Problem 150 - Chemistry

Question

Spinel is a mineral that contains 37.9\(\%\) aluminum, 17.1\(\%\) magnesium, and 45.0\(\%\) oxygen, by mass, and has a density of 3.57 \(\mathrm{g} / \mathrm{cm}^{3} .\) The edge of the cubic unit cell measures 809 \(\mathrm{pm} .\) How many of each type of ion are present in the unit cell?

Step-by-Step Solution

Answer

In the cubic unit cell of Spinel mineral, there are 6 aluminum ions, 3 magnesium ions, and 12 oxygen ions.

Step 1: Calculate the mass of one unit cell

To find the mass of one unit cell, we first need to find the volume of the cubic unit cell, and then multiply it by the density of the mineral. The edge of the cubic unit cell measures 809 pm, which is equal to \(8.09 \times 10^{-8}\) cm. The volume of a cube is given by the formula: \(V = a^3\), where \(a\) is the edge length of the cube. The volume of the unit cell is: \(V = (8.09 \times 10^{-8} \text{ cm})^3 = 5.305 \times 10^{-22} \text{ cm}^3\) Now, we can calculate the mass of the unit cell using the formula: \(M = V \times d\), where \(M\) is the mass of the unit cell, \(V\) is its volume, and \(d\) is the density of the mineral. The mass of the unit cell is: \(M = 5.305 \times 10^{-22} \text{ cm}^3 \times 3.57\frac{\text{g}}{\text{cm}^3} = 1.893 \times 10^{-21}\) g.

Step 2: Find the total number of ions in one unit cell

To find the total number of ions present in one unit cell, we will use the fact that the ionic compound Spinel is formed by the combination of aluminum ions (\(\text{Al}^{3+}\)), magnesium ions (\(\text{Mg}^{2+}\)), and oxygen ions (\(\text{O}^{2-}\)). Thus, the ratio of charges should be equal, i.e., \(\text{Al}^{3+}\) and \(\text{Mg}^{2+}\) together have a net positive charge equal to the net negative charge of \(\text{O}^{2-}\) ions. Let \(n_{Al}\), \(n_{Mg}\), and \(n_O\) be the number of aluminum, magnesium, and oxygen ions in the unit cell, respectively. We have: \(3n_{Al}+2n_{Mg}=2n_O\) Using the mass percentage of Aluminum (\(37.9\%\)), Magnesium (\(17.1\%\)), and Oxygen (\(45.0\%\)): \(0.379M=27n_{Al}\) \(0.171M=24n_{Mg}\) \(0.450M=16n_O\) Now we will find the coefficients mentioned above by dividing each equation by the mass of one unit cell: \(n_{Al}=\frac{0.379M}{27M}=\frac{0.379}{27}\) \(n_{Mg}=\frac{0.171M}{24M}=\frac{0.171}{24}\) \(n_O=\frac{0.450M}{16M}=\frac{0.450}{16}\) Now let's calculate the sum of \(n_{Al}\), \(n_{Mg}\), and \(n_O\) for one unit cell: \(n_{total}=n_{Al}+n_{Mg}+n_O=\frac{0.379}{27}+\frac{0.171}{24}+\frac{0.450}{16}=1\) Thus, there is a total of 1 ion per unit cell.

Step 3: Determine the number of each type of ion

Now that we know there is a total of 1 ion per unit cell, we can calculate the number of each type of ion using the previously calculated coefficients: \(n_{Al}=1 \times \frac{0.379}{27} = 0.014\) \(n_{Mg}=1 \times \frac{0.171}{24} = 0.007\) \(n_O=1 \times \frac{0.450}{16} = 0.028\) Note that these numbers are not integers, and we need an integer number of ions present in the unit cell. To find that, we'll multiply all the calculated values by their least common multiple (lcm): lcm(27, 24, 16) = 432 \(n'_{Al} = n_{Al} \times 432 = 6\) \(n'_{Mg} = n_{Mg} \times 432 = 3\) \(n'_O = n_O \times 432 = 12\) Therefore, there are 6 aluminum ions, 3 magnesium ions, and 12 oxygen ions present in the cubic unit cell of Spinel mineral.

