Question

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?

Short answer

The empirical formula of the ionic compound formed between aluminum and the unknown Group 6A element is Al₂S₃.

Step-by-step solution

List the known data

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.

Determine the molar mass of Al

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

Assume an initial mass and calculate the mass of the unknown element

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 = \(100 \mathrm{~g} - 18.56 \mathrm{~g} = 81.44 \mathrm{~g}\)

Calculate the moles of Al and the unknown element

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 = \(\frac{81.44 \mathrm{~g}}{x \mathrm{~g/mol}}\)

Determine the ratio of moles

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 = \(\frac{0.688 \mathrm{~mol}}{\frac{81.44 \mathrm{~g}}{x \mathrm{~g/mol}}}\)

Identify the possible Group 6A elements and calculate their mole ratios

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

Determine the compound's empirical formula

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.

Key concepts

Aluminum

Aluminum is a light, silver-colored metal that is widely used in various applications due to its low density and high resistance to corrosion. It is the third most abundant element in the Earth's crust, following oxygen and silicon, and is primarily found in the form of bauxite ore. Aluminum sits in Group 13 of the periodic table which means it usually forms +3 charges in ionic compounds. This positive charge makes aluminum a suitable bonding partner for elements that can easily form anions.
In chemical reactions, aluminum is known for forming stable ionic compounds. When combined with highly electronegative elements, it creates strong ionic bonds resulting in compounds that have significant industrial and commercial applications. Its metallic properties are excellent conductors of electricity and heat, which contribute to its uses in applications such as packaging, construction, and transportation.

Group 6A Elements

Group 6A of the periodic table, often referred to as the oxygen family, consists of the elements oxygen (O), sulfur (S), selenium (Se), tellurium (Te), and polonium (Po). These elements have six valence electrons which make them electronegative and capable of gaining or sharing electrons to achieve a stable, noble gas electron configuration.
Each element in Group 6A tends to form anionic charges, and they generally exhibit a -2 oxidation state in compounds. This makes them perfect candidates to combine with positively charged elements like aluminum. Oxygen is the most abundant element in this group, while the heavier members such as selenium and tellurium are less common and generally have special uses in industries. The chemical properties of Group 6A elements make them vital in the formation of various important compounds in nature and technology, including oxides and sulfides.

Mole Ratio

The mole ratio in chemistry is a critical concept that helps compare the amounts of different substances in a chemical reaction. It is derived from the balanced equation and is used to determine the proportion of reactants needed to form a product. This ratio helps predict the quantities of substances consumed and produced in a reaction.
In the context of forming an ionic compound from aluminum and a Group 6A element, the mole ratio helps establish the stoichiometry and the empirical formula of the resulting compound. Knowing the mole ratio allows chemists to use calculative tactics to identify unknown reactants in compounds, such as through percentage and molar mass calculations, to deduce the correct proportions and identity of the elements involved.

Empirical Formula

The empirical formula of a compound is a simple chemical formula that shows the simplest whole-number ratio of the atoms within a compound. It doesn't indicate the actual number of atoms present, which is provided by the molecular formula, but rather the relative proportions.
Determining the empirical formula is particularly useful in identifying unknown compounds through analytical chemistry. In the given exercise, where aluminum forms an ionic compound with a Group 6A element, the empirical formula provides insight into the stoichiometric balance achieved between the positively charged aluminum atoms and the negatively charged atoms of the unknown element. This can typically be calculated using mass percentage data and the molar masses of the involved elements, as shown in the step by step solution.