A
chemical formula conveys considerable quantitative information about a compound
and its constituent elements. We have already learned how to determine the
molar mass of a compound, and, in this section, we consider some other types of
calculations based on the chemical formula.

The
colorless, volatile liquid halothane has been used as a fire extinguisher and
also as an inhalation anesthetic. Its empirical and molecular formulas are C

_{2}HBrClF_{3}, its molecular mass is 197.382 u, and its molar mass is 197.382 g/mol, as calculated below:
The molecular formula of C

_{2}HBrClF_{3}tells us that per mole of halothane there are two moles of C atoms, one mole each of H, Br, and Cl atoms, and three moles of F atoms. This factual statement can be turned into conversion factors to answer such questions as, "How many C atoms are present per mole of halo thane?" In this case, the factor needed is 2 mol C/mol C_{2}HBrClF_{3}. That is,
In
Example 3-3, we use another conversion factor derived from the formula for
halothane. This factor is shown in blue in the setup, which includes other
familiar factors to make the conversion pathway:

**Calculating Percent Composition from a Chemical Formula**

When
chemists believe that they have synthesized a new compound, a sample is
generally sent to an analytical laboratory where its percent composition is
determined. This experimentally determined percent composition is then compared
with the percent composition calculated from the formula of the expected
compound. In this way, chemists can see if the compound obtained could be the
one expected.

Equation
(3.1) establishes how the mass percent of an element in a compound is
calculated. In applying the equation, as in Example 3-4, think in terms of the
following steps.

1.
Determine the molar mass of the compound. This is the denominator in equation
(3.1).

2.
Determine the contribution of the given element to the molar mass. This product
of the formula subscript and the molar mass of the element appears in the
numerator of equation (3.1).

3.
Formulate the ratio of the mass of the given element to the mass of the
compound as a whole. This is the ratio of the numerator from step 2 to the
denominator from step 1.

4.
Multiply this ratio by 100% to obtain the mass percent of the element.

The
mass composition of a compound is the collection of mass percentages of the
individual elements in the compound.

The
percentages of the elements in a compound should add up to 100.00%, and we can
use this fact in two ways.

1. Check
the accuracy of the computations by ensuring that the percentages total
100.00%. As applied to the results of Example 3-4:

12.17% + 0.51% + 40.48% + 17.96% +
28.88% = 100.00%

2. Determine the percentages of all the
elements but one. Obtain that one
difference (subtraction). From Example 3-4:

% H =
100.00% - % C - % Br - % Cl - % F

=
100.00% - 12.17% - 40.48% - 17.96% - 28.88%

=
0.51%

**Establishing Formulas from the Experimentally Determined Percent Composition of Compounds**

At
times, a chemist isolates a chemical compound—say, from an exotic tropical
plant—and has no idea what it is. A report from an analytical laboratory on the
percent composition of the compound yields data needed to determine its
formula.

Percent
composition establishes the relative proportions of the elements in a compound
on a mass basis. A chemical formula requires these proportions to be on a mole
basis, that is, in terms of numbers of atoms. Consider the following five-step
approach to determining a formula from the experimentally determined percent
composition of the compound 2-deoxyribose, a sugar that is a basic constituent
of DNA (deoxyribonucleic acid). The mass percent composition of 2-deoxyribose
is 44.77% C, 7.52% H, and 47.71% O.

1.
Although we could choose any sample size, if we take one of exactly 100 g, the
masses of the elements are numerically equal to their percentages, that is,
44.77 g C, 7.52 g H, and 47.71 g O.

2.
Convert the masses of the elements in the 100.00 g sample to amounts in moles.

3.
Write a tentative formula based on the numbers of moles just determined.

C

_{3.727}H_{7.46}O_{2.982}
4.
Attempt to convert the subscripts in the tentative formula to small whole
numbers. This requires dividing each of the subscripts by the smallest one
(2.982).

If
all subscripts at this point differ only slightly from whole numbers— which is
not the case here—round them off to whole numbers, concluding the calculation
at this point.

5. If
one or more subscripts is still not a whole number—which is the case
here—multiply all subscripts by a small whole number that will make them all
integral. Thus, multiply by 4 here.

The
formula that we get by the method just outlined, C

_{5}H_{10}O_{4}, is the simplest possible formula—the empirical formula. The actual molecular formula may be equal to, or some multiple of, the empirical formula, such as C_{10}H_{20}O_{8}, C_{15}H_{30}O_{12}, C_{20}H_{40}O_{16}, and so on. To find the multiplying factor, we must compare the formula mass based on the empirical formula with the true molecular mass of the compound. We can establish the molecular mass from a separate experiment (by methods introduced in next blog posts). The experimentally determined molecular mass of 2-deoxyribose is 134 u. The formula mass based on the empirical formula, C_{5}HK_{3}O_{4}, is 134.1 u. The measured molecular mass is the same as the empirical formula mass. The molecular formula is also C_{5}H_{10}O_{4}.
We
outline the five-step approach described above in the flow diagram below and
then apply the approach to Example 3-5, where we will find that the empirical
formula and the molecular formula are not the same.

**Combustion Analysis**

Figure
3-6 illustrates an experimental method for establishing an empirical formula
for compounds that are easily burned, such as compounds containing carbon and
hydrogen with oxygen, nitrogen, and a few other elements. In combustion
analysis, a weighed sample of a compound is burned in a stream of oxygen gas.
The water vapor and carbon dioxide gas produced in the combustion are absorbed
by appropriate substances. The increases in mass of these absorbers correspond
to the masses of water and carbon dioxide. We can think of the matter as shown
below. (The subscripts x, y, and z are integers whose values we do not know
initially.)

After
combustion, all the carbon atoms in the sample are found in the C02. All the H
atoms are in the H20. Moreover, the only source of the carbon and hydrogen
atoms was the sample being analyzed. Oxygen atoms in the C02 and H20 could have
come partly from the sample and partly from the oxygen gas consumed in the
combustion. Thus, the quantity of oxygen in the sample has to be determined
indirectly. These ideas are applied in Example 3-6.

We
have just seen how combustion reactions can be used to analyze chemical
substances, but not all samples can be easily burned. Fortunately, several
other types of reactions can be used for chemical analyses. Also, modern
methods in chemistry rely much more on physical measurements with instruments
than on chemical reactions. We will cite some of these methods later in other
posts (use search box above to look for the relative content).