An unknown element crystallizes in a cubic unit cell with an edge length of 3.30 Å. It has a density of 35.77 g/cm3. The atomic mass of the element is 193.5 amu. The mass of one atom is 3.21e-22 g.
Calculate the mass of all atoms in one unit cell in grams.
How many atoms are contained in one unit cell of of this element?
2007-02-12 05:58:03 · 2 answers · asked by AppleCard! 2 in Science & Mathematics ➔ Chemistry
3.3 A = 3.3 * 10^-10 m = 3.3 * 10^-8 cm.
(3.3 * 10^-8 cm)^3 = 3.59 * 10^-23 cm^3.
volume * density = mass
3.59 * 10^-23 cm^3 * 35.77 g/cm^3 = 1.29 * 10^-21 g.
1.29^-21 g / (3.21^-22 g / atoms) = 4 atoms.
By the way, I'm pretty sure no stable isotope of any element has a density of 36 g/cm^3.
2007-02-12 08:18:01 · answer #1 · answered by davisoldham 5 · 0⤊ 0⤋
(1) Figure out the volume of the unit cell. Convert 3.3 angstroms into centimeter and cube it.
(2) Multiply density of this substance by the volume of the unit cell. This gives you the weight of the unit cell (first part of youq question).
(3) Divide that weight by the mass of one atom to calculate the number of atoms in one unit cell (second part of your question).
I got 4 atoms per unit cell. It's probably a face-center-cubic unit cell.
2007-02-12 06:12:33 · answer #2 · answered by Elisa 4 · 0⤊ 0⤋