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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.

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    $\begingroup$ 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... $\endgroup$
    – matt_black
    Sep 17, 2015 at 0:03
  • $\begingroup$ @matt_black I cannot use any other data other than the ones I was given and the ones I measured. $\endgroup$
    – Behroz Henayat
    Sep 17, 2015 at 1:37

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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):

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

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  • $\begingroup$ 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! $\endgroup$
    – Behroz Henayat
    Sep 17, 2015 at 0:58
  • $\begingroup$ 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. $\endgroup$
    – Todd Minehardt ♦
    Sep 17, 2015 at 1:01
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I did this using one estimate for the number of aluminum atoms in 4.55 grams of aluminum, scaled it up to a kilogram, and divided aluminum's current price per kilo (US\$1.87) by my estimate for the number of atoms in a kilogram of aluminum, yielding a price per atom of about $\\\$8.453 \cdot 10^{-26}$, or around eight sextillionths of a penny.

Of course, I'm only in eighth grade, so my calculations may be way off.

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