Question

When aluminum metal is heated with an element from Group 6A 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 formula of the ionic compound is Al\(_2\)S\(_3\). To find this, we first calculated the mass percentages of aluminum and the unknown Group 6A element in the compound, which were 18.56% and 81.44%, respectively. Then, by converting these percentages to masses and moles, we determined the ratio of moles of Al to the moles of the unknown element. By plugging in the atomic masses of different Group 6A elements, we discovered that sulfur resulted in a whole-number ratio, and thus the empirical formula is Al\(_2\)S\(_3\).

Step-by-step solution

Calculate the mass of the unknown element in the compound

Since the compound is \(18.56 \%\) Al by mass, the unknown element must compose the remaining mass, which is \(100\% - 18.56\% = 81.44 \%\).

Convert mass percentages to mass values

Let's assume that we have a \(100 \thinspace g\) sample of this ionic compound. In this case, the mass of aluminum and the unknown element can be calculated by taking the respective percentages of the total mass: - Mass of Al = \((18.56 \%) \times 100 \thinspace g = 18.56 \thinspace g\) - Mass of unknown element = \((81.44 \%) \times 100 \thinspace g = 81.44 \thinspace g\)

Convert mass to moles

Divide the mass of each element by their respective molar mass to find the moles of each element. The molar mass of aluminum is approximately \(26.98 \thinspace g/mol\), and let the molar mass of the unknown element be X. - Moles of Al = \(\frac{18.56 \thinspace g}{26.98 \thinspace g/mol}\) - Moles of unknown element = \(\frac{81.44 \thinspace g}{X \thinspace g/mol}\)

Find the ratio of moles of Al to moles of unknown element

To find the empirical formula, we need to find the lowest whole number ratio of the moles of aluminum to the moles of the unknown element. Let's call this ratio n. n = \(\frac{\frac{18.56 \thinspace g}{26.98 \thinspace g/mol}}{\frac{81.44 \thinspace g}{X \thinspace g/mol}}\)

Identify the unknown element from Group 6A and find the empirical formula

The elements in Group 6A are oxygen (O, atomic mass = 16), sulfur (S, atomic mass = 32), selenium (Se, atomic mass = 79), tellurium (Te, atomic mass = 128), and polonium (Po, atomic mass = 209). The moles ratio (n) calculated in the previous step will be different for each of these elements based on their atomic masses. However, only one of these elements will yield a whole number ratio with the moles of aluminum. Plug in the atomic mass of each element in the value of X and calculate the value of n. The element that results in a whole number ratio is the unknown element. Once you have identified the unknown element, use the whole number ratio as the subscript for each element to represent the empirical formula of the ionic compound.

Key concepts

Periodic Table

The periodic table is a powerful tool that organizes all known chemical elements according to their atomic numbers, electron configurations, and recurring chemical properties. This layout helps us predict the behaviors of elements and their compounds.

One of its key features is the grouping of elements into vertical columns known as "groups" or "families." Elements within the same group often display similar chemical characteristics.

• For example, elements in Group 6A, also known as Group 16, include oxygen, sulfur, and selenium.
• These elements tend to form compounds with similar properties, which is very useful in predicting the outcomes of reactions, such as the one in our exercise with aluminum.

Understanding the placement of elements in the periodic table is crucial in solving chemical problems, like identifying unknown elements in a compound.

Chemical Formula

A chemical formula represents the composition of a compound, indicating the elements involved and their respective quantities.

In ionic compounds like those formed by aluminum and Group 6A elements, the chemical formula is determined through their charges balancing to neutrality. This involves the combination of cations and anions in whole number ratios.

• The formula illustrates the smallest integer ratio of ions that result in electrically neutral compounds.
• For example, if aluminum makes ions with a +3 charge and the Group 6A element forms -2 charged ions, the ratio might be adjusted to balance the charges.

Thus, finding the chemical formula involves calculating the simplest ratio of ions based on their respective charges.

Mass Percentage

Mass percentage is a way of expressing the concentration of a component in a compound. It calculates how much of the total mass of the compound is due to a specific element or component.

In the given exercise, we know that aluminum accounts for 18.56% of the compound’s mass. This helps us understand how much of the compound's mass the unknown Group 6A element contributes.

• This is calculated by taking 100% minus the known percentage of aluminum, resulting in 81.44% for the unknown element.
• Mass percentages help us determine empirical formulas by allowing us to convert these percentages into grams when considering a specific sample size; often 100 grams for simplicity.

By converting these masses to moles, we can find the mole ratio needed to write the empirical formula of the compound.

Group 6A Elements

Group 6A elements, also known as the chalcogens, include oxygen (O), sulfur (S), selenium (Se), tellurium (Te), and polonium (Po). These elements share certain characteristics due to their electron configuration.

• They tend to form -2 anions in ionic compounds, making them quite reactive, especially with metal elements like aluminum.
• Their atomic masses vary significantly, from oxygen's 16 g/mol to polonium's 209 g/mol.
• These elements can form diverse compounds, including oxides, sulfides, selenides, etc., as they're versatile in forming stable ionic and covalent bonds.

Identifying a Group 6A element in a compound involves using these properties, particularly their specific atomic masses, to calculate mole ratios and deduce the empirical formula.