Question

Given that a 40.14 g sample of hydrated NiSO₄·XH₂O is reduced in mass to 22.14 g upon heating. Determine the value of X.

Answer 1

mass of = H2O = mass of hydrated salt - mass of anhydrous salt
mass of = H2O = 40.14 g - 22.14 g
mass of = H2O = 18 g

Molar mass of H2O,
MM = 2*MM(H) + 1*MM(O)
= 2*1.008 + 1*16.0
= 18.016 g/mol


mass(H2O)= 18 g

number of mol of H2O,
n = mass of H2O/molar mass of H2O
=(18.0 g)/(18.016 g/mol)
= 0.9991 mol

Molar mass of NiSO4,
MM = 1*MM(Ni) + 1*MM(S) + 4*MM(O)
= 1*58.69 + 1*32.07 + 4*16.0
= 154.76 g/mol


mass(NiSO4)= 22.14 g

number of mol of NiSO4,
n = mass of NiSO4/molar mass of NiSO4
=(22.14 g)/(154.76 g/mol)
= 0.1431 mol
X = mol (H2O)/mol (NiSO4)
X = 0.9991 / 0.1431
X = 7