How many mols of aluminum atoms are in #1.42 xx 10^24# atoms?

Am I doing this right?

1.42x1024 x 26.98 = 3.83x1025

1 Answer
Aug 31, 2016

What you did was multiply the molar mass of aluminum atom by the number of atoms, which physically doesn't make sense; it gives you:

#"g"/"mol" xx "atoms" = ("g"cdot"atoms")/"mol"#.

The atoms "unit" has to cancel so that you get into the units of #"mol"#s. So, you have to use a relationship between #"mol"#s and #"atoms"# and achieve:

#cancel("atoms") xx "mol"/cancel("atoms")#

How I would do it is to recall that #"1 mol"# of anything is #6.022xx10^23# of that thing. It could be spoons, watches, atoms, pencils, whatever. It could also be #"Al"# atoms, #"C"# atoms, #"H"# atoms, etc.

So if you have #1.42xx10^24# #"atoms"# of aluminum, practically speaking, you should have between #2# and #3# #"mols"# of aluminum atoms because #(1.42xx10^24)/(6.022xx10^23) ~~ 2#.

That's just an estimate, but it gives you an idea of what we expect to get.

#1.42xx10^24 cancel"atoms" xx "1 mol"/(6.022xx10^23 cancel"atoms") = 14.2/6.022 = bb("2.36 mols")# of aluminum atoms.

A #"mol"# is just like a dozen. A dozen eggs means #12# eggs, and #1# dozen eggs is larger in absolute quantity than #1# egg.

Similarly, #"1 mol"# of eggs is larger in absolute quantity than #"1"# egg. So, if you have a large number of atoms that you are converting to #"mol"#s, you should expect the number of #"mol"#s to be smaller, not larger, than your number of atoms.