# Determination of the formula of hydrated Iron (II) Sulphate crystals (FeSO4xH2O)

Determination of the formula of hydrated Iron (II) Sulphate crystals (FeSO4xH2O)AnalysisMethod 11. Using a balance that weighs to two decimal places, weigh a crucible. Add between 1.30 - 1.50g of hydrated iron (II) sulphate crystals. Record the masses.2. Place the crucible containing the hydrated iron (II) sulphate crystals on the pipe-clay triangle and gently heat for two minutes.3. Allow to cool and weigh the crucible and the iron (II) sulphate.4. Repeat steps 2 and 3 until the masses after heating are consistant.5. Record all the masses.Use the results to calculate the moles of FeSO4 and the moles of water and hence deduce the formula of the hydrated iron(II) sulphate crystals, FeSO4xH2O.MassesMass numberItems being weighedMass (g)1Crucible and lid38.862Crucible, lid and FeSO4xH2O40.343FeSO4xH2O1.484.a)Crucible, lid and FeSO4xH2O after heatingafter first heating39.774.b)second heating39.674.c)third heating39.674.d)fourth heating39.67After heating four times, the weight of the crucible, lid and FeSO4xH2O was consistant showing that all of the H2O has evaporated.Therefore the following masses can be calculated:Mass numberSubstancesMass number calculationMass calculation (g)Mass (g)5H2O2 -
4.d)40.34 - 39.670.676FeSO43 - 51.48 - 0.670.81To work out the formula of the FeSO4xH2O:Moles of FeSO4Mr = 56 + 32 + (16 x 4) = 152number of moles = massMrn = 0.81 = 0.00533 moles152Moles of H2OMr = 18n = mass n = 0.67 = 0.0372 molesMr 18Empirical formulaTo work out molar ratio of FeSO4 to H2O (value of x):1 x 0.0372 = 6.980.00533x = 6.98therefore the formula is = FeSO4(H2O)7Method 21. Weight between 2.85 and 3.10 g of hydrated iron(II) sulphate crystals, FeSO4xH2O. Record the mass and then dissolve the crystals in 50.0cm3 of 1 mol dm-3 H2SO4 (aq) ...