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प्रश्न
An organic compound with vapour density = 94 contains C = 12.67%, H = 2.13% and Br = 85.11%. Find the molecule formula.
[Atomic mass: C = 12, H = 1, Br = 80]
उत्तर
| Element | Symbol | Atomic No. | Percentage | Relative no. of atoms | Simplest ratio |
| Hydrogen | H | 1 | 2.13 | `2.13/1` = 2.13 |
`2.13/1.055` = 2 |
| Carbon | C | 12 | 12.67 | `12.67/12` = 1.055 | `1.055/1.055` = 1 |
| Bromine | Br | 80 | 85.11 | `85.11/80` = 1.0638 | `1.0638/1.055` = 1 |
Empirical formula = CH2Br
Given: Vapour density = 94
Molecular mass = 2 × Vapour density
= 2 × 94
= 188
Molecular formula mass = Empirical formula mass × n
188 = 94 × n
n = `188/94`
= 2
Molecular formula = (Empirical formula) × n
= (CH2Br) × 2
= C2H4Br2
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