I. Oxidation Numbers
II. Nomenclature
III. The Mole
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I. Oxidation
Numbers
2
◦
A positive or negative whole number assigned to
an element in a molecule or ion on the basis of a
set of formal rules; to some degree it reflects the
positive or negative character of an atom
1. Keep track of electrons
2. Tell if electrons gained, lost, or unequally shared
3. Allows us to predict formulas of chemical
compound
Not to be confused with oxidation, a process in which
a substance loses one or more electrons….
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Electronegativity
1.
◦
◦
Higher EN = negative oxidation number
Lower EN = positive oxidation number
4

Remember always refer back to:
◦ Electronegativity
◦ Electrostatic forces
◦ Bonding Characteristics
◦ There will be exceptions to the following rules…
5
Oxidation # of free atoms, pure elements,
and polyatomic elements is ZERO
1.
◦
Both atoms have equal EN, no transfer or shift of
electrons
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2. Oxidation # of monatomic ion is equal to the
charge of the ion
If an atom loses 2e-, the other atom(s) must
gain those 2e-. Electrons DO NOT just float
around….
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3. The sum of all oxidation numbers in a
compound must be ZERO.
Compounds are not electrically charged!
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4. Alkali metals always have a +1 oxidation #
when not free
Hydrogen is not an alkali metal although it is in
group 1
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5. Alkaline Earth Metals always have a +2
oxidation number. They form +2 ions when
they bond.
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6a. Certain elements have the same oxidation
# in almost all their compounds.
 Halogens have oxidation number -1 when
bond to metals
 Halogen with higher EN than other bonded
nonmental is assigned the negative number
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6b. Hydrogen assigned a +1 oxidation # in
most compounds
BUT…..

Hydrogen + metal = metallic hydrides
◦ Hydrogen has a -1 oxidation number
◦ Hydrogen more EN than any other metal
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6c. Oxygen has an -2 oxidation number in most
compounds


Oxygen is bonded to highly EN elements does
not have -2 oxidation number
Oxygen is VERY EN and pulls e- from most
other elements
◦ Exception: Perioxide ion
O22- where O has -1 O.N.
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7. Oxidation number of all atoms in a
polyatomic ion add up to the charge of the
ion
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
Follow rules in order
◦ If rules contradict each other, the rule listed first
should be followed


Write an algebraic equation to solve for
unknown oxidation numbers in compounds
Ionic Compounds
◦
◦
◦
◦
“Criss-Cross” method
Use charge of one ion as the subscript for the other
Simplify ratios of atoms
Exception: Peroxide ion O22- example: Na2O2 , not
NaO
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II.
Nomenclature
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




IUPAC developed a systematic way to name
compounds.
Names reveal the composition and qualities
of certain substances
Indicate the types of bonds and
intermolecular attractions
Covalent Compounds
Binary Ionic Compounds
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




Composed of ONLY
nonmetals
Two word names
Use Greek Prefix
System
Least EN element
first, then more EN
element
Ending of last
element is changed
to -ide.
Number
Prefix
1
Mono*
2
Di
3
Tri
4
Tetra
5
Penta
6
Hexa
7
Hepta
8
Octa
9
Nona
10
deca
*omit mono- for first atom
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1.
Antimony tribromide
SbBr3
2.
Hexaboron silicide
B6Si
3.
Chlorine dioxide
4.
Iodide pentafluoride
5.
P4S5
Tetraphorsphorus pentasulfide
6.
Si2Br6
Disilicon hexabromide
7.
CH4
Methane
8.
NF3
Nitrogen trifluoride
ClO2
IF5
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
Compounds formed between metals and nonmetals!

Ionic compounds DO NOT use Greek Prefix System

Named according to the two elements or polyatomic
ions


Positive ions use the same name as their parent
atoms (ex. sodium atoms form sodium ions).
Named FIRST.
Negative ions have an –ide ending.
Named SECOND.
Polyatomic ions
made up of 2+
types of atoms with
covalent bonds!
They act like a
single unit.
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
Oxyanions: anions composed of oxygen and one
other element/ polyatomic ion (table p. 174).
2 Forms of oxyanion:
◦ More oxygens = name ends in “___-ate”
◦ Less oxygens = name ends in “____-ite”
3+ Forms of oxyanion:
◦ Most oxygens = name “per-________-ate”
◦ Least oxygens = name “hypo-______-ite”
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
Polyatomic compounds: compounds that
contain polyatomic ions (see green handout)
Same as binary compounds, but:
◦
◦
◦
◦
◦
Name the cation first
Name the anion second
Replace “-ide” ending with polyatomic ion name
If 2 polyatomic ions, use polyatomic ion names
Do not use Greek prefix system
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1.
2.
3.
4.
5.
6.
7.
8.
MgO
K2 S
Na2SO4
Ba(ClO3)2
NH4Cl
K2Cr2O7
CaSO4
Zn3(PO4)2
Magnesium oxide
Potassium sulfide
Sodium sulfate
Barium chlorate
Ammonium chloride
Potassium dichromate
Calcium sulfate
Zinc phosphate
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
Stock System (Roman Numeral System):
◦ If an element can have more than 1 oxidation state,
a roman numeral is placed in parenthesis after the
element’s name.
◦ Transition metals

