Question

A 2.00 gram sample of Copper (II) acetate produced 2.48 grams of a Copper (II) Formate Hydrate. After heating the sample the final mass of the Copper (II) Formate was 1.69 grams. What was the degree of hydration for the Copper (II) Formate?

Answer 1

Cu(CH3COO)2 ---------> Cu(HCOO)2.xH2O

    2 g                                        2.48 g

Cu(HCOO)2.xH2O -------> Cu(HCOO)2 + xH2O

        2.48                                 1.69

Mass of Copper(II) Formate hydrate = 2.48 g

After heating anhydrous Copper(II) formate is formed = 1.69 g

Mass of H2O = 2.48 - 1.69 = 0.79 g

x = (Mass of H2O / Molar mass of H2O) * Molar mass of Copper(II) formate / Mass of Copper II formate

x = (0.79 / 18) * 153 / 1.69

x = 3.97 = 4

Degree of Hydration for Copper(II) Formate = 4