Question

A 0.551-g sample of hydrated copper(II) sulfate, CuSO₄.xH₂O, was heated carefully until it had changed completely to anydrous copper(II) sulfate (CuSO₄) with a mass of 0.308 g. Determine the value of x. [This number is called the number of waters of hydration of copper(II) sulfate. It specifies the number of water molecules per formula unit of copper(II) sulfate in the hydrated crystal.

Answer 1

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

mass of H2O = 0.551 g - 0.308 g

mass of H2O = 0.243 g

Molar mass of H2O,

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

= 2*1.008 + 1*16.0

= 18.016 g/mol

mass(H2O)= 0.243 g

use:

number of mol of H2O,

n = mass of H2O/molar mass of H2O

=(0.243 g)/(18.02 g/mol)

= 1.349*10^-2 mol

Molar mass of CuSO4,

MM = 1*MM(Cu) + 1*MM(S) + 4*MM(O)

= 1*63.55 + 1*32.07 + 4*16.0

= 159.62 g/mol

mass(CuSO4)= 0.308 g

use:

number of mol of CuSO4,

n = mass of CuSO4/molar mass of CuSO4

=(0.308 g)/(1.596*10^2 g/mol)

= 1.93*10^-3 mol

use:

X = mol (H2O)/mol (CuSO4)

X = 1.349*10^-2 / 1.93*10^-3

X = 7

Answer: 7