Chapter 7 Notes
There are millions of compounds and we must have ways of naming or writing each of
them. In this chapter will learn the most common methods. We must first understand
what a chemical formula is telling us.
C8H18
The letters are the symbols for the elements making up the compound. The subscripts
indicate 8 carbon atoms and 18 hydrogen atoms. Some of these compounds are more
complex. Let’s look at a compound containing a polyatomic ion.
Al2(SO4)3
Note that parantheses are used around the polyatomic ion and it is followed by the
subscript 3. This indicates that 3 polyatomic ions. In this case, there are 2 atoms of Al, 3
(1x3) atoms of S, and 12 (4x3) atoms of O.
NOMENCLATURE is the naming system used to apply names to these compounds.
Naming MONATOMIC IONS are the simplest. Monatomic ions are ions formed from a
single atom. These atoms are found in groups 1, 2, 13 and 15-17. Monatomic ions can
have the charges =1, +2, +3, -1, -2, and -3. Monatomic cations are simply called the
element name. Na+1 is called sodium. A monatomic anion is a little more complicated.
The ending of the element name is replaced with –ide. Flourine in an anionic form, F-1, is
called fluoride. A compound containing sodium and fluorine, NaF, would be called
sodium fluoride. These compounds containing two elements such as NaF, is called a
BINARY COMPOUND. The procedure for naming these compounds is the same
regardless to the number of atoms; the number of elements has to be two. Al2O3 contains
2 atoms of Al and 3 atoms of O. The number of elements in the compound remains at 2;
Al and O. The name for the compound is aluminum oxide.
Writing formulas from the chemical names follow the three simple steps.
Write the symbols for the elements with charge. Cation first.
Al+3 O-2
Cross over the charges (CRISS CROSS) and use the absolute value.
Al+32 O-23
Make sure the subscript is in the simplest whole numbers.
Al2O3
Transition metals pose some difficulty because they exist in more than one ionic form.
Iron, for example, exist as Fe+2 and Fe+3. Regardless, you use the same technigue for
writing formulas. To name compounds with transition metals, you must indicate the
ionic state. This can be done using roman numerals.
Fe+3 O-2
Criss cross the charges.
Fe2O3
You would name the compound, Iron III oxide. Iron III indicates the +3 ionic form of
Iron.
Compounds containing polyatomic ions pose another set of difficulties. POLYATOMIC
IONS are groups of covalently bonded atoms that carry an ionic charge. Most are
negatively charged and contain oxygen. These are sometimes called OXYANIONS.
When naming compounds with polyatomic ions, you use the name of the polyatomic ion.
Ca(OH)2 is called Calcium Hydroxide.
NH3Cl is called Ammonium Chloride.
MOLECULAR COMPOUNDS are covalently bonded compounds. Nomenclature for
molecular compounds is different from that used on ionic compounds. A set of prefixes
are added to the names indicating the number of atoms.
Numbers of atoms
1
2
3
4
5
6
7
8
9
10
Prefix
monoditritetrapentahexaheptaoctanonadeca-
The element in the smaller group number is given first. If in the same group is the same,
use the smaller period. The first element is given a prefix only if it has more than one
atom. The second element is given a prefix and the ending is replaced with –ide. The
letters ‘o’ and ‘a’ at the end of the prefix is dropped if the element name begins with a
vowel. Examples are:
CO
Carbon Monoxide
CO2
Carbon Dioxide
P4O10
Tetraphosphorus Decoxide
OXIDATION NUMBERS indicate how electrons are distributed can be assigned to
atoms of a compound. There is no exact numbers when referring to oxidation numbers,
but there are some guidelines to use.







Pure elements have an oxidation number of zero, 0.
Cations and anions of monatomic elements are equal to the ionic charge.
Fluorine is always -1.
Oxygen is -2 in most compounds. It is -1 in peroxides such as H2O2 and +2 with
fluorine.
Hydrogen is +1 in most compounds except with metals, then it is -1.
The sum of all oxidation numbers must equal zero.