Common Name
◦ Use suffix at the end of the first element (metal)
◦ Smaller oxidation number “-ous”
◦ Larger oxidation number “-ic”
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1.
2.
3.
4.
5.
6.
7.
8.
Hg2I2
CuBr
FeCl2
Co2(C2O4)3
SnO
SnO2
PbSO4
Pb(SO4)2
Mercury (I) iodide
Copper (I) bromide
Iron (II) chloride
Cobalt (III) oxalate
Tin (II) oxide
Tin (IV) oxide
Lead (II) sulfate
Lead (IV) sulfate
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
Hydrates are compounds that have water
molecules in their crystalline structure.
◦ Hold water = “water of hydration”
◦ Formulas written followed by a “dot” and number of
water molecules.
◦ Compounds name + greek prefix + hydrate
 Ex.
Na2CO3
· 7H2O
sodium carbonate heptahydrate
◦ Anhydrates = compounds with NO water in their
structures…
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

Binary acids form when binary compounds
dissolve in water
“hydro”- ___________ acid
◦ HCl: hydrogen chloride  hydrochloric acid
◦ HBr: hydrogen bromide  hydrobromic acid
◦ H2S: hydrogen sulfide  hydrosulfuric acid
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
Ternary acids contain three elements,
generally containing a polyatomic ion, or
combination of hydrogen, oxygen, and a
nonmetal.
◦ Rule1: Addition of 1 oxygen to the acid:
Per__________-ic acid.
◦ Rule 2: Subtraction of 1 oxygen from the acid:
__________-ous acid.
◦ Rule 3: Subtraction of 2 oxygens from the acid:
hypo________-ous acid.
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III. The Mole
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



It is a counting number (like a dozen)
Used to count really, really small things…
Avagadro’s Number (NA)
1 MOLE = 6.022x1023 units (4 sig figs)
A mole is a
amount!!!!!
That is why it is used to count very small things, like
atoms and molecules…
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




1 mole of hockey pucks would equal the mass
of the moon!
1 mole of basketballs would fill a bag the size
of the earth!
1 mole of pennies would cover the Earth ¼
mile deep!
1 mole of sand would fill all the Great Lakes 10
times!
1 mole of popcorn kernels would cover the
United States 9 miles deep!
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Remember……!
1 Mole of ANYTHING is equal to
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6.022 x 10 items
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
We can use the concept of the mole to solve
for problems like:
◦ How many copper atoms are in one penny?
◦ What is the mass of a single atom?
◦ What is the mass of 0.500 moles of helium atoms?
◦ How many atoms are in 33mg of gold?
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
Because a mole is so large, measurements
aren’t counted, they are weighed.
 Molar
mass is the mass
(grams) of 1 mole (NA) of
particles of an element or
compound
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
The same number! Different units!

Look at the periodic table

Scientists chose Avagadro’s # (NA) to:
◦ relate atomic mass units to the larger, more
practical unit of GRAMS.
◦ Represent the # of particles in a mole so that the
atomic mass of an element and mass of a mole of
the element have the same numeric value, just
different units! (Hydrogen experiment)
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
Carbon
12.01 g/mol

Aluminum
26.98 g/mol

Zinc
65.39 g/mol
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
Sodium bicarbonate
◦ NaHCO3
◦ 22.99 + 1.01 + 12.01 + 3(16.00) = 84.01 g/mol
Use atomic mass
from periodic table
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Examples:

H2O
2(1.01) + 16.00 = 18.02 g/mol

NaCl
22.99 + 35.45 = 58.44 g/mol
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
P.184 in textbook flow chart
Mass (g) ↔ Moles ↔ Number of units
(atoms, etc..)
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



How many molecules or atoms are in a
certain amount of a substance?
How many grams are there in a mole of a
substance?
How many moles are there in ??? grams of a
substance?
What is the percent composition of a
substance? (how much do each of the
different types of atoms weigh in the
compound?)
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
Use dimensional analysis to SOLVE
problems…
◦ Sample Problems:








