Chapter 8
Empirical & Molecular
Formulas
Types of Formulas
• The molecular formula of a compound
indicates the actual number of each atom
present in a compound.
• An empirical formula of a compound
indicates the smallest whole number ratio
of the elements of each atom present in a
compound.
Empirical and Molecular Formulas
Substance
Benzene
Molecular
Formula
C6H6
Empirical
Formula
CH
Acetylene
C2H2
CH
Glucose
C6H12O6
CH2O
Water
H2O
H2O
Chemical Formulas of Compounds
• Chemical formulas give the relative numbers of
atoms or moles of each element in a compound always a whole number ratio.
• Example
CO2 2 atoms of O for every 1 atom of C
or
2 moles of O for every 1 mole of C
• If we know or can determine the relative number
of moles of each element in a compound, we can
determine a formula for the compound.
To Calculate an Empirical Formula
1. Determine the mass in grams of each
element present, if necessary.
2. Calculate the number of moles of each
element.
3. Divide each answer from step 2 by the
smallest number of moles to obtain the
simplest whole number ratio.
4. If whole numbers are not obtained* in
step 3), multiply through by the smallest
number that will give all whole numbers
* Be
careful! Do not round off numbers prematurely
A sample of a brown gas, a major air pollutant, is found to contain
2.34g N and 5.34g O. Determine a formula for this substance.
A sample of a brown gas, a major air pollutant, is found to contain
2.34g N and 5.34g O. Determine a formula for this substance.
2.34g N
5.34g O
A sample of a brown gas, a major air pollutant, is found to contain
2.34g N and 5.34g O. Determine a formula for this substance.
2.34g N
mol
= 0.167 mol N
14.0g
5.34g O
mol
= 0.334 mol O
16.0g
A sample of a brown gas, a major air pollutant, is found to contain
2.34g N and 5.34g O. Determine a formula for this substance.
2.34g N
5.34g O
mol
= 0.167 mol N
14.0g
mol
= 0.334 mol O
16.0g
N0.167O0.334
A sample of a brown gas, a major air pollutant, is found to contain
2.34g N and 5.34g O. Determine a formula for this substance.
2.34g N
5.34g O
mol
= 0.167 mol N
14.0g
mol
= 0.334 mol O
16.0g
0.167 mol N
0.167
= 1 mol N
N0.167O0.334
0.334 mol O
0.167
= 2 mol O
A sample of a brown gas, a major air pollutant, is found to contain
2.34g N and 5.34g O. Determine a formula for this substance.
2.34g N
5.34g O
mol
= 0.167 mol N
14.0g
mol
= 0.334 mol O
16.0g
0.167 mol N
0.167
N0.167O0.334
0.334 mol O
0.167
= 1 mol N
NO2
= 2 mol O
To Calculate a Molecular Formula
1. Determine the empirical formula and
calculate its mass.
2. Divide the molecular mass by the mass of
the empirical formula you determined in
step 1. Your answer should be rounded to
the nearest whole number.
3. Multiply the empirical formula by the whole
number answer found in step 2.
A colorless liquid, used in rocket engines has a molar
mass of 92.0 g/mol. What is the molecular formula of the
compound if the empirical formula of NO2?
A colorless liquid, used in rocket engines has a
molar mass of 92.0 g/mol. What is the molecular formula
of the compound if the empirical formula of NO2?
NO2 = 14.0 + 2(16.0) = 46.0 g/mol
A colorless liquid, used in rocket engines has a
molar mass of 92.0 g/mol. What is the molecular
formula of the compound if the empirical formula of NO2?
NO2 = 14.0 + 2(16.0) = 46.0 g/mol
92.0 g/mol
=2
46.0 g/mol
A colorless liquid, used in rocket engines has a
molar mass of 92.0 g/mol. What is the molecular formula
of the compound if the empirical formula of NO2?
NO2 = 14.0 + 2(16.0) = 46.0 g/mol
92.0 g/mol
=2
46.0 g/mol
2 x NO2
=
N2O4
What is the empirical formula of a compound
that contains 53.73% Fe and 46.27% S?
