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# Document - Problem#1 Palladium crystallizes in a...

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Problem #1: Palladium crystallizes in a face-centered cubic unit cell. Its density is 12.023 g/cm 3 . Calculate the atomic radius of palladium. Solution: 1) Calculate the average mass of one atom of Pd: 106.42 g mol¯ 1 ÷ 6.022 x 10 23 atoms mol¯ 1 = 1.767187 x 10¯ 22 g/atom 2) Calculate the mass of the 4 palladium atoms in the face-centered cubic unit cell: 1.767187 x 10¯ 22 g/atom times 4 atoms/unit cell = 7.068748 x 10¯ 22 g/unit cell 3) Use density to get the volume of the unit cell: 7.068748 x 10¯ 22 g ÷ 12.023 g/cm 3 = 5.8793545 x 10¯ 23 cm 3 4) Determine the edge length of the unit cell: [cube root of] 5.8793545 x 10¯ 23 cm 3 = 3.88845 x 10¯ 8 cm 5) Determine the atomic radius: Remember that a face-centered unit cell has an atom in the middle of each face of the cube. The square represents one face of a face-centered cube: Remember that a face-centered unit cell has an atom in the middle of each face of the cube. The square represents one face of a face-centered cube: Problem #2: Nickel crystallizes in a face-centered cubic lattice. If the density of the metal is 8.908 g/cm 3 , what is the unit cell edge length in pm? Solution: This problem is like the one above, it just stops short of determining the atomic radius. 1) Calculate the average mass of one atom of Ni: 58.6934 g mol¯ 1 ÷ 6.022 x 10 23 atoms mol¯ 1 = 9.746496 x 10¯ 23 g/atom 2) Calculate the mass of the 4 nickel atoms in the face-centered cubic unit cell: 9.746496 x 10¯ 23 g/atom times 4 atoms/unit cell = 3.898598 x 10¯ 22 g/unit cell

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3) Use density to get the volume of the unit cell: 3.898598 x 10¯ 22 g ÷ 8.908 g/cm 3 = 4.376514 x 10¯ 23 cm 3 4) Determine the edge length of the unit cell: [cube root of] 4.376514 x 10¯ 23 cm 3 = 3.524 x 10¯ 8 cm 5) Convert cm to pm: cm = 10¯ 2 m; pm = 10¯ 12 m. Consequently, there are 10 10 pm/cm (3.524 x 10¯ 8 cm) (10 10 pm/cm) = 352.4 pm Problem #3: Nickel has a face-centered cubic structure with an edge length of 352.4 picometers. What is the density? This problem is the exact reverse of problem #2. (See problem 4a below for an example set of calculations.) Solution: 1) Convert pm to cm 2) Calculate the volume of the unit cell 3) Calculate the average mass of one atom of Ni 4) Calculate the mass of the 4 nickel atoms in the face-centered cubic unit cell 5) Calculate the density (value from step 4 divided by value from step 2) Problem #4: Krypton crystallizes with a face-centered cubic unit cell of edge 559 pm. a) What is the density of solid krypton? b) What is the atomic radius of krypton? c) What is the volume of one krypton atom? d) What percentage of the unit cell is empty space if each atom is treated as a hard sphere? Solution to a:
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## This note was uploaded on 02/26/2011 for the course CHEM 1000 taught by Professor Hempstead during the Spring '09 term at York University.

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Document - Problem#1 Palladium crystallizes in a...

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