If I know the mass of #"8000 HCl molecules"#, could I find the mass of #"8000 chlorine atoms"#?

See this old answer and this one.

The mole is thus the link between between the micro world of atoms and molecules with the macro world of grams and litres......

These are not the same, as there does NOT exist any atom that is also a molecule.

Avogadro's number is #6.022 xx 10^23 "particles"cdot"mol"^(-1)#. So, if you have a MOLECULAR mass:

#6.022 xx 10^23 "molecules"cdotcancel("mol"^(-1)) xx cancel"1 mol"/"known g of molecule" = "normalized number of molecules"/("1 g molecule")#

If you have some arbitrary number of atoms #N# then given a measured mass of the atoms...

#"known number of atoms" / "known g of atoms"#

#= "normalized number of atoms"/"1 g atoms"#

And thus, these might seem equivalent, since the notion of what kind of particle we are referring to SEEMINGLY cancels out... but they are in fact NOT equivalent. They are dependent on the identity of the substances.

EXAMPLE

If we have #"HCl"#, with molar mass #"36.4609 g/mol"#, then...

#6.022 xx 10^23 "molecules"cdotcancel("mol"^(-1)) xx cancel"1 mol"/"36.4609 g HCl" = color(red)((1.65 xx 10^22 "HCl molecules")/("1 g HCl"))#

Given #"8000 Cl atoms"#, a completely arbitrary number, its mass would be...

#8000 cancel"Cl atoms" xx (cancel"1 mol")/(6.022 xx 10^23 cancel"atoms") xx "35.453 g Cl"/cancel"mol Cl"#

#= 4.71 xx 10^(-19) "g Cl"#

And so, the ratio of the number of #"Cl"# atoms to their mass is:

#"8000 Cl atoms"/(4.71 xx 10^(-19) "g Cl")#

#= color(red)((1.70 xx 10^22 "Cl atoms")/("1 g Cl"))#

Thus, these are NOT numerically equivalent. It was a coincidence that these were similar, but they are not the same thing. Try it with some heavier atom, and it won't be similar.