Question

A hydrate of iron (ii) fluoride has the following formula: FeF2 * xH2O. The water in a 3.41g sample of the hydrate was driven off by heating. The remaining sample had a mass of 1.93g. Find the number of waters of hydration (x) in the hydrate.

Answer 1

mass of = H2O = mass of hydrated salt - mass of anhydrous salt

mass of = H2O = 3.41 g - 1.93 g

mass of = H2O = 1.48 g

Molar mass of H2O,

MM = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

mass(H2O)= 1.48 g

number of mol of H2O,

n = mass of H2O/molar mass of H2O

=(1.48 g)/(18.016 g/mol)

= 8.215*10^-2 mol

Molar mass of FeF2,

MM = 1*MM(Fe) + 2*MM(F)

= 1*55.85 + 2*19.0

= 93.85 g/mol

mass(FeF2)= 1.93 g

number of mol of FeF2,

n = mass of FeF2/molar mass of FeF2

=(1.93 g)/(93.85 g/mol)

= 2.056*10^-2 mol

X = mol (H2O)/mol (FeF2)

X = 8.215*10^-2 / 2.056*10^-2

X = 4