CHEMISTRY UNIT 7
(Chemical Formulas and Compounds)
Monatomic ions are ions formed from a single atom.
Ex: Mg+2, FBinary compounds are compounds composed of two
different elements.
Ex: Mg F2;
H2O
Naming:
To name compounds you simply put the two names
together. The cation is always named first. This
naming system requires that you have knowledge of
ions and balancing.
Ex:
Fe+2 and ClIron II & Chloride becomes
FeCl2 - Iron II chloride
Chemist also use an older form of nomenclature
which takes advantage of prefixes to indicate the
number of each ion present in the compound
Ex: CO2 -Carbon dioxide CO -carbon monoxide
7-2
Prefixes:
mono-1
hexa-6
di-2
tri-3
tetra-4
penta-5
hepta-7 octa-8 nona-9
deca-10
Ex:
carbon tetrachloride
CCl4
dinitrogen monoxide
N2O
dinitrogen pentoxide
N2O5
tetraphosphorus decasulfide P4O10
The Combining of Ions to form Compounds (newest style):
Sodium chloride
Na+ + ClNa+ClSodium chlorate
Na+ + ClO3Na+ClO3Aluminum oxide
Al+3 + O-2
Al2+3 O3-2
Calcium hydroxide
Ca+2 + OHCa+2(OH)2Ammonium phosphate
NH4+1 + PO4-3
(NH4+1)3PO4-3
Mercury (I) phosphate
Hg2+2 + PO4-3
(Hg2+2)3(PO4-3)2
7-3
The Oxidation state of an element is represented by a
signed number called the oxidation number
Oxidation number rules:
1. The oxidation number of a free element is zero.
2. The oxidation number of a mono-atomic ion is
equal to its charge.
3. The algebraic sum of the oxidation numbers of all
the atoms in a compound is zero.
A chemical reaction in which an element attains a more
positive oxidation state (loss of electrons) is called
oxidation
A chemical reaction in which an element attains a more
negative oxidation state (gain of electrons) is called
reduction
The substance that is reduced is called the oxidizing
agent
The substance that is oxidized is called the reducing
agent
7-4
Ex: In the reaction between sodium and sulfur
(Na0 + S0  Na2+1S-2)
Describes the following:
S is reduced by Na
Na is oxidized by S
S is the oxidizing agent Na is the reducing agent
The formula mass of any molecule is the sum of the
average atomic masses of all the atoms represented
in its formula. (that is, add up all the numbers off the
chart.
The molar mass of a substance is the mass of one mole
of atoms (or molecules) of that substance
Ex: The molar mass of carbon is 12.011 g/mole
The gram-molecular weight (G.M.W.) of a molecular
substance is the mass of one mole of molecules of
the substance
Ex: The G.M.W. of water (H2O) is 18.0148 g/mole
H-2 (1.0079) = 2.0158
+ O-1 (15.999) =15.999
18.0158 g/mole
Note: For all practical situations atomic mass, grammolecular weight (G.M.W.), and molar mass are
equivalent.
7-5
The percent composition of a compound is the
percentage by mass of each element in the
compound.
Calculating Percent Composition:
Mass of Element
% Comp. of E =
x 100%
G.M.W of Compound
Ex: Water is made up of Hydrogen and Oxygen. So
is many other compounds.
What percent of water is oxygen?
What percentage of water is hydrogen?
We must first calculate the G.M.W. of water (H2O):
H-2 (1.0079) = 2.0158
+ O-1 (15.999) =15.999
18.0158 g/mole
2.0158
%H =
x 100% = 12.5%
18.0158
15.999
%O =
x 100% = 88.8%
18.0158
7-6
Ex: Find the percent composition for each
constituent of Iron (III) hydroxide.
Iron (III) hydroxide  Fe+3(OH-1)3
Calculating G.M.W.:
Fe-1 (55.847) = 55.847
O-3 (15.999) = 47.997
+ H-3 (1.0079) = 3.0237
106.8677 g/mole
55.847
% Fe =
x 100% = 52.3%
106.8677
47.997
%O =
x 100% = 44.9%
106.8677
3.0237
%H =
x 100% = 2.8%
106.8677
These are the percent compositions for Iron (III) hydroxide.
The percent compositions for Iron (II) hydroxide are different.
7-7
An Empirical formula consists of the symbols for the
elements combined in a compound, with subscripts
showing the smallest whole-number ratio of the
different elements in the compound
Ex: Na17Cl17 is a Molecular formula
NaCl is an Empirical formula
Frequently when evaluating unknown chemicals it is
possible to experimentally determine the percent
composition of constituent elements. This data yields
only the percent composition (or masses) of each
constituent substance in the unknown.
How can we then use this information to predict the
identity (formula) of the unknown compound?
How can we "Calculate the Empirical Formula?


Ex: The decomposition of 11.47g of a compound yields
9.16g of copper and 2.31g of oxygen. What is the
Empirical formula of the compound?
Ex: A compound contains 81.7% carbon and 18.3%
Hydrogen. Find the empirical formula.