   # The radius of gold is 144 pm, and the density is 19.32 g/cm 3 . Does elemental gold have a face-centered cubic structure or a body-centered cubic structure? ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 9, Problem 59E
Textbook Problem
45 views

## The radius of gold is 144 pm, and the density is 19.32 g/cm3. Does elemental gold have a face-centered cubic structure or a body-centered cubic structure?

Interpretation Introduction

Interpretation:

The lattice structure of Gold has to be identified and justified.

Concept introduction:

In packing of atoms in a crystal structure, the atoms are imagined as spheres. The two major types of close packing of the spheres in the crystal are – hexagonal close packing and cubic close packing.

In body-centered cubic unit cell, each of the six corners is occupied by every single atom. Center of the cube is occupied by one atom.

Each atom in the corner is shared by eight unit cells and a single atom in the center of the cube remains unshared. Thus the number of atoms per unit cell in BCC unit cell is,

8×18atomsincorners+1atomatthecenter=1+1=2atoms       The edge length of one unit cell is given bya=4R3where  a=edge length of unit cellR=radiusofatom

In face-centered cubic unit cell, each of the six corners is occupied by every single atom. Each face of the cube is occupied by one atom.

Each atom in the corner is shared by eight unit cells and each atom in the face is shared by two unit cells. Thus the number of atoms per unit cell in FCC unit cell is,

8×18atomsincorners+6×12atomsinfaces=1+3=4atoms The edge length of one unit cell is given bya=2R2where  a=edge length of unit cellR=radiusofatom

### Explanation of Solution

Explanation

Calculate the density of Gold by assuming its structure as BCC.

givendata:R=144pm=144×10-12ma=4R3=4×144×10-12m1.732=332.6pmvolume of unit cell=a3=(332.6×10-12)3m3=3.68×10-23cm3

Average mass of one gold atom=atomicmassofgoldAvogadronumber=197g6.022×1023=32.71×10-23gEachBCCunit cell contains2gold atoms. therefore,Massofaunitcell=2×32.71×10-23g  =65.42×10-23g

density=massvolume=65.42×10-23g3.68×10-23cm3=17.78g/cm3

The atomic radius of Gold is given. The unit cell is assumed as that of body-centered cubic and its edge length is calculated. Accordingly, the volume, mass and density of BCC unit cell are calculated. The obtained value does not agree with the actual value of density of gold.

Calculate the density of gold by assuming its structure as FCC.

givendata:R=144pm=144×10-12ma=2R2=2×144×10-12m×1

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

Find more solutions based on key concepts
A carefully planned diet has which of these characteristics? a. It contains sufficient raw oil. b. It contains ...

Nutrition: Concepts and Controversies - Standalone book (MindTap Course List)

In the following diagram, designate each daughter cell as diploid (2n) or haploid (n).

Human Heredity: Principles and Issues (MindTap Course List)

(a) At the time of this books printing, the U.S. national debt is about 16 trillion. If payments were made at t...

Physics for Scientists and Engineers, Technology Update (No access codes included)

Satellites orbit in space. How can a satellite conduct oceanography research?

Oceanography: An Invitation To Marine Science, Loose-leaf Versin 