Pt crystallizes in a face-centered cubic (fcc) lattice with all atoms at the lattice points. It has a density of 21.45 g/cm³ and an atomic weight of 195.08 amu. Calculate the length of a unit-cell edge. STRATEGY: We can calculate the mass of the unit cell from the atomic weight. Knowing the density and the mass of the unit cell, we can calculate the volume of a unit cell and then the edge length of a unit cell. 195.08 g Pt x 1 mol Pt 1 mol Pt 6.022 x 1023 Pt atoms = 3.239 x 10-22g Pt 1 Pt atom Since there are 4 atoms of Pt per fcc unit cell, the mass of the unit cell = 4 x 3.239 x 10-22 1.296 x 10-21 g Pt = unit cell The Volume of the unit cell is: V = m/d AND V = a³ V = (1.296 x 10-21 g Pt/cell)/(21.45 g/cm³) = 6.042 x 10-23 cm³/cell

Pt crystallizes in a face-centered cubic (fcc) lattice with all atoms at the lattice points. It has a density of 21.45 g/cm³ and an atomic weight of 195.08 amu. Calculate the length of a unit-cell edge. STRATEGY: We can calculate the mass of the unit cell from the atomic weight. Knowing the density and the mass of the unit cell, we can calculate the volume of a unit cell and then the edge length of a unit cell. 195.08 g Pt x 1 mol Pt 1 mol Pt 6.022 x 1023 Pt atoms = 3.239 x 10-22g Pt 1 Pt atom Since there are 4 atoms of Pt per fcc unit cell, the mass of the unit cell = 4 x 3.239 x 10-22 1.296 x 10-21 g Pt = unit cell The Volume of the unit cell is: V = m/d AND V = a³ V = (1.296 x 10-21 g Pt/cell)/(21.45 g/cm³) = 6.042 x 10-23 cm³/cell

Chemistry & Chemical Reactivity

10th Edition

ISBN:9781337399074

Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Chapter12: The Solid State

Section: Chapter Questions

Problem 50GQ: Consider the three types of cubic units cells. (a) Assuming that the spherical atoms or ions in a...

See similar textbooks

Similar questions

Consider the three types of cubic units cells. (a) Assuming that the spherical atoms or ions in a primitive cubic unit cell just touch along the cubes edges, calculate the percentage of occupied space within the unit cell. (Recall that the volume of a sphere is (4/3)r3, where r is the radius of the sphere.) (b) Compare the percentage of occupied space in the primitive cell (pc) with the bcc and fcc unit cells. Based on this, will a metal in these three forms have the same or different densities? If different, in which is it most dense? In which is it least dense?
8.16 Iridium forms a face-centered cubic lattice, and an iridium atom is 271.4 pm in diameter. Calculate the density of iridium.
Lead has a face-centered cubic lattice with all atoms at lattice points and a unit-cell edge length of 495.0 pm. Its atomic mass is 207.2 amu. What is the density of lead?
The CsCl structure is a simple cubic array of chloride ions with a cesium ion at the center of each cubic array (see Exercise 69). Given that the density of cesium chloride is 3.97 g/cm3, and assuming that the chloride and cesium ions touch along the body diagonal of the cubic unit cell, calculate the distance between the centers of adjacent Cs+ and Cl ions in the solid. Compare this value with the expected distance based on the sizes of the ions. The ionic radius of Cs+ is 169 pm, and the ionic radius of Cl is 181 pm.
Aluminum metal crystallizes with a face-centered cubic unit cell. The volume of the cell is 0.0662 nm3. (a) What is the atomic radius of aluminum in cm? (b) What is the volume of a single aluminum atom? (c) What is the density of a single aluminum atom? (d) In face-centered cubic cell packing, the fraction of empty space is 26.0%. When this is factored in, what is the calculated density of aluminum?