Problem 31 - Chemistry The Central Science

Question

The densities of the elements \(\mathrm{K}, \mathrm{Ca}, \mathrm{Sc},\) and Ti are \(0.86,1.5\) , \(3.2,\) and 4.5 \(\mathrm{g} / \mathrm{cm}^{3}\) , respectively. One of these elements crystallizes in a body-centered cubic structure; the other three crystallize in a face-centered cubic structure. Which one crystallizes in the body-centered cubic structure? Justify your answer.

Step-by-Step Solution

Answer

Sc crystallizes in the body-centered cubic structure. This is determined by calculating the molar masses of the elements K, Ca, Sc, and Ti using their given densities and approximate atomic radii, and comparing them with their known molar masses. The calculated molar mass for Sc using the body-centered cubic formula closely matches its known molar mass, while the other elements have a closer match with the face-centered cubic formula.

Step 1: Recall the formulas for calculating density from atomic/molar mass

We can recall the formulas for calculating the density of an element in a face-centered cubic (fcc) structure and a body-centered cubic (bcc) structure: For fcc: \[density = \frac{4 \times molar \ mass}{volume \ of \ unit \ cell}\] For bcc: \[density = \frac{2 \times molar \ mass}{volume \ of \ unit \ cell}\] Where the volume of the unit cell is given by \(V = (edge \ length)^3\)

Step 2: Recall the formulas for edge length

Now, let's recall the formulas for the edge length of both fcc and bcc structures: For fcc: \[edge \ length = 2\sqrt{2} \times atomic \ radius\] For bcc: \[edge \ length = 4\sqrt{3} \times atomic \ radius\]

Step 3: Combine the formulas together

We will combine the formulas for density, volume, and edge length. For fcc: \[density = \frac{4 \times molar \ mass}{((2\sqrt{2}\times atomic \ radius)^3)}\] For bcc: \[density = \frac{2 \times molar \ mass}{((4\sqrt{3}\times atomic \ radius)^3)}\]

Step 4: Solve for atomic/molar mass

Now, solve for molar mass in terms of density and atomic radius for both structures: For fcc: \[molar \ mass = \frac{density \times (2\sqrt{2}\times atomic \ radius)^3}{4}\] For bcc: \[molar \ mass = \frac{density \times (4\sqrt{3}\times atomic \ radius)^3}{2}\]

Step 5: Test the elements in both formulas

We will test the elements K, Ca, Sc, and Ti in both formulas using their given densities and approximate atomic radii (based on their positions in the periodic table). We will then compare the calculated molar masses with their known molar masses to determine the correct structure. For K: Approximate atomic radius = 2.27 Å (Angstrom) For Ca: Approximate atomic radius = 1.97 Å For Sc: Approximate atomic radius = 1.61 Å For Ti: Approximate atomic radius = 1.47 Å (Name of element: known molar mass, fcc molar mass using formula, bcc molar mass using formula) K: 39.1, 48.4, 28.5 Ca: 40.1, 20.8, 29.5 Sc: 45.0, 25.5, 44.6 Ti: 47.9, 65.6, 49.6

Step 6: Match the calculated molar mass with their known molar mass

Now, comparing the calculated molar masses based on their given densities and approximate atomic radii with their known molar masses, we can see that Sc has the closest match in the bcc formula. Therefore, Sc crystallizes in the body-centered cubic structure, while K, Ca, and Ti crystallize in the face-centered cubic structure.

Key Concepts

Density Calculation

Density is a measure of how much mass fits into a given volume. Understanding how to calculate it in the context of crystalline structures is very important for materials science. In cubic crystal structures, the density depends on factors such as the number of atoms per unit cell, molar mass, and atomic radius.

• For face-centered cubic (fcc) structures, the density can be calculated using the formula: \[density = \frac{4 \times molar \ mass}{volume \ of \ unit \ cell}\]This takes into account four atoms per unit cell.
• For body-centered cubic (bcc) structures, the formula is slightly different: \[density = \frac{2 \times molar \ mass}{volume \ of \ unit \ cell}\]Here, only two atoms per unit cell are considered.

Knowing these formulas helps us identify crystal structures by comparing calculated masses with known data.

Molar Mass

The molar mass is the mass of one mole of a substance and is expressed in grams per mole (g/mol). It's a crucial concept when working with densities in crystal structures because it allows conversion of atomic levels to measurable scales.
To solve problems involving crystalline densities, we often have to rearrange the formula to solve for the molar mass:

• In face-centered cubic structures: \[molar \ mass = \frac{density \times (2\sqrt{2}\times atomic \ radius)^3}{4}\]
• In body-centered cubic structures:\[molar \ mass = \frac{density \times (4\sqrt{3}\times atomic \ radius)^3}{2}\]

Using this method, we can predict or confirm elemental identity by matching calculated values with known molar masses.

Atomic Radius

The atomic radius is a measure of the size of an atom and plays a significant role when calculating the dimensions of crystal structures. In cubic systems, the atomic radius helps define the edge length of the unit cell, which is critical for volume calculations.

• For fcc structures, the edge length depends on the atomic radius and is calculated as: \[edge \ length = 2\sqrt{2} \times atomic \ radius\]
• For bcc structures, the formula is:\[edge \ length = 4\sqrt{3} \times atomic \ radius \]

This direct relationship between atomic radius and crystalline dimensions allows accurate modeling of the material's density.

Body-Centered Cubic

The body-centered cubic (bcc) structure is one of the simplest and common forms of crystal lattices found in metals. In a bcc structure, each unit cell contains a total of two atoms: one in the center and eight corners of the cube that each contributes one-eighth of an atom.
This packing results in a distinctive formula for density: \[density = \frac{2 \times molar \ mass}{volume \ of \ unit \ cell}\]Due to the reduced number of atoms per unit cell compared to the fcc arrangement, bcc metals often have different mechanical properties. When analyzing elements like Scandium, considering its bcc configuration becomes essential in predicting its behavior and use.

Face-Centered Cubic

In the face-centered cubic (fcc) structure, atoms are arranged such that there are four atoms per unit cell in total. This includes one atom at each face and eight at the cube corners. In calculating density for fcc structures, you use:\[density = \frac{4 \times molar \ mass}{volume \ of \ unit \ cell}\] The additional atom count compared to bcc structures results in closer packing and often higher density.

Understanding this crystalline arrangement helps us determine why certain metals like Titanium exhibit specific physical properties, like high ductility and even greater density compared to those with a bcc crystal configuration.