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Self-Check Exercise 6.10
The most common form of nylon (Nylon-6) is 63.68% carbon, 12.38% nitrogen, 9.80% hydrogen, and 14.14% oxygen. Calculate the empirical formula for Nylon-6.
Note from Example 6.14 that once the percentages are converted to
masses, this example is the same as earlier examples in which the masses were
given directly.
Calculation of Molecular Formulas
Objective: To learn to calculate the molecular formula of a compound, given
its empirical formula and molar mass.
I
f we know the composition of a compound in terms of the masses (or mass
percentages) of the elements present, we can calculate the empirical formula but not the molecular formula. For reasons that will become clear as
we consider Example 6.15, to obtain the molecular formula we must know
the molar mass. In this section we will consider compounds where both the
percent composition and the molar mass are known.
Connection to Archaeology
Using Chemistry to Understand Lost Worlds
A
rchaeology involves the study of past civilizations by examining the material remains of lost
cultures. Usually it relies on digging. James E. Myster of Hamline University in Minnesota, however,
does much of his archaeology through chemistry.
Myster analyzes the soil to find out where buildings
once stood and what the purposes of the structures
were. This approach works because certain types of
activities leave telltale elements in the soil. For example, a high phosphorus content in a given patch
of soil indicates that animals may have been kept in
that area—animal dung has a high phosphate content. When Myster finds an area characterized by
high calcium and phosphorus content, he concludes
that animals may have been butchered and processed there. Animal flesh contains large amounts of
phosphorus, and bones and teeth contain a great
deal of calcium. On the other hand, areas used for
gardening typically show a depletion of calcium,
magnesium, nitrogen, phosphorus, and sulfur.
Using soil analysis, Myster can even map where
buildings once stood even though no trace of them
remains. Soil inside the former structures contains
elemental evidence of the activities that took place
there. For instance, the former locations of privies
show high phosphate content, whereas the soil under smokehouses has high sodium content from
the salt used for the preservation of meat. The outlines of former buildings are also indicated by the
way that pollutants, such as cadmium, chromium,
lead, and nickel are distributed in the soil. The presence of a building would prevent these elements,
which were scattered by the wind during the early
Industrial Revolution, from being deposited in the
soil.
One of the best things about chemical archaeology is that it is nondestructive. The soil analysis does
not require digging up the site. Coupled with classical methods, chemical archaeology can be very
revealing.
6.8 Calculation of Molecular Formulas
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Example 6.15
Calculating Molecular Formulas
A white powder is analyzed and found to have an empirical formula of
P2O5. The compound has a molar mass of 283.88 g. What is the compound’s molecular formula?
Solution
P
Group
5
O
Group
6
To obtain the molecular formula, we must compare the empirical formula
mass to the molar mass. The empirical formula mass for P2O5 is the mass
of 1 mol of P2O5 units.
Atomic
mass of P
2 mol P: 2 30.97 g 61.94 g
5 mol O: 5 16.00 g 80.00 g
Atomic
mass of O
141.94 g
Mass of 1 mol of P2O5 units
Recall that the molecular formula contains a whole number of empirical formula units. That is,
Molecular formula (empirical formula)n
where n is a small whole number. Now, because
Molecular formula n empirical formula
then
Molar mass n empirical formula mass
Solving for n gives
molar mass
n empirical formula mass
Thus, to determine the molecular formula, we first divide the molar mass
by the empirical formula mass. This tells us how many empirical formula
masses are present in one molar mass.
283.88 g
Molar mass
2
Empirical formula mass
141.94 g
This result means that n 2 for this compound, so the molecular formula consists
of two empirical formula units, and the
molecular formula is (P2O5)2, or P4O10.
The structure of this interesting compound is shown in Figure 6.5.
Figure 6.5
The structure of P4O10 as a “ball-and-stick” model.
This compound has a great affinity for water and is
often used as a desiccant, or drying agent.
184
Chapter 6 Chemical Composition
=P
=O
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Self-Check Exercise 6.11
A compound used as an additive for gasoline to help prevent engine knock
shows the following percentage composition:
71.65% Cl
24.27% C
4.07% H
The molar mass is known to be 98.96 g. Determine the empirical formula
and the molecular formula for this compound.
It is important to realize that the molecular formula is always an integer
multiple of the empirical formula. For example, the sugar glucose (see Figure 6.4)
has the empirical formula CH2O and the molecular formula C6H12O6. In this
case there are six empirical formula units in each glucose molecule:
(CH2O)6 C6H12O6
CHEMISTRY
Molecular formula (empirical
formula)n , where n is an integer.
Focus Questions
In general, we can represent the molecular formula in terms of the empirical
formula as follows:
(Empirical formula)n molecular formula
where n is an integer. If n 1, the molecular formula is the same as the empirical formula. For example, for carbon dioxide the empirical formula (CO2)
and the molecular formula (CO2) are the same, so n 1. On the other hand,
for tetraphosphorus decoxide the empirical formula is P2O5 and the molecular formula is P4O10 (P2O5)2. In this case n 2.
Sections 6.5–6.8
1. Explain how you would find the mass percent of each element in water.
2. Is the empirical formula for a compound ever the same as its molecular
formula? Explain.
3. Why are the subscripts in a chemical formula whole numbers—for
example, N2O5 instead of NO2.5?
4. What critical piece of information must be known (or given in the
problem) to determine the molecular formula for a compound from its
empirical formula?
Focus Questions
185