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प्रश्न
Find the empirical formula of a compound containing 17.64% hydrogen and 82.35% of nitrogen.
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उत्तर १
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
उत्तर २
| 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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0.145 g of X was heated with dry copper (II) oxide and 224 cm3 of carbon dioxide was collected at S.T.P.
Calculate the empirical formula of X by the following step:
Calculate the mass of hydrogen in sample X.
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0.145 g of X was heated with dry copper (II) oxide and 224 cm3 of carbon dioxide was collected at S.T.P.
Calculate the empirical formula of X by the following step:
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