Question

An element crystallizes in a body-centered cubic lattice. The edge of the unit cell is \(2.86 \AA\), and the density of the crystal is \(7.92 \mathrm{~g} / \mathrm{cm}^{3} .\) Calculate the atomic weight of the element.

Short answer

The atomic weight of the element in the given body-centered cubic lattice crystal is approximately 55.8 g/mol.

Step-by-step solution

Determine the number of atoms in a BCC unit cell

In a body-centered cubic lattice, there is one atom at each corner of the unit cell and one atom in the center. There are a total of 8 corners in a cube, and each corner atom is shared with 8 adjacent unit cells. Therefore, each unit cell effectively contains 1/8 of all 8 corner atoms. In addition, there is one full atom in the center. So, the total number of atoms in a BCC unit cell is: \(1 \times Center_{atom} + \left(\frac{1}{8}\times 8\right)Corner_{atoms}= 1+1 = 2\) atoms.

Calculate the volume of the unit cell

We are given that the edge length (a) of the unit cell is 2.86 Å. To calculate the volume (V) of the unit cell, we can use the formula for the volume of a cube: \(V = a^{3}\) Convert the edge length to cm: \(2.86 \AA = 2.86\times 10^{-8} cm\) Now, calculate the volume of the unit cell: \(V = (2.86\times 10^{-8} cm)^{3} = 2.340 \times 10^{-23} cm^{3}\)

Calculate the mass of one unit cell

We are given the density (D) of the crystal, which is 7.92 g/cm³. The mass (M) of one unit cell can be calculated using the formula: \( M = D \times V \) So, the mass of one unit cell is: \( M = (7.92\ g/cm^3) \times (2.340 \times 10^{-23}\ cm^3) = 1.852 \times 10^{-22}\ g \)

Calculate the mass of one atom

As we calculated in Step 1, there are 2 atoms in each unit cell. Therefore, the mass of one atom can be calculated by dividing the mass of the unit cell by the number of atoms: \(Mass_{atom} = \frac{Mass_{unit\ cell}}{Number_{of\ atoms}}\) \(Mass_{atom} = \frac{1.852\times 10^{-22}\ g}{2} = 9.26 \times 10^{-23}\ g\)

Calculate the atomic weight of the element

We can now calculate the atomic weight of the element. The atomic weight is the mass of one mole of atoms, so we need to find how many atoms are in one mole. There are \(6.022 \times 10^{23}\) atoms in one mole (Avogadro's Number). Therefore, the atomic weight (A) can be calculated as: \(A = Mass_{atom} \times Avogadro's\ Number\) \(A = (9.26\times 10^{-23}\ g) \times (6.022\times 10^{23}) = 55.8\ g/mol\)

Conclusion

The atomic weight of the element in the given crystal is approximately 55.8 g/mol.

Key concepts

Crystal Density Calculation

Crystal density is a crucial parameter when studying crystalline materials. It is defined as the mass of the unit cell divided by the volume of the unit cell. For the element crystallizing in a body-centered cubic (BCC) structure, the crystal density can allow us to understand how packed the atoms are in the material's lattice. In this exercise, we start with the given density of the crystal: 7.92 g/cm³.

To find the mass of one unit cell, we use the relationship between density (\(D\)), mass (\(M\)), and volume (\(V\)):

• \( M = D \times V \)

Here, \(M\) represents the mass of the unit cell. By substituting the given density and the calculated volume of the unit cell, we can solve for the mass. This relationship is vital because knowing the density allows us to derive further details about the atomic structure of the crystal.

Atomic Weight Determination

Determining the atomic weight of an element within a crystal, such as a body-centered cubic lattice, involves a few steps based on physical constants and calculated values from the structure. First, we need to understand that the atomic weight represents the mass of one mole of atoms of the element.

• Use Avogadro's Number: \(6.022 \times 10^{23}\) atoms/mol.

The relationship between the mass of a single atom and the atomic weight is found by multiplying the mass of one atom by Avogadro's Number. In our solution, we calculated the atomic weight using:

• \(A = Mass_{atom} \times Avogadro's\ Number\)

Where \(A\) is the atomic weight. This approach gives an atomic weight of approximately 55.8 g/mol for the element.

Unit Cell Volume Calculation

Calculating the volume of a unit cell in a crystal, such as a body-centered cubic (BCC) structure, is straightforward with the knowledge of the edge length. Given that each unit cell is a cube, and the edge length is provided (\(a = 2.86 \, Å)\), calculating the volume involves cubing this length:

• \(V = a^3\)

First, we must convert the edge length from angstroms to centimeters:

• \(1 \AA = 10^{-8} \text{ cm}\)
• \(2.86 \times 10^{-8} \text{ cm}\)

Using these conversions, we find:

• \(V = (2.86 \times 10^{-8} \text{ cm})^3\)
• \(V \approx 2.340 \times 10^{-23} \text{ cm}^3\)

This value gives us the precise spatial volume occupied by the unit cell, useful for calculations like mass and density.