1. Formulas and Percent Composition

2. Percent Composition
The percent composition is the percentage by mass of each element in a compound
This helps distinguish compounds made up of the same elements
FeO or Fe2O3
In Iron (III) Oxide,
Iron is 69.94% Fe and 30.06% O
In Iron (II) Oxide,
Iron is 77.73% Fe and 22.27% O

3. Percent Composition
Calculate the percent composition of copper (I) sulfide.
1. Write the molecular formula
Cu2S
2. Find the molar mass of each element in the compound
(2 mol Cu)(63.55 g/mol Cu) = g Cu
(1 mol S)(32.07 g/mol S) = g S
Molar mass of Cu2S = g/mol
3. Calculate the percent by mass of each element
(127.1g Cu)/(159.2g Cu2S) x 100 = 79.85% Cu
(32.07g S)/(159.2g Cu2S) x 100 = 20.15% S
4. Check to be sure it adds to 100%

4. Percent Composition
Calculate the percent composition of aluminum nitrate.
Al = 12.67%
N = 19.73%
O = 67.60%

5. Empirical Formulas
The Empirical Formula is the simplest whole number ratio among the atoms in a compound
Ammonium Nitrite
NH4NO2
What is the simplest whole number ratio of each element in the compound?
N2H4O2 becomes
NH2O
This is the empirical formula

6. Percent Composition and Empirical Formula
One can determine the Empirical Formula from percent composition
Chemical analysis of a liquid shows that it is 60.0% C, 13.4% H, and 26.6% O by mass. What is the empirical formula for the liquid?
1. Assume you have g of the substance
Then you have 60.0g of C 13.4g of H and 26.6g of O

7. Percent Composition and Empirical Formula
2. Convert from mass to moles
(60.0g C)(1 mol C/12.011g C) = 5.00 mol C
(13.4g H)(1 mol H/1.01g H) = 13.3 mol H
(26.6g O)(1 mol O/16.00g O) = 1.66 mol O
3. Divide all moles by the smallest number to find the ratio
5.00 mol C/1.66 = 3.01 mol C
13.3 mol H/1.66 = 8.01 mol H
1.66 mol O/1.66 = 1.00 mol O
4. This is your Empirical Formula
C3H8O

8. Empirical Formulas
Benzoic Acid contains 68.8% Carbon, 4.95% Hydrogen, and 26.2% Oxygen. Find the empirical formula.
(68.8gC)(1molC/12.01gC) = 5.73 mol C
(4.95gH)(1molH/1.01gH) = 4.90 mol H
(26.2gO)(1molO/16.00gO) = 1.64 mol O
Divide by 1.64 (smallest number)
5.73molC/1.64 = 3.5 mol C
4.90molH/1.64 = 3.0 mol H
1.64molO/1.64 = 1.0 mol O

9. Empirical Formulas Can you have C3.5H3O? No!
You must multiply all by 2 so you can have whole numbers
3.5 x 2, 3 x 2 and 1 x 2
Your final empirical formula is
C7H6O2

10. Molecular Formulas
Molecular formulas are multiples of empirical formulas
CH2O
C2H4O2 is acetic acid (2 x Empirical Formula)
C6H12O6 is Glucose (6 x Empirical Formula)
You can determine the molecular formula from the empirical formula and the mass of the compound

11. Molecular Formulas
The empirical formula for a compound is P2O5. Its experimental molar mass is 284 g/mol. Determine the molecular formula for the compound.
Determine the molar mass of the empirical formula
g/mol
Find the ratio of the compound molar mass to the empirical formula molar mass
(284 g/mol)/( g/mol) = 2.00
Multiply the empirical formula by this number
2(P2O5) = P4O10
Verify the results with a check of the molar mass of the new formula