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A metal has a body-centered cubic lattice with a unit cell edge length of 2.866 Å (1 Å = 10-10m). The density of the metal is 7.87 g/cm2. What is the mass of an atom of this metal in grams? (1 m = 1012 pm). %3D

A metal has a body-centered cubic lattice with a unit cell edge length of 2.866 Å (1 Å = 10-10m). The density of the metal is 7.87 g/cm2. What is the mass of an atom of this metal in grams? (1 m = 1012 pm). %3D

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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**Question:** A metal has a body-centered cubic lattice with a unit cell edge length of 2.866 Å (1 Å = 10^-10 m). The density of the metal is 7.87 g/cm³. What is the mass of an atom of this metal in grams? (1 m = 10^12 pm). **Explanation:** This question involves calculating the mass of an atom in a metal with a body-centered cubic (BCC) structure. ### Key Concepts: - **Body-Centered Cubic (BCC) Lattice:** In a BCC lattice, each unit cell contains two atoms. One atom is at the center of the cube, and the corners of the cube each contribute 1/8th of an atom to the unit cell. - **Unit Cell Edge Length:** The length of the cube's edge is given as 2.866 Å. - **Density:** The density of the metal is 7.87 g/cm³. - **Atomic Mass Calculation:** To find the mass of an atom, use the formula for density and the fact that density = mass/volume. ### Steps to Solve: 1. **Convert Edge Length:** Convert edge length from Ångströms to centimeters. \(2.866 \, \text{Å} = 2.866 \times 10^{-8} \, \text{cm}\) 2. **Calculate Volume of Unit Cell:** Volume = edge length\(^3 = (2.866 \times 10^{-8} \, \text{cm})^3\) 3. **Use Density Formula:** Density = Mass / Volume \(7.87 \, \text{g/cm}^3 = \text{Mass} / \text{Volume of unit cell}\) Solve for Mass of the unit cell. 4. **Calculate Mass of Single Atom:** Since there are two atoms per unit cell in a BCC lattice, divide the mass of the unit cell by 2 to find the mass of a single atom. This process will yield the mass of an individual atom in grams.
Transcribed Image Text:**Question:** A metal has a body-centered cubic lattice with a unit cell edge length of 2.866 Å (1 Å = 10^-10 m). The density of the metal is 7.87 g/cm³. What is the mass of an atom of this metal in grams? (1 m = 10^12 pm). **Explanation:** This question involves calculating the mass of an atom in a metal with a body-centered cubic (BCC) structure. ### Key Concepts: - **Body-Centered Cubic (BCC) Lattice:** In a BCC lattice, each unit cell contains two atoms. One atom is at the center of the cube, and the corners of the cube each contribute 1/8th of an atom to the unit cell. - **Unit Cell Edge Length:** The length of the cube's edge is given as 2.866 Å. - **Density:** The density of the metal is 7.87 g/cm³. - **Atomic Mass Calculation:** To find the mass of an atom, use the formula for density and the fact that density = mass/volume. ### Steps to Solve: 1. **Convert Edge Length:** Convert edge length from Ångströms to centimeters. \(2.866 \, \text{Å} = 2.866 \times 10^{-8} \, \text{cm}\) 2. **Calculate Volume of Unit Cell:** Volume = edge length\(^3 = (2.866 \times 10^{-8} \, \text{cm})^3\) 3. **Use Density Formula:** Density = Mass / Volume \(7.87 \, \text{g/cm}^3 = \text{Mass} / \text{Volume of unit cell}\) Solve for Mass of the unit cell. 4. **Calculate Mass of Single Atom:** Since there are two atoms per unit cell in a BCC lattice, divide the mass of the unit cell by 2 to find the mass of a single atom. This process will yield the mass of an individual atom in grams.
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