Problem 154 - Chemistry and Chemical Reactivity

Question

The "alum" used in cooking is potassium aluminum sulfate hydrate, \(\mathrm{KAl}\left(\mathrm{SO}_{4}\right)_{2} \cdot x \mathrm{H}_{2} \mathrm{O} .\) To find the value of \(x,\) you can heat a sample of the compound to drive off all of the water and leave only \(\mathrm{KAl}\left(\mathrm{SO}_{4}\right)_{2}\). Assume you heat \(4.74 \mathrm{g}\) of the hydrated compound and that the sample loses \(2.16 \mathrm{g}\) of water. What is the value of \(x ?\)

Step-by-Step Solution

Answer

The value of \(x\) is 12.

Step 1: Determine the Mass of Anhydrous Compound

First, we need to find the mass of the anhydrous compound, which is the mass remaining after all the water has been driven off. This can be calculated by subtracting the mass of water lost from the initial mass of the hydrated compound. \[\text{Mass of anhydrous compound} = 4.74\, \text{g} - 2.16\, \text{g} = 2.58 \, \text{g}\]

Step 2: Calculate the Moles of Anhydrous Compound

To find the number of moles of the anhydrous \(\mathrm{KAl(SO}_{4})_{2}\), use its molar mass which is 258\, \text{g/mol}. Using the formula for moles:\[\text{Moles of anhydrous compound} = \frac{\text{mass of anhydrous compound}}{\text{molar mass}} = \frac{2.58\, \text{g}}{258\, \text{g/mol}} = 0.01 \, \text{mol}\]

Step 3: Calculate the Moles of Water Lost

Next, we need to calculate the moles of water lost, knowing its molar mass is 18 \text{g/mol}:\[\text{Moles of water} = \frac{\text{mass of water lost}}{\text{molar mass of water}} = \frac{2.16\, \text{g}}{18\, \text{g/mol}} = 0.12 \, \text{mol}\]

Step 4: Determine the Value of x

Now, the value of \(x\) is found by comparing the ratio of moles of water to moles of the anhydrous compound, since each mole of \(\mathrm{KAl(SO}_{4})_{2}\) corresponds to \(x\) moles of water:\[x = \frac{\text{moles of water}}{\text{moles of anhydrous compound}} = \frac{0.12 \, \text{mol}}{0.01 \, \text{mol}} = 12\]

Step 5: Conclusion: State the Final Result

We've calculated that the value of \(x\), which representsthe number of water molecules per formula unit in the hydrated compound, is 12.

Key Concepts

Stoichiometry in Hydrated Compounds

Stoichiometry is essential when explaining how elements react and combine to form compounds. It's a chemical science that tells us the quantitative relationship between reactants and products in a reaction. In the context of dealing with hydrated compounds like alum, stoichiometry allows us to comprehend how water is incorporated into their structure. Hydrated compounds have water molecules integrated into their crystal lattice. Knowing this helps us calculate how many water molecules, denoted as \(x\), are in each formula unit of the hydrated compound. Here, stoichiometry helps us identify the mole ratio between the compound and its respective components, which is vital for determining \(x\). Understanding this concept will equip you to predict the outcome of chemical reactions and support the calculation of quantities in batch reactions.

Moles Calculation in Chemical Analysis

Chemical analysis often involves calculating moles of substances to understand the nature of chemical reactions. The mole is a fundamental unit in chemistry that quantifies the amount of a substance. Centering around Avogadro's number, which is \(6.022 \times 10^{23}\), the mole allows chemists to count atoms, ions, or molecules in a given sample. For our alum compound, first, we calculated the moles of the anhydrous compound using its mass and molar mass. Then, calculating the moles of water shed light on how much water was originally part of the structure.

• Use the formula: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \).
• This formula helps in identifying how components relate within compound formulas.

Mastering mole calculations is crucial as it forms the foundation for further exploring chemical composition and transformations.

Understanding Thermal Decomposition

Thermal decomposition plays a significant role in deriving pure substances from hydrated compounds. It involves heating a compound to break it down into simpler substances. This process is insightful for removing water from hydrates, leaving behind the anhydrous version of the compound. By heating alum and removing its water content, you can determine the ratio of water in the original hydrate.

• Heating separates the components based on thermal stability.
• The process provides insight into the material's chemical structure.

Understanding thermal decomposition allows chemists to analyze the stability of different compounds and understand how they retain components like water under various conditions.

Alum Chemistry and Its Significance

Alum chemistry delves into compounds composed of sulfates of a univalent metal like potassium and a trivalent metal such as aluminum. Potassium aluminum sulfate, a common alum, is central in industries such as food processing and water purification. As seen in the exercise, alum as a hydrated compound has many water molecules integrated into its structure, which are lost upon heating to form an anhydrous form. This knowledge is instrumental in applications where alum is used, as the loss of water can affect its properties.

• Understanding alum's role is critical in its diverse practical applications.
• Analysis of its hydrate form helps in determining its practicality in industrial processes.

Exploring alum chemistry aids in appreciating its versatile nature and the importance of water in its compound formation.