How to determine the edge length of the aluminium unit cell? I'm having trouble understanding why I can't get the correct answer from this homework problem:


  Solid aluminum forms a face-centered cubic unit cell.  Aluminum has a density of $\pu{2.70 g/cm^3}$. Determine the edge length of the $\ce{Al(s)}$ unit cell in $\pu{cm}$.


Usually if I enter the wrong answer the it will show all the calculations to determine the correct answer. However this one does not show the calculations.

Here is the work I've done:


$\ce{Al }= \pu{26.98 g/mol}$
$\text{Mass} = \frac{26.98}{6.022\times10^{23}} = \pu{4.480E-23 g//atom}$    
$\text{Mass of one atom} = \text{mass of one unit cell}$  
$\text{Mass of one unit cell} = \pu{4.480E-23 g//unit cell}$
$D = \frac{M}{V}$, thus $V = \frac{M}{D}$
$V = \frac{\pu{4.480E-23}}{2.70} = \pu{1.66E-23 cm^3//unit cell}$
For a cube $V = l^3$, thus $l = V^{1/3}$
$\text{Unit cell edge length} = V^{1/3} = {\pu{1.66E-23 cm^3}}^{1/3} = \pu{1.18E-8 cm}$


The correct answer that the book gives is $\pu{4.049E-8 cm}$.

The only example I have to go off of in my book is for a simple cubic unit cell. Since $\ce{Al}$ is a face-centered cubic unit cell and has 4 particles per unit cell I tried multiplying my final answer by $4$, but that gives me $4.72$. I also tried multiplying the unit cell mass by $4$ which gave me a final answer of $1.87$. I just can't seem to get $4.049$. Any help understanding this would be greatly appreciated.
Correct approach is to do two things:

1)  Put four atomic masses into one unit cell.

2)  Properly extract the cube root of the volume you obtain.  That volume should now be about $6.6×10^{-23}\text{cm}^3$ from part 1.  To take the cube root properly, render the exponent on 10 as a multiple of 3, thus  $6.6×10^{-23}$ = $66×10^{-24}$ where $10^{-24}$, not $10^{-23}$, is $(10^{-8})^3$.

Then your numbers should work out.