Hydrate Analysis: Determining Water Content and Formula of a Hydrate

Hydrate Analysis

Introduction to Hydrates

Hydrates are ionic compounds that contain a specific number of water molecules bound within their crystal structure. The water in a hydrate is called water of hydration and is essential to the compound's structure and properties. When hydrates are heated, the water is released, leaving behind the anhydrous salt.

• Hydrate: A compound that includes water molecules within its solid structure (e.g., CuSO4·5H2O).

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

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

Experimental Data Collection

To determine the amount of water in a hydrate, the following masses are measured:

• Mass of empty crucible

• Mass of crucible and hydrate

• Mass of crucible and anhydrous salt (after heating to remove water)

These measurements allow calculation of the mass of the hydrate, the mass of the anhydrous salt, and the mass of water lost during heating.

Sample Data and Calculations

• Mass of empty crucible: 17.95 g

• Mass of crucible and hydrate: 27.16 g

• Mass of hydrate: $27.16\,\text{g} - 17.95\,\text{g} = 10.21\,\text{g}$

• Mass of crucible and anhydrous salt (after heating): 26.07 g

• Mass of anhydrous salt: $26.07\,\text{g} - 17.95\,\text{g} = 8.12\,\text{g}$

• Mass of water lost: $10.21\,\text{g} - 8.32\,\text{g} = 1.89\,\text{g}$

Note: There is a minor discrepancy in the anhydrous salt mass (8.32 g vs. 8.12 g) in the handwritten notes. The calculation above uses the values as written in the image.

Calculating the Percent Water in a Hydrate

The percent by mass of water in the hydrate can be calculated using:

$\text{Percent water} = \frac{\text{Mass of water lost}}{\text{Mass of hydrate}} \times 100\%$

For the sample data:

$\text{Percent water} = \frac{1.89\,\text{g}}{10.21\,\text{g}} \times 100\% = 18.5\%$

Determining the Formula of a Hydrate

To determine the formula of a hydrate (e.g., CuSO4·xH2O), calculate the moles of anhydrous salt and water:

• Moles of anhydrous salt: $\text{moles} = \frac{\text{mass}}{\text{molar mass}}$

• Moles of water: $\text{moles} = \frac{\text{mass}}{18.02\,\text{g/mol}}$

Divide both values by the smaller number of moles to find the simplest whole number ratio, which gives the value of x in the hydrate formula.

Example Calculation

• Mass of anhydrous salt: 8.32 g (assume CuSO4, molar mass = 159.61 g/mol)

• Moles of anhydrous salt: $\frac{8.32\,\text{g}}{159.61\,\text{g/mol}} = 0.0521\,\text{mol}$

• Mass of water lost: 1.89 g

• Moles of water: $\frac{1.89\,\text{g}}{18.02\,\text{g/mol}} = 0.105\,\text{mol}$

• Ratio of moles (water : salt): $\frac{0.105}{0.0521} = 2.02$

• Empirical formula: CuSO4·2H2O

Additional info: The actual hydrate of copper(II) sulfate is pentahydrate (CuSO4·5H2O), but the above calculation is based on the sample data provided.

Summary Table: Key Masses and Calculations

Measurement	Value (g)	Calculation
Mass of empty crucible	17.95	Given
Mass of crucible + hydrate	27.16	Given
Mass of hydrate	10.21	27.16 - 17.95
Mass of crucible + anhydrous salt	26.07	After heating
Mass of anhydrous salt	8.32	26.07 - 17.95
Mass of water lost	1.89	10.21 - 8.32

Key Takeaways

• Heating a hydrate removes water, allowing determination of the percent water and the empirical formula.

• Careful mass measurements before and after heating are essential for accurate results.

• Calculations involve subtraction, percent composition, and mole ratios.