Question

After 0.7 g of CoCl2⋅6H2O is heated, the residue has a mass of 0.321 g . Calculate the % H2O in the hydrate. What was the actual number of moles of water per formula unit CoCl2?

Answer 1

Ans. # Mass of water lost = Mass of (Hydrated sample – Dried sample)

                                    = 0.70 g – 0.321 g = 0.379 g

# Moles of water lost = Mass of water lost / MW of water

                                    = 0.379 g / 18.0 g mol^-1 = 0.021 mol

# Mass of anhydrous salt (CoCl₂) = Mass of dried sample = 0.321 g

Moles of anhydrous salt = 0.321 g / 129.8386 g mol^-1 = 0.0025 mol

# Molar ratio of water to anhydrous salt = Moles of water / moles of anhydrous salt

                                                = 0.021 mol / 0.0025 mol = 8.4 : 1 = 8 : 1

As calculated, the molar ratio of water to anhydrous salt is 5:1. That is, there are 8 water molecules every 1 molecules of the salt.

# % Mass of water in hydrate = (Mass of water / Mass of hydrated sample) x 100

                                                = (0.379 g / 0.70 g) x 100 = 54.14 %