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).

Answer 1

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