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 \(\mathrm{A}\) element, the product is 18.56\(\%\) Al by mass. What is the formula of the compound?

As the ionic compound consists of only Aluminum (Al) and the unknown Group 6A element, we can find the mass percent of the unknown element by subtracting the mass percent of aluminum (Al) from 100%. Mass Percent of the unknown element = 100% - 18.56% = 81.44%.

We know the molar mass of Aluminum (Al) which is 26.98 g/mol. Using the mass percent information, we can calculate the mass of the unknown element in a 100g sample. Mass of Al in 100g sample = (18.56/100) * 100 = 18.56 g Mass of the unknown element in 100g sample = (81.44/100) * 100 = 81.44 g Now, determine the moles of aluminum and the unknown element: Moles of Al = mass of Al / molar mass of Al = 18.56 g / 26.98 g/mol = 0.688 mol Since the ratio of moles of Al to the unknown element must be integers, assume there is 1 mol of Al. Moles of the unknown element = (0.688 mol Al) * (1 mol unknown element / 1 mol Al) = 0.688 mol Molar mass of the unknown element = mass of unknown element / moles of unknown element = 81.44 g / 0.688 mol = 118.28 g/mol

Look up the molar mass of the unknown element among Group 6A (16) elements in the periodic table. The molar masses of group 6A elements are as follows: - Oxygen (O): 16.00 g/mol - Sulfur (S): 32.06 g/mol - Selenium (Se): 78.96 g/mol - Tellurium (Te): 127.60 g/mol - Polonium (Po): 209.00 g/mol The molar mass closest to 118.28 g/mol is 127.60 g/mol for Tellurium (Te). So, the unknown element is Tellurium (Te).

Aluminum (Al) usually forms +3 ions (Al³⁺), while elements from Group 6A form -2 ions (e.g., Te²⁻). To balance the charges and form a neutral ionic compound, we need two Al³⁺ ions and three Te²⁻ ions. The formula of the compound is therefore Al₂Te₃.