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.
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.
1 Answer
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
The volume of a cylinder is the area of the circular base (
#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