Question

How to Calculate the Number of Tin Atoms in a Can?

Thickness of tin coat on a can is 2 μm. Calculate the number of tin atoms that you bring home when you buy a pineapple compote can. The can has a radius of 5.1 cm and is 2.2 cm high. The density of tin is 7.28 g /cm3.

Answer 1

I would imagine the volume of the can as a metallic cylindrical shell, get the can volume using the original radius, then adjust the radius using the thickness to subtract out the inner volume.

I got `7 xx 10^20` `"Sn atoms"`.

---

The volume of a cylinder is the area of the circular base (`pir^2`) times the height (`h`):

`V = pi r^2h`

And so, the entire volume is:

`V_(t ot) = pi cdot ("5.1 cm")^2 cdot "2.2 cm" = ul("179.77 cm"^3)`

From the thickness of the tin, we get that the smaller radius is:

`("5.1 cm") - 2cancel(mu"m") xx (cancel"1 m")/(10^6 cancel(mu"m")) xx ("100 cm")/(cancel"1 m") = "5.0998 cm"`

This is negligibly different, but since we want the difference anyway, this is no longer negligible. The inner volume is:

`V_(i n n er) = pi cdot ("5.0998 cm")^2 cdot "2.2 cm" = ul("179.75 cm"^3)`

The difference in volume gives the volume of the tin shell.

`V_(s h e l l ) = V_(t ot) - V_(i n n er) = "0.02 cm"^3`

Therefore, the mass of it is:

`0.02 cancel("cm"^3) xx "7.28 g"/cancel("cm"^3) = "0.1456 g"`

If we assume it is pure tin, then there are this many mols:

`0.1456 cancel"g Sn" xx "1 mol"/(118.71 cancel"g Sn") = "0.0012 mols Sn"`

And so, there are...

`0.0012 cancel"mols Sn" xx (6.022 xx 10^23 "whatevers")/(cancel"1 mol anything ever")`

`= 7.38 xx 10^20 "Sn atoms"`

But you have only given me one significant figure, so we have only `color(blue)(7 xx 10^20 "Sn atoms")`.