Experimental Determination of Empirical Formulas
Experimental Determination of Empirical Formulas
** The fact
that we can determine the empirical formula of a compound if we know the
percent composition enables us to identify compounds experimentally.
** The
procedure is as follows.
(1) Chemical analysis tells us the number of grams
of each element present in a given amount of a compound.
(2) We convert the
quantities in grams to number of moles of each element.
(3) Using a certain method (we will illustrate it under) , we find the empirical formula of the compound.
** As a specific
example, let us consider the compound ethanol. When ethanol is burned in an
apparatus such as that shown in the following Figure.
Carbon dioxide
(CO2) and water (H2O) are given off. Because neither
carbon nor hydrogen was in the inlet gas, we can conclude that both carbon (C)
and hydrogen (H) were present in ethanol and that oxygen (O) may also be
present. (Molecular oxygen was added in the combustion process, but some of the
oxygen may also have come from the original ethanol sample.)
** The masses
of CO2 and of H2O produced can be determined by measuring
the increase in mass of the CO2 and H2O absorbers,
respectively. Suppose that in one experiment the combustion of 11.5 g of
ethanol produced 22.0 g of CO2 and 13.5 g of H2O. We can
calculate the mass of carbon and hydrogen in the original 11.5-g sample of
ethanol as follows:
** Thus, 11.5 g
of ethanol contains 6.00 g of carbon and 1.51 g of hydrogen. The remainder must
be oxygen, whose mass is:
** The number
of moles of each element present in 11.5 g of ethanol is:
** The formula
of ethanol is therefore C0.50H1.5O0.25 (we
round off the number of moles to two significant figures). Because the number
of atoms must be an integer, we divide the subscripts by 0.25, the smallest
subscript, and obtain for the empirical formula C2H6O.
** Now we can better understand the word (empirical)
which literally means (based only on observation and measurement). The
empirical formula of ethanol is determined from analysis of the compound in
terms of its component elements. No knowledge of how the atoms are linked
together in the compound is required.
Determination of Molecular Formulas
** The formula
calculated from percent composition by mass is always the empirical formula because
the subscripts in the formula are always reduced to the smallest whole numbers.
** To calculate
the actual, molecular formula we must know the approximate molar mass of the compound
in addition to its empirical formula.
** Knowing that
the molar mass of a compound must be an integral multiple of the molar mass of
its empirical formula, we can use the molar mass to find the molecular formula.
Solved problem
A sample of a
compound contains 1.52 g of nitrogen (N) and 3.47 g of oxygen (O). The molar mass
of this compound is between 90 g and 95 g. Determine the molecular formula and
the accurate molar mass of the compound.
Strategy:
** To determine the molecular formula, we first
need to determine the empirical formula. How do we convert between grams and
moles? Comparing the empirical molar mass to the experimentally determined
molar mass will reveal the relationship between the empirical formula and
molecular formula.
Solution:
** We are given grams of N and O. Use molar
mass as a conversion factor to convert grams to moles of each element. Let n represent
the number of moles of each element. We write:
Thus, we arrive
at the formula N0.108O0.217, which gives the identity and
the ratios of atoms present. However, chemical formulas are written with whole
numbers. Try to convert to whole numbers by dividing the subscripts by the
smaller subscript (0.108). After rounding off, we obtain NO2 as the empirical
formula.
** The
molecular formula might be the same as the empirical formula or some integral multiple
of it (for example, two, three, four, or more times the empirical formula). Comparing
the ratio of the molar mass to the molar mass of the empirical formula will
show the integral relationship between the empirical and molecular formulas.
The molar mass of the empirical formula NO2 is:
Next, we
determine the ratio between the molar mass and the empirical molar mass
** The molar
mass is twice the empirical molar mass. This means that there are two NO2
units in each molecule of the compound, and the molecular formula is (NO2)2
or N2O4.
** The actual
molar mass of the compound is two times the empirical molar mass, that is, 2(46.01
g) or 92.02 g, which is between 90 g and 95 g.
Check:
Note that in determining the molecular formula
from the empirical formula, we need only know the approximate molar mass of the
compound. The reason is that the true molar mass is an integral multiple (1×, 2×,
3×, . . .) of the empirical molar mass. Therefore, the ratio (molar mass/empirical
molar mass) will always be close to an integer.
Reference: General Chemistry: The Essential Concepts / Raymond Chang , Jason Overby. (sixth edition).
No comments