   # What fraction of the total volume of a cubic closest packed structure is occupied by atoms? (Hint: V s p h e r e = 4 3 π r 3 . ) What fraction of the total volume of a simple cubic structure is occupied by atoms? Compare the answers. ### 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 61E
Textbook Problem
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## What fraction of the total volume of a cubic closest packed structure is occupied by atoms? (Hint: V s p h e r e = 4 3 π r 3 . ) What fraction of the total volume of a simple cubic structure is occupied by atoms? Compare the answers.

Interpretation Introduction

Interpretation:

• The fraction of total volume of cubic closest packed structure occupied by atoms has to be determined.
• The fraction of total volume of simple cubic structure occupied by atoms has to be determined.

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. Cubic close packing structure has face-centered cubic (FCC) unit cell.

In simple cubic unit cell, each atom in the corner is shared by eight unit cells. Thus, number of atoms per simple cubic unit cell is,

8×18atomsincorners=1atom

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 byl=2r2where  l=edge length of unit cellr=atomicradius.

### Explanation of Solution

Calculate the fraction of total volume occupied by atoms in FCC unit cell.

edgelength,l=2r2Volumeofunitcell,l3=(2r2)3=22.62r3

volumeofatomsinFCCunitcell=4×43πr3=16.76r3

The fraction of total volume occupied by atoms in FCC unit cell is given by,

VatomsVunitcell=16.76r322.62r3=0.7409=74.09%

Cubic closest packed structure has FCC unit cells. Each FCC unit cell has 4 atoms. Thus volume of atoms in unit cell is equivalent to four times the volume of sphere. The atoms touch along the diagonal in FCC unit cell that the edge length is represented by the formula   “ l=2r2 ”.

Figure 1

Volume of the unit cell is obtained by cubing the edge length value. The ratio of volume of atoms in the unit cell to the volume of unit cell gives the fraction of total volume occupied by atoms in FCC unit cell.

Calculate the fraction of total volume occupied by atoms in simple cubic unit cell

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