Page 119 - Theory and Problems of BEGINNING CHEMISTRY

108 FORMULA CALCULATIONS [CHAP. 7

EXAMPLE 7.14. Determine the empirical formula of a sample of a compound which contains 44.21 g Fe and 16.89 g O.
Ans. Since the numbers of grams are given, the step changing percent to grams is not necessary.

1 mol Fe 1 mol O
44.21gFe = 0.7916 mol Fe 16.89gO = 1.056 mol O
55.85gFe 16.00gO
Dividing each by 0.7916 yields 1.000 mol Fe for every 1.333 mol O. This ratio is still not integral. Multiplying these
values by 3 yields integers. (We can round when a value is within 1% or 2% of an integer, but not more.)
3.000 mol Fe and 3.999 or 4 mol O
The formula is Fe 3 O 4 .

7.7. MOLECULAR FORMULAS
Formulas describe the composition of compounds. Empirical formulas give the mole ratio of the various
elements. However, sometimes different compounds have the same ratio of moles of atoms of the same elements.
For example, acetylene, C 2 H 2 , and benzene, C 6 H 6 , each have 1 : 1 ratios of moles of carbon atoms to moles of
hydrogenatoms.Thatis,eachhasanempiricalformulaCH.Suchcompoundshavethesamepercentcompositions.
However, they do not have the same number of atoms in each molecule. The molecular formula is a formula
that gives all the information that the empirical formula gives (the mole ratios of the various elements) plus the
information of how many atoms are in each molecule. In order to deduce molecular formulas from experimental
data, the percent composition and the molar mass are usually determined. The molar mass may be determined
experimentally in several ways, one of which will be described in Chap. 12.
It is apparent that the compounds C 2 H 2 and C 6 H 6 have different molecular masses. That of C 2 H 2 is 26 amu;
that of C 6 H 6 is 78 amu. It straightforward to determine the molecular mass from the molecular formula, but how
can the molecular formula be determined from the empirical formula and the molecular mass? The following
steps are used, with benzene having a molecular mass of 78 amu and an empirical formula of CH serving as an
example.

Step Example
1. Determine the formula mass corresponding to 1 empirical CH has a formula mass of 12 + 1 = 13 amu
formula unit.
2. Divide the molecular mass by the empirical formula mass. Molecular mass = 78 amu
The result must be an integer. 78 amu
Empirical mass = = 6
13 amu
3. Multiply the number of atoms of each element in the em- (CH) 6 = C 6 H 6
pirical formula by the whole number found in step 2.

EXAMPLE 7.15. A compound contains 85.7% carbon and 14.3% hydrogen and has a molar mass of 98.0 g/mol. What is
its molecular formula?
Ans. The first step is to determine the empirical formula from the percent composition data.

1 mol C
85.7gC = 7.14 mol C
12.0gC
1 mol H

14.3gH = 14.2 mol H
1.008gH
The empirical formula is CH 2 .
Now the molecular formula is determined by using the steps outlined above:
1. Mass of CH 2 = 14.0 amu
2. Number of units = (98.0 amu)/(14.0 amu) = 7
3. Molecular formula = (CH 2 ) 7 = C 7 H 14