Empirical & Molecular Formula Notes
Chemistry 4/8/15
Drill
 Calculate the percent composition of
Na2SO4.
 HW: Finish pg. 3 (Empirical Formulas)
 MM=142.05 g/mol
 32.37% Na, 22.58% S, 45.05% O
Objectives
 IWBAT
 Calculate the empirical formula of a
compound, given percent composition or
mass composition.
 Calculate the molecular formula of a
compound, given:
 Empirical formula and molar mass OR
 Percent composition and molar mass
HW Review
 I’ll go over the answers—see if we need to
put any on the board!
Empirical and Molecular Formulas
Empirical Formula

Empirical vs. Molecular Formulas
 Molecular Formula – a formula that
specifies the actual number of atoms of
each element in one molecule of a
compound.
 Empirical Formula – a formula with
the smallest whole-number mole ratio of
the elements that make up a compound.
Empirical Formula
 Empirical Formula
 May or may not be the same as the
molecular formula
 Molecular formula is always a simple
multiple of the empirical formula
 ex. H2O2


Empirical formula is HO
Molecular formula is TWO times the empirical formula
How to calculate an empirical formula
 How to calculate:
 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.
An example:
 STEP 1:
 Compound is 40.05% S and 59.95% O
 STEP 2:
 I assume 100 g of the compound, so I have:
 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:
 1.249 mol S : 3.747 mol O
 Divide each by 1.249 (smallest number in
ratio)
 1 mol S : 3 mol O
 STEP 5:
 SO3
 You are done! The compound is sulfur trioxide.
A way to remember those steps:
 A Poem by Joel Thompson:
 Percent to mass
 Mass to mole
 Divide by small
 Multiply ‘til whole
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:
 You will be given the molar mass of the compound
and the empirical formula.
 STEP 2:
 Calculate the empirical mass (mass of the empirical
formula).
 STEP 3:
 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.
Molecular Formula Example
 STEP 1:
 The empirical formula is CH2O and the molar
mass is 180.18 g.
 STEP 2:
 The empirical mass is 12.01g + 2(1.01g) +
16.00g = 30.03 g
 STEP 3:
 180.18 g/ 30.03 g = 6
 STEP 4:
 CH2O becomes C6H12O6
Empirical Formula Practice
 Calculate the empirical formula for a
compound with a percent composition of
53.73% Fe and 46.27% S.
 Fe2S3
Practice
 The formula mass of a compound is 92 u.
Analysis shows that there are 0.608 g N and
1.388 g O. What is the molecular formula?
 N2O4
Closure
 Acetylene (C2H2) is a gas that is used as a
fuel for welding. Benzene (C6H6) is a liquid
solvent.
 How are they similar?
 How are they different?
 Why are they different?
 Why is one a gas at room temperature and
one a liquid?