Download Regents Review Live

Mark Rosengarten’s Amazing
Chemistry Powerpoint Presentation!
 Aligned to the New York State Standards and
Core Curriculum for “The Physical SettingChemistry”
 Can be used in any high-school chemistry class!
 Please give the link to this file to your chemistry
students! www.markrosengarten.com
 Enjoy it!!! A LOT of work has gone into
bringing you this work, so please credit me when
you use it!
(c) 2006, Mark Rosengarten
Outline for Review
1) The Atom (Nuclear, Electron Config)
2) Matter (Phases, Types, Changes)
3) Bonding (Periodic Table, Ionic, Covalent)
4) Compounds (Formulas, Reactions, IMAF’s)
5) Math of Chemistry (Formula Mass, Gas Laws,
Neutralization, etc.)
6) Kinetics and Thermodynamics (PE Diagrams, etc.)
7) Acids and Bases (pH, formulas, indicators, etc.)
8) Oxidation and Reduction (Half Reactions, Cells, etc.)
9) Organic Chemistry (Hydrocarbons, Families, Reactions)
(c) 2006, Mark Rosengarten
The Atom
1) Nucleons
2) Isotopes
3) Natural Radioactivity
4) Half-Life
5) Nuclear Power
6) Electron Configuation
7) Development of the Atomic Model
(c) 2006, Mark Rosengarten
Nucleons
 Protons: +1 each, determines identity of element, mass of





1 amu, determined using atomic number, nuclear charge
Neutrons: no charge, determines identity of isotope of an
element, 1 amu, determined using mass number - atomic
number (amu = atomic mass unit)
32 S and 33 S are both isotopes of S
16
16
S-32 has 16 protons and 16 neutrons
S-33 has 16 protons and 17 neutrons
All atoms of S have a nuclear charge of +16 due to the 16
protons.
(c) 2006, Mark Rosengarten
Isotopes
 Atoms of the same element MUST contain the same
number of protons.
 Atoms of the same element can vary in their numbers of
neutrons, therefore many different atomic masses can exist
for any one element. These are called isotopes.
 The atomic mass on the Periodic Table is the weightaverage atomic mass, taking into account the different
isotope masses and their relative abundance.
 Rounding off the atomic mass on the Periodic Table will
tell you what the most common isotope of that element is.
(c) 2006, Mark Rosengarten
Weight-Average Atomic Mass
 WAM = ((% A of A/100) X Mass of A) + ((% A of B/100) X Mass of B) + …
 What is the WAM of an element if its isotope masses and
abundances are:
– X-200: Mass = 200.0 amu, % abundance = 20.0 %
– X-204: Mass = 204.0 amu, % abundance = 80.0%
– amu = atomic mass unit (1.66 × 10-27 kilograms/amu)
(c) 2006, Mark Rosengarten
Most Common Isotope
 The weight-average atomic mass of Zinc is
65.39 amu. What is the most common isotope
of Zinc? Zn-65!
 What are the most common isotopes of:
– Co
–S
Ag
Pb
 FACT: one atomic mass unit (1.66 × 10-27
kilograms) is defined as 1/12 of the mass of an
atom of C-12.
 This method doesn’t always work, but it usually
(c) 2006,
Mark
Rosengarten
does. Use
it for
the
Regents exam.
Natural Radioactivity
 Alpha Decay
 Beta Decay
 Positron Decay
 Gamma Decay
 Charges of Decay Particles
 Natural decay starts with a parent
nuclide that ejects a decay particle to
form a daughter nuclide which is more
stable than the parent nuclide was.
(c) 2006, Mark Rosengarten
Alpha Decay
 The nucleus ejects two protons and two neutrons. The
atomic mass decreases by 4, the atomic number decreases
by 2.
 23892U 
(c) 2006, Mark Rosengarten
Beta Decay
 A neutron decays into a proton and an electron. The
electron is ejected from the nucleus as a beta particle. The
atomic mass remains the same, but the atomic number
increases by 1.
 146C 
(c) 2006, Mark Rosengarten
Positron Decay
 A proton is converted into a neutron and a positron. The
positron is ejected by the nucleus. The mass remains the
same, but the atomic number decreases by 1.
 5326Fe 
(c) 2006, Mark Rosengarten
Gamma Decay
 The nucleus has energy levels just like electrons, but the
involve a lot more energy. When the nucleus becomes
more stable, a gamma ray may be released. This is a
photon of high-energy light, and has no mass or charge.
The atomic mass and number do not change with gamma.
Gamma may occur by itself, or in conjunction with any
other decay type.
(c) 2006, Mark Rosengarten
Charges of Decay Particles
(c) 2006, Mark Rosengarten
Half-Life
 Half life is the time it takes for half of
the nuclei in a radioactive sample to
undergo decay.
 Problem Types:
– Going forwards in time
– Going backwards in time
– Radioactive Dating
(c) 2006, Mark Rosengarten
Going Forwards in Time
 How many grams of a 10.0 gram sample of I-131 (half-life
of 8 days) will remain in 24 days?
 #HL = t/T = 24/8 = 3
 Cut 10.0g in half 3 times: 5.00, 2.50, 1.25g
(c) 2006, Mark Rosengarten
Going Backwards in Time
 How many grams of a 10.0 gram sample of I-131 (half-life
of 8 days) would there have been 24 days ago?
 #HL = t/T = 24/8 = 3
 Double 10.0g 3 times: 20.0, 40.0, 80.0 g
(c) 2006, Mark Rosengarten
Radioactive Dating
 A sample of an ancient scroll contains 50% of the original
steady-state concentration of C-14. How old is the scroll?
 50% = 1 HL
 1 HL X 5730 y/HL = 5730y
(c) 2006, Mark Rosengarten
Nuclear Power
 Artificial Transmutation
 Particle Accelerators
 Nuclear Fission
 Nuclear Fusion
(c) 2006, Mark Rosengarten
Artificial Transmutation
 4020Ca + _____ ----->
 9642Mo +
2
40 K
19
+ 11H
1 n + _____
H
----->
1
0
 Nuclide + Bullet --> New Element + Fragment(s)
 The masses and atomic numbers must
add up to be the same on both sides of
the arrow.
(c) 2006, Mark Rosengarten
Particle Accelerators
 Devices that use electromagnetic fields to accelerate
particle “bullets” towards target nuclei to make artificial
transmutation possible!
 Most of the elements from 93 on up (the “transuranium”
elements) were created using particle accelerators.
 Particles with no charge cannot be accelerated by the
charged fields.
(c) 2006, Mark Rosengarten
Nuclear Fission
 9236Kr + 14156Ba + 3 10n + energy
 The three neutrons given off can be reabsorbed by
other U-235 nuclei to continue fission as a chain
reaction

235
92U
+
1
0n
 A tiny bit of mass is lost (mass defect) and converted
2006,
Mark Rosengarten
into a huge amount(c)of
energy.
Chain Reaction
(c) 2006, Mark Rosengarten
Nuclear Fusion
 21H + 21H  42He + energy
 Two small, positively-charged nuclei
smash together at high temperatures and
pressures to form one larger nucleus.
 A small bit of mass is destroyed and
converted into a huge amount of energy,
more than even fission.
(c) 2006, Mark Rosengarten
Electron Configuration
 Basic Configuration
 Valence Electrons
 Electron-Dot (Lewis Dot) Diagrams
 Excited vs. Ground State
 What is Light?
(c) 2006, Mark Rosengarten
Basic Configuration
 The number of electrons is determined from the atomic
number.
 Look up the basic configuration below the atomic number
on the periodic table. (PEL: principal energy level = shell)
 He: 2 (2 e- in the 1st PEL)
 Na: 2-8-1 (2 e- in the 1st PEL, 8 in the 2nd and 1 in the
3rd)
 Br: 2-8-18-7 (2 e- in the 1st PEL, 8 in the 2nd, 18 in the
3rd and 7 in the 4th)
(c) 2006, Mark Rosengarten
Valence Electrons
 The valence electrons are responsible for all chemical




