How many atoms constitute a #3.50*g# mass of #Cu(NO_3)_2#?

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Answer 1 · 2017-02-27T17:53:15

Approx. #1.01xx10^23# individual atoms.

We (i) we work out the molar quantity of #"copper(II) nitrate:"#

#=(3.50*g)/(187.56*g*mol^-1)=1.87xx10^-2*mol#

And (ii), we convert this molar quantity to a number of atoms.

In one formula unit of #Cu(NO_3)_2#, clearly, there are #9# actual atoms:#1xxCu, 2xxN, 6xxO#. Agreed?

So we work out the product, #1.87xx10^-2*molxx9*"atoms"*mol^-1=0.168*molxxN_A#.

Now by definition, in one mole of stuff there are #N_A-=6.022xx10^23# individual items of stuff.

And thus moles of atoms, #=0.168*molxx6.022xx10^23*mol^-1#

#=1.01xx10^23# individual atoms.