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

Short answer

The formula of the ionic compound formed between Aluminum (Al) and the unknown element from Group 6A, which is identified as Tellurium (Te), is Al₂Te₃.

Step-by-step solution

Determine mass percent of the unknown element

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%.

Calculate molar mass of the unknown element

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

Identify the unknown Group 6A element

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).

Determine the formula of the compound

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.

Key concepts

Ionic Compounds

Ionic compounds are formed when metals react with non-metals, resulting in the transfer of electrons from the metal to the non-metal. This electron transfer creates ions; a metal typically forms a positively charged cation, while a non-metal forms a negatively charged anion. The electrostatic force of attraction between these oppositely charged ions holds the compound together in a crystal lattice structure.
For example, in the exercise provided, aluminum (a metal) combines with a non-metal from Group 6A to form an ionic compound. In this case, aluminum loses electrons to form aluminum ions (Al³⁺), while the non-metal gains electrons to form anions (such as Te²⁻ from tellurium).
Understanding ionic compounds involves recognizing this transfer of electrons and the resulting ionic charges, leading to the characteristic high melting and boiling points and electrical conductivity when melted or dissolved in water.

• Electrostatic attraction: Keeps ions in fixed positions within the compound.
• High melting/boiling points: Due to strong ionic bonds.
• Electrical conductivity: In liquid or aqueous state, due to movement of ions.

This foundational knowledge is essential for predicting the properties and behaviors of ionic compounds.

Group 6A Elements

Group 6A of the periodic table is also known as the chalcogens. This group includes the elements oxygen (O), sulfur (S), selenium (Se), tellurium (Te), and polonium (Po). They are characterized by having six electrons in their outermost electron shell. This makes them relatively reactive, as they seek to gain two additional electrons to reach a stable octet configuration.
In chemical reactions, Group 6A elements mostly form negative ions with a charge of 2- (such as S²⁻, Se²⁻, and Te²⁻), due to their high electronegativity. As a result, they frequently form ionic compounds with metals, where the metal loses electrons and the 6A element gains them.
Typically, as you move down the group from oxygen to polonium, elements have increased atomic size and decreased electronegativity. Their metallic character also tends to increase, changing some properties dramatically from the top to the bottom of the group.

• Reactivity: Forms anions typically with a -2 charge.
• Electronegativity: Decreases down the group.
• Uses: For instance, sulfur is widely used in the industry; tellurium is used in electronics.

Comprehending the characteristics of Group 6A elements is crucial for predicting their behavior in compounds.

Aluminum Chemistry

Aluminum is one of the most abundant metals on Earth and plays a critical role in chemical reactions, especially in forming ionic compounds. It is a member of Group 13 in the periodic table and typically displays a +3 oxidation state in its compounds. Aluminum atoms have three outer shell electrons, which it readily loses to achieve a more stable electron configuration.
When reacting with non-metals, aluminum forms cations (Al³⁺) by losing three electrons. This characteristic makes aluminum ideal for forming ionic compounds, as seen in the formation of Al₂Te₃ in the exercise.
Aluminum is known for its ability to form strong, yet lightweight alloys, making it crucial in many industrial applications, from aerospace to packaging.

• Common oxidation state: +3, forming Al³⁺ ions.
• Industrial Uses: Aluminum alloys are crucial in construction and transportation.
• Lightweight: High strength-to-weight ratio beneficial in various applications.

Understanding aluminum's chemical properties enhances comprehension of its crucial role in metallurgy, chemistry, and real-world applications.