bonding.
The valence electrons are the electrons in the outermost
PEL (shell).
He: 2 (2 valence electrons)
Na: 2-8-1 (1 valence electron)
Br: 2-8-18-7 (7 valence electrons)
 The maximum number of valence electrons an atom can
have is EIGHT, called a STABLE OCTET.
(c) 2006, Mark Rosengarten
Electron-Dot Diagrams
 The number of dots equals the number of valence
electrons.
 The number of unpaired valence electrons in a nonmetal
tells you how many covalent bonds that atom can form
with other nonmetals or how many electrons it wants to
gain from metals to form an ion.
 The number of valence electrons in a metal tells you how
many electrons the metal will lose to nonmetals to form an
ion. Caution: May not work with transition metals.
 EXAMPLE DOT DIAGRAMS
(c) 2006, Mark Rosengarten
Example Dot Diagrams
Carbon can also have this dot diagram, which it
has when it forms
compounds.
(c) 2006,organic
Mark Rosengarten
Excited vs. Ground State
 Configurations on the Periodic Table are ground state
configurations.
 If electrons are given energy, they rise to higher energy
levels (excited state).
 If the total number of electrons matches in the
configuration, but the configuration doesn’t match, the
atom is in the excited state.
 Na (ground, on table): 2-8-1
 Example of excited states: 2-7-2, 2-8-0-1, 2-6-3
(c) 2006, Mark Rosengarten
What Is Light?
 Light is formed when electrons drop from
the excited state to the ground state.
 The lines on a bright-line spectrum come
from specific energy level drops and are
unique to each element.
 EXAMPLE SPECTRUM
(c) 2006, Mark Rosengarten
EXAMPLE SPECTRUM
This is the bright-line spectrum of hydrogen. The top
numbers represent the PEL (shell) change that produces the
light with that color and the bottom number is the
wavelength of the light (in nanometers, or 10-9 m).
No other element has the same bright-line spectrum as
hydrogen, so these spectra can be used to identify
elements or mixtures of elements.
(c) 2006, Mark Rosengarten
Development of the Atomic
Model
 Thompson Model
 Rutherford Gold Foil Experiment and
Model
 Bohr Model
 Quantum-Mechanical Model
(c) 2006, Mark Rosengarten
Thompson Model
 The atom is a positively charged diffuse
mass with negatively charged electrons
stuck in it.
(c) 2006, Mark Rosengarten
Rutherford Model
 The atom is made of a small, dense, positively charged
nucleus with electrons at a distance, the vast majority of
the volume of the atom is empty space.
Alpha particles shot
at a thin sheet of gold
foil: most go through
(empty space). Some
deflect or bounce off
(small + charged
nucleus).
(c) 2006, Mark Rosengarten
Bohr Model
 Electrons orbit around the nucleus in energy levels (shells).
Atomic bright-line spectra was the clue.
(c) 2006, Mark Rosengarten
Quantum-Mechanical Model
 Electron energy levels are wave functions.
 Electrons are found in orbitals, regions of space where an
electron is most likely to be found.
 You can’t know both where the electron is and where it is
going at the same time.
 Electrons buzz around the nucleus like gnats buzzing
around your head.
(c) 2006, Mark Rosengarten
Matter
1) Properties of Phases
2) Types of Matter
3) Phase Changes
(c) 2006, Mark Rosengarten
Properties of Phases
 Solids: Crystal lattice (regular geometric pattern),
vibration motion only
 Liquids: particles flow past each other but are still
attracted to each other.
 Gases: particles are small and far apart, they travel in a
straight line until they hit something, they bounce off
without losing any energy, they are so far apart from each
other that they have effectively no attractive forces and
their speed is directly proportional to the Kelvin
temperature (Kinetic-Molecular Theory, Ideal Gas Theory)
(c) 2006, Mark Rosengarten
Solids
The positive and
negative ions
alternate in the
ionic crystal lattice
of NaCl.
(c) 2006, Mark Rosengarten
Liquids
When heated, the ions move
faster and eventually
separate from each other to
form a liquid. The ions are
loosely held together by the
oppositely charged ions, but
the ions are moving too fast
for the crystal lattice to stay
together.
(c) 2006, Mark Rosengarten
Gases
Since all gas molecules spread out
the same way, equal volumes of
gas under equal conditions of
temperature and pressure will
contain equal numbers of
molecules of gas. 22.4 L of any
gas at STP (1.00 atm and 273K)
will contain one mole
(6.02 X 1023) gas molecules.
Since there is space between gas
molecules, gases are affected by
(c) 2006, Mark
Rosengarten
changes
in pressure.
Types of Matter
 Substances (Homogeneous)
– Elements (cannot be decomposed by chemical
change): Al, Ne, O, Br, H
– Compounds (can be decomposed by chemical
change): NaCl, Cu(ClO3)2, KBr, H2O, C2H6
 Mixtures
– Homogeneous: Solutions (solvent + solute)
– Heterogeneous: soil, Italian dressing, etc.
(c) 2006, Mark Rosengarten
Elements
 A sample of lead atoms (Pb). All
atoms in the sample consist of lead,
so the substance is homogeneous.
 A sample of chlorine atoms (Cl).
All atoms in the sample consist of
chlorine, so the substance is
homogeneous.
(c) 2006, Mark Rosengarten
Compounds
 Lead has two charges listed, +2
and +4. This is a sample of lead
(II) chloride (PbCl2). Two or
more elements bonded in a
whole-number ratio is a
COMPOUND.
 This compound is formed from
the +4 version of lead. This is
lead (IV) chloride (PbCl4).
Notice how both samples of lead
compounds have consistent
composition throughout?
(c) 2006, Mark Rosengarten
Compounds are homogeneous!
Mixtures
 A mixture of lead atoms and
chlorine atoms. They exist in no
particular ratio and are not
chemically combined with each
other. They can be separated by
physical means.
 A mixture of PbCl2 and PbCl4
formula units. Again, they are in
no particular ratio to each other
and can be separated without
chemical change.
(c) 2006, Mark Rosengarten
Phase Changes
 Phase Change Types
 Phase Change Diagrams
 Heat of Phase Change
 Evaporation
(c) 2006, Mark Rosengarten
Phase Change Types
(c) 2006, Mark Rosengarten
Phase Change Diagrams
AB: Solid Phase
BC: Melting (S + L)
CD: Liquid Phase
DE: Boiling (L + G)
EF: Gas Phase
Notice how temperature remains constant during a phase
(c) 2006, Mark Rosengarten
change? That’s because the PE is changing, not the KE.
Heat of Phase Change
 How many joules would it take to melt 100. g of H2O (s) at
0oC?
 q=mHf = (100. g)(334 J/g) = 33400 J
 How many joules would it take to boil 100. g of H2O (l) at
100oC?
 q=mHv = (100.g)(2260 J/g) = 226000 J
(c) 2006, Mark Rosengarten
Evaporation
 When the surface molecules of a gas travel upwards at a




great enough speed to escape.
The pressure a vapor exerts when sealed in a container at
equilibrium is called vapor pressure, and can be found on
Table H.
When the liquid is heated, its vapor pressure increases.
When the liquid’s vapor pressure equals the pressure
exerted on it by the outside atmosphere, the liquid can boil.
If the pressure exerted on a liquid increases, the boiling
point of the liquid increases (pressure cooker). If the
pressure decreases, the boiling point of the liquid decreases
(c) 2006, Mark Rosengarten
(special cooking directions for high elevations).
Reference Table H: Vapor
Pressure of Four Liquids
(c) 2006, Mark Rosengarten
Bonding
1) The Periodic Table
2) Ions
3) Ionic Bonding
4) Covalent Bonding
5) Metallic Bonding
(c) 2006, Mark Rosengarten
The Periodic Table
 Metals
 Nonmetals
 Metalloids
 Chemistry of Groups
 Electronegativity
 Ionization Energy
(c) 2006, Mark Rosengarten
Metals
 Have luster, are malleable and ductile, good
conductors of heat and electricity
 Lose electrons to nonmetal atoms to form
positively charged ions in ionic bonds
 Large atomic radii compared to nonmetal atoms
 Low electronegativity and ionization energy
 Left side of the periodic table (except H)
(c) 2006, Mark Rosengarten
Nonmetals
 Are dull and brittle, poor conductors
 Gain electrons from metal atoms to form
negatively charged ions in ionic bonds
 Share unpaired valence electrons with other
nonmetal atoms to form covalent bonds and
molecules
 Small atomic radii compared to metal atoms
 High electronegativity and ionization energy
 Right side of the periodic table (except Group 18)
(c) 2006, Mark Rosengarten
Metalloids
 Found lying on the jagged line between metals and
nonmetals flatly touching the line (except Al and Po).
 Share properties of metals and nonmetals (Si is shiny like a
metal, brittle like a nonmetal and is a semiconductor).
(c) 2006, Mark Rosengarten
Chemistry of Groups
 Group 1: Alkali Metals
 Group 2: Alkaline Earth Metals
 Groups 3-11: Transition Elements
 Group 17: Halogens
 Group 18: Noble Gases
 Diatomic Molecules
(c) 2006, Mark Rosengarten
Group 1: Alkali Metals
 Most active metals, only found in compounds in




nature
React violently with water to form hydrogen gas
and a strong base: 2 Na (s) + H2O (l)  2 NaOH
(aq) + H2 (g)
1 valence electron
Form +1 ion by losing that valence electron
Form oxides like Na2O, Li2O, K2O
(c) 2006, Mark Rosengarten
Group 2: Alkaline Earth Metals
 Very active metals, only found in compounds in




nature
React strongly with water to form hydrogen gas
and a base:
– Ca (s) + 2 H2O (l)  Ca(OH)2 (aq) + H2 (g)
2 valence electrons
Form +2 ion by losing those valence electrons
Form oxides like CaO, MgO, BaO
(c) 2006, Mark Rosengarten
Groups 3-11: Transition Metals
 Many can form different possible charges of ions
 If there is more than one ion listed, give the charge as a
Roman numeral after the name
 Cu+1 = copper (I) Cu+2 = copper (II)
 Compounds containing these metals can be colored.
(c) 2006, Mark Rosengarten
Group 17: Halogens
 Most reactive nonmetals
 React violently with metal atoms to form halide




