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

In 250.0 g of aluminum oxide (Al₂O₃), there are 4.8992 moles of aluminum ions (Al³⁺) present.

Step 1: Find the molar mass of Al₂O₃

To do this, we first need to find the molar masses of aluminum (Al) and oxygen (O) individually. The molar mass of aluminum is 26.98 g/mol, and that of oxygen is 16.00 g/mol. Now we can calculate the molar mass of Al₂O₃: Molar mass of Al₂O₃ = (2 × molar mass of Al) + (3 × molar mass of O) Molar mass of Al₂O₃ = (2 × 26.98 g/mol) + (3 × 16.00 g/mol) Molar mass of Al₂O₃ = 53.96 g/mol + 48.00 g/mol Molar mass of Al₂O₃ = 101.96 g/mol

Step 2: Calculate the moles of Al₂O₃ present in 250.0 g

To do this, we can use the given mass of Al₂O₃ and its molar mass: Moles of Al₂O₃ = (mass of Al₂O₃) / (molar mass of Al₂O₃) Moles of Al₂O₃ = (250.0 g) / (101.96 g/mol) Moles of Al₂O₃ = 2.4496 mol

Step 3: Determine the moles of aluminum ions (Al³⁺) in the given amount of Al₂O₃

In one mole of Al₂O₃, there are 2 moles of Al ions. To find the moles of Al ions present in 2.4496 moles of Al₂O₃, use the below relationship: Moles of Al³⁺ = (2 × moles of Al₂O₃) Moles of Al³⁺ = 2 × 2.4496 mol Moles of Al³⁺ = 4.8992 mol Therefore, there are 4.8992 moles of aluminum ions present in 250.0 g of aluminum oxide.

Key Concepts

Molar Mass Calculation

To calculate the molar mass of a compound, you need to know the atomic masses of the constituent elements. This is essential for converting the mass of a substance to the number of moles, a fundamental concept in chemistry.
In the case of aluminum oxide \((\text{Al}_2\text{O}_3)\), you start by identifying the individual atomic masses:

• Aluminum (Al): 26.98 \(\text{g/mol}\)
• Oxygen (O): 16.00 \(\text{g/mol}\)

Next, multiply the atomic mass of each element by the number of atoms in the formula:

• For Al: \(2 \times 26.98 \text{ g/mol} = 53.96 \text{ g/mol}\)
• For O: \(3 \times 16.00 \text{ g/mol} = 48.00 \text{ g/mol}\)

Finally, sum these values to get the molar mass of Al₂O₃: \(53.96 \text{ g/mol} + 48.00 \text{ g/mol} = 101.96 \text{ g/mol}\).
This molar mass is used to relate the mass of the compound to the number of moles.

Aluminum Oxide

Aluminum oxide is a chemical compound with the formula \(\text{Al}_2\text{O}_3\). It's commonly found in several mineral forms, notably in bauxite, the principal ore of aluminum.
This compound is important in both industrial and manufacturing applications, including metallurgy and ceramics.
In aluminum oxide:

• There are 2 aluminum atoms.
• There are 3 oxygen atoms.

The molecular structure of aluminum oxide causes it to have a high melting point and good electrical insulation properties. This makes it highly valuable as an abrasive and as a component in refractory materials, which can withstand very high temperatures. Understanding its composition is key to calculating chemical quantities like moles, which are vital in chemical reactions.

Moles Calculation

The concept of "moles" in chemistry refers to a standard method for expressing quantities of atoms, molecules, or ions. It's like a bridge between the atomic scale and the macroscopic scale we can measure.
A mole is much like a dozen—it simply means 6.022 x 10²³ of something, which is also known as Avogadro’s number.
To find the number of moles in a given mass of a substance, you use the formula:

• \( \text{Moles} = \frac{\text{Mass of substance}}{\text{Molar mass}} \)

This relationship helps transform a measured mass into the number of "moles," which are more convenient for stoichiometric calculations. For aluminum oxide, as calculated earlier, its molar mass is 101.96 g/mol. Using the formula, the mass of 250.0 g of Al₂O₃ is found to be approximately 2.4496 moles. This calculation is crucial because it allows chemists to relate the amounts of materials used in reactions.

Chemical Composition

Chemical composition refers to the elements that make up a compound and the ratio in which they are present. Understanding this is critical for calculating chemical quantities accurately and predicting the behavior of compounds in various conditions.
The composition of \(\text{Al}_2\text{O}_3\) breaks down into:

• Two atoms of aluminum (Al),
• Three atoms of oxygen (O).

Their proportions by mass can be derived using their molar masses:

• Al contributes \(53.96 \text{ g/mol}\).
• O contributes \(48.00 \text{ g/mol}\).

This precise formula enables the direct calculation of moles for each element in the compound, useful in understanding how the compound will react chemically.
Knowing the chemical composition aids in numerous applications such as determining the necessary amounts of reactants in a chemical reaction or creating desired mixtures in compound synthesis.

Stoichiometry

Stoichiometry is the branch of chemistry involved with calculating the quantities of reactants and products in chemical reactions. This concept applies the law of conservation of mass, which states that mass in a chemical reaction is neither created nor destroyed.
By using balanced chemical equations, stoichiometry allows you to predict the amounts of substances consumed and produced.
In terms of aluminum oxide, stoichiometry comes into play when you understand that one mole of \(\text{Al}_2\text{O}_3\) contains two moles of aluminum ions (Al³⁺). Thus, if you have 2.4496 moles of \(\text{Al}_2\text{O}_3\), it translates to 4.8992 moles of aluminum ions, as calculated by the stoichiometric coefficient.
Understanding stoichiometry is key to scaling laboratory reactions to industrial applications, ensuring that complex reactions can be managed and optimized for various chemical processes.