Question

1.00 g of a hydrated salt contains 0.2014 g of iron, 0.1153 g of sulfur, 0.2301 g of oxygen, and 0.4532 g of water of crystallisation. Find the empirical formula. (At. wt.: Fe = 56; S = 32; O = 16)

Answer 1

Given:

• Mass of hydrated salt = 1.00 g

• Iron (Fe) = 0.2014 g

• Sulfur (S) = 0.1153 g

• Oxygen (O, in salt) = 0.2301 g

• Water of crystallisation (H₂O) = 0.4532 g

• Atomic weights:
  Fe = 56, S = 32, O = 16, H = 1

Iron (Fe): Moles of Fe `= 0.2014/56 = 0.0035946`

Sulfur (S): Moles of S `= 0.1153/32 = 0.0036031`

Oxygen in salt: Moles of O (in salt) `= 0.2301/16 = 0.014381`

Water (H₂O): Molecular mass of H₂O = 18

Moles of H₂O `= 0.4532/18 = 0.02518`

Get the simplest mole ratio

Component	Moles
Fe	0.0035946
S	0.0036031
O (in salt)	0.014381
H₂O	0.02518

Divide all by the smallest value ≈ 0.0036:

Component	Mole Ratio (÷ 0.0036)
Fe	1.00
S	1.00
O (in salt)	4.00
H₂O	7.00

From the mole ratios:

Fe : S : O = 1 : 1 : 4 → This suggests FeSO₄

7 water molecules → ·7H₂O

\[\ce{FeSO4.7H2O}\]