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Question
A Compound on analysis gave Na = 14.31% S = 9.97% H = 6.22% and O = 69.5% calculate the molecular formula of the compound if all the hydrogen in the compound is present in combination with oxygen as water of crystallization. (molecular mass of the compound is 322).
Solution
| Element | % | Relative no. of atoms | Simple ratio |
| Na | 14.31 | `14.31/23` = 0.62 | `0.62/0.31` = 2 |
| S | 9.97 | `9.97/32` = 0.31 | `0.31/0.31` = 1 |
| H | 6.22 | `6.22/1` = 6.22 | `6.22/0.31` = 20 |
| O | 69.5 | `69.5/16` = 4.34 | `4.34/0.31` = 14 |
∴ Empirical formula = Na2SH20O14
η = `"Molar mass"/"Calculated empirical formula mass"`
= `322/322`
= 1
Na2SH20O14 = (2 × 23) + (1 × 32) + (20 × 1) + (14 × 16)
= 46 + 32 + 20 + 224
= 322
Molecular formula = Na2SH20O14
Since all the hydrogen in the compound present as water
∴ Molecular formula is Na2SH20O14
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