A certain compound has the following percent composition: C=40.0% and H=6.7%. The empirical formula is...
Ans :-
| Element | Mass | Atomic mass | Moles | Simplest whole number ratio |
| C | 40 | 12 | 40/12 = 3.33 | 3.33/3.33 = 1 |
| H | 6.7 | 1 | 6.7/1 = 6.7 | 6.7/3.33 = 2 |
Since, the molar ratio of C : H = 1 : 2
Therefore,
The Emperical formula of the compound is = CH2
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