Empirical and Molecular
Formulas
Empirical vs. Molecular Formulas
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Empirical Formula – the simplest formula.
A formula with the smallest whole-number
ratio of the elements that make up a
compound.
Molecular Formula – the true formula.
A formula that specifies the actual number of
atoms of each element in one molecule of a
compound.
Empirical Formula

Empirical Formula


May or may not be the same as the
molecular formula
Molecular formula is always a simple
multiple (ratio) of the empirical formula
 ex. H O
2 2


Empirical formula is HO
Molecular formula is TWO times the
empirical formula
Copy Table
Name
Molecular Empirical Lowest ratio
Formula
Formula of elements
Hydrogen peroxide
H2O2
Glucose
÷2
HO
1:1
C6H12O6 ÷6
CH2O
1:2:1
Benzene
C6H6 ÷6
CH
1:1
Ethyne
C2H2 ÷2
CH
1:1
Aniline
C6H7N
C6H7N
6:7:1
water
H2O
H2O
2:1
How to calculate an empirical formula
STEP 1: You will be given either masses
or percent composition.
 STEP 2: If you are given %
composition, turn it into grams by
assuming a 100.0 g sample. NOTE: If
you are given mass, you do not need to
do this step.
 STEP 3: Convert the masses to the
number of moles of each element.



STEP 4: Figure out the proportion of
moles of each element in the compound
by dividing each by the smallest number
of moles.
STEP 5: If step 4 resulted in whole
numbers, you are done! However, if there
were decimals, you will need to multiply
by small, whole numbers until you have
whole numbers.
A way to remember those
steps: COPY THIS DOWN
A Poem by Joel Thompson:


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
Percent to mass
Mass to mole
Divide by small
Multiply ‘til whole
An example:

STEP 1:


STEP 2:


Compound is 40.05% S and 59.95% O
I assume 100 g of the compound, so it is:
 40.05 g S and 59.95 g O
STEP 3:


40.05 g S•(1 mol S/32.07 g S) = 1.249 mol S
59.95 g O•(1 mol O/16.00 g O) = 3.747 mol O
Continued…

STEP 4:

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STEP 5:
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1.249 mol S : 3.747 mol O
Divide each by 1.249 (smallest number in ratio)
1 mol S : 3 mol O
SO3
You are done! The compound is sulfur
trioxide.

Molecular Formula
Molecular Formula


Molecular Formula – this tells us how
many atoms of each type there really
are in the compound.
Can two substances have the same
empirical formula but be different?


YES! Benzene vs. acetylene: C6H6 vs. C2H2
What is their empirical formula? How is
this different from ionic compounds?
Calculating Molecular Formula

STEP 1:


STEP 2:

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Calculate the empirical mass (mass of the
empirical formula).
STEP 3:


You will be given the molar mass of the
compound and the empirical formula.
Divide the given molar mass by the empirical
mass. You should get a small whole number.
STEP 4:

Multiply the subscripts of the empirical
formula with the number obtained.
Step #3 in more detail
Ratio = Molar Mass (Actual)
Molar Mass (EF)
Your ratio will give you a
whole number ratio.
Example

The empirical formula of a compound is found
to be CH2O. After careful analysis the molar
mass is found to be 180.18 g. Determine the
molecular formula of this compound.
Molecular Formula Example

STEP 1:


STEP 2:
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The empirical mass is 12.01g + 2*1.01g +
16.00g = 30.03 g
STEP 3:


The empirical formula is CH2O and the molar
mass is 180.18 g.
Ratio= 180.18 g/ 30.03 g = 6
STEP 4:

CH2O becomes C6H12O6