Page 1 of 3
Formulas of Compounds
Objective: To understand the meaning of the empirical formulas of compounds.
A
ssume that you have mixed two solutions, and a solid product forms.
How can you find out what the solid is? What is its formula? Although
an experienced chemist can often predict the product expected in a chemical reaction, the only sure way to identify the product is to perform experiments. Usually we compare the physical properties of the product to the properties of known compounds.
Sometimes a chemical reaction gives a product that has never been obtained before. In such a case, a chemist determines what compound has been
formed by determining which elements are present and how much of each
is there. These data can be used to obtain the formula of the compound. In
Section 6.5 we used the formula of the compound to determine the mass of
Connection to Histor y
“Mauvelous” Chemistry
S
ometimes a little chemical knowledge can lead
to a new and profitable career—as happened to
a young chemistry student named William Perkin.
In one of his chemistry lectures, Perkin’s professor
said it would be very desirable to make quinine artificially (in 1856, the only source for quinine—a
drug effective against malaria—was the bark of the
cinchona tree). Perkin, who was only 18 at the time,
decided to make quinine artificially in his home laboratory as a project over Easter vacation.
To do so, he looked at the known formulas for
quinine and toluidine (his starting reactant). Perkin
recognized that he simply needed to add some carbon and hydrogen atoms and then several oxygens
to get quinine. He had made some large oversimplifications and his reactions produced an ugly reddish-brown sludge. But Perkin didn’t give up—he
changed to a new starting material called aniline.
This time he got an even worse-looking black solid.
He was ready to throw the compound out when he
noticed that it turned a beautiful purple color in water or alcohol.
Perkin liked the color so much that he became
distracted from his original quest to make quinine.
In testing the purple solutions, he found he could
use them to dye cloth. He sent his synthetic dye to
a dye works, which found that it worked well on
silk and pretreated cotton.
Perkin decided to quit school (against the advice
of his professors) and convinced his wealthy father
to back his venture. He patented his dye and built
a factory to produce it on a large scale. His dye
eventually became know as mauve.
174
Chapter 6 Chemical Composition
Until Perkin’s discovery, all permanent purple or
lavender dyes were outrageously expensive. The
only source for the natural dye was a mollusk in the
Mediterranean Sea. It took 9000 mollusks for 1g of
dye—no wonder purple became the color associated with royalty. Perkin changed everything by producing a purple dye from coal tar that virtually
everyone could afford and wear.
Perkin went on to develop another popular dye
called alizarin in 1869. This red dye was being produced by the ton in his own factory by 1871.
Perkin sold his factory in 1874 and at age 36 had
enough money to spend the rest of his life doing
pure research. He bought a new house and continued working in his old laboratory. Eventually he
made courmarin, the first perfume from coal tar and
cinnamic acid. Perkin’s work started an explosion in
the manufacture of organic
chemicals and research
into their chemistry.
Image not available for this
CD-ROM. Please refer to
the image in the textbook.
A portrait of Pliny
wearing royal purple robes.
Page 2 of 3
each element present in a mole of the compound. To obtain the formula of
an unknown compound, we do the opposite. That is, we use the measured
masses of the elements present to determine the formula.
Recall that the formula of a compound represents the relative numbers
of the various types of atoms present. For example, the molecular formula
CO2 tells us that for each carbon atom there are two oxygen atoms in each
molecule of carbon dioxide. So to determine the formula of a substance we
need to count the atoms. As we have seen in this chapter, we can do this by
weighing. Suppose we know that a compound contains only the elements
carbon, hydrogen, and oxygen, and we weigh out a 0.2015-g sample for analysis. Using methods we will not discuss here, we find that this 0.2015-g sample of compound contains 0.0806 g of carbon, 0.01353 g of hydrogen, and
0.1074 g of oxygen. We have just learned how to convert these masses to
numbers of atoms by using the atomic mass of each element. We begin by
converting to moles.
