Question

An element crystallizes in a face centered cubic lattice and the edge of the unit cell is \(0.559\) \(\mathrm{nm}\). The density is \(3.19 \mathrm{~g} / \mathrm{cm}^{3}\). What is the atomic weight? (a) \(87.6\) (b) \(79.9\) (c) \(85.5\) (d) \(83.9\)

Short answer

(a) 87.6

Step-by-step solution

Calculate the volume of the unit cell

The edge length of the unit cell is given as 0.559 nm (nanometers). To calculate the volume of the unit cell, which is a cube, use the formula for the volume of a cube, which is side length cubed, \( V = a^3 \). First, convert the edge length from nanometers to centimeters to make it compatible with the density units: \(1 nm = 10^{-7} cm\). So, the edge length in centimeters is \(0.559 nm \times 10^{-7} cm/nm = 5.59 \times 10^{-8} cm\). Now, calculate the volume: \( V = (5.59 \times 10^{-8} cm)^3 = (5.59)^3 \times 10^{-24} cm^3\).

Calculate the number of atoms per unit cell

In a face-centered cubic (fcc) lattice, there are 4 atoms per unit cell. This is because each of the 8 corner atoms contributes 1/8th of an atom to the unit cell, and each of the 6 face atoms contributes 1/2 of an atom. So, the total number of atoms per unit cell in an fcc lattice is \(8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4 \).

Calculate the mass of the unit cell

The density (\(\rho\)) of the substance is given, which can be used to calculate the mass of the unit cell (\(m\)) using the formula \(\rho = \frac{m}{V}\), where \(V\) is the volume. Rearrange the formula to solve for mass: \(m = \rho \times V\). Density is given as 3.19 g/cm^3, and the volume of the unit cell has been calculated in step 1. Plugging in the values gives the mass.

Calculate the atomic weight of the element

The atomic weight can be found by dividing the mass of the unit cell by the number of atoms per unit cell, and then using Avogadro's number to convert it to grams/mole. The mass of the unit cell is calculated in step 3, and the number of atoms per unit cell was determined in step 2 to be 4. Use Avogadro's number, which is approximately \(6.022 \times 10^{23} atoms/mole\), to find the atomic weight: \( \text{Atomic Weight} = \frac{m}{4} \times \frac{1 mole}{6.022 \times 10^{23} atoms}\).

Perform the calculations

First calculate the volume of the unit cell in step 1, then calculate the mass using the density in step 3. Divide this mass by the number of atoms per unit cell from step 2. Finally, multiply by Avogadro's number to find the atomic weight of the element. Remember to keep track of units and to convert them as necessary.

Key concepts

Unit Cell Volume Calculation

Understanding the volume of a unit cell is essential in material science as it represents the smallest repeating unit of a crystal lattice. In a face-centered cubic (fcc) lattice, the unit cell is a cube. To calculate its volume, you simply raise the edge length to the third power. Here, the edge length is converted from nanometers to centimeters before calculation to correspond with density measurements, typically given in g/cm³.

It's crucial to recognize that all length measurements should be in the same unit when calculating volume, thus ensuring that the mass derived from the density will provide an accurate atomic weight later on. Remember, conversion factors, such as 1 nm = 10^-7 cm, are key in these calculations.

Atomic Weight Calculation

Calculating the atomic weight of an element from its crystal structure involves several intertwined concepts. Once we have the mass of a unit cell from the density and volume, we divide it by the number of atoms in the unit cell to get the mass of a single atom. In a face-centered cubic lattice, there are effectively 4 atoms per unit cell. With this individual atomic mass, we can then find the atomic weight by accounting for Avogadro's number, which allows conversion of mass per atom to molar mass (grams per mole).

This step requires precise handling of the units throughout the process. It's about transforming the scale from an individual atomic level within the lattice to the macroscopic level we use for measuring and comparing substances in chemistry - grams per mole.

Crystal Lattice Density

Key Factors Affecting Density

The density of a crystal lattice like that of a face-centered cubic (fcc) system is a measure of how tightly packed the atoms are within the unit cell. It's a critical property that influences the material's overall characteristics. Density is calculated using the mass of the unit cell and its volume.

This is why understanding the unit cell's geometry is so vital: the density gives us direct insights into the material's structure. Whether for practical applications like calculating how much force a material can withstand, or for theoretical pursuits such as determining atomic arrangements, the density is a foundational parameter in the study of crystalline materials.

Avogadro's Number Application

From Atoms to Moles

Avogadro's number — approximately 6.022 x 10^23 atoms per mole — is a bridge between the atomic scale and the observable world. When computing the atomic weight based on the mass of a unit cell and the number of atoms within that cell, Avogadro's number is used to convert from the atomic mass unit to grams per mole, the standard unit for molar mass.

Applying Avogadro's number is not just a mathematical exercise; it's an application of one of the fundamental principles of chemistry. By using this number, scientists can work with amounts of substance in a way that's relevant to the scales required for experimentation and industrial applications.