# Hydrates

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Topics: Water, Ion
CHEMISTRY EXPERIMENT NUMBER 8

PERCENTAGE OF WATER IN A HYDRATE

Objectives

1. To determine the percentage of water in barium chloride dihydrate.

2. To determine the percentage of water in an unknown hydrate salt.

3. To calculate the water of crystallization for the unknown hydrate salt.

Discussion

A hydrate salt is composed of anions (negative ions) and cations (positive ions) which are surrounded by and weakly bonded water molecules. Each hydrate salt has a fixed number of water molecules associated with it, called waters of hydration or water of crystallization. When a salt holds waters of hydration, we call it a hydrated salt or a hydrate (hydrate from hydor, the Greek word for water).

Barium chloride dihydrate, BaCl2(2H2O, has two waters of crystallization, or two waters of hydration. Other hydrates have waters of hydration ranging from one to twelve. Upon heating, a hydrate decomposes and produces an anhydrous salt and water (in the form of steam).

BaCl2(2H2O (s ) ( BaCl2 (s) + 2 H2O (g)

Examples and Problems

Problem 1-Theoretical Percentage

Calculate the theoretical percentage of water in barium chloride dihydrate.

Solution: The formula mass of BaCl2(2H2O is

Ba 1 x 137.3 = 137.3 amu Cl 2 x 35.5 = 71.0 amu H2O 2 x 18.0 = 36.0 amu 244.3 amu

The theoretical percentage of water is found by dividing the water of crystallization mass, 36.0 amu, by the hydrate mass, 244.3 amu.

36.0 amu x 100 = 14.7 % 244.3 amu

Problem 2 – Experimental percentage

The experimental percentage of water in a hydrate is found by comparing the mass of water driven off to the total mass of the compound, expressed as a percentage.

A 1.250 g sample of barium chloride dihydrate has a mass of 1.060 g after heating. Calculate the experimental percentage of water.

Solution: The mass of water lost is found by difference:

1.250 g – 1.060 g = 0.190 g hydrate

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