Carbon
1 mol C atoms
(0.0806 g C) 0.00671 mol C atoms
12.01 g C
Hydrogen
1 mol H atoms
(0.01353 g H) 0.01342 mol H atoms
1.008 g H
Oxygen
1 mol O atoms
(0.1074 g O) 0.006713 mol O atoms
16.00 g O
Let’s review what we have established. We now know that 0.2015 g of
the compound contains 0.00671 mol of C atoms, 0.01342 mol of H atoms,
and 0.006713 mol of O atoms. Because 1 mol is 6.022 1023, these quantities can be converted to actual numbers of atoms.
Carbon
(6.022 1023 C atoms)
(0.00671 mol C atoms) 4.04 1021 C atoms
1 mol C atom
Hydrogen
(6.022 1023 H atoms)
(0.01342 mol H atoms) 8.08 1021 H atoms
1 mol H atom
Oxygen
(6.022 1023 O atoms)
(0.006713 mol O atoms) 4.043 1021 O atoms
1 mol O atom
These are the numbers of the various types of atoms in 0.2015 g of compound. What do these numbers tell us about the formula of the compound?
Note the following:
1. The compound contains the same number of C and O atoms.
2. There are twice as many H atoms as C atoms or O atoms.
We can represent this information by the formula CH2O, which expresses the
relative numbers of C, H, and O atoms present. Is this the true formula for
the compound? In other words, is the compound made up of CH2O molecules? It may be. However, it might also be made up of C2H4O2 molecules,
6.6 Formulas of Compounds
175
Page 3 of 3
H
H
H
C
C
OH
O
H
OH
C
HO
OH
H
C
C
H
OH
C
H
Figure 6.4
The glucose molecule. The molecular formula is C6H12O6, as
can be verified by counting the
atoms. The empirical formula for
glucose is CH2O.
C3H6O3 molecules, C4H8O4 molecules, C5H10O5 molecules, C6H12O6 molecules, and so on. Note that each of these molecules has the required 121
ratio of carbon to hydrogen to oxygen atoms (the ratio shown by experiment
to be present in the compound).
When we break a compound down into its separate elements and “count”
the atoms present, we learn only the ratio of atoms—we get only the relative
numbers of atoms. The formula of a compound that expresses the smallest
whole-number ratio of the atoms present is called the empirical formula or
simplest formula. A compound that contains the molecules C4H8O4 has the
same empirical formula as a compound that contains C6H12O6 molecules. The
empirical formula for both is CH2O. The actual formula of a compound—the
one that gives the composition of the molecules that are present—is called
the molecular formula. The sugar called glucose is made of molecules with
the molecular formula C6H12O6 (Figure 6.4). Note from the molecular formula for glucose that the empirical formula is CH2O. We can represent the
molecular formula as a multiple (by 6) of the empirical formula:
C6H12O6 (CH2O)6
In the next section, we will explore in more detail how to calculate the
empirical formula for a compound from the relative masses of the elements
present. As we will see in Sections 6.7 and 6.8, we must know the molar mass
of a compound to determine its molecular formula.
Example 6.10
Determining Empirical Formulas
In each case below, the molecular formula for a compound is given. Determine the empirical formula for each compound.
a. C6H6. This is the molecular formula for benzene, a liquid commonly
used in industry as a starting material for many important products.
b. C12H4Cl4O2. This is the molecular formula for a substance commonly called dioxin, a powerful poison that sometimes occurs as a
by-product in the production of other chemicals.
c. C6H16N2. This is the molecular formula for one of the reactants used
to produce nylon.
Solution
a. C6H6 (CH)6; CH is the empirical formula. Each subscript in the
empirical formula is multiplied by 6 to obtain the molecular
formula.
PROBLEM SOLVING
When a subscript is outside of
the parentheses, it applies to
everything in the parentheses.
Each subscript inside is multiplied by the subscript outside
the parentheses.
176
b. C12H4Cl4O2; C12H4Cl4O2 (C6H2Cl2O)2; C6H2Cl2O is the empirical
formula. Each subscript in the empirical formula is multiplied by 2
to obtain the molecular formula.
c. C6H16N2 (C3H8N)2; C3H8N is the empirical formula. Each subscript in the empirical formula is multiplied by 2 to obtain the molecular formula.
Chapter 6 Chemical Composition