When 2.246 grams of a calcium chloride hydrate was heated, 0.728 grams of water were
released. determine the formula of the hydrate (hint: take the same approach as for the
copper(ii) sulfate hydrate).

CaCl₂.3H₂O

We are given;

We are supposed to determine the formula of the hydrate;

Mass of anhydrous salt = Mass of the hydrate - mass of the water of crystallization

Mass of anhydrous CaCl₂ = 2.246 g - 0.728 g

                                          = 1.518 g

Moles = Mass ÷ Molar mass

Moles of anhydrous salt

Molar mass of anhydrous CaCl₂ = 110.98 g/mol

Moles of CaCl₂ = 1.518 g ÷ 110.98 g/mol

                         = 0.0137 moles

Moles of water

Molar mass of water = 18.02 g/mol

Moles of water = 0.728 g ÷ 18.02 g/mol

                        = 0.0404 moles

CaCl₂ : H₂O

0.0137 moles : 0.0404 moles

0.0137/0.0137  : 0.0404/0.0137

                      1 : 2.94

                      1 :  3

We get; CaCl₂.3H₂O

Therefore; the formula of the hydrate is CaCl₂.3H₂O

with this one i know the answer in craters one reason i looked it up and i learned about it there is my evidence

fe 55.85 g/mol and h2o 18.02 g/mol.