Question

Calcium crystallizes in a face-centered cubic unit cell at room temperature that has an edge length of \(558.8 \mathrm{pm}\). (a) Calculate the atomic radius of a calcium atom. (b) Calculate the density of Ca metal at this temperature.

Short answer

The atomic radius of a calcium atom is approximately 192.5 pm, and the density of Ca metal at room temperature is approximately 1.54 g/cm³.

Step-by-step solution

Calculate the atomic radius using the edge length

In a face-centered cubic unit cell, the atoms are located at the corners and face centers of the cube. The relationship between the atomic radius (r) and the edge length (a) can be determined by using simple geometry. For a face-centered cubic structure, the diagonal of the cube can be defined in terms of atomic radius: \[Diagonal = 4r\] Moreover, the diagonal of the cube can also be defined using Pythagorean theorem: \[Diagonal = \sqrt{a^2 + a^2 + a^2} = \sqrt{3a^2}\] Now, equating both expressions for the diagonal and solving for the atomic radius: \[4r = \sqrt{3a^2}\] \[r = \frac{\sqrt{3a^2}}{4}\] Given the edge length a = 558.8 pm, we can calculate the atomic radius: \[r = \frac{\sqrt{3(558.8 \mathrm{pm})^2}}{4}\]

Calculate the atomic radius of calcium

Calculate r using the equation found in step 1: \[r \approx \frac{\sqrt{3(558.8 \mathrm{pm})^2}}{4} = 192.5 \mathrm{pm}\] The atomic radius of a calcium atom is approximately 192.5 pm.

Calculate the density of Ca metal

To calculate the density of calcium metal, we need the mass of one unit cell and the volume of the unit cell. - The mass of one unit cell can be found using the atomic mass of calcium, the Avogadro's number, and the number of atoms in a face-centered cubic unit cell (which is equal to 4). \[Mass_{unit \thinspace cell} = \frac{4 \times Atomic \thinspace Mass_{Ca}}{Avogadro's \thinspace Number}\] - The volume of the unit cell can be calculated using the cube edge length (a): \[Volume_{unit \thinspace cell} = a^3\] Finally, the density can be computed as follows: \[Density = \frac{Mass_{unit \thinspace cell}}{Volume_{unit \thinspace cell}}\]

Calculate the mass of one unit cell

The atomic mass of calcium (Ca) is 40.08 g/mol, and the Avogadro's number is \(6.022 \times 10^{23}\) atoms/mol. Calculate the mass of one unit cell: \[Mass_{unit \thinspace cell} = \frac{4 \times 40.08 \thinspace g/mol}{6.022 \times 10^{23}\thinspace atoms/mol} \approx 2.672 \times 10^{-22} \thinspace g\]

Calculate the volume of the unit cell

Calculate the volume of the unit cell using the cube edge length a = 558.8 pm (note that 1 pm = \(10^{-12}\) m): \[Volume_{unit \thinspace cell} = (558.8 \times 10^{-12} \thinspace m)^3 = 1.74 \times 10^{-28} \thinspace m^3\]

Calculate the density of Ca metal

Calculate the density of Ca metal using the mass and volume of the unit cell: \[Density = \frac{2.672 \times 10^{-22} \thinspace g}{1.74 \times 10^{-28} \thinspace m^3} \approx 1.54 \thinspace g/cm^3\] (Don't forget to convert the units from g/m³ to g/cm³ with the conversion factor \(10^6\)) The density of Ca metal at room temperature is approximately 1.54 g/cm³.

Key concepts

Atomic Radius in a Face-Centered Cubic Unit Cell

Calcium crystallizes in what's called a face-centered cubic (FCC) unit cell. These unit cells are like tiny building blocks that have atoms positioned at each cube corner and in the middle of each cube face. In an FCC unit cell, the relationship between the edge length, denoted as \(a\), and the atomic radius \(r\) is cleverly determined through geometric principles involving the cube's diagonal.
This diagonal is equivalent to four times the atomic radius, or mathematically expressed as:

• \(Diagonal = 4r\)

Similarly, using the Pythagorean theorem, the diagonal also equals \(\sqrt{3a^2}\). By equating these two formulas, we derive:

• \(r = \frac{\sqrt{3a^2}}{4}\)

Given that calcium's FCC edge length \(a\) is 558.8 picometers (pm), we can plug this value into the equation:

• \(r \approx \frac{\sqrt{3(558.8\, \text{pm})^2}}{4} = 192.5\, \text{pm}\)

This helps us find the atomic radius of a calcium atom, which is approximately 192.5 pm.

Density Calculation of Calcium Metal

Density is an intriguing property that tells us how much mass is present in a certain volume. For calcium metal crystallized in an FCC unit cell, calculating its density requires two essential pieces: the mass of a unit cell and its volume.
The number of atoms per unit cell in an FCC structure is 4. The atomic mass of calcium is 40.08 grams per mole, and Avogadro’s number, the famous constant that relates moles to individual atoms, is \(6.022 \times 10^{23}\) atoms/mol. By combining these values, we find:

• \(\text{Mass}_{\text{unit cell}} = \frac{4 \times 40.08 \, \text{g/mol}}{6.022 \times 10^{23} \, \text{atoms/mol}} \approx 2.672 \times 10^{-22} \, \text{g}\)

Next, calculate the volume using the unit cell's edge length \(a\) of 558.8 pm or \(558.8 \times 10^{-12} \, \text{m}\):

• \(\text{Volume}_{\text{unit cell}} = (558.8 \times 10^{-12} \, \text{m})^3 = 1.74 \times 10^{-28} \, \text{m}^3\)

Finally, determine the density by dividing the mass by the volume:

• \(\text{Density} = \frac{2.672 \times 10^{-22} \, \text{g}}{1.74 \times 10^{-28} \, \text{m}^3} \approx 1.54 \, \text{g/cm}^3\)

Thus, the density of calcium metal at room temperature is about \(1.54\, \text{g/cm}^{3}\).

Unique Properties of Calcium Metal

Calcium is an abundant element that plays many roles both in technology and in biological systems. As a metal, calcium is soft with a silvery appearance, offering distinct characteristics that set it apart from other metals.

• Calcium is an alkaline earth metal with atomic number 20, denoting its position in the periodic table.
• Its role in the formation of face-centered cubic structures is a typical trait among metals due to efficient packing.
• Calcium exhibits low density compared to other metals; it is lightweight yet possesses notable strength.
• Beyond its mechanical properties, calcium's reactivity is significant, important for applications like metallurgy and in the synthesis of other compounds.

Understanding calcium's structural attributes and its calculative density offers insights into its usefulness in various industrial processes and biochemistry. Its FCC unit cell arrangement enhances its utility as a lightweight metal, vital in different technological developments.