Question

An aluminum can has a height of 4.83 inches, a diameter of 2.60 inches, and a uniform wall/end thickness of 0.0112 inches. If the can costs exactly five cents to purchase, determine the cost (with appropriate significant figures) of one single aluminum atom from the can.

Answer 1

Strategy: with the measures you can determine the volume of the plate of aluminum. Then you can use the density of aluminum to calculate the mass.

With the mass of aluminum and its atomic mass you can find the number of moles and thereafter the number of atoms.

Finally divide the cost by the number of atoms to find the cost of one single atom.

Let's do it.

Volume of aluminum plate, V: 0.0112 in* 4.83 in* 2.60 in * [2.54 cm/in]^3 = 2.305 cm^3

Density of aluminum (from Wikipedia), d = 2.70 g/cm^3

mass, m = d*V = 2.305 cm^3 * 2.70 g/ cm^3 = 6.22 g

Atomic mass of aluminum (from Wikipedia), am = 27 g / mol

Number of moles, n = m/am = 6.22 g / 27 g / mol =   0.23 mol

Number of atoms = n*Avogadro constant = 0.23 mol * 6.022 * 10^23 atoms/mol = 1.39*10^23

Cost per atom = cost of the can / number of atoms =$ 0.05 /1.39*10^23 atoms  =  3.60 * 10^ - 25 $/atom