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mple Exercise 6: Empirical Formula and Density: Ionic Solid The unit cell of a binary compound of copper and oxygen is shown here. Given this image and the ionic radii rcut = 0.74 Å and ro² = 1.26 Å, (a) determine the empirical formula of this compound, (b) determine the coordination numbers of copper and oxygen, (c) estimate the length of the edge of the cubic unit cell, and (d) estimate the density of the compound. Solution (a) To determine the empirical formula we need to determine how many of each type of ion there are per unit cell. O: 8(1/8) + 1 = 2 02-ions in the unit cell. Cu: 4(1) = 4 Cut ions in the unit cell. Empirical Formula: Cu₂O O Cu

mple Exercise 6: Empirical Formula and Density: Ionic Solid The unit cell of a binary compound of copper and oxygen is shown here. Given this image and the ionic radii rcut = 0.74 Å and ro² = 1.26 Å, (a) determine the empirical formula of this compound, (b) determine the coordination numbers of copper and oxygen, (c) estimate the length of the edge of the cubic unit cell, and (d) estimate the density of the compound. Solution (a) To determine the empirical formula we need to determine how many of each type of ion there are per unit cell. O: 8(1/8) + 1 = 2 02-ions in the unit cell. Cu: 4(1) = 4 Cut ions in the unit cell. Empirical Formula: Cu₂O O Cu

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Sample Exercise 6: Empirical Formula and Density: Ionic Solid The unit cell of a binary compound of copper and oxygen is shown here. Given this image and the ionic radii rcut = 0.74 Å and ro² = 1.26 Å, (a) determine the empirical formula of this compound, (b) determine the coordination numbers of copper and oxygen, (c) estimate the length of the edge of the cubic unit cell, and (d) estimate the density of the compound. Solution (a) To determine the empirical formula we need to determine how many of each type of ion there are per unit cell. O: 8(1/8) + 1 = 2 0²-ions in the unit cell. Cu: 4(1) = 4 Cut ions in the unit cell. Empirical Formula: Cu₂O (b) If we can visually determine the coordination number of one of the ions, we can use Equation 12.1 to determine the coordination number of the other ion. Number of cations per formula unit Number of anions per formula unit anion coordination number cation coordination number Cation coordination number = anion coordination number Copper coordination number = 4 # of anions per formula unit # of cations per formula unit ( 2 [12.1] = 2 Cu
Transcribed Image Text:Sample Exercise 6: Empirical Formula and Density: Ionic Solid The unit cell of a binary compound of copper and oxygen is shown here. Given this image and the ionic radii rcut = 0.74 Å and ro² = 1.26 Å, (a) determine the empirical formula of this compound, (b) determine the coordination numbers of copper and oxygen, (c) estimate the length of the edge of the cubic unit cell, and (d) estimate the density of the compound. Solution (a) To determine the empirical formula we need to determine how many of each type of ion there are per unit cell. O: 8(1/8) + 1 = 2 0²-ions in the unit cell. Cu: 4(1) = 4 Cut ions in the unit cell. Empirical Formula: Cu₂O (b) If we can visually determine the coordination number of one of the ions, we can use Equation 12.1 to determine the coordination number of the other ion. Number of cations per formula unit Number of anions per formula unit anion coordination number cation coordination number Cation coordination number = anion coordination number Copper coordination number = 4 # of anions per formula unit # of cations per formula unit ( 2 [12.1] = 2 Cu
(c) To estimate the length of the unit cell edge we must first determine the direction along which the ions touch. We can then use ionic radii and trigonometry to estimate the size of the unit cell. V Cell edge, y = √a² + x² y = √a² + (√2a)² = √3a Body diagonal, y Face diagonal, x x = √a² + a² = √2a P = Σ, Z₁ * M₁ V cell * N P (Part of the diagram has been cut away for clarity). Along the Body diagonal (y) we have: y = r0²- + 2rCu+ + 2r02- + 2rCu+ + r0²- y = 4r0²- + 4rCu+ Also: y = 3¹/2 a Vcell (4.619 x 10-8 cm)³ = Vcell 9.86 x 10-23 cm³ (d) Use The Crystallographer's Formula to calculate the density of Cu₂0: = y = 4(1.26 Å) + 4(0.74 Å) = 8.00 Å Thus: a = 8.00 Å a = 4.619 Å 31/2 a = 4.619 Å x 1 x 10-8 cm = 4.619 x 10-8 cm Å MM Cu = 63.5463 g/mol Z = 4 MM O = 15.9994 g/mol Z = 2 (4 x 63.5463) + (2 x 15.9994 g/mol) (9.86 x 10-23 cm³)* (6.022 x 10²3) p = 4.82 g/cm³
Transcribed Image Text:(c) To estimate the length of the unit cell edge we must first determine the direction along which the ions touch. We can then use ionic radii and trigonometry to estimate the size of the unit cell. V Cell edge, y = √a² + x² y = √a² + (√2a)² = √3a Body diagonal, y Face diagonal, x x = √a² + a² = √2a P = Σ, Z₁ * M₁ V cell * N P (Part of the diagram has been cut away for clarity). Along the Body diagonal (y) we have: y = r0²- + 2rCu+ + 2r02- + 2rCu+ + r0²- y = 4r0²- + 4rCu+ Also: y = 3¹/2 a Vcell (4.619 x 10-8 cm)³ = Vcell 9.86 x 10-23 cm³ (d) Use The Crystallographer's Formula to calculate the density of Cu₂0: = y = 4(1.26 Å) + 4(0.74 Å) = 8.00 Å Thus: a = 8.00 Å a = 4.619 Å 31/2 a = 4.619 Å x 1 x 10-8 cm = 4.619 x 10-8 cm Å MM Cu = 63.5463 g/mol Z = 4 MM O = 15.9994 g/mol Z = 2 (4 x 63.5463) + (2 x 15.9994 g/mol) (9.86 x 10-23 cm³)* (6.022 x 10²3) p = 4.82 g/cm³
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