How to solve empirical Formulae and Molecular Formula

Empirical and Molecular Formulas can be used to represent compounds and there are simple ways to do this calculation.

Compounds can be represented with empirical, and structural formulas.

The empirical formula of a compound is the simplest formula of a compound that shows the atoms in their simplest ratio.

The molecular formula is the formula of a compound that shows the actual atoms of the elements contained in the compound.

The structural formula of a compound however is more descriptive because the Structural formulae of a compound show all the atoms and all the bonds present n the compound.

I want to show you how to deduce the empirical and molecular formula of compounds and also how to easily solve simple problems involving the formula of compounds.

How to deduce Empirical and Molecular Formula of compounds

Example I

An organic compound contains 92.3% Carbon and 7.7% Hydrogen. If the molecular mass is 78 Calculate the empirical and molecular formula of the compound.

Solution

    CarbonHydrogen
Percentage of elements   92.3%  7.7%  
  Divide by the relative atomic mass   92.3/12    7.7  7.7/1 7.7
Divide by the smallest ratio 7.7/7.7      17.7/7.7     1

        Therefore the empirical formula gotten is CH,

       Molecular mass is 78, then equate the molecular mass to the empirical formula.

               [CH]n = 78

               [12 +1]n = 78

                13n = 78

         n= 78/13 = 6

        The molecular formula is  C6H6, the compound is Benzene.

  Example 2

An organic compound on analysis contains 2.0g Carbon, 0.34g Hydrogen, and 2.67g of Oxygen. If the molecular mass is 60, calculate the empirical and molecular formula of the compound.

Solution

    CarbonHydrogenOxygen
Mass on analysis   2.00.34  2.67
  Divide by the relative atomic mass  2.0/12    0.170.34/1 0.34 2.67/16 0.17
Divide by the smallest ratio 0.17/0.17      12/0.17     20.17/0.17      1

       Therefore the empirical formula gotten is CH2O

       Molecular mass is 60, then equate the molecular mass to the empirical formula.

               [CH2O]n = 60

               [12 +1X2 +16]n = 60

               30n = 60, Then  n = 60/30 = 2

              [CH2O]6 =  C2H4O2  or CH3COOH

Example 3

Determine the empirical and molecular formula of an oxide of Nitrogen containing 70% oxygen.. if the relative molecular mass is 92. Deduce the molecular formula.

Solution

    NitrogenOxygen  
Percentage of elements   30 %  70%  
  Divide by the relative atomic mass    30/14    7.7 70/16 7.7
Divide by the smallest ratio  2.1/2.1      14.38/2.1     2

      Therefore the empirical formula gotten is NO2

       Molecular mass is 92, then equate the molecular mass to the empirical formula.

               [NO2]n = 92

               [14  +16 X 2]n = 92

               46n = 92, Then  n = 92/46 = 2

              [NO2]2 =  N2O4

In conclusion, the empirical and molecular formula can be deduced with dat given in the questions but most times, you will need to get the empirical formula first before the molecular formula.

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