Question

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.

Answer 1

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