compounds: 2 Na + Cl2  2 NaCl
Only found in compounds in nature
Have 7 valence electrons
Gain 1 valence electron from a metal to form -1
ions
Share 1 valence electron with another nonmetal
atom to form one covalent bond.
(c) 2006, Mark Rosengarten
Group 18: Noble Gases
 Are completely nonreactive since they have eight
valence electrons, making a stable octet.
 Kr and Xe can be forced, in the laboratory, to give
up some valence electrons to react with fluorine.
 Since noble gases do not naturally bond to any
other elements, one atom of noble gas is
considered to be a molecule of noble gas. This is
called a monatomic molecule. Ne represents an
atom of Ne and a molecule of Ne.
(c) 2006, Mark Rosengarten
Diatomic Molecules
 Br, I, N, Cl, H, O and F are so reactive that they exist in a
more chemically stable state when they covalently bond
with another atom of their own element to make two-atom,
or diatomic molecules.
 Br2, I2, N2, Cl2, H2, O2 and F2
 The decomposition of water: 2 H2O  2 H2 + O2
(c) 2006, Mark Rosengarten
Electronegativity
 An atom’s attraction to electrons in a chemical bond.
 F has the highest, at 4.0
 Fr has the lowest, at 0.7
 If two atoms that are different in EN (END) from each
other by 1.7 or more collide and bond (like a metal atom
and a nonmetal atom), the one with the higher
electronegativity will pull the valence electrons away from
the atom with the lower electronegativity to form a (-) ion.
The atom that was stripped of its valence electrons forms a
(+) ion.
 If the two atoms have an END of less than 1.7, they will
(c) 2006, Mark Rosengarten
share their unpaired valence electrons…covalent bond!
Ionization Energy
 The energy required to remove the most loosely held
valence electron from an atom in the gas phase.
 High electronegativity means high ionization energy
because if an atom is more attracted to electrons, it will
take more energy to remove those electrons.
 Metals have low ionization energy. They lose electrons
easily to form (+) charged ions.
 Nonmetals have high ionization energy but high
electronegativity. They gain electrons easily to form (-)
charged ions when reacted with metals, or share unpaired
valence electrons with other nonmetal atoms.
(c) 2006, Mark Rosengarten
Ions
 Ions are charged particles formed by the gain or loss of
electrons.
– Metals lose electrons (oxidation) to form (+) charged
cations.
– Nonmetals gain electrons (reduction) to form (-)
charged anions.
 Atoms will gain or lose electrons in such a way that they
end up with 8 valence electrons (stable octet).
– The exceptions to this are H, Li, Be and B, which are
not large enough to support 8 valence electrons. They
must be satisfied with 2 (Li, Be, B) or 0 (H).
(c) 2006, Mark Rosengarten
Metal Ions (Cations)
 Na: 2-8-1
 Na+1: 2-8
 Ca: 2-8-8-2
Note that when the atom
loses its valence electron,
the next lower PEL
becomes the valence PEL.
 Ca+2: 2-8-8
 Al: 2-8-3
 Al+3: 2-8
Notice how the dot
diagrams for metal ions
lack dots! Place brackets
around the element symbol
and put the charge on the
upper right outside!
(c) 2006, Mark Rosengarten
Nonmetal Ions (Anions)
 F: 2-7
 F-1: 2-8
 O: 2-6
 O-2: 2-8
Note how the ions all have 8
valence electrons. Also note the
gained electrons as red dots.
Nonmetal ion dot diagrams show
8 dots, with brackets around the
dot diagram and the charge of
the ion written to the upper right
side outside the brackets.
 N: 2-5
 N-3: 2-8
(c) 2006, Mark Rosengarten
Ionic Bonding
 If two atoms that are different in EN (END) from each
other by 1.7 or more collide and bond (like a metal atom
and a nonmetal atom), the one with the higher
electronegativity will pull the valence electrons away from
the atom with the lower electronegativity to form a (-) ion.
The atom that was stripped of its valence electrons forms a
(+) ion.
 The oppositely charged ions attract to form the bond. It is
a surface bond that can be broken by melting or dissolving
in water.
 Ionic bonding forms ionic crystal lattices, not molecules.
(c) 2006, Mark Rosengarten
Example of Ionic Bonding
(c) 2006, Mark Rosengarten
Covalent Bonding
 If two nonmetal atoms have an END of 1.7 or less, they
will share their unpaired valence electrons to form a
covalent bond.
 A particle made of covalently bonded nonmetal atoms is
called a molecule.
 If the END is between 0 and 0.4, the sharing of electrons is
equal, so there are no charged ends. This is NONPOLAR
covalent bonding.
 If the END is between 0.5 and 1.7, the sharing of electrons
is unequal. The atom with the higher EN will be d- and the
one with the lower EN will be d+ charged. This is a
(c) 2006, Mark Rosengarten
POLAR covalent bonding.
(d means “partial”)
Examples of Covalent Bonding
(c) 2006, Mark Rosengarten
Metallic Bonding
 Metal atoms of the same element bond with each other by
sharing valence electrons that they lose to each other.
 This is a lot like an atomic game of “hot potato”, where
metal kernals (the atom inside the valence electrons) sit in
a crystal lattice, passing valence electrons back and forth
between each other).
 Since electrons can be forced to travel in a certain direction
within the metal, metals are very good at conducting
electricity in all phases.
(c) 2006, Mark Rosengarten
Compounds
1) Types of Compounds
2) Formula Writing
3) Formula Naming
4) Empirical Formulas
5) Molecular Formulas
6) Types of Chemical Reactions
7) Balancing Chemical Reactions
8) Attractive Forces
(c) 2006, Mark Rosengarten
Types of Compounds
 Ionic: made of metal and nonmetal ions. Form an ionic
crystal lattice when in the solid phase. Ions separate when
melted or dissolved in water, allowing electrical
conduction. Examples: NaCl, K2O, CaBr2
 Molecular: made of nonmetal atoms bonded to form a
distinct particle called a molecule. Bonds do not break
upon melting or dissolving, so molecular substances do not
conduct electricity. EXCEPTION: Acids [H+A- (aq)]
ionize in water to form H3O+ and A-, so they do conduct.
 Network: made up of nonmetal atoms bonded in a
seemingly endless matrix of covalent bonds with no
(c) 2006, Mark
Rosengarten
distinguishable molecules.
Very
high m.p., don’t conduct.
Ionic Compounds
Ionic Crystal Structure, then adding heat (or dissolving in water) to break
up the crystal into a liquid composed of free-moving ions.
(c) 2006, Mark Rosengarten
Molecular Compounds
(c) 2006, Mark Rosengarten
Network Solids
Network solids are made of nonmetal atoms covalently
bonded together to form large crystal lattices. No individual
molecules can be distinguished. Examples include C
(diamond) and SiO2 (quartz). Corundum (Al2O3) also forms
these, even though Al is considered a metal. Network solids
are among the hardest materials known. They have
extremely high melting points and do not conduct electricity.
(c) 2006, Mark Rosengarten
Formula Writing
 The charge of the (+) ion and the charge of the (-) ion must





cancel out to make the formula. Use subscripts to indicate
how many atoms of each element there are in the
compound, no subscript if there is only one atom of that
element.
Na+1 and Cl-1 = NaCl
Ca+2 and Br-1 = CaBr2
Al+3 and O-2 = Al2O3
Zn+2 and PO4-3 = Zn3(PO4)2
Try these problems!
(c) 2006, Mark Rosengarten
Formulas to Write
 Ba+2 and N-3
 NH4+1 and SO4-2
 Li+1 and S-2
 Cu+2 and NO3-1
 Al+3 and CO3-2
 Fe+3 and Cl-1
 Pb+4 and O-2
 Pb+2 and O-2
(c) 2006, Mark Rosengarten
Formula Naming
 Compounds are named from the elements or




polyatomic ions that form them.
KCl = potassium chloride
Na2SO4 = sodium sulfate
(NH4)2S = ammonium sulfide
AgNO3 = silver nitrate
 Notice all the metals listed here only have
one charge listed? So what do you do if a
metal has more than one charge listed? Take
a peek!
(c) 2006, Mark Rosengarten
The Stock System
 CrCl2 = chromium (II) chloride
 CrCl3 = chromium (III) chloride
 CrCl6 = chromium (VI) chloride
 FeO = iron (II) oxide
Try
Co(NO3)2 and
Co(NO3)3
MnS = manganese (II) sulfide
 Fe2O3 = iron (III) oxide
MnS2 = manganese (IV) sulfide
 The Roman numeral(c)is2006,
theMark
charge
of the metal ion!
Rosengarten
Empirical Formulas
 Ionic formulas: represent the simplest whole number mole
ratio of elements in a compound.
 Ca3N2 means a 3:2 ratio of Ca ions to N ions in the
compound.
 Many molecular formulas can be simplified to empirical
formulas
– Ethane (C2H6) can be simplified to CH3. This is the
empirical formula…the ratio of C to H in the molecule.
 All ionic compounds have empirical formulas.
(c) 2006, Mark Rosengarten
Molecular Formulas
 The count of the actual number of atoms of each element




