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Dimension and Density: ength for Ag was determined to be 408.6 pm. Ag crystallizes in a fcc lattice with all atoms at the lattice points. Ag has a density of 10.50 g/cm³. Calculate the mass of a single Ag atom. STRATEGY: We can calculate the unit cell volume. Then, from the density, we can find the mass of the unit cell and then the mass of a single Ag atom. The volume of the cell = a³: V = (408.6 x 10-12 m)3 = 6.822 x 10-29 m³ The density of silver (in g/m³) is: d=10.50 g/cm³ x (1 cm/10-2m)3 = 1.050 x 107 g/m³ The mass of a unit cell is: m = d*V m = 1.050 x 107 g/m³ (6.822 x 10-29 m³) m = 7.163 x 10-22 g Ag/cell Because there are 4 atoms in a fcc cell, the mass of a single Ag atom in the cell m = 7.163 x 10-22 g Ag/cell/ (4 atoms/cell) OR: m = 1.791 x 10-22 g Ag/atom M = rho*V*N/Z = (10.50 g/cm³)(6.822 x 10-29 m³)(10°cm³/m³)(6.022x1023)/4 M = 197.84 g/mol Therefore, 197.84 g/mol/6.022x23 atoms/mol = 1.791 x 10-22 g Ag/atom

Dimension and Density: ength for Ag was determined to be 408.6 pm. Ag crystallizes in a fcc lattice with all atoms at the lattice points. Ag has a density of 10.50 g/cm³. Calculate the mass of a single Ag atom. STRATEGY: We can calculate the unit cell volume. Then, from the density, we can find the mass of the unit cell and then the mass of a single Ag atom. The volume of the cell = a³: V = (408.6 x 10-12 m)3 = 6.822 x 10-29 m³ The density of silver (in g/m³) is: d=10.50 g/cm³ x (1 cm/10-2m)3 = 1.050 x 107 g/m³ The mass of a unit cell is: m = d*V m = 1.050 x 107 g/m³ (6.822 x 10-29 m³) m = 7.163 x 10-22 g Ag/cell Because there are 4 atoms in a fcc cell, the mass of a single Ag atom in the cell m = 7.163 x 10-22 g Ag/cell/ (4 atoms/cell) OR: m = 1.791 x 10-22 g Ag/atom M = rho*V*N/Z = (10.50 g/cm³)(6.822 x 10-29 m³)(10°cm³/m³)(6.022x1023)/4 M = 197.84 g/mol Therefore, 197.84 g/mol/6.022x23 atoms/mol = 1.791 x 10-22 g Ag/atom

Chemistry & Chemical Reactivity
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
Section12.1: Crystal Lattices And Unit Cells
Problem 12.1CYU: (a) Determining an Atom Radius from Lattice Dimensions: Gold has a face-centered unit cell, and its...
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Dimension and Density: ength for Ag was determined to be 408.6 pm. Ag crystallizes in a fcc lattice with all atoms at the lattice points. Ag has a density of 10.50 g/cm³. Calculate the mass of a single Ag atom. STRATEGY: We can calculate the unit cell volume. Then, from the density, we can find the mass of the unit cell and then the mass of a single Ag atom. The volume of the cell = a³: V = (408.6 x 10-12 m)3 = 6.822 x 10-29 m³ The density of silver (in g/m³) is: d=10.50 g/cm³ x (1 cm/10-2m)3 = 1.050 x 107 g/m³ The mass of a unit cell is: m = d*V m = 1.050 x 107 g/m³ (6.822 x 10-29 m³) m = 7.163 x 10-22 g Ag/cell Because there are 4 atoms in a fcc cell, the mass of a single Ag atom in the cell m = 7.163 x 10-22 g Ag/cell/ (4 atoms/cell) OR: m = 1.791 x 10-22 g Ag/atom M = rho*V*N/Z = (10.50 g/cm³)(6.822 x 10-29 m³)(10°cm³/m³)(6.022x1023)/4 M = 197.84 g/mol Therefore, 197.84 g/mol/6.022x23 atoms/mol = 1.791 x 10-22 g Ag/atom
Transcribed Image Text:Dimension and Density: ength for Ag was determined to be 408.6 pm. Ag crystallizes in a fcc lattice with all atoms at the lattice points. Ag has a density of 10.50 g/cm³. Calculate the mass of a single Ag atom. STRATEGY: We can calculate the unit cell volume. Then, from the density, we can find the mass of the unit cell and then the mass of a single Ag atom. The volume of the cell = a³: V = (408.6 x 10-12 m)3 = 6.822 x 10-29 m³ The density of silver (in g/m³) is: d=10.50 g/cm³ x (1 cm/10-2m)3 = 1.050 x 107 g/m³ The mass of a unit cell is: m = d*V m = 1.050 x 107 g/m³ (6.822 x 10-29 m³) m = 7.163 x 10-22 g Ag/cell Because there are 4 atoms in a fcc cell, the mass of a single Ag atom in the cell m = 7.163 x 10-22 g Ag/cell/ (4 atoms/cell) OR: m = 1.791 x 10-22 g Ag/atom M = rho*V*N/Z = (10.50 g/cm³)(6.822 x 10-29 m³)(10°cm³/m³)(6.022x1023)/4 M = 197.84 g/mol Therefore, 197.84 g/mol/6.022x23 atoms/mol = 1.791 x 10-22 g Ag/atom
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