4.5 EMPIRICAL
FORMULAS
Formulas
• Law of Constant Composition - a compound contains
elements in a certain fixed proportions (ratios),
regardless of how the compound is prepared or where
it is found in nature.
• If you have one molecule of methane gas, you will
always have 1 carbon atoms and 4 hydrogen atoms.
3
3 Steps for determining Chemical
Formulas
1. Determine the percent composition of all elements.
2. Convert this information into an empirical formula
3. Find the true number of atoms/ elements in the
compound (Molecular Formula)
4.5 Empirical Formula
2. Empirical Formula is the formula that gives the lowest
ratio of atoms in a compound. It does not necessarily tell
you the exact number of each type of atom.
Steps to Determine Empirical Formula
1. List given values.
2. Write the mass percent as mass (mass percent is just
mass per 100g sample).
3. Convert each mass to moles.
4. State the mole ratios.
5. Calculate lowest whole number ratio.
Example 1: The percent composition of a compound is
69.9 % iron and 30.1% oxygen. What is the empirical
formula of a compound?
Step 1: List the given values
Fe = 69.9% and O = 30.1%
Step 2: Calculate the mass (m) of each element in
a 100g sample.
mFe= 69.9 x 100g = 69.9g
100
mO= 30.1 x 100g = 30.1g
100
Step 3: Convert Mass (m) into moles (n)
nFe= 69.9g x 1mol/ 55.86g = 1.25 mol Fe
nO= 30.1g x 1mol/ 16.00g = 1.88 mol O
Step 4: State the Amount Ratio
nFe : nO
1.25mol : 1.88 mol
Step 5: Calculate lowest whole number ratio
1.25mol : 1.88 mol
1.25mol 1.25 mol
1 : 1.5
2 : 3
When you don’t get a whole number,
multiply entire ratio by 2, 3, 4 etc. until
you get a whole number
Empirical Formula
is Fe2O3
Example 2: The percentage
composition of a compound is 21.6%
sodium, 33.3% chlorine, and 45.1%
oxygen. What is the empirical
formula of the compound?
Step 1: List the given values
Cl=33.3%, Na = 21.6% and O = 45.1%
Step 2: Calculate the mass (m) of each
element in a 100g sample.
mNa= 21.6 x 100g = 21.6g Na
100
mCl= 33.3 x 100g = 33.3g Cl
100
mO= 45.1 x 100g = 45.1g O
100
Step 3: Convert Mass (m) into moles (n)
nNa= 21.6g x 1mol/ 23.0g = 0.94 mol Na
nCl= 33.3g x 1mol/ 35.5g = 0.94 mol Cl
nO= 45.1g x 1mol/ 16.00g = 2.82 mol O
Step 4: State the Amount Ratio
nNa
: nCl
: nO
0.94mol : 0.94mol : 2.82 mol
Step 5: Calculate lowest whole number ratio
0.94mol : 0.94mol : 2.82 mol
0.94mol : 0.94mol : 0.94 mol
1 : 1: 3
Empirical Formula
is NaClO3
Example 3: A barium salt is found
to contain 21.93 g of barium, 5.12 g
of sulfur, and 10.24 g of oxygen.
What is the simplest formula of this
compound?
Step 1: List the given values
Ba = 21.93g
S = 5.12g O = 10.24g
Step 2: Calculate the mass (m) of each
element.
***Mass is given, so proceed to step 3!
Assignment
• Proceed with the assignment that follows
Empirical Formula Check Point
1. Determine the empirical formula for the
following compound: 22.1% aluminum, 25.4 %
phosphorus and 52.5% oxygen
2. Name the compound! ;)