Determine the radius of an Al atom (in pm) if the density of aluminum is 2.71 g/cm3 . Aluminum crystallizes in a face centered cubic structure with an edge leng

143pm is the radius of an Al atom

Explanation:

In a face centered cubic structure, FCC, there are 4 atoms per unit cell.

First, you need to obtain the mass of an unit cell using molar mass of Aluminium  and thus, obtain edge length and knowing Edge = √8R you can find the radius, R, of an Al atom.

Mass of an unit cell

As 1 mole of Al weighs 26.98g. 4 atoms of Al weigh:

4 atoms × (1mole / 6.022x10²³atoms) × (26.98g / mole) = 1.792x10⁻²²g

Edge length

As density of aluminium is 2.71g/cm³, the volume of an unit cell is:

1.792x10⁻²²g × (1cm³ / 2.71g) = 6.613x10⁻²³cm³

And the length of an edge of the cell is:

∛6.613x10⁻²³cm³ = 4.044x10⁻⁸cm = 4.044x10⁻¹⁰m

Radius:

As in FCC structure, Edge = √8 R, radius of an atom of Al is:

4.044x10⁻¹⁰m = √8 R

1.430x10⁻¹⁰m = R.

In pm:

1.430x10⁻¹⁰m ₓ (1x10¹²pm / 1m) =

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