How many atoms constitute a 3.50*g3.50⋅g mass of Cu(NO_3)_2Cu(NO3)2?

1 Answer
anor277
Feb 27, 2017

Approx. 1.01xx10^231.01×1023 individual atoms.

Explanation:

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

=(3.50*g)/(187.56*g*mol^-1)=1.87xx10^-2*mol=3.50⋅g187.56⋅g⋅mol−1=1.87×10−2⋅mol

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

In one formula unit of Cu(NO_3)_2Cu(NO3)2, clearly, there are 99 actual atoms:1xxCu, 2xxN, 6xxO1×Cu,2×N,6×O. Agreed?

So we work out the product, 1.87xx10^-2*molxx9*"atoms"*mol^-1=0.168*molxxN_A1.87×10−2⋅mol×9⋅atoms⋅mol−1=0.168⋅mol×NA.

Now by definition, in one mole of stuff there are N_A-=6.022xx10^23NA≡6.022×1023 individual items of stuff.

And thus moles of atoms, =0.168*molxx6.022xx10^23*mol^-1=0.168⋅mol×6.022×1023⋅mol−1

=1.01xx10^23=1.01×1023 individual atoms.

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