Empirical fomulas are the simplified versions of compounds.
For example C3H9 can be simplified to CH3 and on the otherhand molecular formulas are the unsimplified versions of empirical compounds like C3H9
Empirical and molecular formulas can be expressed as percent compositions.
For exmaple water is 89% oxygen and 11% hydrogen. Though more hydrogen atoms are presesnt in the formula H2O the percent compisition is based on the masses of atoms divided by the total mass of the compound.
For H2O the percent composition is determined by:
Oxygen = 16 g/mol
Hydrogen = 1g/mol
Total mass = 16 + 2 (2 hydrogen atoms are present)
Oxygen % = 16/18 or 89%
Hydrogen %= 2/18 11%
With organic compounds that are burned it is possible to find out how much of each atom was present in the beginning of the reaction and after.
If a 6.5 gram sample of C and H burn to produce 20.5 grams of CO2 and 8.4 grams of H2O
You can figure out the empircal or molecular formula by finding how many moles are present of each compound.
There is .466 mol of CO2 (remember mole conversions)
And .467 mol of H2OIf you divide by the lowest mole amount you can get how many of each compound are present.
.467/4.66 = 1
.466/.466 = 1
So theres about 1 of H2O and CO2
And knowing this simple ratio of xCHy + O2 = xCO2 + (y/2)H2O
In this case it would be 1CH2 + 2O2 = 1CO2 + 1H2O
And to find the oxygen is as simple as adding the oxygens present on the the right side of the formula and dividing by two.
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