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Empirical FormulasEmpirical Formulas
Chapter 7Chapter 7--4, p. 229 4, p. 229 -- 231231
Empirical FormulaEmpirical Formula
•• Defined as:Defined as:
the symbols for the elements the symbols for the elements
combined in a chemical compound, combined in a chemical compound,
with subscripts showing with subscripts showing the smallest the smallest
whole number mole ratiowhole number mole ratio of the of the
different atoms in the compound.different atoms in the compound.
ExampleExample
CC22HH6 6 = = dicarbondicarbon hexahydridehexahydride (ethane)(ethane)
has an has an empirical formulaempirical formula of CHof CH33
(notice the common multiple is 2, (notice the common multiple is 2,
which you use to divide to get the which you use to divide to get the
smallest whole number ratio)smallest whole number ratio)
Ethane’s Ethane’s molecular formulamolecular formula is is CC22HH66
Determining Empirical Formula Determining Empirical Formula
from Percent Compositionfrom Percent Composition
1.1.Assume you have 100.00 g of the Assume you have 100.00 g of the
compound.compound.
2.2.Use the number in percent of each Use the number in percent of each
element and change it to grams. For element and change it to grams. For
example, the percent oxygen in NaOH example, the percent oxygen in NaOH
is 32.6%. If you have a 100.00 g is 32.6%. If you have a 100.00 g
sample of NaOH, 32.6 g of it is oxygen.sample of NaOH, 32.6 g of it is oxygen.
3.3.Divide by the molar mass of each Divide by the molar mass of each
element from the periodic table.element from the periodic table.
4.4.Compare the mole ratios by dividing Compare the mole ratios by dividing
all values by the smallest mole value.all values by the smallest mole value.
ExampleExample
A compound is 32.38% Na, 22.65% S A compound is 32.38% Na, 22.65% S
and 44.99% O. Determine the and 44.99% O. Determine the
empirical formula.empirical formula.
32.38 g Na 1 mol Na = 1.408 mol Na32.38 g Na 1 mol Na = 1.408 mol Na
22.99 g Na22.99 g Na
22.65 g S 1 mol S = 0.7063 mol S22.65 g S 1 mol S = 0.7063 mol S
32.07 g S32.07 g S
44.99 g O 1 mol O = 2.812 mol O44.99 g O 1 mol O = 2.812 mol O
16.00 g O16.00 g O
Which molar value is the smallest?Which molar value is the smallest?
ExampleExample
A compound is 32.38% Na, 22.65% S A compound is 32.38% Na, 22.65% S
and 44.99% O. Determine the and 44.99% O. Determine the
empirical formula.empirical formula.
1.408 mol Na1.408 mol Na = 1.993 mol Na = 2 Na= 1.993 mol Na = 2 Na
0.70630.7063
0.7063 mol S0.7063 mol S = 1.000 mol S = 1 S= 1.000 mol S = 1 S
0.70630.7063
2.812 mol O 2.812 mol O = 3.981 mol O = 4 O= 3.981 mol O = 4 O
0.70630.7063
The empirical formula is NaThe empirical formula is Na22SOSO44
