Chapter 7: Chemical Formulas and Chemical Compounds
I. CHEMICAL NAMES AND FORMULAS
A.
Significance of a Chemical Formula
1.
A chemical formula shows the relative number of atoms of each kind in a chemical
compound, with subscripts indicating the number of atoms of one type of element.
2.
In a molecular formula, the chemical formula shows the number of atoms of each
element in one molecule of the compound.
a)
For example, H2O, the chemical formula for water, indicates the presence of 2
hydrogen atoms and 1 oxygen atom.
3.
In an ionic compound, the chemical formula represents one formula unit, which is the
simplest ratio of the compound‚ positive ions (cations) and negative ions (anions).
B.
Monatomic Ions
1.
Many main-group elements can become ions by gaining or losing electrons to have
noble-gas configurations.
2.
Monatomic ions are ions formed from a single atom.
a)
Group 1 metals lose one electron; Group 2 electrons lose two electrons. They
will have +1 and +2 charges, respectively.
b)
The nonmetals in Group 15, 16, and 17 gain electrons and form anions.
c)
Metals with d orbitals are slightly different, as they can form +2, +3, or even +1
and +4. Some of them can even form ions of different charges.
(1)
Refer to the chart on page 221 of Modern Chemistry published by Holt,
Rinehart, and Winston.
C.
Naming Monatomic Ions
1.
Monatomic cations are identified by their normal name.
2.
Monatomic anions, however, have their ending dropped and the ending ‚ -ide is added
to the root name.
D.
Binary Ionic Compounds
1.
Binary compounds are compounds composed of two elements.
2.
In a binary ionic compound, the total numbers of positive charges and negative
charges must be equal.
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a)
For example, Mg (magnesium) has a +2 charge and Br (bromine bromide) has a
-1 charge. In order for a neutral charge, there must be 2 bromide ions bonded to the
Mg. Thus, +2 + (-2) = 0. The chemical formula will be written as MgBr2.
E.
Naming Binary Ionic Compounds
1.
The nomenclature of binary ionic compounds involves combining the names of the
compound (positive and negative ions).
2.
The positive ion goes first, followed by the name of the anion.
a)
The ratio of the ions does not influence its name. For example, Al2O3 is called
aluminum oxide.
3.
F.
Practice:
The Stock System of Nomenclature
1.
Some elements form two or more cations with different charges.
2.
Scientists use the Stock system of nomenclature to distinguish the ions formed by
such elements.
3.
The system consists of a Roman numeral placed in parentheses after the metal name.
a)
Examples are iron (II) and iron (III). Metals that commonly only form one cation
do not include a roman numeral. A compound such as CuCl2 is called copper (II)
chloride.
G.
Compounds Containing Polyatomic Ions
1.
Most polyatomic ions are negatively charged and most are oxyanions which means
they contain oxygen.
2.
The name a compound receives depends on the number of oxygen atoms.
a)
The name with a greater number ends in ‚ -ate, while the smaller number gets
an ‚-ite.
b)
Some elements can also form more than two types of oxyanions.
(1)
Anions with one fewer oxygen than the ‚ -ite anion has the prefix hypo-.
(2)
Atoms with one more oxygen than the -ate anion has the prefix per-.
(a)
Refer to page 226 of Modern Chemistry published by Holt,
Rinehart, and Winston for a chart of common polyatomic ions.
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H.
Naming Binary Molecular Compounds
1.
Molecular compounds are composed of individually covalently bonded units.
2.
Prefixes are used in naming molecular compounds.
a)
For example, tri- indicates three of one type of atom.
b)
The second element is named by combining the prefix indicative of the number
of atoms, the root name, and then the ending -ide.
(1)
The o or a at the end of a prefix is usually dropped if there is another
vowel following it.
(a)
For example, monoxide and pentoxide instead of mono-oxide and
penta-oxide.
3.
I.
Practice:
Acids and Salts
1.
Binary acids consist of two elements, usually hydrogen and a halogen.
a)
To name a binary acid, the hydrogen is named hydro and the name of the
anion ending in -ic.
2.
Oxyacids contain hydrogen, oxygen, and a third element (usually a nonmetal).
a)
They are named based on how many oxygen atoms are in the acid. The
common acid ends in -ic.
b)
The acid with one more oxygen has the prefix per- and ends in - ic. The acid
with one less oxygen ends in -ous and the one with two less oxygen atoms ends in -ous
and has the hypo- prefix.
c)
Practice:
3.
An ionic compound composed of a cation and the anion from an acid ois often
referred to as a salt.
a)
Examples:
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II. USING CHEMICAL FORMULAS
A.
Formula Masses
1.
To find the formula mass of any molecule, formula unit, or ion, add the average
atomic masses of all atoms represented in its formula.
2.
For example, to find the formula mass of water, add the masses of two hydrogen
atoms and one oxygen atom: 1.01 + 1.01 + 16.0 = 18.02 g.
3.
Formula masses are numerically equivalent to molar masses, the mass in grams of one
mole of the substance (g/mol).
B.
Percentage Composition
1.
The percentage by mass of each element in a compound is known as the percentage
composition of the compound.
a)
In hydrates, where there are water molecules attached, the percent
composition of the water molecules are calculated together. In other words, you
would find the percent H2O instead of separating out the Hs and Os.
III. DETERMINING CHEMICAL FORMULAS
A.
An empirical formula consists of the symbols for the elements combined in a
compound, with subscripts showing the smallest whole-number mole ratio of the different
atoms in the compound.
B.
Calculation of Empirical Formulas
1.
To determine empirical formula from percentage composition, assume that there is a
100 g sample. The percentages will then indicate how many grams are in a 100 g sample.
2.
For example, 78.1% B in a 100 g sample means that there are 78.1 grams B in that
sample and 21.9% H means there are 21.9 grams H.
3.
Next, convert the mass composition into a mole composition by dividing by the molar
mass.
a)
B has a molar mass of 10.8, so divide 78.1 by 10.8 = 7.23 mol B.
b)
H has a molar mass of 1.01, so divide 21.9 by 1.01 to get 21.7 mol H.
c)
Then divide by the smallest number to get a ratio of smallest whole numbers.
7.23 divided by 7.23 equals 1. 21.9 divided by 7.23 equals 3.03. Round to get a ratio of
1 B: 3 H.
d)
C.
The empirical formula is BH3.
Calculation of Molecular Formulas
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1.
The molecular formula is the actual formula of a molecular compound. Compounds
may share the same ratio (empirical formula) but may be very different from having distinct
molecular formulas. A molecular formula is x (empirical formula), where x is a whole-number
multiple indicating the factor the subscripts of the empirical formula are multiplied by.
2.
To determine the molecular formula, it is necessary to have the empirical formula and
the compound’s formula mass/molar mass.
a)
First, find the mass of the empirical formula. Then divide the experimental
formula mass (given) by the empirical formula mass to get x. Then multiply the
subscripts in the empirical formula by x.
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