If a polyatomic ion, the sum of all oxidation numbers must equal the charge on
the polyatomic ion.
Lets assign oxidation numbers to UF6. Remember, fluorine
is always -1.
UF6-1
The net charge is -6, because there are 6 fluorine atoms. So,
-1 x 6 is -6. In order for the sum of the oxidation numbers
to equal zero, the oxidation number for uranium has to be
+6.
Lets look at a compound that has more than
two elements such as H2SO4. Remember that
hydrogen is +1 and oxygen is -2.
H2+1SO4-2
The net oxidation number for hydrogen is +2
and oxygen is -8. So the oxidation number for
sulfur has to be +6. Therefore +2+6-8=0.
FORMULA MASS is the sum of the average
atomic masses of all the elements in the
compound. The formula mass is actually the
MOLAR MASS for the compound. To
calculate the molar mass of a compound, use
the periodic table. For example, the chemical
formula for water is H2O. Water has 2 atoms of hydrogen and 1 atom of oxygen.
Hydrogen
2 atoms x 1.01 grams/mole =
Oxygen
1 atom x 16.00 grams/mole =
Combined average atomic mass
2.02 grams/mole
16.00 grams/mole
18.02 grams/mole
The formula mass for water is 18.02 grams/mole. It is important to remember the
relationship between molar mass, the mole, and Avogrado’s number.
MOLAR MASS = MOLE = 6.022 x 1023 MOLECULES
You can now do molar conversions with compounds just as we did with elements.
m = mass of substance
M = Molar Mass of substance
v = volume of substance
V = Concentration
n = Number of moles
PERCENT COMPOSITION is a way to compare the masses of each component of a
compound. First, you must calculate the molar mass for the compound.
Cu2S
2 atoms Cu x 63.6 grams/mole =
1 atom S x 32.1 grams/mole =
127.1 grams/mole
32.1 gams/mole
169.2 grams/mole
Then divide the mass of each component by the molar mass of the compound and
multiply by 100%.
% Composition of Cu
% Composition of S
(127.1 grams/169.2 grams)x100% = 79.9% Cu
(32.1 grams/169.2 grams)x100% = 20.1% S
The sum of the % compositions will be 100%, 79.9%+20.1%=100%.
The percent composition is a vital part of analytical chemistry. The percent composition
can allow for the determination of chemical formulas using only composition data. The
EMPIRICAL FORMULA is the chemical formula expressed with subscripts in the
smallest whole number ratio of the elements in the compound.
For example, the composition of a compound is
determined to contain 78.1% boron and 21.9%
hydrogen. What is the empirical formula for the
compound?
Because the composition is given in percentages,
it can be expressed as 78.1 grams of boron and
21.9 grams of hydrogen. You can then convert
the grams to moles.
78.1 grams B/10.81 grams per mole =
7.22 moles Boron
21.9 grams H/1.01 grams per mole =
21.7 moles Hydrogen
The moles are then placed in the form of a ratio.
7.22 moles B: 21.7 moles H
Divide by the smallest.
7.22 moles B: 21.7 moles H
7.22
7.22
1 mole B: 3.01 moles H
The empirical formula of the compound can be written as
BH3.
Remember the empirical formula gives the atomic ratio in the smallest whole numbers.
This formula may not represent the actual molecular formula. To convert the empirical
formula to the molecular formula, one must know the molar mass of the actual
compound. The relationship between the empirical formula and molecular formula is
expressed inn terms of mass.
X(empirical formula mass) = molecular formula mass
Or
X = molecular formula mass/empirical formula mass
With the empirical formula above, the molecular mass is measured to be 27.67 grams.
What is the actual molecular formula for the compound?
Calculate the empirical formula mass.
1 atom of boron x 10.81 grams/mole =
3 atoms of hydrogen x 1.01 grams/mole =
10.81 grams/mole
3.03 grams/mole
13.84 grams/mole
Plug the empirical formula into the formula.
X = 27.67 grams/13.84 grams = 2.00
So each element of the compound is increase by a factor of 2.
2(BH3) = B2H6
B2H6 is the actual molecular formula.