# Empirical vs Molecular notes ```Empirical vs. Molecular
Formulas
Empirical Formula
The lowest whole
number ratio
(subscripts) of elements
in a compound.
Cannot be reduced!!!
not empirical
empirical
Ex: C6H12O6 
CH2O
Molecular Formula
Actual number of atoms in a chemical compound
molecular
EX: C12H24O12
Molecular Formulas can be reduced to Empirical Formulas
molecular
EX: C12H24O12  CH2O
empirical
Different molecular formulas can have similar empirical formulas
molecular
EX:
N3O9
N12O36 
empirical

NO3
PRACTICE:
1. Identify
each as empirical (can’t be reduced)
or molecular (can be reduced)
2. If its molecular – write the empirical
1.
2.
3.
4.
5.
C2H4
NO3
S9Cl12
C3Cl9
N4S9
molecular
CH2
- empirical
empirical
molecular
molecular
empirical
S3Cl4
- empirical
CCl3 - empirical
Finding Empirical Formula from Percent Composition
• Ex: A compound was found to be 54.53% Carbon, 9.15%
Hydrogen, and 36.32% Oxygen. Find its Empirical Formula.
Steps:
1. Assume a 100g sample (change %  g)
2. &divide; by the molar mass of that element to find moles (sig fig it!)
3. Identify the lowest # of moles and &divide; them all by that number
4. Round each to the nearest whole # (if .0 or .9) (if not - special)
5. The resulting whole #are the subscripts for that element in the
empirical formula
Calculating Empirical Formula
63.5% Silver
8.2% Nitrogen
28.3% Oxygen
63.5 g Ag
107.87
8.2 g N
14.01
28.3 g O
16.00
.589 mole Ag
.589
.59 mole N
.589
1.77 mole O
.589
1
1
AgNO3
3
Calculating Empirical Formula (special)
60.00%C
60.00g C
12.01
4.48%H
4.48g H
1.01
4.996 mole C
2.221
4.44 mole H
2.221
2.249
2
35.53%O
35.53g O
16.00
2.221 mole N
2.221
1
x4
x4
x4
9
8
4
C9H8O4
Calculating Molecular Formula
1.Find the empirical formula
2.Calculate the molar mass of your empirical formula
3.Identify the molar mass of your molecular (GIVEN in the
problem every time!)
4.Divide the molecular mass / empirical mass
5.Round to the nearest whole #
6.Multiply the whole # by the subscripts in the Empirical
formula
Practice
If a compound has an empirical formula of NO3 and a
molecular mass of 186g – what is the molecular formula?
Empirical formula: NO3
Molecular mass (given)
empirical mass
3 x NO3
=
N3O9
molar mass:
62.01g
186g
62.01
```