Question

Barium has a body-centered cubic structure. If the atomic radius of barium is $222\mathrm{pm}$, calculate the density of solid barium.

Short answer

The density of solid barium can be calculated using the given atomic radius and following these steps: 1. Calculate the edge length 'a' from the atomic radius (r) using the formula $4r=\sqrt{3a^2}$, which results in $a=\sqrt{\frac{(4(222){)}^2}{3}}\approx 505.77\mathrm{pm}$. 2. Calculate the volume of the unit cell (V) as $V=a^3\approx (505.77{)}^3\approx 1.293\times {10}^8{\mathrm{pm}}^3$. 3. In a BCC structure, there are 2 atoms per unit cell, and the molar mass of barium (Ba) is 137.33 g/mol. Find the mass of one Ba atom using Avogadro's Number: $\frac{137.33\mathrm{g}/\mathrm{mol}}{6.022\times {10}^{23}\mathrm{atoms}/\mathrm{mol}}\approx 2.28\times {10}^{-22}\mathrm{g}/\mathrm{atom}$. 4. Calculate the mass of atoms in the unit cell: $2.28\times {10}^{-22}\mathrm{g}/\mathrm{atom}\times 2\approx 4.56\times {10}^{-22}\mathrm{g}$. 5. Calculate the density of solid barium (ρ) using the formula $\rho =\frac{\text{Mass}}{\text{Volume}}$, which results in $\rho =\frac{4.56\times {10}^{-22}\mathrm{g}}{1.293\times {10}^8{\mathrm{pm}}^3}\approx 3.53\mathrm{g}/{\mathrm{cm}}^3$. Therefore, the density of solid barium is approximately $3.53\mathrm{g}/{\mathrm{cm}}^3$.

Step-by-step solution

Determine the unit cell edge length from the atomic radius

Given the atomic radius of barium as $r=222\mathrm{pm}$, we need to find the edge length (a) of the unit cell for the body-centered cubic structure. In a bcc lattice, the body diagonal of the unit cell is equal to 4 times the atomic radius. The body diagonal can be found using the Pythagorean theorem in a 3D space, which is given by: $d=\sqrt{a^2+a^2+a^2}=\sqrt{3a^2}$ We know that $d=4r$, therefore $4r=\sqrt{3a^2}$ Now, we can calculate the value of the edge length 'a'.

Calculate the edge length 'a'

Given $4r=\sqrt{3a^2}$ and $r=222\mathrm{pm}$, we can find the value of edge length 'a' by substituting the value of r in the equation: $4(222)=\sqrt{3a^2}$ We now need to solve for 'a'. Square both sides of the equation: $(4(222){)}^2=3a^2$ Divide the equation by 3: $\frac{(4(222){)}^2}{3}=a^2$ Now calculate the value of 'a': $a=\sqrt{\frac{(4(222){)}^2}{3}}$

Calculate the volume of the unit cell

Now that we have the edge length 'a', we can calculate the volume of the unit cell as follows: Volume of the unit cell (V) = $a^3$

Determine the number of atoms per unit cell and their mass

In a body-centered cubic structure, there are 2 atoms per unit cell. The molar mass of barium (Ba) is 137.33 g/mol. To find the mass of one barium atom, we can use Avogadro's Number (N_a = $6.022\times {10}^{23}$ atoms/mol): Mass of one Ba atom = $\frac{137.33\mathrm{g}/\mathrm{mol}}{6.022\times {10}^{23}\mathrm{atoms}/\mathrm{mol}}$ To find the mass of atoms in the unit cell, multiply the mass of one barium atom by the number of atoms per unit cell (2). Mass of atoms in the unit cell = Mass of one Ba atom * 2

Calculate the density of solid barium

Finally, we can calculate the density of solid barium (ρ) using the mass of atoms in the unit cell and the volume of the unit cell. The density formula is given by: $\rho =\frac{\text{Mass}}{\text{Volume}}=\frac{\text{Mass of atoms in the unit cell}}{\text{Volume of the unit cell}}$ Substitute the values found in the previous steps and calculate the density of solid barium.

Key concepts

Atomic Radius

The atomic radius refers to the size of an atom, typically represented as the distance from the nucleus of the atom to the outermost electron orbit. It is an important value in material science as it determines how atoms pack together in a solid structure.

• For barium, the given value is 222 picometers (pm).
• This measurement helps us estimate the size of the unit cells in a crystal lattice.

Understanding atomic radius aids in calculating the dimensions of a unit cell in different crystalline structures, such as the body-centered cubic structure. By indicating how closely the atoms are packed, it affects the material's density and physical properties.

Body-Centered Cubic Structure

The body-centered cubic (bcc) structure is a type of crystal arrangement found in metals like barium. In this structure, each unit cell consists of an atom at each corner and one atom positioned at the center.

• Each corner atom is shared by eight neighboring cells, and the center atom belongs entirely to the cell.
• The body-centered cubic structure is characterized by high packing density, increasing the possible density of the material.
• There are two atoms in every bcc unit cell.

The bcc structure impacts how we calculate density since it dictates the arrangement of atoms, with symmetry across the unit cell helping determine the volume. This structure is vital in determining mechanical properties like strength and durability.

Unit Cell

A unit cell is the smallest repeating unit in a crystal lattice that embodies the entire structure's symmetry and properties. Understanding the dimensions and volume of the unit cell is critical for calculating density.

• The edge length ('a') is key to defining the unit cell's boundaries.
• Calculating the edge length in a bcc structure involves using the atomic radius and body diagonal relations.
• The volume is determined as the cube of the edge length: $$V=a^3$$

By comprehending the unit cell, you can derive the density by comparing the total atom mass to the cell volume, an essential step in material science.

Avogadro's Number

Avogadro's Number, $6.022\times {10}^{23}$, is one of chemistry's fundamental constants, representing the number of atoms or molecules in one mole of a substance. This constant is crucial when converting between atomic scale and macroscopic quantities.

• It allows for the calculation of the mass of a single atom, which is essential for determining the mass in a unit cell.
• For barium, given the molar mass, using Avogadro's number helps to find the mass of individual atoms.

Utilizing Avogadro's number, you can bridge atomic-level measurements with the bulk material properties, enabling the calculation of densities in crystal structures like bcc.