An element crystallizes in a body-centered cubic lattice. Theedge of the unit ce

Question

An element crystallizes in a body-centered cubic lattice. Theedge of the unit cell is 2.82A(A with a circle on top), and thedensity of the crystal is 7.92g/cm^3. Calculate the atomic weightof the element. Thank you so much! PS: Could you tell me how to set up the problem, how toapproach it, rather than just an answer. An element crystallizes in a body-centered cubic lattice. Theedge of the unit cell is 2.82A(A with a circle on top), and thedensity of the crystal is 7.92g/cm^3. Calculate the atomic weightof the element. Thank you so much! PS: Could you tell me how to set up the problem, how toapproach it, rather than just an answer.

Explanation / Answer

A body-centered cubic lattice has an atom at each of 8 corners ofthe cell and an atom in the center of the cell. The fraction of anatom that occupies a corner of a unit cell is 1/8 and the fractionof the center atom in the unit cell is, of course, 1. So, 8 corner atoms * (1/8) + 1 center atom = 2 atoms in the unitcell Then, density = mass / volume density = 7.92 g/cm^3 mass = x amu (what you are looking for) volume = (2.86 Angstroms)^3 (length times width times height ofunit cell) amu = density/volume = (7.92 g/cm^3)/(2.86 A)^3 Then to convert amu to grams, divide the number of amu from abovestep by Avogadro's number, 6.02 x 10^23 amu. This gives the answerin grams. ---------------------------------------------------------------------------------------------------------------------- l = 2.86 x 10^-8 cm V = ( 2.86 x 10^-8)^3 = 2.34 x 10^-23 cm^3 mass = 7.92 g/cm^3 x 2.34 x 10^-23 cm^3 = 1.85 x 10^-22 g number atoms per cell = 2 1.85 x 10^-22 = 2 x molar mass / 6.02 x 10^23 molar mass = 55.8 g/mol ( iron)

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