Emergency Chemistry Help! Gold cube atoms and diameter?

Question

A cube of gold that is 1.00 cm on a side has a mass of 19.3 g. A single gold atom has a mass of 197.0 amu.

(a) How many gold atoms are in the cube?

(b) From the information given, estimate the diameter in angstroms of a single gold atom.

Answer 1

(a)

197.0 amu / atom = 197.0 g/mole

19.3 g x (1 mole / 197.0g) x (6.022 x 10²³ atoms / mole) = 5.90 x 10²²

atoms

(b)

if we assume the volume of the cube = volume of each atom x number of atoms then...

the volume of the cube is (1.00 cm)³ = 1.00 cm³

and 100 cm³  =  5.90 x 10²² x volume of each atom

volume of an atom = 100 cm³  /  5.90 x 10²²  = 1.69 x 10^-21 cm³

since V = 4/3 x pi x r³ for a sphere...

1.69 x 10^-21 cm³ = 4/3 x pi x r³

r = ³√(1.69 x 10^-21 cm³ x 3 / (4 x pi)) = ³√(4.05 x 10^-22 cm³)

r = 7.40 x 10^-8 cm =  7.40 x 10^-10 m = 7.40 Å

that is of course the radius.. so the diameter would be 2x that or

diameter = 14.80 Å

*******************

of course the volume of the cube is never exactly = volume of the atoms x number of atoms due to packing density.   if we assume that the spheres take up 74.08% of the total volume (see the link under source)..

then....

volume of an atom = 74.08 cm³  /  5.90 x 10²²  = 1.26 x 10^-21 cm³

r = ³√(1.26 x 10^-21 cm³ x 3 / (4 x pi)) = ³√(3.00 x 10^-22 cm³)

r = 6.69 x 10^-8cm = 6.69 Å

and

diameter = 13.4 Å