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प्रश्न
A compound has O = 61.32%, S = 11.15%, H = 4.88% and Zn = 22.65%. The relative molecular mass of the compound is 287 amu. Find the molecular formula of the compound, assuming that all the hydrogen is present as a water of crystallization.
उत्तर
| Element | Percentage | Atomic Weight | Ratio | Simplest Ratio |
| O | 61.32 | 16 | `61.32/16` = 3.83 | `3.83/0.35` = 10.9 ≈ 11 |
| S | 11.15 | 32 | `11.15/32` = 0.35 | `0.35/0.35` = 1 |
| H | 4.88 | 1 | `4.88/1` = 4.88 | `4.88/0.35` = 13.9 ≈ 14 |
| Zn | 22.65 | 65.37 | `22.65/65.97` = 0.345 | `0.345/0.35` = 1 |
Empirical formula of the given compound = ZnSH14O11
Empirical formula mass = 65.37 + 32 + 141 + 11 + 16 = 287.37
Molecular mass = 287
n = `"Molecular mass"/"Empirical formula mass" = 287/287=1`
Molecular formula = ZnSO11H14
=ZnSO4.7H2O
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