Question

A student heats a sample of Copper (II) sulfate in a crucible and records the data shown in the table. What is the complete formula and name for the compound before heating?
Mass of empty crucible 128.10 g
Mass of crucible + sample before heating 152.00 g
Mass of crucible + sample after heating 147.60 g

Answer 1

Answer:

Copper(II) Sulphate (CuSO4.7H2O) is generally present in a hydrated form.

Mass of the CuSO4.7H2O sample before heating

(152.00-128.10)g = 23.90 grams

Mass of the water loss due to heating

(152.00-147.60)g = 4.40 grams

Answer 2

Explanation:

Copper (II) sulfate is usually present as a hydrous state, which is of the form CuSO4 * nH2O, where n is a whole number.

Mass of sample (CuSO4 * nH2O)

= 152.00g - 128.10g = 23.90g.

Mass of water loss during heating

= 152.00g - 147.60g = 4.40g.

Molar mass of H2O = 18g/mol

Moles of H2O in sample

= 4.40g / (18g/mol) = 0.244mol.

Mass of anhydrous sample (CuSO4)

= 23.90g - 4.40g = 19.50g

Molar mass of CuSO4 = 159.61g/mol

Moles of CuSO4 in sample

= 19.50g / (159.61g/mol) = 0.122mol.

Since mole ratio of CuSO4 to H2O

= 0.122mol : 0.244mol = 1:2, n = 2.

Hence we have CuSO4 * 2H2O.