# 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 could always look up the bulk price of aluminium per tonne (this will be a lot less than the foil but will still be valid). Aluminium is a commonly traded commodity and will be sold by weight. You already know the weight of the sample... – matt_black Sep 17 '15 at 0:03
• @matt_black I cannot use any other data other than the ones I was given and the ones I measured. – Behroz Henayat Sep 17 '15 at 1:37

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):

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

• There is an error with your answer. The molar mass is wrong, it is supposed to be 26.98. Can you do your calculation again so I can confirm that my answer is right? I would appreciate that. Thanks! – Behroz Henayat Sep 17 '15 at 0:46
• @BProgram - Thanks for catching that, fixed and recalculated. – Todd Minehardt Sep 17 '15 at 0:51
• One more thing, could you explain to me why in the equation it is 232.5/1.35 and above the equation it was 1.35/232.5? Why was it flipped? I can't seem to understand that. Thanks! – Behroz Henayat Sep 17 '15 at 0:58
• It's just to make the units come out (dimensional analysis). It can be as I wrote it initially or inverted: it's the ratio that matters. How it's written depends on what units cancel and make it through to the end. Same with Avogadro's number here. – Todd Minehardt Sep 17 '15 at 1:01