When aluminum metal is heated with an element from Group \(6 \mathrm{~A}\) of the periodic table, an ionic compound forms. When the experiment is performed with an unknown Group 6 A element, the product is \(18.56 \%\) Al by mass. What is the formula of the compound?

We know the following information: 1. Aluminum metal (Al) combines with an element from Group 6A of the periodic table. 2. The product is 18.56% Al by mass.

The molar mass of aluminum (Al) is approximately 26.98 g/mol.

Assuming we have 100 grams of the compound, since the percentage of aluminum is given as 18.56% of the total mass, then the mass of Al would be \(18.56 \mathrm{~g}\). This means that the mass of the unknown Group 6A element would be the difference between the total mass and the mass of Al: Mass of the unknown element = \(100 \mathrm{~g} - 18.56 \mathrm{~g} = 81.44 \mathrm{~g}\)

Now, we will determine the moles of aluminum by dividing the mass by the molar mass: Moles of Al = \(\frac{18.56 \mathrm{~g}}{26.98 \mathrm{~g/mol}} ≈ 0.688 \mathrm{~mol}\) Let x represent the molar mass of the unknown element. We can calculate its moles using the mass: Moles of the unknown element = \(\frac{81.44 \mathrm{~g}}{x \mathrm{~g/mol}}\)

We then calculate the mole ratio of aluminum to the unknown element by dividing the moles of aluminum by the moles of the unknown element: Mole ratio = \(\frac{0.688 \mathrm{~mol}}{\frac{81.44 \mathrm{~g}}{x \mathrm{~g/mol}}}\)

There are four main Group 6A elements in the periodic table: oxygen (O), sulfur (S), selenium (Se), and tellurium (Te). Their molar masses are approximately 16 g/mol (O), 32 g/mol (S), 79 g/mol (Se), and 128 g/mol (Te). We will calculate the mole ratios for each element and identify the one with the closest integer ratio: 1. For oxygen (O): Mole ratio = \(\frac{0.688 \mathrm{~mol}}{\frac{81.44 \mathrm{~g}}{16 \mathrm{~g/mol}}}\) ≈ 1.37 2. For sulfur (S): Mole ratio = \(\frac{0.688 \mathrm{~mol}}{\frac{81.44 \mathrm{~g}}{32 \mathrm{~g/mol}}}\) ≈ 2.72 3. For selenium (Se): Mole ratio = \(\frac{0.688 \mathrm{~mol}}{\frac{81.44 \mathrm{~g}}{79 \mathrm{~g/mol}}}\) ≈ 6.88 4. For tellurium (Te): Mole ratio = \(\frac{0.688 \mathrm{~mol}}{\frac{81.44 \mathrm{~g}}{128 \mathrm{~g/mol}}}\) ≈ 10.9

From the calculated mole ratios, aluminum and sulfur have the closest integer ratio (approximately 1:3). Therefore, the empirical formula of the compound is Al₂S₃.