Question

An organic compound has the following percentage composition C= 12.36%, H= 2.13%, Br= 85%. Its vapour density is 94.find its molecular formula
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Answer 1

★Given:-

Percentage composition:

• C= 12.36%, H= 2.13%, Br= 85%.
• Vapour density = 94.

★To find:-

• The molecular formula.

Steps to find the molecular formula:-

1.Calculate the relative no.of.moles.

2.Find the simplest ratio by dividing mole by smallest molar quantity,to get the whole ratio.

3.From the whole ratio,write the empirical formula.

4.Calculate the empirical formula mass.

5.Find the molecular mass by the formula:

✦Molecular mass = 2×vapour density

6.Find value of 'n' by the formula:

✦n = molecular mass / empirical mass

7.To find molecular formula,use:

✦Molecular formula = n×empirical formula.

★Solution:-

We know,

Atomic mass:

• C=12
• H=1
• Br=80

Given percentage:

• C=12.36%
• H=2.13%
• Br=85%

Relative no.of.moles:-

✦No.of.moles = Given weight/molar mass

No of moles:

=> C = 12.76/12

        = 1.06

=> H=2.13/1

       =2.13

=>Br=85.11/80

       =1.06

As we can see,1.06 is the smallest mole quantity.

Simplest mole ratio:-

=>C=1.06/10.6 = 1

=>H=2.13/1.06 = 2

=>Br=1.06/1.06 = 1

The whole ratio is 1:2:1.Therefore,

The empirical formula is CH2Br.

Now,

✦Empirical mass =Sum of atomic masses of C,H,Br.

→EM= 12×(1×2)+80

          =94

✦Molecular mass=2×vapour density

→Molecular mass = 2×94

                           =188.

✦n = Molecular mass/empirical mass

→ n = 188/94

      = 2

✦Molecular formula = n×empirical formula

→ MF = 2×CH2Br

         =C2H4Br2

                               

Hence,the molecular formula of the compund is C2H4Br2.

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