Cadmium sulfide (CdS) has a cubic unit cell, and from x-ray diffraction data it is known that the cell edge length is \(0.582 \mathrm{~nm}\). If the measured density is \(4.82 \mathrm{~g} / \mathrm{cm}^{3}\), how many \(\mathrm{Cd}^{2+}\) and \(\mathrm{S}^{2-}\) ions are there per unit cell?

X-ray diffraction (XRD) is a powerful tool used to study the structure of crystalline materials. By measuring the angles and intensities of X-rays scattered by a crystal, we can determine the distances between planes of atoms within the crystal, known as lattice spacings. The data obtained from XRD can be used to calculate the size of the unit cell, which is the smallest repeating unit within the crystal structure that defines its symmetry and structure.

In the case of Cadmium sulfide (CdS), XRD data revealed that its unit cell is cubic and have a cell edge length, which we then use in our calculations. Our understanding of the crystalline structure directly impacts our ability to determine how many ions of each type are present within a single unit cell of CdS. By assessing the X-ray diffraction data, we gathered essential information about the atomic arrangement, which sets the stage for further calculations regarding the crystal's density and composition.

Breaking Down Crystal Density

The density of a crystal is a measure of how much mass is contained in a given volume of the material. For crystalline solids like Cadmium sulfide, the mass of a single unit cell can be calculated using the density along with the volume of the unit cell.

To perform a crystal density calculation, first, the volume of the unit cell (V) must be obtained using the unit cell edge length acquired from X-ray diffraction data. The formula for the volume of a cubic unit cell is simple:
V = a^3, where a is the cell edge length. Then, the mass of the unit cell can be found by multiplying its volume by the measured density of the substance.

Applying the Calculation

Once the mass is determined, it can reveal how many moles of the substance compose a single unit cell, taking us one step closer to unlocking the number of ions within that space. This step is crucial for students to understand because it directly correlates the macroscopic property of density to the microscopic arrangement of ions in a crystal.

Avogadro's number, which is approximately 6.022 x 10^23, represents the number of particles in one mole of a substance. Applying Avogadro's number allows us to transition from the scale of moles, which is convenient for chemical reactions and bulk measurements, to actual counts of ions or atoms, providing a link between the macroscopic and microscopic worlds.

In our Cadmium sulfide unit cell example, we start with the measured mass of the unit cell obtained from the crystal density calculation. To pinpoint the number of ions in each unit cell, we divide this mass by the molar mass of CdS. This step yields the number of moles of CdS within the unit cell. We then multiply the total number of moles by Avogadro's number to get the number of ions, which needs to be rounded to the nearest whole number since it’s not possible to have a fraction of an ion.

It is essential for students to grasp the concept that Avogadro's number serves as a bridge between a quantity that we can measure in the laboratory (moles) and the actual number of discrete ions in a substance. This concept is not only fundamental for solving problems involving unit cell calculations but also is foundational to all of chemistry.