Empirical
Formula and
Molecular
Formula
Empirical Formula
 Empirical
formula is the simplest whole
number ratio of atoms in a formula
 Empirical
lab data
information is always based on
Molecular Formula
 Molecular
formula is the actual ratio of
the atoms that form a molecule of a
substance.
 Sometimes
the empirical and molecular
are the same but they do not have to be!
More molecular….

Example H2O for water empirical and molecular
formula are the same.

H 2 : O 1 is the simplest whole number ratio of atoms in
the formula and it is also how the molecule actually
forms.

Example glucose C6H12O6 for glucose the molecule is
6:12:6 This is not the simplest whole number ratio of
atoms in the formula.
Molecular and Empirical
Formulas
 If
glucose were analyzed in the laboratory
its empirical ratio would be 1:2:1.
Additional test would have to be run to
determine the mass of the substance
Calculations
Step one= List individual elements in the
formula. Change the given information
from grams or percent into moles
Step two= Decide which mole value is the
smallest on the list.
Step three= Divide all mole values by that
smallest value
more Calculations……
 Step
four= If a decimal of .5 or .3 appear
in the ratio you obtained do not round up
or off!
Ratios with .5 must be multiplied by two.
Ratios with .3 must be multiplied by three.
Formula of Hydrates
 The
three basic steps are the same for
hydrates. But with hydrates……
 The two parts being compared are:

Moles of anhydrous in the sample vs.
Moles of water in the sample

Elements are not listed separately!

Example Empirical /Molecular
 An
organic compound was found to
contain 92.25 % carbon and 7.75 %
hydrogen. If the molecular mass is 78.0
grams, please calculate both the
empirical and molecular formulas for this
substance.
Step 1
C
92.25g/12.01 g = 7.681 moles
 H 7.75g/1.01 g= 7.67 moles
 Divide through by the smaller value!
 1 : 1 ratio C1H1 is empirical formula
 Now we must check for molecular
formula!
Step 2
 C1H1
add up the empirical mass of this
substance.
 12.01+ 1.01g = 13.02grams
 Divide the given molecular mass by the
empirical mass.
 78.0/ 13.02 = 6
 Molecular formula is C6H6
Final Answer
 Empirical
formula C1H1
 Molecular
formula C6H6
Modern Chemistry
example p.246
 Quantitative
analysis shows that a
compound contains 32.38% sodium,
22.65%sulfur, and 44.99% oxygen. Please
find the empirical formula for this
compound.
Hydrate Example
 Strontium
chloride hexahydrate
 158.52g plus 108.12g = 266.64g
 108.12/266.64 x 100 = 40.5 % H2O
Hydrate Calculation
 CuSO4*
? H2O
 Crucible&cover
 Crucible& cov& hydrate
 Crucible/cov& anhydrous
111.55g
131.53g
124.32g
Hydrates
 Subtract
C&C from anhydrous (after
heating) this will equal grams of CuSO4.
 Change
grams to moles.
Hydrates
 If
you subtract hydrate from anhydrous
you will find the mass of the water
removed from the sample.
 Change grams water to moles.
Hydrates
 Look
at mole values.
 Decide which mole value is smaller
 Divide both values by the smaller to get a
one to something ratio
 Taa!
Daa! you are done.
The End!
 Go
forth and calculate!