How do I calculate the cost of a single Aluminum atom? I am doing a Chemistry assignment in which I was given a piece of aluminum foil which I had to measure and use for my calculations. I am supposed to calculate the cost of a single aluminum atom using the given data from my teacher and the data which I have measured from my piece of aluminum foil. I understand the basic concept of how I am supposed to do this, but I am having some trouble. Here is the data:

Data given by teacher:  
  Cost of a roll           $5.53  
  Roll size                30.5 cm x 30.5 m  
  Density of Al            2.70 g/cm^3  
  Molar Mass Al            26.98 g/mol

Data measured from my piece of Aluminum foil:  
  Mass     1.35 g  
  Size     18.6 cm x 12.5 cm


I have determined a way to do this, but I am having trouble with it. First of all, I have determined the amount in moles which is $\mathrm{0.0500~mol}$. Then I multiplied the amount in moles with Avogadro's constant ($\mathrm{6.02 \cdot 10^{23}~mol^{-1}}$) to determine the number of aluminum atoms in the piece of aluminum foil I received which is $\mathrm{3.01 \cdot 10^{22}}~\text{atoms}$. Now I just have to find the cost of a single aluminum atom, but this is where I am stuck. I cannot use the cost of the aluminum foil roll to divide it by the number of atoms since I was given a piece of aluminum foil so I need to find the cost of the piece of aluminum foil and then divide it by the number of atoms.

So basically my question is: How do I find the cost of the piece of aluminum foil? If I can find the cost then I can divide it by the number of atoms to determine the cost of a single aluminum atom.

However, if there is a way to find the number of atoms in the roll of aluminum foil, I could then use the cost of the aluminum roll to find the cost of a single aluminum atom. Please explain if there is a way to do this using the data given by the teacher.
You can compute the area of aluminum foil in the roll as:

$$30.5\;\rm{cm}\cdot 3050\;\rm{cm} = 93025\;\rm{cm}^{2}$$

You have measured the mass per square centimeter as:

$${1.35\;\rm{g}\over 18.6\;\rm{cm}\cdot 12.5\;\rm{cm}} = {1.35\;\rm{g}\over 232.5\;\rm{cm}^{2}}$$

You have been given the cost of the roll (and thus the cost per area):

$${\$5.53\over 93025\;\rm{cm}^{2}}$$ 

Using those numbers, the molar mass for aluminum, and Avogadro's constant (here, it's just the number of atoms so I designate it as such):

$$\rm{cost\;per\;atom} = \left({\$5.53\over 93025\;\rm{cm}^{2}}\right)
\left({232.5\;\rm{cm}^{2}\over 1.35\;\rm{g}}\right)
\left({26.98\;\rm{g}\over 1\;\rm{mol}}\right)
\left({1\;\rm{mol}\over 6.022\cdot 10^{23}\;\rm{atoms}}\right)
$$
$$\rm{cost\;per\;atom} = \$4.59\cdot 10^{-25}$$