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).

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.

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