P.
P.
P.
P.
P.
P.
P.
P.
184
185
186
187
189
190
192
193
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 Example:
How many moles are in
22 grams of copper metal?
In all problems like this, you need to go
through four steps to find a solution.
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
Step 1: Figure out how many parts in your
calculation you will have by using this
diagram
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
Step 2: Make a T-chart, and put whatever
information the problem gave you in the top
left. After that, put the units of whatever you
were given in the bottom right of the T, and
the units of what you want to find in the top
right.
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
Step 3: Put the conversion factors into the Tchart in front of the units on the right.
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
Step 4: Cancel out the units from the top left
and bottom right, then find the answer by
multiplying all the stuff on the top together
and dividing it by the stuff on the bottom.
Pau!
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
Continue by adding another section in the
T-chart… repeat steps…
…and there you go.
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

Molarity is the amount of a substance
dissolved in one liter of solution.
Molarity (M) = moles/ liters of solution
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


Chemical compounds contain two or more
atoms chemically combined to behave as one
unit.
Masses of compound units can be found by
adding the masses of the atoms contained in
them.
Formula unit = a single unit of a compound
◦ NaCl = one unit of sodium chloride
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
The mass of a mole of a substance:
◦ Gram-atomic mass = mass of a mole of atoms
◦ Gram-molecular mass = mass of a mole of
molecules
◦ Gram-formula mass= mass of a mole of formula
units in an ionic compound
 All have the units g/mol
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

Formulas tell us the proportions of atoms in a
molecule or ionic compound
We can relate # of atoms  # of moles
◦ Example: gram-molecular mass of NH3?
◦ 1 mole of nitrogen = (1) 14.01 = 14.01 g/mol
◦ 3 moles of hydrogen = (3) 1.008 = 3.024 g/mol
◦ 14.01 + 3.024 = 17.03 g/mol
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


Example: gram-formula mass of Al2(SO4)3?
Each formula unit contains two Al, three S, and
12 O. A mole of Al2(SO4)3 consists of 2 moles of
Al atoms, 3 moles of S atoms, and 12 moles of
O atoms.
Find the molar mass….. 342.1 g/mol Al2(SO4)3
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
Structural Formulas –
◦ Show the types of atoms
◦ Exact composition of each molecule
◦ Arrangement of chemical bonds
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
Molecular Formula – “TRUE FORMULA”
◦ Shows the types of atoms
◦ Exact composition of ATOMS in each molecule
◦ Does not show shape, location of bonds, or bond
type
H2O = water
C2H4 = ethene
Cl2 = chlorine
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
Empirical Formulas –
◦ Tell what elements are present in simple ratios
◦ Used for both ionic compounds and molecules
◦ Careful when writing empirical formulas for molecules…
 May be the actual molecular composition OR
 May only show the simplest ratio of atoms in the molecule
H2O = water  Empirical formula = H20
C2H4 = ethene  Empirical formula = CH2
Cl2 = chlorine  Empirical formula = Cl
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
Empirical Formulas – REDUCE SUBSCRIPTS
Example:
◦
◦
◦
◦
C2H6  CH3
1. FIND MASS (OR %) OF EACH ELEMENT
2. Find moles of each element
3. Divide moles by the smallest # to find subscripts
4. When necessary, multiply subscripts by 2,3, or 4
to get whole #’s
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
Find the empirical formula for a sample of
25.9% N and 74.1% O.
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
Empirical formula:
N1O2.5
Need to make subscripts whole numbers 
x2
N2O5
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

Mole ratio in the EF  mass ratio
Example:
◦ Water formula – H20  2 moles hydrogen for every 1
mole of oxygen
◦ Express in mass – 1 mole of water contains 2.016g of
hydrogen atoms and 16.00g of oxygen atoms
◦ Convert moles to mass (grams)
 2 mole H (1.008g/1 mole H) = 2.016g H
 1 mole O (16.00g/ 1 mole O) = 16.00g O
 Total mass of water = 2.016g + 16.00g = 18.02g
 Now find % composition…..
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
Molecular Formula:
1.
2.
3.
4.
Find the empirical formula
Find the empirical formula mass
Divide the molecular mass by the empirical mass
Multiply each subscript by the answer from step 3
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

The empirical formula for ethene is CH2. Find
the molecular formula is the molecular mass
is 28.1 g/mol.
Empirical mass = 14.03 g/mol
(28.1 g/mol)/ (14.03 g/mol) = 2.00
(CH2)2  C2H4
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



Percent composition = the mass composition
of a compound
All other formulas describe # of atoms
%Comp. describes masses of atoms
Which atoms make up the most mass in a
compound or molecule?
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
Percent = “per hundred”

General setup:
◦ Part/whole x 100%

Example:
◦ Lab analysis of 30.00g Al2(SO4)3
 4.731g Al  (4.731/30.00) x 100% = 15.77% Al
 8.433g S  (8.433/30.00) x 100% = 28.11% S
 16.836g O  (16.836/ 30.00) x 100% = 56.12% O
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
Sample Problem p. 192
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