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How do I calculate the cost of a single Aluminum atom?
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Marcus Johnson
How do I calculate the cost of a single Aluminum atom?
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.
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.
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 cant seem to understand that. Thanks!More
Its just to make the units come out (dimensional analysis). It can be as I wrote it initially or inverted: its the ratio that matters. How its written depends on what units cancel and make it through to the end. Same with Avogadros number here.More
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.
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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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):
$$\rm{cost\;per\;atom} = \left({\$5.53\over 93025\;\rm{cm}^{2}}\right) \left({232.5\;\rm{cm}^{2}\over 1.35\;\rm{g}}\right) \left({26.98\;\rm{g}\over 1\;\rm{mol}}\right) \left({1\;\rm{mol}\over 6.022\cdot 10^{23}\;\rm{atoms}}\right) $$ $$\rm{cost\;per\;atom} = \$4.59\cdot 10^{-25}$$
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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