How many ATOMS in a #0.400*mol# quantity of #N_2O_5#?

1 Answer
Mar 27, 2017

Do you mean #"how many molecules"#, or #"how many atoms?"#

Explanation:

Clearly, there are #N_Axx0.400# #N_2O_5# molecules in such a molar quantity. Here we use #N_A#, #"Avogadro's number"#, just as we would use any other collective number, like a #"dozen"#, or a #"score"#, or a #"gross"#.

And so there are #0.400*molxx6.022xx10^23*mol^-1#, approx. #2.40xx10^23# individual #N_2O_5# molecules. What is the mass of this molar quantity?

And as regard to #"atoms"# there are #0.400*molxx7*"atoms"*mol^-1# #=# #2.8xxN_A# atoms.

The Lewis structure of #N_2O_5# is non-trivial, and charge separation with quaternized nitrogen occurs to give: #""^(-)O(O=)N^(+)-O-^(+)N(=O)O^-#. What is the overall charge on this molecular representation?