Empirical and Molecular
Formulas
Empirical vs. Molecular
• Empirical Formula: represents the smallest
ratio of atoms in a formula.
– In other words it represents the simplest
chemical formula for a particular compound.
• Molecular Formula: the actual number of
each type of atom (a multiple of the
empirical formula)
Steps to Solve!
• You need to follow these basic steps to
determine your answer.
• 1) Make sure you are starting in grams
• 2) Convert all grams to moles (Molar Mass)
• 3) Divide each by the smallest number of
moles.
• 4) Create whole number ratio of subscripts
• 5) Multiply if necessary
Example 1
• In an unknown molecule, there are
4.15 grams of Carbon and 1.38
grams of Hydrogen. Determine the
empirical formula for the substance!
• Step 1) Make sure you are starting in
grams
• Step 2) Convert each mass in grams
to moles.
Now What?
• Step 3) Divide each by the smallest
number of moles. Set up the Ratio of
elements
Solution!
• The 1 and the 4 represent the whole
number ratio of the elements in the
chemical formula.
• The 1 represents that there is only one
Carbon atom
• The 4 represents that there are four
Hydrogen atoms
• The Solution: The empirical formula is
CH4
• Phenol, a general disinfectant, is 76.57
% Carbon, 6.43 % Hydrogen, and 17.00
% Oxygen. Determine it’s empirical
formula.
• Step 1) Make sure you are starting in
grams…….
• How can we go from percent to grams?
• Easy! Assume we have a 100 gram
sample so there would be how much of
each element?
• Step 2) Convert each mass in grams to moles.
Try this!
• Element “A” is 78.1% abundant and has a
molar mass of 10.81 g/mol. Element “B” is
21.9% abundant and has a molar mass of
1.01 g/mol. Determine the empirical formula.
•
• 78.1g A X 1mol/10.81g = 7.22 mol B
• 21.9g B X 1mol/1.01g = 21.7 mol H
• 7.22 mol ÷ 7.22mol =1
• 21.7 mol ÷ 7.22mol = 3.01
• Ratio of 1:3
• AB3
Molecular Formula
• The molecular formula is related, but
different to the empirical formula.
• Remember that Empirical Formula
represents the lowest ratio of the
atoms in a compound.
• The Molecular Formula is the actual,
or true ratio of the elements in the
compound.
Needs and Steps!
• 1. In order to solve you need the
molar mass of the molecular formula
for a compound in g/mol.
• 2. Determine the empirical formula.
• 3. Calculate the molar mass of the
empirical formula
• 4. Divide the molar mass of the
molecular formula by the molar mass
of the empirical formula. Apply ratio to
all subscripts.
Example
• The empirical formula for
hydroquinone, a chemical used in
photography, is C3H3O. The
molecular weight of the
compound Is 110 g/mol.
Determine the molecular formula.
• Step 1) Determine the molecular
weight of the compound: Given at
110 g/mol.
• Step 2) Determine the empirical
formula: Given C3H3O.
• Step 3) Determine the molar mass of
the empirical formula.
Think…I Know it’s hard…
• Step 4) Think about the molar
mass of the molecular formula
compared to the molar mass of
the empirical formula.
• Mol. Formula
Emp. Formula
• 110 g/mol
55.06 g/mol
• Divide and get the ratio
• 110 g/mol / 55.06 g/mol = 2
So….
• That 2 represents what you will
multiply all the subscripts by to
determine the correct molecular
formula.
• C 3 H3 O
x 2
=
C6H6O2
Molecular Formula
Try this
• The empirical formula for a compound
is C3H7. If the molecular weight is 86
g/mol, then what is the molecular
formula?
• 1. The molecular weight is given:
– 86 g/mol
• 2. The empirical formula is given:
- C3H7
• 3. Calculate the molar mass for
the empirical formula.
Compare the two….
• The Molecular weight was 86
g/mol
• The Empirical weight was 43.1
g/mol
• Divide Molecular by Empirical to
determine the ratio!
• 86 g/mol / 43.1 g/mol = 2
• So 2 is the ratio, multiply all
the subscripts by 2
• C3H7 x 2 = C6H14 is the
Mol. Formula
Practice!
A) Empirical Formula is S
Molecular weight is 256 grams
per mol
B) Empirical Formula is NO2
Molecular Weight is 46 g/mol