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Question
Find the empirical formula of a compound containing 17.64% hydrogen and 82.35% of nitrogen.
Solution 1
Given is the % of H n N
So atomic weight of H is 1
the atomic weight of N is 14
Divide % of H with its atomic weight. weight and similarly N % with its atomic weight I.e. 14
You will get the relative no of atoms.
to get the simplest ratio. Divide the smaller value of relative no of atoms with the two. you will get the simplest ratio
NH3
Solution 2
| Element | Percentage | Atomic Weight | Ratio | Simplest Ratio |
| Hydrogen | 82.35% | 14 | `82.35/14` = 5.88 | `5.88/5.88` = 1 |
| Nitrogen | 17.64% | 1 | `17.46/1` = 17.46 | `17.46/5.88` = 3 |
∴ Empirical Formula= N1H3 = NH3
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