1. The problem statement, all variables and given/known data "Calcium crystallizes in a body-centered cubic structure. (a) How many Ca atoms are contained in each unit cell? (b) How many nearest neighbors does each Ca atom possess? (c) Estimate the length of the unit cell edge, a, from the atomic radius of calcium (1.97A). (d) Estimate the density of Ca metal." 2. Relevant equations (a) two Ca atoms (b) eight neighbors Pythagorean theorem: a2 + b2 = c2 http://education.mrsec.wisc.edu/SlideShow/images/unit_cells/body_centered_cubic2.jpg r = 1.97 A DCa = 1.54 g/cm3 3. The attempt at a solution Okay, I am sure I messed up at part (c), where I'm calculating the length of the unit cell's edges in relation to the atomic radius. The longest distance inside the cell from one end to another is 4r. I started out with the edge of the cube, which I presumed to be 2r + x (some unknown distance I needed to figure out). Then I would have to figure out the distance between two corners diagonally opposite each other on any given face of a cube, given by the square root of 2(2r +x)2. I would use the Pythagorean theorem to equate the squares of those two distances to 4r, which is the longest distance inside a cube. Then I would work backwards from there. This is how my distance calculations turned out, and I'm pretty sure I screwed something up. (c) (7.88)2 = (3.94 + x)2 + √2(3.94 + x)2) 62.0944 = 15.5236 + 7.88x + x2 + (3.94 +x)√(2) x2 +7.88x - 46.5708 + 5.572 + x√(2) = 0 x2 + 9.3x - 41 = 0 x = (-9.3 ± √(86.49 + 164))/2 x = (-9.3 ± 15.827)/2 x = 3.264 (ignoring negative result) d = 3.264 + 3.94 = 7.2 A Next, I did the volume and mass calculations, the errors of which I am sure are the result of my distance calculations. (d) 7.2 A = 7.2 ⋅ 10-8 cm (7.2 ⋅ 10-8 cm)3 = 373.2 ⋅ 10-24 cm3 (40.078 g/mol)(1 mol/6.022 ⋅ 1023 atoms) = 6.66 ⋅ 10-23 g/atom (2 atoms)(6.66 ⋅ 10-23 g/atom) = 13.32 ⋅ 10-23 g D = 13.32 ⋅ 10-23 g/373.2 ⋅ 10-24 cm3 = 133.2 g/373.2 cm3 = 0.36 g/cm3 Now I know this isn't the correct answer. I checked, and the density of solid calcium is 1.54 g/cm3. Can someone tell me what I'm doing wrong with my calculations?