Chemistry Notes
Empirical & Molecular
Formulas
Empirical Formula
The empirical formula gives you
the lowest, whole number ratio
of elements in the compound.
The empirical formula may or
may not be the same as the
molecular formula.
For example:
For carbon dioxide, the molecular
formula is CO2, and the empirical
formula is CO2. One carbon and
two oxygens are the lowest ratio of
atoms.
The molecular formula for dinitrogen
tetrahydride is N2H4, but the
empirical formula is NH2.
What is the empirical formula for:
C6H12O6
What is the empirical formula for:
C6H12O6
CH2O
What is the empirical formula for:
C6H12O6
C6H12O2
CH2O
What is the empirical formula for:
C6H12O6
C6H12O2
CH2O
C3H6O
What is the empirical formula for:
C6H12O6
C6H12O2
N2H2
CH2O
C3H6O
What is the empirical formula for:
C6H12O6
C6H12O2
N2H2
CH2O
C3H6O
NH
What is the empirical formula for:
C6H12O6
C6H12O2
N2H2
CH4
CH2O
C3H6O
NH
What is the empirical formula for:
C6H12O6
C6H12O2
N2H2
CH4
CH2O
C3H6O
NH
CH4
To Find the Empirical
Formula from % Composition:
If given the percentages, assume
there are 100.0 grams of the
compound.
Convert the grams of each
element to moles.
To Find the Empirical Formula
from % Composition:
Divide by the smaller amount of
moles, then manipulate the ratio
so that all numbers are whole.
Two Examples:
What is the empirical formula of a
compound that is 27.3% carbon
and 72.7% oxygen?
Two Examples:
What is the empirical formula of a compound that is 27.3% carbon and 72.7%
oxygen?
27.3 g Carbon x 1 mole C = 2.28 mol C / 2.28 = 1 mole C
12 g C
72.7 g Oxygen x 1 mole O = 4.54 mol O / 2.28 = 2 moles O
16 g O
The empirical formula would be: CO2
Two Examples:
What is the empirical formula of a
compound that is 25.9%
nitrogen and 74.1% oxygen?
Two Examples:
What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen?
25.9 g Nitrogen x 1 mole N = 1.85 mol N / 1.85 = 1 mole N x 2 = 2 moles N
14 g N
74.1 g Oxygen x 1 mole O = 4.63 mol O / 1.85 = 2.5 moles O x 2 = 5 moles O
16 g O
The empirical formula would be: N2O5
More to try…
Calculate the empirical formula of
a compound that is 94.1%
oxygen, 5.9% hydrogen.
More to try…
Calculate the empirical formula of a compound that is 94.1% oxygen, 5.9%
hydrogen.
94.1 g Oxygen x 1 mole O = 5.88 mol O / 5.88 = 1 mole O
16 g O
5.9 g Hydrogen x 1 mole H = 5.9 mol H / 5.88 = 1 mole H
1gH
The empirical formula would be: OH
More to try…
Calculate the empirical formula of
a compound that is 79.8%
carbon, 20.2% hydrogen.
More to try…
Calculate the empirical formula of a compound that is 79.8% carbon, 20.2%
hydrogen.
79.8 g Carbon x 1 mole C = 6.65 mol C / 6.65 = 1 mole C
12 g C
20.0 g Hydrogen x 1 mole H = 20.2 mol H / 6.65 = 3 mole H
1gH
The empirical formula would be: CH3
From there, find it’s molecular
formula:
If given the molar mass (how
many g/mol of the compound)
then you can calculate the
molecular formula from the
empirical formula.
From there, find it’s molecular
formula:
Take the compound’s empirical
formula mass and compare to
the molecular mass. The
molecular mass will be a
multiple of the empirical
formula’s mass.
Example:
If the molecular mass of the first
example (N2O5) problem is 216
g/mol, then what is the
molecular formula for the
compound?
Example:
If the molecular mass of the first example (N2O5)
problem is 216 g/mol, then what is the molecular
formula for the compound?
216 g /mol = 2
108 g/mol
The molecular formula is: N4O10
Try these:
What is the molecular formula of
a compound whose molar mass
is 60.0 g and whose empirical
formula is CH4N?
Try these:
What is the molecular formula of a compound whose molar
mass is 60.0 g and whose empirical formula is CH4N?
60.0 g/mol = 2
30.0 g/mol
The molecular formula is: C2H8N2
Try these:
What is the molecular formula for
a compound whose molar mass
is 78 g and whose empirical
formula is CH?
Try these:
What is the molecular formula for a compound whose molar mass is 78 g
and whose empirical formula is CH?
78 g/mol = 6
13 g/mol
The molecular formula is: C6H6