Question

Aluminum reacts with oxygen to yield aluminum oxide. If \(5.0 \mathrm{~g}\) of \(\mathrm{Al}\) reacts with \(4.45 \mathrm{~g}\) of \(\mathrm{O}_{2}\), what is the empirical formula of aluminum oxide?

Short answer

The empirical formula of aluminum oxide is Al2O3.

Step-by-step solution

Write the equation of reaction

Aluminum (Al) reacts with oxygen (O2) to form aluminum oxide (Al2O3). The chemical equation for this reaction is: \(4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3\).

Calculate moles of aluminum

The molar mass of aluminum (Al) is approximately \(27\text{ g/mol}\). To find the number of moles of aluminum, use the formula: \(\text{Moles of Al} = \frac{5.0\text{ g}}{27\text{ g/mol}}\). This gives approximately \(0.185\text{ mol}\) of Al.

Calculate moles of oxygen

The molar mass of oxygen gas (O2) is approximately \(32\text{ g/mol}\). To find the number of moles of oxygen, use the formula: \(\text{Moles of } O_2 = \frac{4.45\text{ g}}{32\text{ g/mol}}\). This gives approximately \(0.139\text{ mol}\) of O2.

Find the simplest molar ratio

The reaction requires breaking down the moles into a ratio that reflects the empirical formula. We have \(0.185\text{ mol of Al}\) and \(0.139\text{ mol of } O_2\). Multiply the moles of \(O_2\) by 2 (since each molecule provides 2 atoms), to reflect individual oxygen atoms: \(0.139 \times 2 = 0.278\). Divide by the smallest amount to find the simplest ratio: \(\frac{0.185}{0.139} \approx 1.33\) for Al and \(\frac{0.278}{0.139} \approx 2\) for O.

Determine the empirical formula

For a clearer formula, multiply the ratio values by 3 to adjust the fraction representing Al: \(1.33 \times 3 = 4\) for Al and \(2 \times 3 = 6\) for O. Simplifying gives the empirical formula: Al2O3.

Key concepts

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, are transformed into new substances called products. In the context of our exercise, aluminum (Al) reacts with oxygen (O2) to form aluminum oxide (Al2O3). Understanding the chemical equation is key because it shows the transformation and the stoichiometric proportions involved.
A balanced chemical equation ensures that the same number of atoms for each element appear in both reactants and products. This is crucial as it reflects the conservation of mass in a chemical reaction:

• It illustrates how atoms rearrange.
• Relates masses of substances in a reaction.

In our specific reaction:

• 4 Al atoms react with 3 O2 molecules to form 2 Al2O3 molecules.
• This balanced equation tells us the mole ratio of aluminum to oxygen and the formula of aluminum oxide.

By understanding this equation, we lay the groundwork for calculating moles involved and determining the empirical formula of the compound formed.

Moles Calculation

Calculating moles is a fundamental step in determining the proportions of elements in a chemical reaction. A mole is a unit that measures the amount of substance, similar to how a dozen measures quantity. It's important as it allows chemists to count atoms by weighing them.

• The molar mass is the mass of one mole of a substance, given in grams per mole (g/mol).
• For aluminum, the molar mass is approximately 27 g/mol, and for oxygen gas (O2), it is 32 g/mol.

To calculate moles, use:

• Formula: \(\text{Moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}\)

In our case:

• 5.0 g of aluminum translates into approximately 0.185 moles.
• 4.45 g of oxygen translates into approximately 0.139 moles.

These calculations are pivotal as they help determine the number of atoms involved in the reaction to derive the empirical formula.

Aluminum Oxide

Aluminum oxide is the product formed when aluminum reacts with oxygen. It’s a compound with significant industrial applications due to its properties, such as high melting point and resistance to corrosion.
Understanding the empirical formula of aluminum oxide (Al2O3) is essential:

• The empirical formula represents the simplest whole-number ratio of elements in a compound.
• For our reaction, breaking down the moles of each element helps find this ratio.

Through mole calculations:

• The ratio of aluminum to oxygen comes out to approximately 1.33:2, which simplifies to Al2O3 upon multiplying to clear the fraction.

This formula not only tells us the composition of aluminum oxide but also its stable atomic arrangement, helping us understand its chemical behavior and usage in various domains.