in a molecule.
H2O: a molecule made of two H atoms and one O atom
covalently bonded together.
C2H6O: A molecule made of two C atoms, six H atoms and
one O atom covalently bonded together.
Molecular formulas are whole-number multiples of
empirical formulas:
– H2O = 1 X (H2O)
– C8H16 = 8 X (CH2)
Calculating Molecular
Formulas
(c) 2006,
Mark Rosengarten
Types of Chemical Reactions
 Redox Reactions: driven by the loss (oxidation) and gain
(reduction) of electrons. Any species that does not change
charge is called the spectator ion.
– Synthesis
– Decomposition
– Single Replacement
 Ion Exchange Reaction: driven by the formation of an
insoluble precipitate. The ions that remain dissolved
throughout are the spectator ions.
– Double Replacement
(c) 2006, Mark Rosengarten
Synthesis
 Two elements combine to form a compound
 2 Na + O2  Na2O
 Same reaction, with charges added in:
– 2 Na0 + O20  Na2+1O-2
 Na0 is oxidized (loses electrons), is the reducing agent
 O20 is reduced (gains electrons), is the oxidizing agent
 Electrons are transferred from the Na0 to the O20.
 No spectator ions, there are only two elements here.
(c) 2006, Mark Rosengarten
Decomposition
 A compound breaks down into its original elements.
 Na2O  2 Na + O2
 Same reaction, with charges added in:
– Na2+1O-2  2 Na0 + O20
 O-2 is oxidized (loses electrons), is the reducing agent
 Na+1 is reduced (gains electrons), is the oxidizing agent
 Electrons are transferred from the O-2 to the Na+1.
 No spectator ions, there are only two elements here.
(c) 2006, Mark Rosengarten
Single Replacement
 An element replaces the same type of element in a




compound.
Ca + 2 KCl  CaCl2 + 2 K
Same reaction, with charges added in:
– Ca0 + 2 K+1Cl-1  Ca+2Cl2-1 + 2 K0
Ca0 is oxidized (loses electrons), is the reducing agent
K+1 is reduced (gains electrons), is the oxidizing agent
 Electrons are transferred from the Ca0 to the K+1.
(c) 2006, Mark Rosengarten
 Cl-1 is the spectator ion, since it’s charge doesn’t change.
Double Replacement
 The (+) ion of one compound bonds to the (-) ion of





another compound to make an insoluble precipitate. The
compounds must both be dissolved in water to break the
ionic bonds first.
NaCl (aq) + AgNO3 (aq)  NaNO3 (aq) + AgCl (s)
The Cl-1 and Ag+1 come together to make the insoluble
precipitate, which looks like snow in the test tube.
No species change charge, so this is not a redox reaction.
Since the Na+1 and NO3-1 ions remain dissolved throughout
the reaction, they are the spectator ions.
How do identify the(c)precipitate?
2006, Mark Rosengarten
Identifying the Precipitate
 The precipitate is the compound that is insoluble. AgCl is
a precipitate because Cl- is a halide. Halides are soluble,
(c) 2006, Mark Rosengarten
except when combined
with Ag+ and others.
Balancing Chemical Reactions
 Balance one element or ion at a time
 Use a pencil
 Use coefficients only, never change formulas
 Revise if necessary
 The coefficient multiplies everything in the formula by that
amount
– 2 Ca(NO3)2 means that you have 2 Ca, 4 N and 12 O.
 Examples for you to try!
(c) 2006, Mark Rosengarten
Reactions to Balance
 ___NaCl  ___Na + ___Cl2
 ___Al + ___O2  ___Al2O3
 ___SO3  ___SO2 + ___O2
 ___Ca + ___HNO3  ___Ca(NO3)2 + ___H2
 __FeCl3 + __Pb(NO3)2  __Fe(NO3)3 + __PbCl2
(c) 2006, Mark Rosengarten
Attractive Forces
 Molecules have partially charged ends. The d+ end of one
molecule attracts to the d- end of another molecule.
 Ions are charged (+) or (-). Positively charged ions attract
other to form ionic bonds, a type of attractive force.
 Since partially charged ends result in weaker attractions
than fully charged ends, ionic compounds generally have
much higher melting points than molecular compounds.
 Determining Polarity of Molecules
 Hydrogen Bond Attractions
(c) 2006, Mark Rosengarten
Determining Polarity of
Molecules
-----------------------------------------------------------------------------
(c) 2006, Mark Rosengarten
Hydrogen Bond
Attractions
A hydrogen bond attraction is a
very strong attractive force
between the H end of one polar
molecule and the N, O or F end
of another polar molecule. This
attraction is so strong that water
is a liquid at a temperature
where most compounds that are
much heavier than water (like
propane, C3H8) are gases. This
also gives water its surface
tension and its ability to form a
(c) 2006, Mark Rosengarten
meniscus in a narrow glass tube.
Math of Chemistry
1) Formula Mass
2) Percent Composition
3) Mole Problems
4) Gas Laws
5) Neutralization
6) Concentration
7) Significant Figures and Rounding
8) Metric Conversions
9) Calorimetry (c) 2006, Mark Rosengarten
Formula Mass
 Gram Formula Mass = sum of atomic masses of all




elements in the compound
Round given atomic masses to the nearest tenth
H2O: (2 X 1.0) + (1 X 16.0) = 18.0 grams/mole
Na2SO4: (2 X 23.0)+(1 X 32.1)+(4 X 16.0) = 142.1 g/mole
Now you try:
– BaBr2
– CaSO4
– Al2(CO3)3
(c) 2006, Mark Rosengarten
Percent Composition
What is the % composition, by mass,
of each element in SiO2?
%Si =
(28.1/60.1) X 100 = 46.8%
%O = (2 X 16.0 = 32.0), (32.0/60.1) X 100 = 53.2%
The mass of part is the number of atoms of that element in the
compound. The mass of whole is the formula mass of the
compound. Don’t forget to take atomic mass to the nearest
tenth! This is a problem
for Mark
youRosengarten
to try.
(c) 2006,
Practice Percent
Composition Problem
 What is the percent by mass of each element in Li2SO4?
(c) 2006, Mark Rosengarten
Mole Problems
 Grams <=> Moles
 Molecular Formula
 Stoichiometry
(c) 2006, Mark Rosengarten
Grams <=> Moles
 How many grams will 3.00 moles of NaOH (40.0 g/mol)
weigh?
 3.00 moles X 40.0 g/mol = 120. g
 How many moles of NaOH (40.0 g/mol) are represented
by 10.0 grams?
 (10.0 g) / (40.0 g/mol) = 0.250 mol
(c) 2006, Mark Rosengarten
Molecular Formula
 Molecular Formula = (Molecular Mass/Empirical Mass) X Empirical Formula
 What is the molecular formula of a compound with an
empirical formula of CH2 and a molecular mass of 70.0
grams/mole?
 1) Find the Empirical Formula Mass: CH2 = 14.0
 2) Divide the MM/EM: 70.0/14.0 = 5
 3) Multiply the molecular formula by the result:
5 (CH2) = C5H10
(c) 2006, Mark Rosengarten
Stoichiometry
 Moles of Target = Moles of Given X (Coefficent of Target/Coefficient of given)
 Given the balanced equation N2 + 3 H2  2 NH3, How
many moles of H2 need to be completely reacted with
N2 to yield 20.0 moles of NH3?
 20.0 moles NH3 X (3 H2 / 2 NH3) = 30.0 moles H2
(c) 2006, Mark Rosengarten
Gas Laws
 Make a data table to put the numbers so you can eliminate




