Introduction to Fundamental Concepts of Chemistry -Emperical Formula

9/24/2015 CBSE

Empirical Formula



Definition

It is the simplest formula of a chemical compound which represents the
element present of the compound and also represent the simplest ratio between
the elements of the compound.




Examples

�The empirical formula of benzene is "CH". It indicates that the benzene molecule
is composed of two elements carbon and
hydrogen and the ratio between these two elements is 1:1.

�The empirical formula of glucose is "CH2O". This formula represents that glucose
molecule is composed of three elements carbon, hydrogen and oxygen. The ratio
between carbon and oxygen is equal but hydrogen is double.





Determination of Empirical Formula



�To determine the empirical formula of a compound following steps are required.


�1. To detect the elements present in the compound.

�2. To determine the masses of each
element.

�3. To calculate the
percentage of each element.

�4. Determination of mole composition of each element.

�5. Determination of simplest ratio between the element of the compound.





Illustrated Example of Empirical Formula



�Consider an unknown compound whose empirical formula is to be determined is
given to us. Now we will use the above five steps in order to calculate the
empirical formula.



Step I - Determination of the Elements

�By performing test it is found that the compound contains magnesium and oxygen
elements.



Step II - Determination of the Masses

�Masses of the elements are experimentally determined which are given below.

�Mass of Mg = 2.4 gm

�Mass of Oxygen = 1.6 gm



Step III - Estimation of the Percentage

�The percentage of an element may be determined by using the formula.

�% of element = Mass of element / Mass of compound x 100

�In the given compound two elements are present which are magnesium and oxygen,
therefore mass of compound is equal to the sum of the mass of magnesium and mass
of oxygen.

�Mass of compound = 2.4 + 1.6 = 4.0 gm

�% Mg = Mass of Mg / Mass of Compound x 100

�= 2.4 / 4.0 x 100

�= 60%

�% O = Mass of Oxygen / Mass of Compound x 100

�= 1.6 / 4.0 x 100

�= 40%



Step IV - Determination of Mole Composition

�Mole composition of the elements is obtained by dividing percentage of each
element with its atomic mass.

�Mole ratio of Mg = Percentage of Mg / Atomic Mass of Mg

�= 60 / 24

�= 2.5

�Mole ratio of Mg = Percentage of Oxygen / Atomic Mass of Oxygen

�= 40 / 16

�= 2.5



Step V - Determination of Simplest Ratio

�To obtain the simplest ratio of the atoms the quotients obtained in the step IV
are divided by the smallest quotients.

�Mg = 2.5 / 2.5 = 1

�O = 2.5 / 2.5 = 1

�Thus the empirical formula of the compound is MgO



Note

�If the number obtained in the simplest
ratio is not a whole number then multiply this number with a smallest number
such that it becomes a whole number maintain their proportion.

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