help with covalent radius/aluminum

[Baykko]

The covalent radius of an aluminum atom is 1.18 x 10^-8 cm. Assuming that in aluminum foil,the atoms are packed one on top of each other, how many atoms thick is a sheet of aluminum foil? (if aluminum thickness= 0.0017)

 

for this.. ive tried dividing the thickness with the radius/diameter, giving me answer of around 14k and 7k but not sure if that is correct or what is the right way to solve it.

 

then i have no idea what to do for this problem,

A cube of aluminum of side 2.00cm has a mass of 21.60g. Using the covalent radius of an aluminum atom, calculate the volume of the "empty space" inside the cube.

[UC]

For the first one, you are doing it right, but which measurement (diameter or radius) makes sense? If you stack spheres, isn't their total height going to be the diameters of each added or their radii?

 

For the second one, you need to look up what crystal structure aluminum crystallizes in. Assume that the aluminum atoms are solid spheres with a radius equal to the covalent radius. Wikipedia will tell you that it is face-centered cubic. Then, draw a unit cell and calculate what's called the atomic packing factor. http://en.wikipedia.org/wiki/Atomic_packing_factor This is amount of space that the aluminum atoms occupy per unit volume. FCC crystal structure has an atomic packing factor of 0.74, although I suspect that your homework should show the derivation. This means that 26% of space in any given volume of crystal is empty. From the size of your cube, calculating the volume of "empty space" should be easy. The mass of the cube is irrelevant.

[hermanntrude]

For the first one, you are doing it right, but which measurement (diameter or radius) makes sense? If you stack spheres, isn't their total height going to be the diameters of each added or their radii?

 

For the second one, you need to look up what crystal structure aluminum crystallizes in. Assume that the aluminum atoms are solid spheres with a radius equal to the covalent radius. Wikipedia will tell you that it is face-centered cubic. Then, draw a unit cell and calculate what's called the atomic packing factor. http://en.wikipedia.org/wiki/Atomic_packing_factor This is amount of space that the aluminum atoms occupy per unit volume. FCC crystal structure has an atomic packing factor of 0.74, although I suspect that your homework should show the derivation. This means that 26% of space in any given volume of crystal is empty. From the size of your cube, calculating the volume of "empty space" should be easy. The mass of the cube is irrelevant.

 

UC, read the question. it says "assuming that the atoms are stacked on top of each other"

 

that means we don't need to confuse people with crystal lattices and packing spheres

[UC]

UC, read the question. it says "assuming that the atoms are stacked on top of each other"

 

that means we don't need to confuse people with crystal lattices and packing spheres

 

Reread the original post. Those are two seperate questions being asked. The stacking refers to the foil thickness only. How would that help to calculate "empty space" anyway? For this, examination of the lattice structure is necessary.

 

The given mass and volume produce the theoretical density for aluminum, showing it is a "solid" metal cube with no macroscopic voids.