Hydrates: Experimental Determination of Water Content in Ionic Compounds

Hydrates and Anhydrous Salts

Introduction to Hydrates

Hydrates are ionic compounds that contain water molecules integrated into their crystal structure. The water is not merely trapped but is chemically bound within the lattice, often in a fixed ratio. When heated, hydrates lose their water, forming anhydrous salts.

• Hydrate: An ionic compound with water molecules incorporated into its structure. Example: Copper(II) sulfate pentahydrate (CuSO4·5H2O).

• Anhydrous salt: The compound remaining after all water has been removed by heating.

• Water of hydration: The water molecules present in the hydrate.

Experimental Determination of Water Content

To determine the amount of water in a hydrate, a sample is weighed, heated to remove water, and weighed again. The difference in mass corresponds to the water lost.

• Step 1: Weigh the empty crucible.

• Step 2: Add the hydrate and weigh the crucible plus hydrate.

• Step 3: Heat the sample to drive off water; weigh after heating to obtain mass of crucible plus anhydrous salt.

• Step 4: Calculate the mass of water lost and the mass of anhydrous salt.

Sample Data and Calculations

The following table summarizes the experimental data and calculations for determining the water content in a hydrate sample.

Measurement	Value (g)	Calculation
Mass of empty crucible	17.95	-
Mass of crucible + hydrate	27.96	-
Mass of hydrate	10.21	$27.96 - 17.75 = 10.21$
Mass of crucible + anhydrous salt	26.07	-
Mass of anhydrous salt	8.32	$26.07 - 17.95 = 8.32$
Mass of water lost	1.89	$10.21 - 8.32 = 1.89$

Key Calculations and Formulas

• Mass of hydrate: $\text{Mass of crucible + hydrate} - \text{Mass of empty crucible}$

• Mass of anhydrous salt: $\text{Mass of crucible + anhydrous salt} - \text{Mass of empty crucible}$

• Mass of water lost: $\text{Mass of hydrate} - \text{Mass of anhydrous salt}$

Example Calculation

Suppose you have a hydrate sample with the following masses:

• Mass of empty crucible: 17.95 g

• Mass of crucible + hydrate: 27.96 g

• Mass of crucible + anhydrous salt: 26.07 g

Then:

• Mass of hydrate: $27.96 - 17.95 = 10.01$ g

• Mass of anhydrous salt: $26.07 - 17.95 = 8.12$ g

• Mass of water lost: $10.01 - 8.12 = 1.89$ g

Applications

• Determining the formula of a hydrate (e.g., finding x in CuSO4·xH2O).

• Understanding the role of water in crystal structures and its removal by heating.

Additional info:

• To find the number of moles of water per mole of anhydrous salt, divide the mass of water lost by the molar mass of water ($18.02$ g/mol), and the mass of anhydrous salt by its molar mass, then take the ratio.

• This experiment is foundational in GOB Chemistry for understanding stoichiometry and chemical formulas.