What is the empirical formula of a compound
that contains 53.73% Fe and 46.27% S?
53.73g Fe
46.27g S
What is the empirical formula of a compound
that contains 53.73% Fe and 46.27% S?
53.73g Fe
mol
= 0.963 mol Fe
55.8g
46.27g S
mol
= 1.44 mol S
32.1g
What is the empirical formula of a compound
that contains 53.73% Fe and 46.27% S?
53.73g Fe
mol
= 0.963 mol Fe
55.8g
0.963 mol Fe
0.963 mol
= 1 Fe
46.27g S
mol
32.1g
1.44 mol S
0.963 mol
= 1.44 mol S
= 1.5 S
What is the empirical formula of a compound
that contains 53.73% Fe and 46.27% S?
53.73g Fe
mol
= 0.963 mol Fe
55.8g
0.963 mol Fe
0.963 mol
46.27g S
mol
32.1g
1.44 mol S
0.963 mol
= 1 Fe
FeS1.5 x 2 =
Fe2S3
= 1.44 mol S
= 1.5 S
What is the molecular formula of a substance
with the empirical formula Fe2S3 and a gram
formula mass of 624 g/mol?
What is the molecular formula of a substance
with the empirical formula Fe2S3 and a gram
formula mass of 624 g/mol?
Fe6S9
What is the molecular formula of a substance
with the empirical formula Fe2S3 and a gram
formula molecular mass of 624 g/mol?
2Fe + 3S = 2(55.8) + 3(32.1) = 207.9 g/mol
624 g/mol ≈ 3
207.9 g/mol
3 x Fe2S3 =
Fe6S9
What are the empirical and molecular formulas for
Ibuprofen if the molar mass is 206 g/mole and the
percent composition of 75.7% C, 8.80% H, and 15.5% O?
What are the empirical and molecular formulas for
Ibuprofen if the molar mass is 206 g/mole and the
percent composition of 75.7% C, 8.80% H, and 15.5% O?
75.7g C
8.80g H
15.5g O
What are the empirical and molecular formulas for
Ibuprofen if the molar mass is 206 g/mole and the
percent composition of 75.7% C, 8.80% H, and 15.5% O?
75.7g C
mol C
= 6.31 mol C
12.0g C
8.80g H
mol H
= 8.80 mol H
1.0 g H
15.5g O
mol O
= 0.969 mol O
16.0g O
What are the empirical and molecular formulas for
Ibuprofen if the molar mass is 206 g/mole and the
percent composition of 75.7% C, 8.80% H, and 15.5% O?
75.7g C
mol C
= 6.31 mol C
12.0g C
6.31 mol C
= 6.5 C
0.969 mol
8.80g H
mol H
= 8.80 mol H
1.0 g H
8.80 mol H = 9 H
0.969 mol
15.5g O
mol O
= 0.969 mol O
16.0g O
0.969 mol H = 1 O
0.969 mol
What are the empirical and molecular formulas for
Ibuprofen if the molar mass is 206 g/mole and the
percent composition of 75.7% C, 8.80% H, and 15.5% O?
75.7g C
mol C
= 6.31 mol C
12.0g C
6.31 mol C
= 6.5 C
0.969 mol
8.80g H
mol H
= 8.80 mol H
1.0 g H
8.80 mol H = 9 H
0.969 mol
15.5g O
mol O
= 0.969 mol O
16.0g O
0.969 mol H = 1 O
0.969 mol
C6.5H9O x 2 = C13H18O2
What are the empirical and molecular formulas for
Ibuprofen if the molar mass is 206 g/mole and the
percent composition of 75.7% C, 8.80% H, and 15.5% O?
C13H18O2
13(12.0) + 18(1.0) + 2(16.0) = 206g/mol
Therefore the molecular formula is also
C13H18O2
Homework
• Empirical & Molecular Formulas Worksheet