Bell Work: Empirical
1. __________ formulas show the actual ratio
of elements found in a compound in nature.
2. _________ formulas show the simplified ratio
of elements in a compound.
3. Name three compounds that have identical
molecular and empirical formulas.
4. Percent to ____, mass to ___, _____ by
small, multiply ‘til _____.
Bell Work: Empirical vs. Molecular
1. Draw a Venn diagram to compare and
contrast Empirical and Molecular
formulas.
2. Draw a Venn diagram to compare and
contrast moles and grams.
3. Draw a Venn diagram to compare and
contrast formula units and molecules.
Molecular Formula
 The molecular formula needs to be found by
going one step further
 Molecular formula = (Empirical formula) x scaling
factor
 To find the scaling factor
1) Determine the empirical mass (total mass of all
elements in empirical formula)
2) Divide molecular mass by empirical mass
Finding a Molecular Formula
Example 3:
Chemical analysis of succinic acid indicates it is
composed of 40.68% C, 5.08% H, and 54.24 % O, and
has a molar mass of 118.1 g/mol. Determine the
empirical and molecular formulas for succinic acid.
1)Convert the percent for each element into moles (use
the percent given as the amount in grams for each
element in 100 g of the compound)
40.68 g C x (1 mol C/12.0 g C) = 3.39 mol C
5.08 g H x (1 mol H/1.0 g H) = 5.08 mol H
54.24 g O x (1 mol O/16.0 g O) = 3.39 mol O
Molecular Formula
2)Next, divide each mol amount by the smallest mol amount.
3.39 mol C/ 3.39 = 1 mol C
5.08 mol H/ 3.39 = 1.5 mol H
3.39 mol O/ 3.39 = 1 mol O
Ratio of C : H : O = 1 : 1.5 : 1
3) Write the Empirical Formula:
You can’t have half-moles, so multiply everything by 2.
Empirical Formula: C2H3O2
Molecular Formula
4) We need to find the empirical mass using the masses
of each element.
2 mol C x (12.0 g C/1 mol C) = 24.0 g C.
3 mol H x (1.0 g H/1 mol H) = 3.0 g H.
2 mol O x (16.0 g O/1 mol O) = 32.0 g O.
Empirical Mass: 59.0 g/ mol C2H3O2
Molecular Formula
5) Now, divide the molar mass by the empirical mass to
determine the scaling factor.
118.1 / 59.0 = 2.00
Multiply the subscripts of the empirical formula by 2 to
find the molecular formula.
Molecular Formula: C4H6O4
Learning Check EF-3
A compound has a formula mass of 176.0
and an empirical formula of C3H4O3. What
is the molecular formula?
1) C3H4O3
2) C6H8O6
3) C9H12O9
Timberlake LecturePLUS
8
Solution EF-3
A compound has a formula mass of 176.0
and an empirical formula of C3H4O3. What is
the molecular formula?
2) C6H8O6
C3H4O3 = 88.0 g/EF
176.0 g
=
2.00
88.0
Timberlake LecturePLUS
9
Learning Check EF-5
Aspirin is 60.0% C, 4.5 % H and 35.5 O.
Calculate its simplest formula. In 100
g of aspirin, there are 60.0 g C, 4.5 g
H, and 35.5 g O.
Timberlake LecturePLUS
10
Solution EF-5
60.0 g C x
4.5 g H
___________= ______ mol C
x ___________ = _______mol H
35.5 g O x ___________ = _______mol O
Timberlake LecturePLUS
11
Solution EF-5
60.0 g C x
1 mol C
=
5.00 mol C
=
4.5 mol H
=
2.22 mol O
12.0 g C
4.5 g H
x
1 mol H
1.01 g H
35.5 g O x
1mol O
16.0 g O
Timberlake LecturePLUS
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Divide by the smallest # of moles.
5.00 mol C =
________________
______ mol O
4.5 mol H
=
______ mol O
________________
2.22 mol O =
________________
______ mol O
Are are the results whole numbers?_____
Timberlake LecturePLUS
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Divide by the smallest # of moles.
5.00 mol C =
___2.25__
2.22 mol O
4.5 mol H
2.22 mol O
=
___2.00__
2.22 mol O =
___1.00__
2.22 mol O
Are are the results whole numbers?_____
Timberlake LecturePLUS
14
Practice Problems
Pg 335: 52
A Handy Flowchart