Question

An element crystallizes in a face-centered cubic lattice. The edge of the unit cell is 4.078$\hat{A}$ , and the density of the crystal is 19.30 $\mathrm{g}/{\mathrm{cm}}^3.$ Calculate the atomic weight of the element and identify the element.

Short answer

In a face-centered cubic lattice, there are 4 atoms per unit cell. The volume of the unit cell is 6.797 x 10^(-23) cm³, and its mass is 1.311 x 10^(-21) g. The mass of a single atom is 3.278 x 10^(-22) g, and its atomic weight is approximately 197.38 g/mol. This atomic weight corresponds to gold (Au).

Step-by-step solution

Determine the number of atoms per unit cell for an fcc lattice

In a face-centered cubic lattice, there is one atom at each corner and one atom at the center of each face. In total, the number of atoms per unit cell is given by: AtomsPerUnitCell = 8(corner atoms) x 1/8 (atoms shared by adjacent unit cells) + 6(face atoms) x 1/2 (atoms shared by neighboring cells) AtomsPerUnitCell = 1 + 3 = 4 Therefore, in an fcc lattice, there are 4 atoms per unit cell.

Calculate the volume of the unit cell

We are given that the edge length of the unit cell is 4.078Å. To get the volume, we simply need to cube the edge length: Volume = (EdgeLength)³ Volume = (4.078 Å)³ Since we need the answer in cubic centimeters, we have to convert Å to cm: 4.078 Å * (1 x 10^(-8) cm) = 4.078 x 10^(-8) cm Volume = (4.078 x 10^(-8) cm)³ = 6.797 x 10^(-23) cm³

Calculate the mass of the unit cell

We know the density (ρ) and volume (V) of the unit cell, so we can find its mass (m) with the following equation: m = ρ x V The density of the crystal is given as 19.30 g/cm³. m = (19.30 g/cm³) x (6.797 x 10^(-23) cm³) = 1.311 x 10^(-21) g

Calculate the mass of a single atom

Now that we have the mass of the unit cell, we can divide the mass by the number of atoms per unit cell to find the mass of a single atom: MassOfSingleAtom = MassOfUnitCell / AtomsPerUnitCell MassOfSingleAtom = (1.311 x 10^(-21) g) / 4 = 3.278 x 10^(-22) g

Calculate the atomic weight and identify the element

To find the atomic weight, we first need to calculate the moles (n) of the single atom. The Avogadro's number (N_A) is 6.022 x 10^(23) atoms/mol. Therefore: n = MassOfSingleAtom x N_A n = (3.278 x 10^(-22) g) x (6.022 x 10^(23) atoms/mol) = 197.38 g/mol The atomic weight of the element is approximately 197.38 g/mol. This atomic weight corresponds to gold (Au), with a standard atomic weight of 197.0 g/mol.

Key concepts

Crystal Density Calculation

Crystal density is a critical concept in materials science and solid-state physics. It refers to the mass of the crystal per unit volume. The density can reveal a lot about the crystal's composition and structure. To calculate the density, one can use the formula:

$$\rho =\frac{m}{V}$$
Where $\rho$ is the density, $m$ is the mass, and $V$ is the volume. For a face-centered cubic lattice, this calculation becomes particularly interesting because of the efficient packing of atoms.

In the provided exercise, we were given the density, and our goal was to reverse-engineer the problem to find the atomic weight. However, usually, if the atomic weight is known, the density can also be calculated by manipulating the previous equation to accommodate the number of atoms in a unit cell and their individual masses.

To improve understanding of this concept, imagine filling a box with uniform spheres until no additional spheres can fit without changing the box's size. The density calculation for a crystal mirrors this idea, except instead of spheres, the box, our unit cell, is filled with atoms.

Atomic Weight Determination

The atomic weight, or atomic mass, is fundamentally important in chemistry and physics. It is a value that tells us the average mass of atoms of an element, measured in atomic mass units (amu). To find the atomic weight from the density of a crystal, we first need to calculate the mass of a single atom within the crystal lattice.

As illustrated in our problem, once we have the mass of a single atom, we can use Avogadro's number to convert this mass to atomic weight. Avogadro's number, $6.022\times {10}^{23}$, is the number of particles in one mole of substance. The atomic weight is then represented as grams per mole (g/mol).

Understanding the route from the mass of a single atom to atomic weight involves grasping the concept of moles and Avogadro's constant. Simply put, if you know the mass of one atom and consider that one mole of any element contains the same number of atoms, you can extrapolate to find the mass of one mole of the element—its atomic weight.

Unit Cell Volume

The volume of the unit cell in a crystal lattice is the three-dimensional space that a single cell occupies. For a face-centered cubic lattice, the unit cell has a distinctive cubic shape, and hence the volume is found by cubing the edge length.

Let's recall the formula:$$V=a^3$$
Where $V$ is the volume and $a$ is the edge length of the cube. For our exercise, the unit cell's edge was given as $4.078\hat{A}$, or angstroms. Since volume is essentially 'how much space is taken up', understanding this calculation is like knowing how much water would fit inside a small cube-shaped container with each side of the length $a$.

To further improve comprehension, one must remember to convert the edge length from angstroms, a common unit for atomic scales, to centimeters, which is more convenient for volume calculations in a laboratory setting. Each angstrom is one ten-billionth of a meter ($1\hat{A}=1\times {10}^{-10}$ meters), a conversion necessary to align units when using density in grams per cubic centimeter (g/cm³).