Empirical Formula
October 23, 2017
Empirical formula of a chemical compound is the simplest whole number ratio of atoms of each element present in a compound. The molecular formula specifies the actual number of atoms of each element in a molecule.
The data may be given in grams (2.4 grams of magnesium combines with 1.6 grams of oxygen) or percentage (A compound contains 60% Mg and 40% Oxygen) of different elements.
They are divided separately by the respective atomic masses of the elements. All the results are calculated correct to three decimal points.
The answers in the above step are then divided by the smallest answer.
Some answers in the above step will be integer number or very close to integer number. Some may be in decimal form such as 2.499. This should never be written as 3 but 2.5. In the final step, all answers are then multiplied by a suitable number to get all integer numbers. For example, if the ratio is 1:1.5, the final ratio is 2:3. However, if the ratio is 1:1.333, the final ratio is 3:4.
 Magnesium  Oxygen 
Percentage  60  40 
Atomic mass  24  16 
Divide percentage by the atomic mass  60/24 = 2.5  40/16 = 2.5 
Divide all the results by the smallest result  2.5/2.5  2.5/2.5 = 1 
Empirical Formula  MgO 
Molecular formula
Once the empirical formula is established, their atomic masses are added. This is known as empirical formula mass. The given molecular mass is then divided by the empirical formula mass. If the answer is 1, empirical and molecular formula is the same. However, if the answer is 2,3 or 4, or any other number, empirical formula is then multiplied to get the molecular formula.
Empirical formula  Molecular mass  Molecular formula 



CH  78  C_{6}H_{6} 
C_{2}H_{5}  58  C_{4}H_{10} 



Resources others found helpful
Iambic Pentameter: Iambic WHAT!?
To help students of Shakespeare understand the power of iambic pentameter.
A comprehensive review of Spanish tenses