Determine the empirical formula for a compound that has the following analysis: 8.09% carbon, 0.34% hydrogen, 10.78% oxygen, and 80.78% bromine.

C₂HO₂Br₃

Explanation:

Data given

Carbon = 8.09%

Hydrogen = 0.34%

Oxygen = 10.78%

bromine = 80.78%

Empirical formula = ?

Solution:

First find the masses of each component

Consider total compound is 100g

As we now

mass of element = % of component

So,

8.09 g of C = 8.09% of Carbon

0.34 g og H = 0.34% of  Hydrogen

10.78 g of O =  10.78% of Oxygen

80.78 g of bromine = 80.78% of bromine

Now convert the masses to moles

For Carbon

Molar mass of C = 12 g/mol

            no. of mole = mass in g / molar mass

Put value in above formula

            no. of mole = 8.09 g/ 12 g/mol

            no. of mole = 0.674

mole of C = 0.6742

For Hydrogen

Molar mass of H = 1 g/mol

              no. of mole = mass in g / molar mass

Put value in above formula

              no. of mole = 0.34 g/ 1 g/mol

               no. of mole = 0.34

mole of H = 0.34 mole

For Oxygen

Molar mass of O = 16 g/mol

              no. of mole = mass in g / molar mass

Put value in above formula

              no. of mole = 10.78 g/ 16 g/mol

               no. of mole = 0.674

mole of O = 0.674 mole

For Br

Molar mass of Br = 80 g/mol

             no. of mole = mass in g / molar mass

Put value in above formula

              no. of mole = 80.78 g/ 80 g/mol

              no. of mole = 1.0098

mole of Br = 1.0098 moles

Now we have values in moles as below

C = 0.6742

H = 0.34

O = 0.674

Br = 1.0098

Divide the all values on the smallest values to get whole number ratio

C = 0.6742 / 0.34 = 1.983 ≅ 2

H = 0.34 / 0.34 = 1

O = 0.674 /0.34 = 1.983 ≅ 2

Br = 1.0098 / 0.34 = 2.97 ≅ 3

So all have round value 1 mole

C = 2

H = 1

O = 2

Br = 3

So the empirical formula will be C₂HO₂Br₃ i.e. all 3 atoms in simplest small ratio

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