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Chapter 12: Problem 40

An element crystallizes in a face-centered cubic lattice. The edge of the unit cell is \(0.408 \mathrm{nm}\), and the density of the crystal is $10.49 \mathrm{~g} / \mathrm{cm}^{3}$. Calculate the atomic weight of the element and identify the element.

Short Answer

Expert verified
The element has an atomic weight of approximately \(1.781 \times 10^{-22} \mathrm{~g/atom}\). By comparing this value with atomic weights of elements in the periodic table, we identify the element as Copper (Cu).

Step by step solution

01

Determine the number of atoms in face-centered cubic lattice unit cell

In a face-centered cubic lattice, there is one atom at each corner and one at the center of each face of the unit cell. There are 8 corners and 6 faces. Each corner atom is being shared by 8 unit cells, so each unit cell has 8*(1/8) = 1 corner atom completely inside it. Each face-centered atom is shared between 2 unit cells, so each unit cell has 6*(1/2) = 3 face-centered atoms inside it. Thus, we have a total of 1 (from corners) + 3 (from face-centers) = 4 atoms in each face-centered cubic unit cell.
02

Calculate the volume of the unit cell

The edge length of the unit cell is given as 0.408 nm (nanometers), but we need to convert it to cm (centimeters) because the given density is in g/cm³. 1 nm = 1x10⁻⁹ m = 1x10⁻⁷ cm So, 0.408 nm = 0.408x10⁻⁷ cm The volume V of the unit cell is the cube of the edge length: V = (0.408x10⁻⁷ cm)³ = 6.795x10⁻²³ cm³
03

Calculate the mass of the unit cell

Using the given density (D) and volume (V) of the unit cell, we can calculate the mass (M) of the unit cell: D = M / V M = D * V M = (10.49 g/cm³) * (6.795x10⁻²³ cm³) = 7.125x10⁻²² g
04

Calculate the atomic weight of the element

We know there are 4 atoms in the unit cell, and we already calculated the mass of the unit cell. Now, we can find the atomic weight of the element (W) by dividing the mass by the number of atoms: W = M / number of atoms W = (7.125x10⁻²² g) / 4 = 1.781x10⁻²² g/atom
05

Identify the element

The calculated atomic weight of the element is approximately 1.781x10⁻²² g/atom. To identify the element, we will compare this value with the atomic weights of the elements in the periodic table. The closest match we find is Copper (Cu), which has an atomic weight of approximately 63.55 g/mol. Keep in mind that the result may not be exactly equal due to some rounding errors in calculations, but it's close enough to identify the element as Copper.

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Most popular questions from this chapter

Which of the three-dimensional primitive lattices has a unit cell where none of the internal angles is \(90^{\circ}\) ? (a) Orthorhombic, (b) hexagonal, (c) rhombohedral, (d) triclinic, (e) both rhombohedral and triclinic.

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