Bell Work: % Comp Review
**Turn in late Folder Checks**
A 100 gram sample of carbon dioxide is
27.3% carbon.
1. How many grams of C are in the sample?
2. How many grams of O are in the sample?
3. What’s the molar mass of CO2?
Empirical and Molecular
Formulas
Section 11.4
Empirical formulas are like reduced
fractions.
10
 can
12
be reduced to . . . . . .
5
6
Mg2O2 can be reduced to . . . . . . MgO
C6H12O6 can be reduced to . . . . . . CH2O
Types of Formulas
The formulas for compounds can be
expressed as an empirical formula and as a
molecular(true) formula.
Empirical
Molecular (true)
Name
CH
C 2H 2
acetylene
CH
C 6H 6
benzene
CO2
CO2
CH2O
C5H10O5
carbon dioxide
Timberlake LecturePLUS
ribose
4
Empirical Formulas
Write your own one-sentence definition for
each of the following:
Empirical formula
Molecular formula
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An empirical formula represents the
simplest whole number ratio of the
atoms in a compound.
The molecular formula is the true or
actual ratio of the atoms in a
compound.
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Learning Check EF-1
A. What is the empirical formula for C4H8?
1) C2H4
2) CH2
3) CH
B. What is the empirical formula for C8H14?
1) C4H7
2) C6H12
3) C8H14
C. Which are possible molecular formulas for
CH2O?
1) CH2O
2) C2H4O2
Timberlake LecturePLUS
3) C3H6O3
7
Solution EF-1
A. What is the empirical formula for C4H8?
2) CH2
B. What is the empirical formula for C8H14?
1) C4H7
C. What is a molecular formula for CH2O?
1) CH2O
2) C2H4O2
Timberlake LecturePLUS
3) C3H6O3
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Empirical Formulas
 What if you don’t know the chemical formula?
 We can use the percent composition to find the empirical
formula.
 An empirical formula represents the simplest whole
number ratio of the atoms in a compound.
 HF5
Empirical Formula
 The empirical formula may or may not be the same as the
actual molecular formula.
 Molecular Formula: the actual number of atoms of each
element in one molecule or formula unit of a substance.
Empirical Formula
 Example: glucose
 Molecular formula = C6H12O6
 Empirical formula = CH2O
What is their common element ratio?
Ratio of one C to 2 H to 1 O
 What is the scaling factor?
6
Finding Empirical Formulas from
%Composition
 If you have the percent composition given to you, you can
determine the empirical formula.
 You need to assume a few things.
 The total mass of the compound is 100.0g.
 The percent composition of the element is equal to the mass
in grams of the element.
 Example 1:
An oxide of sulfur has a percent composition of 40.05% S
and 59.95% O.
In 100 g of the compound, 40.05 g are S and 59.95 g are O.
Next, find the amount of mol for each element.
Empirical Formula
 Next, find the amount of mol for each element.
 1 mol S 
40.05g S x 
  1.249 mol S
 32.07g S 
 1 mol O 
59.95g O x 
  3.747 mol O.
 16.00g O 
 OK, Now what?
 These numbers will help us determine the subscripts
for the empirical formula.
 You cannot use the exact numbers that we just found
because they are not whole numbers.
Empirical Formula
 Divide all numbers by the smallest number.
 So, S has a subscript of one:
1.249 mol S / 1.249 = 1 mol S
 Then, divide the mol O by the same number to find its
subscript in the empirical formula.
3.747 mol O/ 1.249 = 3 mol O
Then write your empirical formula using your
subscripts: SO3
Example Problem 2
Pg. 333: 46
1 𝑚𝑜𝑙 𝑁
36.84𝑔𝑁𝑥
= 2.63 𝑚𝑜𝑙 𝑁
14.0 𝑔 𝑁
1 𝑚𝑜𝑙 𝑂
63.16𝑔𝑁𝑥
= 3.95 𝑚𝑜𝑙 𝑂
16.0 𝑔 𝑂
N: 2.63 / 2.63 = 1 mol
O: 3.95 / 2.63 = 1.5 mol
Have to double moles to get whole # subscripts.
N2O3
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
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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
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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.
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Solution EF-5
60.0 g C x
4.5 g H
___________= ______ mol C
x ___________ = _______mol H
35.5 g O x ___________ = _______mol O
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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
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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?_____
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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?_____
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Practice Problems
Pg 335: 52
A Handy Flowchart