Empirical and Molecular
Formulas
For compounds:
•How to calculate Empirical Formula
•How to calculate Molecular Formula
Warmup
• Calculate the percent composition of these compounds:
1.
C2H2 (Acetylene)
2. C8H8 (Styrene)
Acetylene Cutting Torch
Styrene Foam
Learning Objectives
• I can describe the difference between
molecular and empirical formulas
• I can determine the empirical formula for a
compound
• I can determine the molecular formula for a
compound given its empirical formula and
the compound’s molar mass.
Calculating Empirical Formulas
• The molecular formula of a compound shows the actual
number and kinds of atoms present in a molecule.
– example : H2O2 is hydrogen peroxide
• The empirical formula just gives the lowest whole-number
ratio of atoms of the elements in a compound.
– If you were to decompose hydrogen peroxide into its elements, you
would find one hydrogen for every oxygen, a 1:1 ratio.
– therefore the empirical formula for hydrogen peroxide is HO
Calculating Empirical Formulas
• An empirical may or may not be the same as a molecular
formula.
• If the formulas are different, the molecular formula is a
simple multiple of the empirical formula.
hydrogen
peroxide
molecular
formula
empirical
formula
molar mass of
H2O2
H2O2
HO
34.0 g/mol
Note: A compound’s molar mass depends on its molecular
formula: H2O2 = 2x1.0 + 2x16.0 = 34.0g/mol
Calculating Empirical Formulas
• Example: What is the empirical formula of a compound
that is 25.9% nitrogen and 74.1% oxygen?
• The % composition tells the ratio of the masses of
nitrogen atoms to oxygen atoms.
• fact: percent = grams per 100 grams
• Step 1: Convert the % composition into mass ratio.
– We could say: “In 100g of the compound, there are ______g N
and _______g O”
Calculating Empirical Formulas
• Example: What is the empirical formula of a compound
that is 25.9% nitrogen and 74.1% oxygen?
• The % composition tells the ratio of the masses of
nitrogen atoms to oxygen atoms.
• fact: percent = grams per 100 grams
• Step 1: Convert the % composition into mass ratio.
– We could say: “In 100g of the compound, there are 25.9g N and
74.1g O”
Calculating Empirical Formulas
• Example: What is the empirical formula of a compound
that is 25.9% nitrogen and 74.1% oxygen?
Step 2: Use percent composition values to convert to
moles. In 100g of the compound, there are 25.9g N and
74.1g O.
1 mol N
25.9 g N 
 1.85 mol N
14.0 g N
74.1 g O 
1 mol O
 4.63 mol O
16.0 g O
Calculating Empirical Formulas
• Example: What is the empirical formula of a compound
that is 25.9% nitrogen and 74.1% oxygen?
Step 3: Divide both molar quantities by the smaller number
of moles. This will give 1 for the element with smaller
moles.
1.85 mol N
 1 mol N
1.85
4.63 mol O
 2.50 mol O
1.85
Is the final answer
N1O2.5 ?????
Obviously not!!
Calculating Empirical Formulas
• Example: What is the empirical formula of a compound
that is 25.9% nitrogen and 74.1% oxygen?
Step 4: Multiply each part of the ratio by a number to
convert the fraction to a whole number:
1 mol N  2  2 mol N
2.50 mol O  2  5 mol O
The empirical formula is N2O5
--- dinitrogen pentoxide ---
You try it now…
• Calculate the empirical formula of a compound that is 94.1% O and
5.9% H
Step 1: Use percent composition values to convert to moles. In 100g of
the compound, there are ______ g O and _______g H.
Step 2: Use percent composition values to convert to moles.
mol O = ______ mol O
mol H = ______ mol H
Step 3: Divide both molar quantities by the smaller number of moles.
This will give 1 for the element with smaller moles.
______ mol O
______ mol H
Step 4: Multiply each part of the ratio by a number to convert the
fraction to a whole number.
empirical formula : ___________
You try it now…
• Calculate the empirical formula of a compound that is 94.1% O and
5.9% H
Step 1: Use percent composition values to convert to moles. In 100g of
the compound, there are 94.1 g O and 5.9 g H.
Step 2: Use percent composition values to convert to moles.
mol O = 5.88 mol O
mol H = 5.9 mol H
Step 3: Divide both molar quantities by the smaller number of moles.
This will give 1 for the element with smaller moles.
.9966 = 1 mol O
1 mol H
Step 4: Multiply each part of the ratio by a number to convert the
fraction to a whole number.
since the numbers in step 3 are whole already, the empirical formula is HO
CALCULATING MOLECULAR
FORMULAS
If we know a compound’s empirical formula and its molar mass, we
can determine the molecular formula
Calculating Molecular Formulas p194
• Different compounds can have the same
empirical formula.
• The empirical formula gives the lowest wholenumber ratio of atoms of the elements in a
compound.
example:
C2H2 (Acetylene)
molar mass
26.0 g/mol
empirical formula
CH
C8H8 (Styrene)
104.0 g/mol
CH
Calculating Molecular Formulas p194
• Different compounds can have the same
empirical formula.
• The empirical formula gives the lowest wholenumber ratio of atoms of the elements in a
compound.
example:
C2H2 (Acetylene)
molar mass
26.0 g/mol
empirical formula
CH
C8H8 (Styrene)
104.0 g/mol
CH
Calculating Molecular Formulas
• You can determine the molecular formula of a compound if
you know its empirical formula and its molar mass.
• step 1: Calculate the empirical formula molar mass. (efm).
• step 2: Divide the efm into the molar mass to get a whole
number.
• step 3: multiply the empirical formula subscripts to get the
molecular formula.
Calculating Molecular Formulas
• Example: Calculate the molecular formula for the
compound whose molar mass is 60.0g and empirical
formula CH4N
step 1: empirical formula mass (efm)
12.0 + 4x1.0 + 14.0 = 30.0 g/mol
step 2: divide efm into the molar mass to get a whole number:
60.0 / 30.0 = 2
step 3: multiply the empirical formula subscripts to get the
molecular formula.
CH4N subscripts x 2 is molecular formula C2H8N2
Calculating Molecular Formulas
• Example: Calculate the molecular formula for the
compound whose molar mass is 60.0g and empirical
formula CH4N
step 1: empirical formula mass (efm)
12.0 + 4x1.0 + 14.0 = 30.0 g/mol
step 2: divide efm into the molar mass to get a whole number:
60.0 / 30.0 = 2
step 3: multiply the empirical formula subscripts to get the
molecular formula.
CH4N subscripts x 2 is molecular formula C2H8N2
You try it: Calculating Molecular Formulas
The empirical formula for ethylene glycol used in car
antifreeze is CH3O. Its molar mass is 62.0 g/mol.
Find the molecular formula:
step 1: Determine the empirical formula molar mass (efm)
step 2: divide efm into the molar mass to get a whole number:
step 3: multiply the empirical formula subscripts to get the
molecular formula.