Key Concepts

Aluminum Ions

Aluminum ions are a significant component in many ionic compounds and are characterized by their charge of +3, denoted as \(\text{Al}^{3+}\). This charge results from aluminum losing three electrons to achieve a stable electron configuration similar to the noble gas neon. Aluminum's role in ionic compounds, such as Spinel, is crucial because of its ability to balance the overall charge. In the context of the mineral Spinel, aluminum ions contribute significantly to the stability of the crystal lattice by providing positive charges that counterbalance the negative charges from oxygen ions. This balance is essential for maintaining the integrity of the compound's structure. When considering the ionic structure, it's important to calculate the proportion of aluminum ions based on the compound's composition and the mass percentages provided. When solving problems involving aluminum ions in unit cells, ensure to correctly determine their number using their mass percentage and the overall mass of the unit cell.

Magnesium Ions

Magnesium ions, denoted as \(\text{Mg}^{2+}\), play a fundamental role in forms of ionic compounds like Spinel. With a charge of +2, magnesium ions lose two electrons, favoring a stable electronic configuration similar to neon. This charge consistency supports their interaction with negatively charged ions, such as oxygen, to form neutral compounds. In the context of Spinel, magnesium ions contribute to the overall positive charge balance required to stabilize the mineral's crystalline structure. Calculating the number of magnesium ions involves considering their relative mass percentage and the unit cell's total mass. Understanding this concept is crucial when determining how ions are distributed within a crystalline unit cell. Accurate quantification based on provided chemical properties helps understand the spatial arrangement and mineral density, affecting how Spinel's structure is represented physically.

Oxygen Ions

Oxygen ions, often carrying a charge of -2 (\(\text{O}^{2-}\)), are pivotal in forming ionic compounds by providing necessary negative charges. This electron gain facilitates stability by mirroring the electron configuration of neon. Within mineral Spinel, oxygen ions balance the positive charges from aluminum and magnesium ions, maintaining neutrality. Oxygen's role and proportion need to be understood quantitatively; it’s often presented based on a compound's composition and mass percentages. The precise calculation of oxygen ions provides insights into the mineral's unit cell formation and aids in understanding its physical and chemical properties. Each oxygen ion shares robust bonds with surrounding cations, helping maintain the crystal lattice structure key to mineral stability.

Unit Cell Calculation

Unit cell calculation in crystallography is crucial for understanding the atomic arrangement in minerals like Spinel. A unit cell represents the smallest section of a lattice that visually and compositionally repeats across the crystal. The process starts by determining the geometric size of the unit cell, typically using the edge length. For cubic unit cells, such as those in Spinel, the volume \(V\) is calculated using \(a^3\), where \(a\) is the edge length. Understanding the volume is essential because it directly affects the material's density and the number of constituent ions. With the volume known, one can then apply the density to find the total mass, facilitating a complete analysis of the ionic composition within the cube. Through calculations involving mass percentages and relationships between different ions, one can discern the precise number of each ion type in the specified unit cell. This understanding is critical for predicting how minerals behave physically and chemically.

Density and Volume in Chemistry

Density and volume are foundational concepts in chemistry, especially when analyzing ionic compounds and their structures. Density \((d)\) relates mass \((m)\) to volume \((V)\) via the formula \(d = \frac{m}{V}\), providing insights into how tightly matter is compacted. This relationship is key when interpreting mineral structures like Spinel's, as density reflects how ions are packed into the crystal lattice. Volume, particularly in crystallography, describes the three-dimensional space occupied by the unit cell, essential for calculating the number of ions within it. Translating edge lengths into volume, using formulas like \(a^3\) for cubic cells, allows for precise measurement of properties on an atomic level. By understanding these calculations, students can predict the behavior of materials under various conditions and correlate ionic arrangements with physical properties such as hardness, stability, and reactivity in different environments.