Empirical and Molecular
Formulas
This presentation has been
brought to you by the CSI Lab
located in N. Haven, Ct.
Empirical and Molecular

Molecular Formula:



Empirical Formula:



The formula for a compound that exist as the
nbr. of atoms.
Ex. H202  Hydrogen Peroxide
Indicate the lowest whole number ratio of
atoms of each element in a compound.
Ex. HO  Hydrogen Peroxide
Ho, Ho, Ho!!!





The molecular formula of glucose is…
C6H12O6, , its empirical formula is…
CH2O …
So what’s my point?
Empirical formulas are lowest whole
number ratios of atoms of each type …to
determine them experimentally we
need numbers of atoms!
Consider the following:






Analysis of a compound at CSI North
Haven Labs indicates that it contains
32.38 g Na
22.65 g S
44.99 g O
How can we use this data to determine
the formula?...Think…
Mass  number of atoms!
Remember doing mole
conversions?




Use the molar mass of element convert
to moles.
A mole is a number of atoms … you now
have a ratio of atoms in the compound.
Divide each mole amount by the smallest
mole amount… whole # ratio???
Use ratio to write subscripts

1.408 mol Na : 0.7063 mol S : 2.812mol O

Divide each by smallest 

1.993 mol Na : 1 mol S : 3.981 mol O
 Na2SO4
Try this..



Analysis indicates that a compound
contains 78.1% Boron and 21.9%
hydrogen determine the empirical formula.
Why is this one any different from the last
example?
Assume you have a 100 gram
sample…now how many grams of boron
do you have?




7.22 mol B :
21.7 mol H
Divide through by smallest 
1 mol B
:
3.01 mol H
BH3
Try this one …

Analysis of a 10.150 g sample of a
compound known to contain only
phosphorus and oxygen indicates a
phosphorus content of 4.433 g. What is
the empirical formula?

Grams of P = 4.433g Grams of O = ?

Now try it…

0.1431 mol P : 0.3573 mol O

Divide by smallest 

1 mol P : 2.497 mol O …

Now what?...Multiply through by two to
obtain whole numbers… 2 : 5
P2O5
HW

Study for quiz next class

Read pgs 229-233

Answer questions #1-3 pg 233
Find empirical formulas and the
empirical masses…
1. 36.48% Na, 25.41% S, 38.11 % O…
Na2SO3
126 g/mol
2. 53.70 % iron and 46.30% sulfur…
Fe2S3
3.
208 g/mol
1.04 g K, 0.70 g Cr and 0.86 g O…
K2CrO4
194 g/mol
4. 36.51% C, 6.20% H and 57.29 % O …
CH2O
30 g/mol
Molecular Formulas


You have already solved for the Empirical
Formula
Now find the mass of the Empirical
Formula



Ex. CH2O  ~30 g/mol
You will be given a molecular weight to
start with (~184.62 g/mol)
Now divide the molecular mass by the
Empirical mass

This will tell you the ratio, or the amount
of times that the Empricial Formula goes
into the Molecular formula


Answer should be close to a whole
number


(184.62 g/mol)/(30 g/mol)
6.154  6
Multiply the Empirical Formula by that
whole nbr. and you have the molecular
formula

C6H12O6
Quantitative Analysis

Various methods are used to determine
the mass or percent composition of each
element in a given compound…more
about those later…
Analysis by Combustion
Percent Composition
Believe it or not you’ve already
done it.
% comp.


Tells you the percent of the mass made up
by each element in the compound
The mass of each element in a compound
compared to the entire mass of the
compound and multiplied by 100 %

Find the percent composition of copper (I)
sulfide, Cu2S

A sample of an unkn. Compound with a
mass of 0.2370 g is extracted from the
roots of a plant. Decomposition of the
sample produces 0.09480 g of carbon,
0.1264 g of oxgyen, and 0.0158 g of
hydrogen. What is the percentage comp.
of the compound?
Hydrates



inorganic salts which contain a specific
number of water molecules that are
loosely attached
Written as: CuSO4 • 5H2O
Copper Sulfate with 5 water molecules
Formula for a hydrate


Finding the empirical formula for a hydrate
involves finding the mass difference of the
hydrate and anhydrous salt by driving off
the excess water:
DEMO:
http://www.chem.iastate.edu/group/Green
bowe/sections/projectfolder/flashfiles/stoic
hiometry/empirical.html










Example you conduct an experiment using
CuWO4 • xH2O
Mass of hydrate = 2.286 g
Mass of anhydrous salt = 2.050 g
Mass of H2O driven off = 0.236 g
CuWO4 = 311.4 g/mol
Mol salt = 2.050g (1mol/311.4g) = 0.00658
mol
Mol H2O = 0.236g (1mol/18g) = 0.0131 mol
Divide by the smallest mol amount
0.0131 mol/0.00658 mol =~2 ; 2 water
molecules
CuWO4 • 2H2O
One interesting note!







A compound with a molecular mass of 59.7 g is
determined to be made up of 40% Na and 60%
Cl. What is the Empirical Formula? What is the
Molecular Formula?
Empirical Formula – NaCl
Molecular Formula – NaCl
What are the charges (oxidation nbrs) on the Na
cation and Cl anion?
+1 and -1
What is the overall charge of the cmpd?
ZERO
Oxidation Numbers by Group
0
1+
2+
3+
Tend to have more
than one oxidation
number (+1 to +7)
3+
3+ or 4+
varies
3-
2-
1-


The sum of the oxidation nbrs in an ionic
cmpd must be zero.
Ex. Magnesium Iodide
Mg+2 I-1
 Goes to
MgI2
 Again the charge on the cmpd is 0

Aluminum Oxide Al2O3?