the words.
Make sure that any Celsius temperatures are converted to
Kelvin (add 273).
Rearrange the equation before substituting in numbers. If
you are trying to solve for T2, get it out of the denominator
first by cross-multiplying.
If one of the variables is constant, then eliminate it.
Try these problems!
(c) 2006, Mark Rosengarten
Gas Law Problem 1
 A 2.00 L sample of N2 gas
at STP is compressed to
4.00 atm at constant temperature. What is the new
volume of the gas?
 V2 = P1V1 / P2
 = (1.00 atm)(2.00 L) /
(4.00 atm)
 = 0.500 L
(c) 2006, Mark Rosengarten
Gas Law Problem 2
 To what temperature must a 3.000 L sample of O2 gas at
300.0 K be heated to raise the volume to 10.00 L?
 T2 = V2T1/V1
 = (10.00 L)(300.0 K) / (3.000 L) = 1000. K
(c) 2006, Mark Rosengarten
Gas Law Problem 3
 A 3.00 L sample of NH3 gas at 100.0 kPa is cooled from
500.0 K to 300.0 K and its pressure is reduced to 80.0 kPa.
What is the new volume of the gas?
 V2 = P1V1T2 / P2T1
 = (100.0 kPa)(3.00 L)(300. K) / (80.0 kPa)(500. K)
 = 2.25 L
(c) 2006, Mark Rosengarten
Neutralization
 10.0 mL of 0.20 M HCl is neutralized by 40.0 mL of
NaOH. What is the concentration of the NaOH?
 #H MaVa = #OH MbVb, so Mb = #H MaVa / #OH Vb
 = (1)(0.20 M)(10.0 mL) / (1) (40.0 mL) = 0.050 M
 How many mL of 2.00 M H2SO4 are needed to completely
neutralize 30.0 mL of 0.500 M KOH?
(c) 2006, Mark Rosengarten
Concentration
 Molarity
 Parts per Million
 Percent by Mass
 Percent by Volume
(c) 2006, Mark Rosengarten
Molarity
 What is the molarity of a 500.0 mL solution of NaOH (FM
= 40.0) with 60.0 g of NaOH (aq)?
– Convert g to moles and mL to L first!
– M = moles / L = 1.50 moles / 0.5000 L = 3.00 M
 How many grams of NaOH does it take to make 2.0 L of a
0.100 M solution of NaOH (aq)?
– Moles = M X L = 0.100 M X 2.0 L = 0.200 moles
– Convert moles to grams: 0.200 moles X 40.0 g/mol = 8.00 g
(c) 2006, Mark Rosengarten
Parts Per Million
 100.0 grams of water is evaporated and analyzed for lead.
0.00010 grams of lead ions are found. What is the
concentration of the lead, in parts per million?
 ppm = (0.00010 g) / (100.0 g) X 1 000 000 = 1.0 ppm
 If the legal limit for lead in the water is 3.0 ppm, then the water
sample is within the legal limits (it’s OK!)
(c) 2006, Mark Rosengarten
Percent by Mass
 A 50.0 gram sample of a solution is evaporated and found
to contain 0.100 grams of sodium chloride. What is the
percent by mass of sodium chloride in the solution?
 % Comp = (0.100 g) / (50.0 g) X 100 = 0.200%
(c) 2006, Mark Rosengarten
Percent By Volume
 Substitute “volume” for “mass” in the above equation.
 What is the percent by volume of hexane if 20.0 mL of
hexane are dissolved in benzene to a total volume of 80.0
mL?
 % Comp = (20.0 mL) / (80.0 mL) X100 = 25.0%
(c) 2006, Mark Rosengarten
Sig Figs and Rounding
 How many Significant Figures does a number have?
 What is the precision of my measurement?
 How do I round off answers to addition and subtraction
problems?
 How do I round off answers to multiplication and division
problems?
(c) 2006, Mark Rosengarten
How many Sig Figs?
 Start counting sig figs at the first non-zero.
 All digits except place-holding zeroes are sig figs.
Measurement
# of Sig Figs
Measurement
# of Sig Figs
0.115 cm
3
234 cm
3
0.00034 cm
2
67000 cm
2
0.00304 cm
3
_
45000 cm
4
0.0560 cm
3
560. cm
3
0.00070700 cm
5
560.00 cm
5
(c) 2006, Mark Rosengarten
What Precision?
 A number’s precision is determined by the furthest
(smallest) place the number is recorded to.
 6000 mL : thousands place
 6000. mL : ones place
 6000.0 mL : tenths place
 5.30 mL : hundredths place
 8.7 mL : tenths place
 23.740 mL : thousandths place
(c) 2006, Mark Rosengarten
Rounding with addition and
subtraction
 Answers are rounded to the least precise place.
1) 4.732 cm
16.8
cm
+ 0.781 cm
---------22.313 cm
22.3 cm
2)
17.440 mL
3.895 mL
+ 16.77 mL
-------------38.105 mL
38.11 mL
(c) 2006, Mark Rosengarten
3)
32.0
MW
+ 0.0059 MW
--------------32.0059 MW
32.0 MW
Rounding with multiplication
and division
 Answers are rounded to the fewest number of significant
figures.
1)
37.66 KW
x 2.2 h
---------82.852 KWh
83 KWh
2)
14.922 cm
x 2.0 cm
----------2
29.844 cm
2
30. cm
(c) 2006, Mark Rosengarten
3) 98.11 kg
x 200 m
---------19 622 kgm
20 000 kgm
Metric Conversions
 Determine how many powers of ten
difference there are between the two
units (no prefix = 100) and create a
conversion factor. Multiply or divide
the given by the conversion factor.
How many kg are in 38.2 cg?
(38.2 cg) /(100000 cg/kg) = 0.000382 km
How many mL in 0.988 dL?
(0.988 dg) X (100 mL/dL) = 98.8 mL
(c) 2006, Mark Rosengarten
Calorimetry
 This equation can be used to determine any of the variables
here. You will not have to solve for C, since we will
always assume that the energy transfer is being absorbed
by or released by a measured quantity of water, whose
specific heat is given above.
 Solving for q
 Solving for m
(c) 2006, Mark Rosengarten
 Solving for DT
Solving for q
 How many joules are absorbed by 100.0 grams of water in
a calorimeter if the temperature of the water increases from
20.0oC to 50.0oC?
 q = mCDT = (100.0 g)(4.18 J/goC)(30.0oC) = 12500 J
(c) 2006, Mark Rosengarten
Solving for m
 A sample of water in a calorimeter cup increases from
25oC to 50.oC by the addition of 500.0 joules of energy.
What is the mass of water in the calorimeter cup?
 q = mCDT, so m = q / CDT = (500.0 J) / (4.18 J/goC)(25oC) = 4.8 g
(c) 2006, Mark Rosengarten
Solving for DT
 If a 50.0 gram sample of water in a calorimeter cup
absorbs 1000.0 joules of energy, how much will the
temperature rise by?
 q = mCDT, so DT = q / mC = (1000.0 J)/(50.0 g)(4.18 J/goC) = 4.8oC
 If the water started at 20.0oC, what will the final
temperature be?
– Since the water ABSORBS the energy, its temperature will
INCREASE by the DT: 20.0oC + 4.8oC = 24.8oC
(c) 2006, Mark Rosengarten
Kinetics and Thermodynamics
1) Reaction Rate
2) Heat of Reaction
3) Potential Energy Diagrams
4) Equilibrium
5) Le Châtelier’s Principle
6) Solubility Curves
(c) 2006, Mark Rosengarten
Reaction Rate
 Reactions happen when reacting particles collide with
sufficient energy (activation energy) and at the proper
angle.
 Anything that makes more collisions in a given time will
make the reaction rate increase.
– Increasing temperature
– Increasing concentration (pressure for gases)
– Increasing surface area (solids)
 Adding a catalyst makes a reaction go faster by removing
steps from the mechanism and lowering the activation
(c) 2006,
Markup
Rosengarten
energy without getting
used
in the process.
Heat of Reaction
 Reactions either absorb PE (endothermic, +DH) or release
PE (exothermic, -DH)
Exothermic, PEKE, Temp
Endothermic, KEPE, Temp
Rewriting the equation with heat included:
4 Al(s) + 3 O2(g)  2 Al2O3(s) + 3351 kJ
N2(g) + O2(g) +182.6 kJ  2 NO(g)
(c) 2006, Mark Rosengarten
Potential Energy Diagrams
 Steps of a reactions:
– Reactants have a certain amount of PE stored in their
bonds (Heat of Reactants)
– The reactants are given enough energy to collide and
react (Activation Energy)
– The resulting intermediate has the highest energy that
the reaction can make (Heat of Activated Complex)
– The activated complex breaks down and forms the
products, which have a certain amount of PE stored in
their bonds (Heat of Products)
– Hproducts - Hreactants(c)=2006,
DHMark Rosengarten EXAMPLES
Making a PE Diagram
 X axis: Reaction Coordinate (time, no units)
 Y axis: PE (kJ)
 Three lines representing energy (Hreactants, Hactivated complex,
Hproducts)
 Two arrows representing energy changes:
– From Hreactants to Hactivated complex: Activation Energy
– From Hreactants to Hproducts : DH
 ENDOTHERMIC PE DIAGRAM
 EXOTHERMIC PE DIAGRAM
(c) 2006, Mark Rosengarten
Endothermic PE Diagram
If a catalyst is added? (c) 2006, Mark Rosengarten
Endothermic with Catalyst
The red line represents(c)the
catalyzed reaction.
2006, Mark Rosengarten
Exothermic PE Diagram
(c) 2006, Mark Rosengarten
What does it look like with a catalyst?
Exothermic with a Catalyst
The red line represents the catalyzed reaction. Lower
2006, Mark Rosengarten
A.E. and faster reaction(c)time!
Equilibrium
When the rate of the forward
reaction equals the rate of the
(c) 2006, Mark Rosengarten
reverse reaction.
Examples of Equilibrium
 Solution Equilibrium: when a solution is saturated, the
rate of dissolving equals the rate of precipitating.
– NaCl (s)  Na+1 (aq) + Cl-1 (aq)
 Vapor-Liquid Equilibrium: when a liquid is trapped with
air in a container, the liquid evaporates until the rate of
evaporation equals the rate of condensation.
– H2O (l)  H2O (g)
 Phase equilibrium: At the melting point, the rate of solid
turning to liquid equals the rate of liquid turning back to
solid.
– H2O (s)  H2O (l)
(c) 2006, Mark Rosengarten
Le Châtelier’s Principle
 If a system at equilibrium is stressed, the equilibrium




will shift in a direction that relieves that stress.
A stress is a factor that affects reaction rate. Since
catalysts affect both reaction rates equally, catalysts have
no effect on a system already at equilibrium.
Equilibrium will shift AWAY from what is added
Equilibrium will shift TOWARDS what is removed.
This is because the shift will even out the change in
reaction rate and bring the system back to equilibrium
» NEXT
(c) 2006, Mark Rosengarten
Steps to Relieving Stress
 1) Equilibrium is subjected to a STRESS.
 2) System SHIFTS towards what is removed from the
system or away from what is added.
 The shift results in a CHANGE OF CONCENTRATION
for both the products and the reactants.
– If the shift is towards the products, the concentration of
the products will increase and the concentration of the
reactants will decrease.
– If the shift is towards the reactants, the concentration of
the reactants will increase and the concentration of the
products will decrease.
(c) 2006, Mark Rosengarten
» NEXT
Examples
 For the reaction N2(g)
+ 3H2(g)  2 NH3(g) + heat
– Adding N2 will cause the equilibrium to shift RIGHT, resulting in
an increase in the concentration of NH3 and a decrease in the
concentration of N2 and H2.
– Removing H2 will cause a shift to the LEFT, resulting in a
decrease in the concentration of NH3 and an increase in the
concentration of N2 and H2.
– Increasing the temperature will cause a shift to the LEFT, same
results as the one above.
– Decreasing the pressure will cause a shift to the LEFT, because
there is more gas on the left side, and making more gas will bring
the pressure back up to its equilibrium amount.
– Adding a catalyst will
have no effect, so no shift will happen.
(c) 2006, Mark Rosengarten
Solubility Curves
 Solubility: the maximum quantity of solute that can be




