1. Percent Composition Percent = part / whole Example: MgO
Find molar mass of whole compound:
MgO = = 40.3 grams
% Mg (by mass) = mass of Mg = g x 100 = 60.3%
molar mass of MgO g
% O (by mass) = mass of O = g x 100 = 39.7%
molar mass of MgO g

2. Empirical Formulas
Gives the lowest whole-number ratio of elements and compounds in a formula
4 Steps:
Change the percent to grams
Convert grams to moles
Divide each of the moles by the smallest number to find the ratio
Round ratio to a whole number. If the ratio is not a whole number (ex. 1.5) multiply each element by 2 to get a whole number

3. Examples
A compound contains 94.1% Oxygen and 5.9% Hydrogen. What is its empirical formula?
94.1% = 94.1g O
5.9% = 5.9g H
5.9g H
1 mol H
1 mol O
94.1g O
= 5.9 mol H
= 5.88 mol O
16 g O
1 g H
5.9
5.88
5.88
= 1.003
(can round to 1.0)
= 1.0
4) Ratio = 1:1 Formula = OH

4. Examples
A compound contains 67.6% Mercury and 10.8% Sulfur and 21.6% Oxygen. What is its empirical formula?
67.6% = 67.6g Hg
10.8% = 10.8g S
21.6% = 21.6g O
67.6g Hg = mol Hg
10.8g S = mol S
21.6g O = 1.35 mol O
0.336
0.338
0.336
1.35
0.336
= 1.00 Hg
= 1.00 S
= O
4) Ratio = 1:1:4 Formula = HgSO4

5. Examples
What is the empirical formula for a compound containing 70.0% Fe and 30.0% O?
70.0% = 70.0g Fe
30.0% = 30.0g O
70.0g Fe = 1.25 mol Fe
30.0g O = mol O
1.875
1.25
1.25
= 1.5 O
= 1.00 Fe
4) Ratio = 1 : 1.5
5) Mult ratio by 2 = 2 : 3
Formula = Fe2O3

6. Molecular Formulas
Same as empirical formula, or a simple whole-number multiple of it
Steps:
Calculate the mass in grams of the empirical formula provided
Divide the molar mass by the mass of the empirical formula
Multiply this whole number ratio by the empirical formula

7. Examples
Calculate the molecular formula of the compound whose molar mass is 60.0g and empirical formula is CH4N.
1) CH4N = (1) + 14 = 30.0 g
60.0g
30.0g
= 2.0
3) (CH4N)
= C2H8N2

8. Examples
What is the molecular formula of ethylene glycol (CH3O), used in antifreeze. It has a molar mass of 62 g/mol.
1) CH3O = 31.0 g
3) 2.0 (CH3O)
= C2H6O2
62.0g
31.0g
= 2.0
Find the molecular formula of C3H2Cl, which is mothballs. Its molar mass is 147 g/mol.
1) C3H2Cl = 73.0 g
3) 2.0 (C3H2Cl)
= C6H4Cl2
147.0g
73.0g
= 2.0