Question

Calcium has a cubic closest packed structure as a solid. Assuming that calcium has an atomic radius of \(197 \mathrm{pm},\) calculate the density of solid calcium.

Short answer

In a cubic closest packed structure, atomic radius (r) is related to the unit cell edge length (a) by \(a = 2\sqrt{2}r = 556.16 \, \mathrm{pm}\). The volume of the unit cell (V) is \(a^3 = 1.719 \times 10^{8}\, \mathrm{pm}^3 = 1.719 \times 10^{-6} \, \mathrm{cm}^3\). In this arrangement, there are 4 atoms per unit cell, with a total mass of \(2.662 \times 10^{-22} \, \mathrm{g}\). From the mass and volume, the density (ρ) of solid calcium is \(\rho = \frac{M}{V} = 1.548 \, \mathrm{g/cm^3}\).

Step-by-step solution

Calculate the length of a unit cell

In a ccp structure, the atoms are arranged in a face-centered cubic (fcc) arrangement, which means that the atoms are present at the corners and the center of each face of the cube. The relationship between the atomic radius (r) and the unit cell edge length (a) in an fcc structure is given by the following equation: \[a = 2\sqrt{2}r\] We are given the atomic radius of calcium (r) as 197 pm. Now, we can calculate the length of a unit cell (a) for calcium: \[a = 2\sqrt{2}(197 \, \mathrm{pm}) = 556.16 \, \mathrm{pm}\]

Calculate the volume of the unit cell

Now that we have the edge length (a) of the unit cell, we can calculate the volume (V) of the unit cell using the formula for the volume of a cube: \[V = a^3\] Substituting the edge length (556.16 pm) calculated in Step 1: \[V = (556.16 \, \mathrm{pm})^3 = 1.719 \times 10^8 \, \mathrm{pm}^3\]

Calculate the number of atoms per unit cell

In an fcc arrangement, there are 4 atoms per unit cell. This is because each of the 8 corners of the cube contains 1/8 of an atom, and each of the 6 faces contains 1/2 of an atom: \[\text{Number of atoms per unit cell} = 8 \times (\frac{1}{8}) + 6 \times (\frac{1}{2}) = 4\]

Calculate the mass of the atoms in the unit cell

To calculate the mass of the atoms in the unit cell, we need to find the molar mass of calcium and convert it to the mass of one atom. The molar mass of calcium is 40.08 g/mol. To find the mass of one atom, we can use Avogadro's number (6.022 x 10^23 atoms/mol): \[\text{Mass of one calcium atom} = \frac{40.08 \, \mathrm{g/mol}}{6.022 \times 10^{23} \, \mathrm{atoms/mol}} = 6.655 \times 10^{-23} \, \mathrm{g}\] Therefore, the mass of the atoms in the unit cell (4 calcium atoms) is: \[\text{Total mass in one unit cell} = 4 \times (6.655 \times 10^{-23} \, \mathrm{g}) = 2.662 \times 10^{-22} \, \mathrm{g}\]

Calculate the density of solid calcium

Finally, we can calculate the density (ρ) of solid calcium using the mass (M) and volume (V) of the unit cell: \[\rho = \frac{M}{V}\] Substituting the values obtained in Steps 2 and 4: \[\rho = \frac{2.662 \times 10^{-22} \, \mathrm{g}}{1.719 \times 10^8 \, \mathrm{pm}^3}\] To convert the volume from \(pm^3\) to \(cm^3\), we use the conversion factor \(1\, cm = 1\times 10^{10}\, pm\), and so: \[V = 1.719 \times 10^8 \, \mathrm{pm}^3 \times (\frac{1 \, \mathrm{cm}}{1\times 10^{10} \, \mathrm{pm}})^3 = 1.719 \times 10^{-6} \, \mathrm{cm}^3\] Now we can calculate the density: \[\rho = \frac{2.662 \times 10^{-22}\, \mathrm{g}}{1.719 \times 10^{-6}\, \mathrm{cm}^3} = 1.548 \, \mathrm{g/cm^3}\] Thus, the density of solid calcium is approximately 1.548 g/cm³.

Key concepts

Cubic Closest Packed Structure

The cubic closest packed (CCP) structure is a highly efficient way to arrange atoms in a solid state. It is also known as the face-centered cubic (FCC) structure.
In this arrangement, each atom is surrounded by 12 other atoms, forming a highly dense and stable configuration. The CCP structure ensures that the maximum amount of space is utilized within the crystal lattice.
Such structures are common in metals, because they offer a balance of density and structural integrity. Not only does this configuration support the strong metallic bonds, but it also allows more flexibility and ductility, enabling metals to be shaped without breaking.

• The atoms in a CCP structure are arranged at the corners and the face centers of the cube.
• This structure is characterized by its ability to pack atoms in a way that minimizes wasted space.

Understanding CCP is crucial when analyzing metal density and properties.

Atomic Radius

The atomic radius is an essential concept in solid-state chemistry. It defines the size of an atom from its nucleus to the outer boundary of its electron cloud. This measurement is crucial for understanding how atoms pack together in solid structures.
For calcium, the atomic radius is given as 197 pm (picometers). This radius influences how closely calcium atoms can approach each other, affecting the overall density and packing structure of calcium in its solid form.
The atomic radius can vary slightly depending on the number of surrounding atoms (or coordination number), because atoms can slightly expand or contract in response to their environment. When calculating the unit cell dimensions in a CCP structure, the atomic radius is fundamental.

• Atomic radius affects the edge length of the unit cell in crystalline structures.
• It provides insight into the space an atom occupies in a lattice.

Knowing the atomic radius helps predict interactions and properties in materials science.

Molar Mass of Calcium

The molar mass of calcium refers to the weight of one mole of calcium atoms. It is expressed in grams per mole (g/mol) and is critical for converting between atomic-scale and macroscopic measurements.
For calcium, the molar mass is 40.08 g/mol. This value is used to calculate the mass of a specific number of atoms, such as those in a unit cell of a crystalline solid.
To find the mass of a single atom, we divide the molar mass by Avogadro's number, which is approximately 6.022 x 10^23 atoms/mol. This conversion is vital when determining how much an individual unit of the crystal weighs, helping to calculate density.

• Molar mass allows us to bridge atomic-scale measurements with bulk material properties.
• It is necessary for calculating the mass of atoms within a crystal lattice.

Molar mass is thus a key step in understanding how individual atoms contribute to the overall material.

Face-Centered Cubic Arrangement

In a face-centered cubic (FCC) arrangement, atoms are located at each of the eight corners as well as at the center of each face of the cube. This results in a highly efficient and densely packed structure.
Calcium, as an FCC structure, features 4 equivalent atoms per unit cell. This structure is achieved by considering that each corner atom is shared among 8 adjacent unit cells, and each face-centered atom is shared between 2, effectively contributing a total of one atom from the faces.
The FCC arrangement is noticeable for its high packing efficiency, utilized by many metals to maintain strength while being malleable. It results in 74% of the space within the unit cell being occupied by atoms, leaving minimal unoccupied volume.

• The FCC arrangement provides a balance between high packing efficiency and good mechanical properties.
• It allows metals like calcium to be relatively soft and ductile.

Understanding this arrangement provides insight into why some metals are so adaptable to different forms and conditions.