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Nickel has a face-centered unit cell with all atoms at lattice points and an edge length of \(352.4 \mathrm{pm} .\) The density of metallic nickel is \(8.91 \mathrm{~g} / \mathrm{cm}^{3} .\) What is the mass of a nickel atom? From the atomic mass, calculate Avogadro's number.

Short Answer

Expert verified
Mass of a nickel atom: \(9.75 \times 10^{-23} \text{ g}\); Avogadro's number: \(6.02 \times 10^{23} \text{ atoms/mol}\).

Step by step solution

01

Determine Volume of the Unit Cell

The edge length of a face-centered cubic (FCC) unit cell is given as \( a = 352.4 \text{ pm} \). Convert this to cm: \( 352.4 \text{ pm} = 3.524 \times 10^{-8} \text{ cm} \). The volume of the cubic unit cell is \( V = a^3 = (3.524 \times 10^{-8} \text{ cm})^3 = 4.38 \times 10^{-23} \text{ cm}^3 \).
02

Calculate Mass of Unit Cell

The density (\( \rho \)) of nickel is \( 8.91 \text{ g/cm}^3 \). The mass of the unit cell is calculated using \( \text{mass} = \text{density} \times \text{volume} = 8.91 \times 4.38 \times 10^{-23} = 3.90 \times 10^{-22} \text{ g} \).
03

Calculate Number of Atoms per Unit Cell

In a face-centered cubic lattice, there are 4 atoms per unit cell.
04

Calculate Mass of a Single Nickel Atom

Divide the mass of the unit cell by the number of atoms per unit cell: \( \text{mass of a nickel atom} = \frac{3.90 \times 10^{-22}}{4} = 9.75 \times 10^{-23} \text{ g} \).
05

Determine Moles of Nickel

From the periodic table, the atomic mass of nickel is roughly \( 58.69 \text{ g/mol} \). This represents the mass of one mole of nickel atoms.
06

Calculate Avogadro's Number

Using the relation \( \text{molar mass} = \text{mass of one mole} = \text{mass of one atom} \times N_A \), where \( N_A \) is Avogadro鈥檚 number, we solve: \( 58.69 \text{ g/mol} = 9.75 \times 10^{-23} \text{ g} \times N_A \). Thus, \( N_A = \frac{58.69}{9.75 \times 10^{-23}} \approx 6.02 \times 10^{23} \text{ atoms/mol} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Face-Centered Cubic Lattice
In crystallography, the face-centered cubic (FCC) lattice is a common arrangement where atoms are positioned at each of the corners and on the center of each face of the cube. This lattice is structured in such a way that it has a high packing efficiency, with 4 atoms per unit cell.
Each corner atom is shared among eight unit cells, and each face-centered atom is shared between two unit cells. This makes for an efficient and densely packed structure. FCC lattices are prevalent in many metals like nickel, aluminum, and copper, contributing to their well-known properties like high density and strong bonds.
Understanding how atoms are arranged helps us predict and explain the physical properties of materials such as strength, flexibility, and conductivity.
Density of Metals
Density is a fundamental physical property of metals, often described as mass per unit volume. It is usually expressed in grams per cubic centimeter (g/cm鲁).
The density of a metal indicates how tightly atoms are packed in its crystallographic structure. Nickel, for instance, has a density of 8.91 g/cm鲁, signifying that its atoms are closely packed in its FCC lattice structure.
To determine the density, you can use the mass and volume of the unit cell, where density (\( \rho \)) equals mass divided by volume. Density affects many of the metal鈥檚 properties such as its strength, melting point, and how it will interact with other materials.
Atomic Mass
Atomic mass is the mass of an atom, typically expressed in atomic mass units (amu). It is sometimes referred to as atomic weight and represents the weighted average mass of an element's isotopes. For elements in a solid material, like metals, the atomic mass is vital in converting between the mass of a sample and the number of atoms it contains.
For example, the atomic mass of nickel is approximately 58.69 g/mol, meaning one mole of nickel atoms has a mass of 58.69 grams. This average is indispensable when calculating other chemical properties, such as the number of atoms present in a certain mass of the material.
Understanding atomic mass helps chemists and material scientists relate microscopic properties of atoms to the macroscopic properties of substances.
Unit Cell Volume
The unit cell volume is a measure of the size of the smallest repetitive volume unit of a crystal lattice. For a cubic unit cell, like those in FCC lattices, the volume is determined by cubing the edge length of the cell.
For example, in the problem mentioned, the unit cell of nickel has an edge length of 352.4 picometers (pm), which when converted to centimeters is 3.524 x 10鈦烩伕 cm. The volume is then calculated as \( V = a^3 = (3.524 \times 10^{-8} \text{ cm})^3 = 4.38 \times 10^{-23} \text{ cm}^3 \).
Unit cell volume is crucial for calculating the density and also for determining the arrangements of atoms within the material. By knowing the volume, you can deduce how compact the crystal lattice is and how the metallic bonds are distributed.

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Most popular questions from this chapter

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