Percent of Water in an Unknown Hydrated Salt: Principles, Calculations, and Laboratory Procedure

Percent of Water in an Unknown Hydrated Salt

Objectives

• Determine the percent (by mass) of water in an unknown hydrated salt.

• Determine the number of water molecules per formula unit of the unknown hydrated salt.

Introduction to Hydrated Salts

Hydrated salts are ionic compounds that crystallize with a specific number of water molecules incorporated into their structure. These water molecules are called water of hydration and are essential to the physical properties of the salt.

• Hydrate: A compound that contains water molecules within its crystal structure.

• Anhydrous salt: The compound remaining after the water of hydration is removed, typically by heating.

Common examples of hydrated salts include:

• CuSO4·5H2O (copper(II) sulfate pentahydrate)

• BaCl2·2H2O (barium chloride dihydrate)

• Na2CO3·10H2O (sodium carbonate decahydrate)

Properties and Stability of Hydrates

Hydrates can lose water spontaneously when exposed to air, a process called efflorescence. Some hydrates require heating to drive off the water of hydration. The stability of a hydrate depends on the strength of the interaction between the salt and water molecules.

• Efflorescent substances: Lose water to the atmosphere easily (e.g., CuSO4·5H2O).

• Deliquescent substances: Absorb water from the atmosphere and may dissolve in it.

Reaction of Hydrated Salts Upon Heating

When a hydrated salt is heated, it undergoes a reversible reaction:

• Hydrated salt (solid) → Anhydrous salt (solid) + Water of hydration

The general equation is:

• $\text{CuSO}_4 \cdot 5\text{H}_2\text{O} \xrightarrow{\text{heat}} \text{CuSO}_4 + 5\text{H}_2\text{O}$

This reaction is used to determine the amount of water present in a hydrate by measuring the mass lost upon heating.

Calculating Percent Water in a Hydrate

Percent Water Formula

The percent of water in a hydrate is calculated using the following formula:

• $\%\ \text{water} = \frac{\text{mass of water lost}}{\text{mass of hydrate}} \times 100$

Example: If a sample of hydrate loses 1.89 g of water and the original mass of hydrate was 10.21 g:

• $\%\ \text{water} = \frac{1.89}{10.21} \times 100 = 18.5\%$

Determining the Number of Water Molecules per Formula Unit

To find the number of water molecules (n) per formula unit in the hydrate, use the following steps:

1. Calculate moles of water lost:

   • $\text{moles of water} = \frac{\text{mass of water lost}}{\text{molar mass of H}_2\text{O}}$

2. Calculate moles of anhydrous salt:

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

3. Find the ratio:

   • $n = \frac{\text{moles of water}}{\text{moles of anhydrous salt}}$

Round n to the nearest whole number to determine the formula of the hydrate (e.g., CuSO4·nH2O).

Laboratory Procedure Overview

Preparation of the Crucible

• Clean and heat the crucible to remove moisture.

• Weigh the empty crucible.

Heating the Hydrate

• Weigh the crucible with the unknown hydrate.

• Heat the crucible to drive off water of hydration.

• Cool and reweigh the crucible and contents.

• Repeat heating and weighing until mass is constant (indicating all water has been removed).

Data Collection and Calculations

• Record all masses: empty crucible, crucible + hydrate, crucible + anhydrous salt.

• Calculate mass of hydrate, mass of anhydrous salt, and mass of water lost.

• Use the formulas above to determine percent water and number of water molecules per formula unit.

Sample Data Table

The following table summarizes the key measurements and calculations performed in the experiment:

Measurement	Trial 1	Trial 2
Mass of empty crucible (g)	17.95	(Additional info: Not provided)
Mass of crucible + hydrate (g)	27.16	(Additional info: Not provided)
Mass of hydrate (g)	10.21	(Additional info: Not provided)
Mass of crucible + anhydrous salt (g)	26.07 (after first heating)	(Additional info: Not provided)
Mass of anhydrous salt (g)	8.329	(Additional info: Not provided)
Mass of water lost (g)	1.89	(Additional info: Not provided)
Percent water	18.5%	(Additional info: Not provided)

Key Equations

• $\%\ \text{water} = \frac{\text{mass of water lost}}{\text{mass of hydrate}} \times 100$

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

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

• $n = \frac{\text{moles of water}}{\text{moles of anhydrous salt}}$

Reproducibility and Accuracy

To ensure accuracy, repeat the heating and weighing process until the mass of the crucible and its contents is constant. This confirms that all water has been removed. If results from repeated trials agree within 0.05 g, the measurement is considered reliable.

Summary Table: Types of Hydrated Salts

Hydrated Salt	Formula	Common Name
Copper(II) sulfate pentahydrate	CuSO4·5H2O	Blue vitriol
Barium chloride dihydrate	BaCl2·2H2O	(Additional info: Common laboratory reagent)
Sodium carbonate decahydrate	Na2CO3·10H2O	Washing soda

Conclusion

Determining the percent of water in a hydrated salt and the number of water molecules per formula unit is a fundamental experiment in GOB Chemistry. It illustrates the principles of stoichiometry, chemical formulas, and the properties of ionic compounds. Mastery of these calculations is essential for understanding the composition and behavior of hydrates in chemical and biological systems.