Problem 51 - Chemistry: An Atoms-Focused Approach

Question

A 3.556 g sample of a pure aluminum oxide decomposes under high heat to produce \(1.674 \mathrm{g}\) of oxygen in addition to pure aluminum metal. What is the empirical formula of the aluminum oxide?

Step-by-Step Solution

Answer

Answer: The empirical formula of aluminum oxide is Al2O3.

Step 1: Determine the moles of oxygen produced

Divide the mass of oxygen produced (1.674 g) by its molar mass (16.00 g/mol) to find the number of moles of oxygen. Moles of Oxygen = \(\frac{Mass\ of\ Oxygen}{Molar\ Mass\ of\ Oxygen}\) Moles of Oxygen = \(\frac{1.674\ g}{16.00\ g/mol}\) Moles of Oxygen = \(0.1046\ mol\)

Step 2: Determine the mass of aluminum

Subtract the mass of oxygen from the mass of the aluminum oxide sample to find the mass of aluminum. Mass of Aluminum = (Mass of Aluminum Oxide) - (Mass of Oxygen produced) Mass of Aluminum = 3.556 g - 1.674 g Mass of Aluminum =1.882 g

Step 3: Determine the moles of aluminum

Divide the mass of aluminum (1.882 g) by its molar mass (26.98 g/mol) to find the number of moles of aluminum. Moles of Aluminum = \(\frac{Mass\ of\ Aluminum}{Molar\ Mass\ of\ Aluminum}\) Moles of Aluminum = \(\frac{1.882\ g}{26.98\ g/mol}\) Moles of Aluminum = \(0.0697\ mol\)

Step 4: Determine the mole ratio of aluminum to oxygen

Divide the moles of aluminum and oxygen by their smallest value to find the simplest whole number ratio between the elements. Mole Ratio of Aluminum to Oxygen = \(\frac{Moles\ of\ Aluminum}{Moles\ of\ Oxygen}\) ` Mole Ratio of Aluminum to Oxygen = \(\frac{0.0697\ mol}{0.1046\ mol}\) Mole Ratio of Aluminum to Oxygen = \(\frac{2}{3}\)

Step 5: Find the empirical formula

Based on the mole ratio found in step 4, the empirical formula of aluminum oxide is Al2O3.

Key Concepts

Aluminum Oxide

Aluminum oxide, often referred to as alumina, is a compound made up of aluminum and oxygen. It is commonly found in minerals such as bauxite and corundum. In this context, it's important to determine its empirical formula, which represents the simplest ratio of atoms in a compound. Understanding the empirical formula helps chemists comprehend the composition and properties of the substance. For example, the empirical formula for aluminum oxide is Al₂O₃. This indicates that in aluminum oxide, two aluminum atoms are present for every three oxygen atoms. Knowing this formula is pivotal when analyzing reactions and properties of the substance.

Mole Ratio

The mole ratio is an essential concept when determining empirical formulas. It reflects the simplest whole number relationship between the elements in a compound. Calculating the mole ratio involves dividing the number of moles of each element by the smallest number of moles calculated. For instance, if you have 0.0697 moles of aluminum and 0.1046 moles of oxygen, the mole ratio calculation involves:

• Divide the moles of aluminum: \(\frac{0.0697\ mol}{0.0697\ mol} = 1\)
• Divide the moles of oxygen: \(\frac{0.1046\ mol}{0.0697\ mol} \approx 1.5\)

By rounding to the nearest whole numbers, we maintain the ratio as 2:3, crucially informing us of the compound's formula. Understanding mole ratios is vital in stoichiometry and helps in predicting product yields from reactions.

Moles Calculation

In chemistry, converting mass to moles is a fundamental technique to find concentrations and calculate molecular interactions. A mole represents 6.022 × 10²³ particles, whether atoms, molecules, or ions. To convert mass to moles, you divide the mass of the element by its molar mass.To find the moles of oxygen produced in the aluminum oxide decomposition, the calculation is:

• \(\text{Moles of Oxygen} = \frac{1.674\ g}{16.00\ g/mol} = 0.1046\ mol\)

Similarly, for aluminum, it is:

• \(\text{Moles of Aluminum} = \frac{1.882\ g}{26.98\ g/mol} = 0.0697\ mol\)

These calculations form the basis for determining the empirical formula and guide further chemical analyses.

Elemental Composition

Elemental composition refers to the percentage of each element by mass in a compound. By decomposing a compound and measuring the masses of the resultant elements, we gain insight into its composition. For aluminum oxide, understanding its elemental composition means knowing how the total mass is distributed between aluminum and oxygen.In our exercise, we noted that:

• 3.556 g of aluminum oxide contains 1.674 g of oxygen and 1.882 g of aluminum.

The results inform us about the makeup of the compound.To find the mass percentage of each element:

• \( \text{Percentage of Aluminum} = \frac{1.882\ g}{3.556\ g} \times 100\% = 52.9\% \)
• \( \text{Percentage of Oxygen} = \frac{1.674\ g}{3.556\ g} \times 100\% = 47.1\% \)

Grasping these percentages aids in analyzing the compound's properties and predicting its behavior during reactions.