If I have #9.71xx10^22# #"platimum atoms"#, what molar quantity, and what mass of metal are present?

1 Answer
Sep 1, 2016

Answer:

You have approx #0.16*mol# of platinum atoms, with a mass of approx. #30*g#.

Explanation:

#"Moles of platinum"# #=# #(9.71xx10^22*"platinum atoms")/(6.022xx10^23*"platinum atoms"*mol^-1)# #=# #??"moles"#?

Now #1# #mol# of #Pt# atoms has a mass of #195.08*g#.

So you should have a mass of approx. #30*g#.

I appreciate that it is hard to work with such unfeasibly large numbers. But remember that the mole is just another number, like a score, or a dozen, or 1 gross - it's just that a mole is *&&??! LARGE. #"Avogadro's number"# of platinum atoms specifies #1# #mol# and has a mass of #195.08*g#. Likewise #"Avogadro's number"# of #""^1H# atoms has a mass of #1*g# precisely.

This idea of molar and mass equivalence is something that is fundamental to chemistry, and you should take some effort to appreciate it.