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Question
An organic compound has the following percentage composition: C = 12.76%, H = 2.13%, Br = 85.11%. The vapour density of the compound is 94. Find out its molecular formula.
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Solution
| Element | Atomic mass | Percentage | Relative number of moles | Simplest mole ratio |
Whole number ratio |
| C | 12 | 12.76 | 12.76/12 = 1.06 | 1.06/1.06 = 1 | 1 |
| H | 1 | 2.13 | 2.13/1 =2.13 | 2.13/1.06 = 2 | 2 |
| Br | 80 | 85.11 | 85.11/80 =1.06 | 1.06/1.06 = 1 | 1 |
So the Empirical formula of the compound will be CH2Br.
Now the Empirical formula mass will be = Atomic mass of C + Atomic mass of H + Atomic mass of Br
= 12 + 1 x 2 + 80 = 94
Now as Molecular mass = 2 x vapour density
= 2 x 94 = 188.
So n = Molecular mass / empirical formula mass
= 188/ 94 = 2
Molecular formula of the compound is = n x empirical formula
= 2 x (CH2Br) = C2H4Br2
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