Iron (ii) sulfate forms a blue-green hydrate with the formula feso4 ·n h2o (s). if this hydrate is heated to a high enough temperature, h2o (g) can be driven off, leaving the dirty yellow anhydrous salt feso4 (sa 18.300-g sample of the hydrate was heated to 300 °c. the resulting feso4 (s) had a mass of 9.9993 g. calculate the value of n in feso4 ·n h2o (s).

Mass of FeSO4 = 9.9993g

Mass of Water = Anhydrous FeSO4 (18.300g) - FeSO4 (9.9993g) = 8.3007g Molar Mass of water = H (2 x 1.01) + O (16) = 18.02 g/mol

Molar Mass of FeSO4 = Fe (55.845) + S (32.065) + O (4 x 16) = 151.91 g/mol Moles of Water = 8.3007 / 18.02 = 0.46063

Moles of FeSO4 = 9.9993 / 151.91 = 0.06582

n = Moles of Water / Moles of FeSO4 = 0.46063 / 0.06582 = 7

So the compound is FeSO4 7 H2O

Answer is: n is seven (FeSO₄·7H₂O).

m (FeSO₄·nH₂O) = 18,300 g.

m (FeSO₄) = 9,9993 g.

m (H₂O) = 18,300 g - 9,9993 g = 8,3007 g.

M (FeSO₄) = 151,9 g/mol.

n (FeSO₄) = 9,9993 g : 151,9 g/mol = 0,0658 mol.

n (H₂O) = 8,3007 g : 18 g/mol = 0,461 mol.

n (FeSO₄) : n (H₂O) = 0,0658 mol : 0,461 mol.

n (FeSO₄) : n (H₂O) = 1 : 7.