dissolved in a given quantity of solvent at a given
temperature to make a saturated solution.
Saturated: a solution containing the maximum quantity of
solute that the solvent can hold. The limit of solubility.
Supersaturated: the solution is holding more than it can
theoretically hold OR there is excess solute which
precipitates out. True supersaturation is rare.
Unsaturated: There are still solvent molecules available to
dissolve more solute, so more can dissolve.
How ionic solutes dissolve in water: polar water
(c) 2006, Mark Rosengarten
molecules attach to the ions and tear them off the crystal.
Solubility
Solubility: go to the temperature
and up to the desired line, then
across to the Y-axis. This is how
many g of solute are needed to
make a saturated solution of that
solute in 100g of H2O at that
particular temperature.
At 40oC, the solubility of KNO3 in
100g of water is 64 g. In 200g of
water, double that amount. In 50g
of water, cut it in half.
(c) 2006, Mark Rosengarten
Supersaturated
If 120 g of NaNO3 are added to
100g of water at 30oC:
1) The solution would be
SUPERSATURATED, because
there is more solute dissolved
than the solubility allows
2) The extra 25g would
precipitate out
3) If you heated the solution up
by 24oC (to 54oC), the excess
solute would dissolve.
(c) 2006, Mark Rosengarten
Unsaturated
If 80 g of KNO3 are added to
100g of water at 60oC:
1) The solution would be
UNSATURATED, because there
is less solute dissolved than the
solubility allows
2) 26g more can be added to
make a saturated solution
3) If you cooled the solution
down by 12oC (to 48oC), the
solution would become saturated
(c) 2006, Mark Rosengarten
How Ionic Solutes Dissolve in
Water
Water solvent molecules attach to the
Water solvent holds the ions apart and
(c)
2006,
Mark
Rosengarten
ions (H end to the Cl-, O end to the Na+) keeps the ions from coming back together
Acids and Bases
1) Formulas, Naming and Properties of Acids
2) Formulas, Naming and Properties of Bases
3) Neutralization
4) pH
5) Indicators
6) Alternate Theories
(c) 2006, Mark Rosengarten
Formulas, Naming and
Properties of Acids
 Arrhenius Definition of Acids: molecules that dissolve in
water to produce H3O+ (hydronium) as the only positively
charged ion in solution.
 HCl (g) + H2O (l)  H3O+ (aq) + Cl Properties of Acids
 Naming of Acids
 Formula Writing of Acids
(c) 2006, Mark Rosengarten
Properties of Acids
 Acids react with metals above H2 on Table J to





form H2(g) and a salt.
Acids have a pH of less than 7.
Dilute solutions of acids taste sour.
Acids turn phenolphthalein CLEAR, litmus RED
and bromthymol blue YELLOW.
Acids neutralize bases.
Acids are formed when acid anhydrides (NO2,
SO2, CO2) react with water for form acids. This is
how acid rain forms from auto and industrial
emissions.
(c) 2006, Mark Rosengarten
Naming of Acids
 Binary Acids (H+ and a nonmetal)
– hydro (nonmetal) -ide + ic acid
• HCl (aq) = hydrochloric acid
 Ternary Acids (H+ and a polyatomic ion)
– (polyatomic ion) -ate +ic acid
• HNO3 (aq) = nitric acid
– (polyatomic ion) -ide +ic acid
• HCN (aq) = cyanic acid
– (polyatomic ion) -ite +ous acid
• HNO2 (aq) = nitrous acid
(c) 2006, Mark Rosengarten
Formula Writing of Acids
 Acids formulas get written like any other. Write the H+1







first, then figure out what the negative ion is based on the
name. Cancel out the charges to write the formula. Don’t
forget the (aq) after it…it’s only an acid if it’s in water!
Hydrosulfuric acid: H+1 and S-2 = H2S (aq)
Carbonic acid: H+1 and CO3-2 = H2CO3 (aq)
Chlorous acid: H+1 and ClO2-1 = HClO2 (aq)
Hydrobromic acid: H+1 and Br-1 = HBr (aq)
Hydronitric acid:
Hypochlorous acid:
(c) 2006, Mark Rosengarten
Perchloric acid:
Formulas, Naming and
Properties of Bases
 Arrhenius Definition of Bases: ionic compounds that
dissolve in water to produce OH- (hydroxide) as the only
negatively charged ion in solution.
 NaOH (s)  Na+1 (aq) + OH-1 (aq)
 Properties of Bases
 Naming of Bases
 Formula Writing of Bases
(c) 2006, Mark Rosengarten
Properties of Bases
 Bases react with fats to form soap and glycerol. This





process is called saponification.
Bases have a pH of more than 7.
Dilute solutions of bases taste bitter.
Bases turn phenolphthalein PINK, litmus BLUE and
bromthymol blue BLUE.
Bases neutralize acids.
Bases are formed when alkali metals or alkaline earth
metals react with water. The words “alkali” and “alkaline”
mean “basic”, as opposed to “acidic”.
(c) 2006, Mark Rosengarten
Naming of Bases
 Bases are named like any ionic
compound, the name of the metal ion
first (with a Roman numeral if
necessary) followed by “hydroxide”.
Fe(OH)2 (aq) = iron (II) hydroxide
Fe(OH)3 (aq) = iron (III) hydroxide
Al(OH)3 (aq) = aluminum hydroxide
NH3 (aq) is the same thing as NH4OH:
NH3 + H2O  NH4OH
(c) 2006, Mark Rosengarten
Also called ammonium hydroxide.
Formula Writing of Bases
 Formula writing of bases is the same as for any ionic







formula writing. The charges of the ions have to cancel
out.
Calcium hydroxide = Ca+2 and OH-1 = Ca(OH)2 (aq)
Potassium hydroxide = K+1 and OH-1 = KOH (aq)
Lead (II) hydroxide = Pb+2 and OH-1 = Pb(OH)2 (aq)
Lead (IV) hydroxide = Pb+4 and OH-1 = Pb(OH)4 (aq)
Lithium hydroxide =
Copper (II) hydroxide =
Magnesium hydroxide =
(c) 2006, Mark Rosengarten
Neutralization
 H+1 + OH-1  HOH
 Acid + Base  Water + Salt (double replacement)
 HCl (aq) + NaOH (aq)  HOH (l) + NaCl (aq)
 H2SO4 (aq) + KOH (aq)  2 HOH (l) + K2SO4 (aq)
 HBr (aq) + LiOH (aq) 
 H2CrO4 (aq) + NaOH (aq) 
 HNO3 (aq) + Ca(OH)2 (aq) 
(c) 2006,
Mark Rosengarten
 H3PO4 (aq) + Mg(OH)
2 (aq) 
pH
 A change of 1 in pH is a tenfold increase in acid or base
strength.
 A pH of 4 is 10 times more acidic than a pH of 5.
(c) 2006, Mark Rosengarten
 A pH of 12 is 100 times more basic than a pH of 10.
Indicators
At a pH of 2:
Methyl Orange = red
Bromthymol Blue = yellow
Phenolphthalein = colorless
Litmus = red
Bromcresol Green = yellow
Methyl orange is red at a pH of
3.2 and below and yellow at a pH
Thymol Blue
of 4.4 and higher. In between the
two numbers, it is an intermediate
color that is not listed on this
table.
(c) 2006, Mark Rosengarten
= yellow
Alternate Theories
 Arrhenius Theory: acids and bases must be in aqueous
solution.
 Alternate Theory: Not necessarily so!
– Acid: proton (H+1) donor…gives up H+1 in a reaction.
– Base: proton (H+1) acceptor…gains H+1 in a reaction.
 HNO3 + H2O  H3O+1 + NO3-1
– Since HNO3 lost an H+1 during the reaction, it is an acid.
– Since H2O gained the H+1 that HNO3 lost, it is a base.
(c) 2006, Mark Rosengarten
Oxidation and Reduction
1) Oxidation Numbers
2) Identifying OX, RD and SI Species
3) Agents
4) Writing Half-Reactions
5) Balancing Half-Reactions
6) Activity Series
7) Voltaic Cells
8) Electrolytic Cells
9) Electroplating
(c) 2006, Mark Rosengarten
Oxidation Numbers
 Elements have no charge until they bond to other elements.




