Problem 146 - Chemistry Matter and Change

Question

Calculate the moles of aluminum ions present in 250.0 \(\mathrm{g}\) of aluminum oxide \(\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right) .\)

Step-by-Step Solution

Answer

There are approximately 4.90 moles of aluminum ions in 250.0 g of \( \mathrm{Al}_2 \mathrm{O}_3 \).

Step 1: Determine Molar Mass of Aluminum Oxide

First, we find the molar mass of aluminum oxide, \( \mathrm{Al}_2 \mathrm{O}_3 \). This compound contains 2 aluminum (\(\mathrm{Al}\)) atoms and 3 oxygen (\(\mathrm{O}\)) atoms. Calculate the molar mass using the periodic table, where the atomic mass of \(\mathrm{Al}\) is approximately 27.0 g/mol, and the atomic mass of \(\mathrm{O}\) is approximately 16.0 g/mol. Thus, the molar mass of \( \mathrm{Al}_2 \mathrm{O}_3 \) is: \[ 2(27.0) + 3(16.0) = 54.0 + 48.0 = 102.0 \, \mathrm{g/mol}. \]

Step 2: Calculate Moles of Aluminum Oxide

Next, we use the molar mass to find the number of moles of \( \mathrm{Al}_2 \mathrm{O}_3 \) present in 250.0 g of the compound. This is done using the formula: \[ \text{moles} = \frac{\text{mass}}{\text{molar mass}}. \] Substitute the known values: \[ \text{moles of } \mathrm{Al}_2 \mathrm{O}_3 = \frac{250.0 \, \mathrm{g}}{102.0 \, \mathrm{g/mol}} \approx 2.45 \, \text{moles}. \]

Step 3: Calculate Moles of Aluminum Ions

Finally, we determine the number of moles of aluminum ions, \(\mathrm{Al}^{3+}\). Each formula unit of \( \mathrm{Al}_2 \mathrm{O}_3 \) contains 2 \( \mathrm{Al}^{3+} \) ions. Thus, for the moles of \( \mathrm{Al}_2 \mathrm{O}_3 \), the moles of \( \mathrm{Al}^{3+} \) are: \[ \text{moles of } \mathrm{Al}^{3+} = 2 \times \text{moles of } \mathrm{Al}_2 \mathrm{O}_3 = 2 \times 2.45 \approx 4.90 \, \text{moles}. \]

Key Concepts

Aluminum oxide

Aluminum oxide, also known by its chemical formula \( \mathrm{Al}_2 \mathrm{O}_3 \), is a compound formed by aluminum and oxygen. It is often found in nature as corundum and is critically important in materials science, especially in the production of aluminum metal. Each molecule of aluminum oxide consists of:

• 2 aluminum (\( \mathrm{Al} \)) atoms
• 3 oxygen (\( \mathrm{O} \)) atoms

This composition makes it a stable compound with many industrial applications such as an abrasive due to its hardness, and as an electrical insulator. Understanding its chemical makeup is essential for mole calculations, such as determining the number of particles present in a given mass of the substance. Using aluminum oxide in mole-related problems often revolves around its unique chemical properties and the stoichiometry of its formation.

Molar mass

Molar mass is a crucial concept in chemistry when dealing with reactions and conversions between mass and moles. It is the mass of one mole of a given substance, usually expressed in grams per mole (g/mol). To calculate the molar mass of a compound like aluminum oxide, \( \mathrm{Al}_2 \mathrm{O}_3 \), follow these steps:

• Find the atomic masses from the periodic table: Aluminum ≤ \( 27.0 \, \mathrm{g/mol} \)
• Oxygen ≤ \( 16.0 \, \mathrm{g/mol} \)
• Calculate the sum for the compound: \( 2 \times 27.0 + 3 \times 16.0 = 102.0 \, \mathrm{g/mol} \)

This molar mass becomes the denominator when converting grams to moles using the formula \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). Understanding how to accurately compute molar mass is fundamental to study chemical reactions and predict the quantities involved in various processes.

Aluminum ions

Aluminum ions, specifically \( \mathrm{Al}^{3+} \), are positively charged ions formed when aluminum atoms lose three electrons. In the compound aluminum oxide \( (\mathrm{Al}_2 \mathrm{O}_3) \), these ions play a vital role in its chemical structure and properties. To find the moles of aluminum ions in a sample, consider these factors:

• Each molecule of \( \mathrm{Al}_2 \mathrm{O}_3 \) contains 2 \( \mathrm{Al}^{3+} \) ions due to its composition.
• If you determine there are 2.45 moles of \( \mathrm{Al}_2 \mathrm{O}_3 \), apply simple multiplication: \( 2 \times 2.45 \approx 4.90 \) moles of \( \mathrm{Al}^{3+} \).

Counting ions in a compound involves recognizing its basic formula unit and multiplying by the respective number of ions each unit contributes. Accurately finding the number of ions is important in applications involving electrochemistry and reaction stoichiometry.