EMPIRICAL FORMULA: Simplest whole number ratio of atoms of each Element in the Compound
s- Calculated from knowledge of the ratio of masses of each Element in the Compound
Example:
sA compound that contains 10 g of Hydrogen and 80 g of Oxygen has an empirical formula of H2O. This can be shown by the following calculations:
sAmount of Hydrogen Atoms = Mass in Grams ÷ Ar of Hydrogen = (10 ÷ 1) = 10 Mol Amount of Oxygen Atoms = Mass in Grams ÷ Ar of Oxygen = (80 ÷ 16) = 5 Mol The Ratio of Moles of Hydrogen Atoms to Moles of Oxygen Atoms: Hydrogen Oxygen Moles 10 : 5 Ratio 2 : 1 Ratio of Hydrogen Atoms to Oxygen Atoms is 2: 1 Hence, the Empirical Formula is H2O |
MOLECULAR FORMULA: Exact number of atoms of each Element present in the formula of the Compound
s- Divide the Relative Formula Mass of the Molecular Formula by the Relative Formula Mass of the Empirical Formula
- Multiply this to each number of Elements
RELATIONSHIP BETWEEN EMPIRICAL AND MOLECULAR FORMULA:
sNAME OF COMPOUND | EMPIRICAL FORMULA | MOLECULAR FORMULA |
METHANE | CH4 | CH4 |
ETHANE | CH3 | C2H6 |
ETHENE | CH2 | C2H4 |
BENZENE | CH1 | C6H6 |
Example:
sThe Empirical Formula of X is C4H10S1 and the Relative Formula Mass of X is 180
sWhat is the Molecular Formula of X?
sRELATIVE FORMULA MASS: Carbon : 12 Hydrogen : 1 Sulfur : 32 STEP 1 - Calculate Relative Formula Mass of Empirical Formula ( C x 4 ) + ( H x 10 ) + ( S x 1) = ( 12 x 4 ) + ( 1 x 10 ) + ( 32 x 1) = 90 STEP 2 - Divide Relative Formula Mass of X by Relative Formula Mass of Empirical Formula 180 / 90 = 2 STEP 3 - Multiply Each Number of Elements by 2 ( C 4 x 2 ) + ( H 10 x 2 ) + ( S 1 x 2 ) = ( C 8 ) + ( H 20 ) + ( S 2 ) Molecular Formula of X = C8H20S2 |
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