Problem 112 - General Chemistry

Question

The edge length of the unit cell of tantalum metal, Ta, is \(330.6 \mathrm{pm} ;\) the unit cell is body-centered cubic (one atom at each lattice point). Tantalum has a density of \(16.69 \mathrm{~g} / \mathrm{cm}^{3} .\) What is the mass of a tantalum atom? Use Avogadro's number to calculate the atomic mass of tantalum.

Step-by-Step Solution

Answer

The atomic mass of tantalum is approximately 181.2 g/mol.

Step 1: Calculate Volume of Unit Cell

The edge length of the unit cell is given as 330.6 pm. Convert this to centimeters, since density is given in g/cm³. 330.6 pm = 330.6 x 10^{-12} m = 330.6 x 10^{-10} cmThe volume of the cubic unit cell can be calculated using the formula for the volume of a cube, \(V = a^3\), where \(a\) is the edge length.\[V = (330.6 \times 10^{-10})^3\]\[V = 3.609 \times 10^{-23} \text{ cm}^3\]

Step 2: Determine Mass of Unit Cell

Since density \(\rho\) is mass per unit volume, use the formula \(\text{Mass} = \rho \times \text{Volume}\) to find the mass of the unit cell.Given \(\rho = 16.69 \text{ g/cm}^3\), and using the volume calculated earlier:\[\text{Mass of unit cell} = 16.69 \times 3.609 \times 10^{-23}\]\[= 6.023 \times 10^{-22} \text{ g}\]

Step 3: Understand Composition of the Unit Cell

A body-centered cubic (BCC) unit cell contains a total of 2 atoms: one at the center and one-eighth of an atom at each of the 8 corners (since an atom is shared by 8 unit cells).\[\text{Total atoms per unit cell} = 1 + \frac{1}{8} \times 8 = 2\]

Step 4: Calculate Mass of One Tantalum Atom

From Step 2, we know the mass of the unit cell is \(6.023 \times 10^{-22} \text{ g}\). As found in Step 3, the unit cell contains 2 tantalum atoms.Thus, the mass of one tantalum atom is:\[\text{Mass of one atom} = \frac{6.023 \times 10^{-22}}{2}\]\[= 3.0115 \times 10^{-22} \text{ g}\]

Step 5: Convert Mass of Atom to Molar Mass Using Avogadro's Number

Avogadro's number is \(6.022 \times 10^{23}\) atoms/mol. To convert the mass of one atom to their molar mass:\[\text{Molar mass} = 3.0115 \times 10^{-22} \text{ g/atom} \times 6.022 \times 10^{23} \text{ atoms/mol}\]\[= 181.2 \text{ g/mol}\]

Step 6: Final Step: Verification

Double-check each calculation to ensure accuracy and correct any rounding errors. Confirm that the steps follow logically from one step to the next for consistency.

Key Concepts

Body-Centered Cubic Structure

A fascinating aspect of crystallography is understanding how atoms are arranged in space. One common structure is the body-centered cubic (BCC).
In a BCC lattice, an atom is positioned at each corner of a cube, while an additional atom sits centrally within the cube. This unique positioning ensures stability and density attributes peculiar to BCC structures.
This specific configuration allows us to calculate the number of atoms within the unit cell—there are effectively 2 atoms per unit cell. Specifically, the center atom contributes entirely to the single unit cell, while the 8 corner atoms contribute collectively as one (since each is shared among 8 adjoining cells).
The BCC structure is therefore crucial in calculating properties such as mass when such lattice parameters as density and edge length are known.

Unit Cell Volume

The unit cell's volume is a fundamental concept in solid-state chemistry, providing insights into atomic spacing and density. Calculating the volume of a cubic unit cell involves using its edge length. For tantalum, the given edge length is 330.6 pm (picometers).
To convert it into a more usable unit, such as centimeters, we apply the conversion factor (1 pm = 1x10^{-10} cm). This conversion helps maintain consistency, especially when density is expressed in grams per cubic centimeter (g/cm³).
The formula for volume is straightforward: \(V = a^3\), where \(a\) is the edge length. For tantalum, \(V = (330.6 \times 10^{-10})^3\), resulting in a volume of approximately \(3.609 \times 10^{-23} \, \text{cm}^3\).
Accurately calculating this volume is essential for further computations, such as determining the mass and density of the metal.

Density Calculation

Understanding density is crucial for material characterization. Density is defined as mass per unit volume. It's the key parameter that links the atomistic structure to measurable macroscopic properties like weight and volume.
In our scenario with tantalum, the density is given as \(16.69 \, \text{g/cm}^3\). Using the unit cell volume calculated earlier, the formula for mass \(\text{mass} = \rho \times \text{volume}\) becomes indispensable.
For tantalum, multiplying the density \(16.69\) by the unit cell volume \(3.609 \times 10^{-23} \, \text{cm}^3\) gives us the mass of the unit cell, found to be \(6.023 \times 10^{-22} \, \text{g}\).
This calculated mass can then be divided by the number of atoms in the cell to deduce the atomic mass, providing a systematic approach to understanding material density in solid-state applications.

Avogadro's Number

Avogadro's number is a fundamental constant in chemistry, symbolizing the number of constituent particles, usually atoms or molecules, that are contained in one mole of a substance. This constant is valued at approximately \(6.022 \times 10^{23}\).
It helps link the atomic world to the macroscopic world we can measure directly. In the context of tantalum, Avogadro's number is used to calculate the molar mass from the atomic mass derived earlier.
By taking the atomic mass of a single tantalum atom \(3.0115 \times 10^{-22} \, \text{g/atom}\) and multiplying it by Avogadro's number, we convert to molar mass: \(\text{molar mass} = 3.0115 \times 10^{-22} \, \text{g/atom} \times 6.022 \times 10^{23} \, \text{atoms/mol}\).
The outcome, \(181.2 \, \text{g/mol}\), offers a comprehensive picture of tantalum's atomic scale and its relation to bulk properties observed in the laboratory.