– Na0, Li0, H20. S0, N20, C600
The formula of a compound is such that the charges of the
elements making up the compound all add up to zero.
The symbol and charge of an element or polyatomic ion is
called a SPECIES.
Determine the charge of each species in the following
compounds:
NaCl
KNO3
CuSO4
Fe2(CO3)3
(c) 2006, Mark Rosengarten
Identifying OX, RD, SI
Species
 Ca0 + 2 H+1Cl-1  Ca+2Cl-12 + H20
 Oxidation = loss of electrons. The species becomes more
positive in charge. For example, Ca0  Ca+2, so Ca0 is the
species that is oxidized.
 Reduction = gain of electrons. The species becomes more
negative in charge. For example, H+1  H0, so the H+1 is
the species that is reduced.
 Spectator Ion = no change in charge. The species does not
gain or lose any electrons. For example, Cl-1  Cl-1, so
the Cl-1 is the spectator
ion.
(c) 2006, Mark Rosengarten
Agents
 Ca0 + 2 H+1Cl-1  Ca+2Cl-12 + H20
 Since Ca0 is being oxidized and H+1 is being reduced, the
electrons must be going from the Ca0 to the H+1.
 Since Ca0 would not lose electrons (be oxidized) if H+1
weren’t there to gain them, H+1 is the cause, or agent, of
Ca0’s oxidation. H+1 is the oxidizing agent.
 Since H+1 would not gain electrons (be reduced) if Ca0
weren’t there to lose them, Ca0 is the cause, or agent, of
H+1’s reduction. Ca0 is the reducing agent.
(c) 2006, Mark Rosengarten
Writing Half-Reactions
 Ca0 + 2 H+1Cl-1  Ca+2Cl-12 + H20
 Oxidation: Ca0  Ca+2 + 2e Reduction: 2H+1 + 2e-  H20
The two electrons lost
by Ca0 are gained by
the two H+1 (each H+1
picks up an electron).
PRACTICE SOME!
(c) 2006, Mark Rosengarten
Practice Half-Reactions
 Don’t forget to determine the charge of each species first!
 4 Li + O2  2 Li2O
 Oxidation Half-Reaction:
 Reduction Half-Reaction:
 Zn + Na2SO4  ZnSO4 + 2 Na
 Oxidation Half-Reaction:
 Reduction Half-Reaction:
(c) 2006, Mark Rosengarten
Balancing Half-Reactions
 Ca0 + Fe+3  Ca+2 + Fe0
– Ca’s charge changes by 2, so double the Fe.
– Fe’s charge changes by 3, so triple the Ca.
– 3 Ca0 + 2 Fe+3  3 Ca+2 + 2 Fe0
 Try these:
 __Na0 + __H+1  __Na+1 + __H20
– (hint: balance the H and H2 first!)
 __Al0 + __Cu+2  __Al+3 + __Cu0
(c) 2006, Mark Rosengarten
Activity Series
 For metals, the higher up the chart the
element is, the more likely it is to be
oxidized. This is because metals like to
lose electrons, and the more active a
metallic element is, the more easily it can
lose them.
 For nonmetals, the higher up the chart the
element is, the more likely it is to be
reduced. This is because nonmetals like to
gain electrons, and the more active a
nonmetallic element is, the more easily it
can gain them.
(c) 2006, Mark Rosengarten
Metal Activity
3 K0 + Fe+3Cl-13
REACTION
Fe0 + 3 K+1Cl-1
NO REACTION
 Metallic elements start out with a charge
of ZERO, so they can only be oxidized to
form (+) ions.
 The higher of two metals MUST undergo
oxidation in the reaction, or no reaction
will happen.
 The reaction 3 K + FeCl3  3 KCl + Fe
WILL happen, because K is being
oxidized, and that is what Table J says
should happen.
 The reaction Fe + 3 KCl  FeCl3 + 3 K
(c) 2006, Mark Rosengarten
will
NOT happen.
Voltaic Cells
 Produce electrical current using a spontaneous redox
reaction
 Used to make batteries!
 Materials needed: two beakers, piece of the oxidized metal
(anode, - electrode), solution of the oxidized metal, piece
of the reduced metal (cathode, + electrode), solution of the
reduced metal, porous material (salt bridge), solution of a
salt that does not contain either metal in the reaction, wire
and a load to make use of the generated current!
 Use Reference Table J to determine the metals to use
– Higher = (-) anode
Lower = (+) cathode
(c) 2006, Mark Rosengarten
Making Voltaic Cells
More
Info!!!
Create
Your
Own
(c) 2006, Mark Rosengarten
Cell!!!!
How It Works
Since Zn is listed above Cu, Zn0 will be
oxidized when it reacts with Cu+2. The
reaction: Zn + CuSO4  ZnSO4 + Cu
 The Zn0 anode loses 2 e-, which go up the wire and
through the load. The Zn0 electrode gets smaller as the Zn0
becomes Zn+2 and dissolves into solution. The e- go into
the Cu0, where they sit on the outside surface of the Cu0
cathode and wait for Cu+2 from the solution to come over
so that the e- can jump on to the Cu+2 and reduce it to Cu0.
The size of the Cu0 electrode increases. The negative ions
in solution go over the salt bridge to the anode side to
complete the circuit.(c) 2006, Mark Rosengarten
You Start At The Anode
Vital to make a battery
Is this electrochemistry
You take two half-cells
And connect them up so well
With a load to power in between
You need to have electrodes you see
Full of that metallicity
Let electrons flow
Across the salt bridge we go!
Allowing us to make electricity
We start the anode
Electrons are lost there
And go through the wire
And through the load on fire
They get to the cathode
And reduce the cations
And the anions go through the salt bridge
Back to where…
(c) 2006, Mark Rosengarten
WHERE?
Make Your Own Cell!!!
(c) 2006, Mark Rosengarten
Electrolytic Cells
 Use electricity to force a nonspontaneous redox reaction to
take place.
 Uses for Electrolytic Cells:
– Decomposition of Alkali Metal Compounds
– Decomposition of Water into Hydrogen and Oxygen
– Electroplating
 Differences between Voltaic and Electrolytic Cells:
– ANODE:
Voltaic (-)
Electrolytic (+)
– CATHODE:
Voltaic (+)
Electrolytic (-)
– Voltaic: 2 half-cells, a salt bridge and a load
(c) 2006,
– Electrolytic: 1 cell,
no Mark
saltRosengarten
bridge, IS the load
Decomposing Alkali
Metal Compounds
2 NaCl  2 Na + Cl2
The Na+1 is reduced at
the (-) cathode,
picking up an e- from
the battery
(c) 2006, Mark Rosengarten
The Cl-1 is oxidized at
the (+) anode, the ebeing pulled off by the
battery (DC)
Decomposing Water
2 H2O  2 H2 + O2
The H+ is reduced at
the (-) cathode,
yielding H2 (g), which
is trapped in the tube.
The O-2 is oxidized at
the (+) anode, yielding
O2 (g), which is
(c) 2006, Mark Rosengarten
trapped in the tube.
Electroplating
The Ag0 is oxidized to Ag+1
when the (+) end of the
battery strips its electrons
off.
The Ag+1 migrates through
the solution towards the (-)
charged cathode (ring),
where it picks up an electron
from the battery and forms
Ag0, which coats on to the
(c) 2006, Mark Rosengarten
ring.
Organic Chemistry
1) Hydrocarbons
2) Substituted Hydrocarbons
3) Organic Families
4) Organic Reactions
(c) 2006, Mark Rosengarten
Hydrocarbons
 Molecules made of Hydrogen and Carbon
 Carbon forms four bonds, hydrogen forms one bond
 Hydrocarbons come in three different homologous series:
– Alkanes (single bond between C’s, saturated)
– Alkenes (1 double bond between 2 C’s, unsaturated)
– Alkynes (1 triple bond between 2 C’s, unsaturated)
 These are called aliphatic, or open-chain, hydrocarbons.
 Count the number of carbons and add the appropriate
suffix!
(c) 2006, Mark Rosengarten
Alkanes
 CH4 = methane
 C2H6 = ethane
 C3H8 = propane
 C4H10 = butane
 C5H12 = pentane
 To find the number of hydrogens,
double the number of carbons and add
2.(c) 2006, Mark Rosengarten
Methane
Meth-: one carbon
-ane: alkane
The simplest organic
molecule, also known as
natural gas!
(c) 2006, Mark Rosengarten
Ethane
Eth-: two carbons
-ane: alkane
(c) 2006, Mark Rosengarten
Propane
Prop-: three carbons
-ane: alkane
Also known as “cylinder gas”, usually stored under pressure
and used for gas grills and stoves. It’s also very handy as a
fuel for Bunsen burners!
(c) 2006, Mark Rosengarten
Butane
But-: four carbons
-ane: alkane
Liquefies with moderate pressure, useful for gas lighters. You
have probably lit your gas grill with a grill lighter fueled with
butane!
(c) 2006, Mark Rosengarten
Pentane
Pent-: five carbons
-ane: alkane
Your Turn!!!
Draw Hexane:
Draw Heptane:
(c) 2006, Mark Rosengarten
Alkenes
 C2H4 = Ethene
 C3H6 = Propene
 C4H8 = Butene
 C5H10 = Pentene
 To find the number of hydrogens, double
the number of carbons.
(c) 2006, Mark Rosengarten
Ethene
Two carbons, double bonded.
Notice how each carbon has
four bonds? Two to the other
carbon and two to hydrogen
atoms.
Also called “ethylene”, is used for the production of
polyethylene, which is an extensively used plastic. Look for
the “PE”, “HDPE” (#2 recycling) or “LDPE” (#4 recycling)
on your plastic bags and containers!
(c) 2006, Mark Rosengarten
Propene
Three carbons, two of them
double bonded. Notice how
each carbon has four bonds?
If you flipped this molecule so that the double bond was on
the right side of the molecule instead of the left, it would still
be the same molecule. This is true of all alkenes.
Used to make polypropylene (PP, recycling #5), used for
dishwasher safe containers and indoor/outdoor carpeting!
(c) 2006, Mark Rosengarten
Butene
