How many mols of #"Cu"# are in #4.5 xx 10^22 "atoms"#?

You have the number of #"atoms"# of #"Cu"#. You need the #"mol"#s of #"Cu"#.

Recall that Avogadro's number is #6.022_1413xx10^(23) "atoms/mol"#, which tells you the relationship you need (the subscript numbers indicate extra decimal places you could use if you want but may not be necessary).

Basically what you want to do is do a unit conversion that cancels out #\mathbf"atoms"# and gives you #\mathbf"mol"#s. The beauty of Avogadro's number is that it is always the same number, even for different elements.

#(4.5xx10^(22) cancel("atoms Cu"))("1 mol atoms Cu"/(6.0221413xx10^(23) cancel("atoms Cu")))#

#= color(blue)(0.074"7 mol atoms Cu")#

If you don't get this result, check your parentheses. It should be solved with the following parentheses placements to avoid scaling the result by #10^23#:

#(4.5xx10^(22))*("1"/(6.0221413xx10^(23)))#
(try no parentheses and see what you get if you don't use #"E"# notation for your exponentiations)

Note that by virtue of using the unit of a #"mol"#, the number ideally should be between about #\mathbf0.01# and #\mathbf10#. Anything outside that range looks strange, because the unit is intended to make the numbers look nicer.

(We say #"1 mile"# instead of #"5280 ft"#, for instance. Or, for example, saying you have #"0.001 mol"# can be hard to read, so you can instead say you have #"1 mmol"#.)