Chemical Composition Review
Conversions
This part is the logic that will help you remember
Moles are a chemist’s dozen:
 3 dozen eggs x 12 = 36 eggs
 Moles x (conversion factor) = new unit

Conversion factors are
o Atomic mass or molar mass
 Units of grams per mole (g / mol)
 Used for converting between grams and moles
o Avogadro’s Number
 Units of atoms per mole (atoms / mol)
 Used for converting between moles and atoms
Here are the equations
Moles x AM = grams
moles x Av# = atoms
Grams / AM = moles
atoms / Av# = moles
Don’t forget Av# = 6.022 x 1023
Percent Composition
Designated as a fraction 
Part
Whole
Example: CH3O
AM = 31.034
The part is each element times its quantity
The whole is the AM for the compound
C
CH 3O
3H
CH 3O
O
CH 3O



Make sure you use the mass of each
12.011
(100) = 38.703% Carbon
31.034
3.0237
(100) = 9.7432% Hydrogen
31.034
15.999
(100) = 51.553% Oxygen
31.034
Determining Empirical Formulas
Given either mass of percentages of constituent elements it works the same
Example: An unknown sample was found to have a percent composition as follows:
47.0% K, 14.5% C, and 38.5% O. Find the empirical formula.
Given %, assume a 100 g sample so:
47.0 g K
14.5 g C
38.5 g O
This is a mass ratio, but all analysis is done by mole ratio so convert each into moles
47.0
14.5
38.5
= 1.20 mol K
= 1.21 mol C
= 2.41 mol O
39.098
12.011
15.999
These are the right units, but we need to reduce our numbers
Divide all by the smallest reducing the least down to one while holding the ratio
1.20
1.21
2.41
= 1K
= 1C
= 2O
1.20
1.20
1.20
These numbers should be almost exactly integers
Be careful when numbers are near 0.5, for all may need to be doubled
If these numbers are not near integers or halfs, then check your math
So the final compound is KCO2
Determining Molecular Formulas
Once you have the empirical formula the molecular formula can be calculated
Given the empirical formula above, determine the molecular formula if the atomic
mass of the compound is 166.214 g/mol
First find the empirical mass (atomic mass of the empirical formula)
39.098 + 12.011 + 2 (15.999) = 83.107 g/mol
Now divide the atomic mass by the empirical mass
166.214 / 83.107 = 2
This is your multiplier (n) value
KCO2 x 2 = K2C2O4