (C XHYOZ etc…)
3. Divide each of the individual moles by the smallest
number of moles to gain the molar ratios for each
element in the compound.
5.397 mol C 1.349 mol O 13.48 mol H
These are the formula subscripts. (C XHYOZ etc…)
Subscript for C
X=
5.397
= 4.001
1.349
Example: A compound is found to be 64.82 %
Example: A compound is found to be 64.82 %
carbon, 21.59 % oxygen and 13.59 % hydrogen.
What is the empirical formula for this compound?
carbon, 21.59 % oxygen and 13.59 % hydrogen.
What is the empirical formula for this compound?
3. Divide each of the individual moles by the smallest
number of moles to gain the molar ratios for each
element in the compound.
5.397 mol C 1.349 mol O 13.48 mol H
These are the formula subscripts. (C XHYOZ etc…)
3. Divide each of the individual moles by the smallest
number of moles to gain the molar ratios for each
element in the compound.
5.397 mol C 1.349 mol O 13.48 mol H
These are the formula subscripts. (C XHYOZ etc…)
Subscript for C
X=
Subscript for H
13.48
5.397
= 4.001 Y =
= 9.992
1.349
1.349
Subscript for C
X=
Subscript for H
Subscript for O
13.48
1.349
5.397
= 4.001 Y =
= 9.992 Z =
= 1.000
1.349
1.349
1.349
Chapter 2 — Atoms, Molecules, Ions
23
Example: A compound is found to be 64.82 %
Example: A compound is found to be 64.82 %
carbon, 21.59 % oxygen and 13.59 % hydrogen.
What is the empirical formula for this compound?
carbon, 21.59 % oxygen and 13.59 % hydrogen.
What is the empirical formula for this compound?
**If the ratios are fractional (0.5, 1.5 or 0.333) multiply
each ratio by a whole number to get even number
Rounding to the nearest whole numbers:
CXHYOZ
formula subscripts.
X = 4.001 = 4
Y = 9.992 =10
Z = 1.000 = 1
Examples:
0.5  2 = 1
0.25  4 = 1
0.333  3 = 1
C4H10O
The empirical formula is:
The results of this calculation tells us only about the
empirical formula of the compound.
Example: A compound is found to be 64.82 %
Example: A compound is found to be 64.82 %
carbon, 21.59 % oxygen and 13.59 % hydrogen.
What is the empirical formula for this compound?
carbon, 21.59 % oxygen and 13.59 % hydrogen.
What is the empirical formula for this compound?
4. Determining the Molecular Formula
4. Determining the Molecular Formula
For some compounds, the molecular formula is a
multiple of the empirical formula:
empirical formula
n x CXHYOZ
Since the molecular formula is a multiple is scaled
by a factor “n”, the molecular and empirical
molar masses must also scale by the same ratio.
molecular formula
=
( )
( )
molar formula mass g
CnXHnYOnZ
n = 2,3,4…
mol
=n
empirical formula mass g
mol
Example: A compound is found to be 64.82 %
Example: A compound is found to be 64.82 %
carbon, 21.59 % oxygen and 13.59 % hydrogen.
What is the empirical formula for this compound?
carbon, 21.59 % oxygen and 13.59 % hydrogen.
What is the empirical formula for this compound?
4. Determining the Molecular Formula
4. Determining the Molecular Formula
• The empirical formula weight for C4H10O is 74.12
g/mol
• In a separate experiment, the molar mass of the
compound was determined to be 222.1 g/mol.
• What is the molecular formula?
222.1 g / mol
= 2.996 or 3
74.12 g / mol
(C4H10O )´ 3 = C12H30O3
The molecular formula of the compound is:
C12H30O3
Chapter 2 — Atoms, Molecules, Ions
Determine the Formula of a
Compound of Sn & I
24
Example: 1.056g of Sn is reacted with 1.947g of I2.
After the reaction is complete, 0.601g of Sn is
recovered. What is the formula for SnIx?
Sn(s) + I2(s)  SnIx
unknown formula
Example: 1.056g of Sn is reacted with 1.947g
of I2. After the reaction is complete, 0.601g of
Sn is recovered. What is the formula for SnIx?
Example: 1.056g of Sn is reacted with 1.947g
of I2. After the reaction is complete, 0.601g of
Sn is recovered. What is the formula for SnIx?
What do we know?
• The mass of I2 reacted.
• The mass of Sn reacted.
(mass initial – mass left over)
What do we know?
• The mass of I2 reacted.
• The mass of Sn reacted.
(mass initial – mass left over)
What does that tell us?
• Moles of Sn and I2 reacted.
Example: 1.056g of Sn is reacted with 1.947g
of I2. After the reaction is complete, 0.601g of
Sn is recovered. What is the formula for SnIx?
Example: 1.056g of Sn is reacted with 1.947g
of I2. After the reaction is complete, 0.601g of
Sn is recovered. What is the formula for SnIx?
What do we know?
• The mass of I2 reacted.
• The mass of Sn reacted.
(mass initial – mass left over)
What does that tell us?
• Moles of Sn and I2 reacted.
What does that yield?
• The compound’s formula!
mass of Sn that combined
• Mass of Sn initially
• Mass of Sn recovered
• Mass of Sn reacted
moles of Sn used:
0.455 g Sn ´
with the I2:
=
1.056 g
=
0.601 g
=
0.455 g
1 mol Sn
= 3.83 x 10-3 mol Sn
118.7 g Sn
Chapter 2 — Atoms, Molecules, Ions
25
Example: 1.056g of Sn is reacted with 1.947g
of I2. After the reaction is complete, 0.601g of
Sn is recovered. What is the formula for SnIx?
Example: 1.056g of Sn is reacted with 1.947g
of I2. After the reaction is complete, 0.601g of
Sn is recovered. What is the formula for SnIx?
mols of I that combined with the Sn:
Now find the ratio of number of moles of moles of I
and Sn that combined.
1.947 g I2 ´
1 mol I2
2 mol I
´
= 1.534 ´ 10-2 mol I
253.81 g I2 1 mol I2
1.534 x 10-2 mol I
3.83 x 10-3 mol Sn
=
4.01 mol I
1.00 mol Sn
Empirical formula is SnI4
Hydrates
Naming Hydrates
A hydrate is a substance composed of an inorganic salt
and physically bound water.
MXnH2O
salt
prefix:
water
n = is the ratio of moles of water to 1 mole of the
salt
n=
Salt name + prefix hydrate
mols H 2O
mols MX
Heating Hydrates
mono, di, tri etc…
BaCl22H2O barium chloride dihydrate
sodium sulfate pentahydrate
Hydrate Formulas
When a hydrate is heated, the water is liberated.
Ä
® BaCl2 (s) +2H2O(g)
BaCl2 × 2H2O(s) ¾¾
• For every one mole of hydrate, n moles (in this case
2) of water are liberated.
• After heating, any mass that remains is due to the
salt residue.
• Therefore, any mass loss in heating is due to the
water loss.
mass sample – mass residue = mass water lost
The solution to the
problem involves
the following series
of steps.
Na2SO45H2O
Chapter 2 — Atoms, Molecules, Ions
Hydrate Formulas
From the mass of the water lost, one can find the mols of
water associated with the hydrate.
Ä
® BaCl2 (s) +2H2O(g)
BaCl2 × 2H2O(s) ¾¾
Knowing the mass and chemical identity of the anhydrous
salt…
One can determine the mols of the anhydrous salt.
The ratio of mols of water to mols of salt (n) is found by:
n=
mols H2O from hydrate
mols anhydrous salt
26