This is 1-butene, because the double
bond is between the 1st and 2nd
carbon from the end. The number 1
represents the lowest numbered
carbon the double bond is touching.
This is 2-butene. The double bond
is between the 2nd and 3rd carbon
from the end. Always count from
the end the double bond is closest
to.
ISOMERS: Molecules that share the same molecular
formula, but have different
structural formulas.
(c) 2006, Mark Rosengarten
Pentene
This is 1-pentene. The double bond is
on the first carbon from the end.
This is 2-pentene. The double bond is
on the second carbon from the end.
This is not another isomer of pentene.
This is also 2-pentene, just that the
double bond is closer to the right end.
(c) 2006, Mark Rosengarten
Alkynes
 C2H2 = Ethyne
 C3H4 = Propyne
 C4H6 = Butyne
 C5H8 = Pentyne
 To find the number of hydrogens, double
the number of carbons and subtract 2.
(c) 2006, Mark Rosengarten
Ethyne
Now, try to draw propyne! Any isomers? Let’s see!
Also known as “acetylene”, used by miners by dripping
water on CaC2 to light up mining helmets. The “carbide
lamps” were attached to miner’s helmets by a clip and had
a large reflective mirror that magnified the acetylene
flame.
Used for welding and cutting applications, as ethyne
oC!
burns at temperatures
over
3000
(c) 2006, Mark Rosengarten
Propyne
This is propyne! Nope! No
isomers.
OK, now draw butyne. If there are any isomers, draw
them too.
(c) 2006, Mark Rosengarten
Butyne
Well, here’s 1-butyne!
And here’s 2-butyne!
Is there a 3-butyne? Nope! That would be 1-butyne. With
four carbons, the double bond can only be between the 1st
and 2nd carbon, or between the 2nd and 3rd carbons.
Now, try pentyne!
(c) 2006, Mark Rosengarten
Pentyne
1-pentyne
2-pentyne
Now, draw all of the possible isomers for hexyne!
(c) 2006, Mark Rosengarten
Substituted Hydrocarbons
 Hydrocarbon chains can have three kinds of “dingly-
danglies” attached to the chain. If the dingly-dangly is
made of anything other than hydrogen and carbon, the
molecule ceases to be a hydrocarbon and becomes another
type of organic molecule.
– Alkyl groups
– Halide groups
– Other functional groups
 To name a hydrocarbon with an attached group, determine
which carbon (use lowest possible number value) the
group is attached to. Use di- for 2 groups, tri- for three.
(c) 2006, Mark Rosengarten
Alkyl Groups
(c) 2006, Mark Rosengarten
Halide Groups
(c) 2006, Mark Rosengarten
Organic Families
 Each family has a functional group to identify it.
– Alcohol (R-OH, hydroxyl group)
– Organic Acid (R-COOH, primary carboxyl group)
– Aldehyde (R-CHO, primary carbonyl group)
– Ketone (R1-CO-R2, secondary carbonyl group)
– Ether (R1-O-R2)
– Ester (R1-COO-R2, carboxyl group in the middle)
– Amine (R-NH2, amine group)
– Amide (R-CONH2, amide group)
 These molecules are alkanes with functional groups
(c) 2006, Mark Rosengarten
attached. The name is based on the alkane name.
Alcohol
On to DI and TRIHYDROXY
ALCOHOLS
(c) 2006, Mark Rosengarten
Di and Trihydroxy Alcohols
(c) 2006, Mark Rosengarten
Positioning of
Functional Group
PRIMARY (1o): the functional group is
bonded to a carbon that is on the end of
the chain.
SECONDARY (2o): The functional
group is bonded to a carbon in the middle
of the chain.
TERTIARY (3o): The functional group
is bonded to a carbon that is itself
directly bonded to three other carbons.
(c) 2006, Mark Rosengarten
Organic Acid
These are weak acids. The H on the right side is the one
that ionized in water to form H3O+. The -COOH
(carboxyl) functional group is always on a PRIMARY
carbon.
Can be formed from the oxidation of primary alcohols
using a KMnO4 catalyst.
(c) 2006, Mark Rosengarten
Aldehyde
Aldehydes have the CO (carbonyl) groups ALWAYS on a
PRIMARY carbon. This is the only structural difference
between aldehydes and ketones.
Formed by the oxidation of primary alcohols with a catalyst.
Propanal is formed from the oxidation of 1-propanol using
pyridinium chlorochromate
(PCC)
catalyst.*
(c) 2006, Mark
Rosengarten
Ketone
Ketones have the CO (carbonyl) groups ALWAYS on a
SECONDARY carbon. This is the only structural difference
between ketones and aldehydes.
Can be formed from the dehydration of secondary alcohols
with a catalyst. Propanone is formed from the oxidation of 2propanol using KMnO4 or PCC catalyst.*
(c) 2006, Mark Rosengarten
Ether
Ethers are made of two alkyl groups surrounding one oxygen
atom. The ether is named for the alkyl groups on “ether” side
of the oxygen. If a three-carbon alkyl group and a fourcarbon alkyl group are on either side, the name would be
propyl butyl ether. Made with an etherfication reaction.
(c) 2006, Mark Rosengarten
Ester
Esters are named for the alcohol and organic acid that
reacted by esterification to form the ester. If the alcohol was
1-propanol and the acid was hexanoic acid, the name of the
ester would be propyl hexanoate. Esters contain a COO
(carboxyl) group in the middle of the molecule, which
differentiates them from organic acids.
(c) 2006, Mark Rosengarten
Amine
- Component of amino acids, and therefore proteins, RNA and
DNA…life itself!
- Essentially ammonia (NH3) with the hydrogens replaced by
one or more hydrocarbon chains, hence the name “amine”!
(c) 2006, Mark Rosengarten
Amide
Synthetic Polyamides: nylon, kevlar
Natural Polyamide: silk!
For more information on polymers, go here.
(c) 2006, Mark Rosengarten
Organic Reactions
 Combustion
 Fermentation
 Substitution
 Addition
 Dehydration Synthesis
– Etherification
– Esterification
 Saponification
 Polymerization
(c) 2006, Mark Rosengarten
Combustion
 Happens when an organic molecule reacts with oxygen gas
to form carbon dioxide and water vapor. Also known as
“burning”.
(c) 2006, Mark Rosengarten
Fermentation
 Process of making ethanol by having yeast digest simple
sugars anaerobically. CO2 is a byproduct of this reaction.
 The ethanol produced is toxic and it kills the yeast when
the percent by volume of ethanol gets to 14%.
(c) 2006, Mark Rosengarten
Substitution
 Alkane + Halogen  Alkyl Halide + Hydrogen Halide
 The halogen atoms substitute for any of the hydrogen
atoms in the alkane. This happens one atom at a time. The
halide generally replaces an H on the end of the molecule.
C2H6 + Cl2  C2H5Cl + HCl
The second Cl can then substitute for another H:
C2H5Cl + HCl  C2H4Cl2 + H2
(c) 2006, Mark Rosengarten
Addition
 Alkene + Halogen  Alkyl Halide
 The double bond is broken, and the halogen adds at either
side of where the double bond was. One isomer possible.
(c) 2006, Mark Rosengarten
Etherification*
 Alcohol + Alcohol  Ether + Water
 A dehydrating agent (H2SO4) removes H from one
alcohol’s OH and removes the OH from the other. The
two molecules join where there H and OH were removed.
Note: dimethyl ether and diethyl ether are also produced from
this reaction, but can be separated out.
(c) 2006, Mark Rosengarten
Esterification
 Organic Acid + Alcohol  Ester + Water
 A dehydrating agent (H2SO4) removes H from the organic
acid and removes the OH from the alcohol. The two
molecules join where there H and OH were removed.
(c) 2006, Mark Rosengarten
Saponification
The process of making soap from glycerol esters (fats).
Glycerol ester + 3 NaOH  soap + glycerol
Glyceryl stearate + 3 NaOH  sodium stearate + glycerol
The sodium stearate is the soap! It emulsifies
grease…surrounds globules with its nonpolar ends, creating
micelles with - charge that water can then wash away. Hard
water replaces Na+ with Ca+2 and/or other low solubility ions,
which forms a precipitate called “soap scum”.
Water softeners remove these hardening ions from your tap
water, allowing the soap to dissolve normally.
(c) 2006, Mark Rosengarten
Polymerization
 A polymer is a very long-chain molecule made up of many




monomers (unit molecules) joined together.
The polymer is named for the monomer that made it.
– Polystyrene is made of styrene monomer
– Polybutadiene is made of butadiene monomer
Addition Polymers
Condensation Polymers
Rubber
(c) 2006, Mark Rosengarten
Addition Polymers
Joining monomers together by breaking double
bonds
Polyvinyl chloride (PVC): vinyl siding, PVC pipes, etc.
Vinyl chloride
polyvinyl chloride

n C2H3Cl
-(-C2H3Cl-)-n
Polytetrafluoroethene (PTFE, teflon):
TFE
n C2F4
PTFE

-(-C2F4-)-n
(c) 2006, Mark Rosengarten
Condensation Polymers
Condensation polymerization is just dehydration synthesis,
except instead of making one molecule of ether or ester, you
make a monster molecule of polyether or polyester.
(c) 2006, Mark Rosengarten
Rubber
The process of toughing rubber by cross-linking the polymer
strands with sulfur is called...
(c) 2006, Mark Rosengarten
VULCANIZATION!!!
(c) 2006, Mark Rosengarten
THE END
(c) 2006, Mark Rosengarten