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

Feb 27, 2017

Approx. $1.01 \times {10}^{23}$ individual atoms.

#### Explanation:

We (i) we work out the molar quantity of $\text{copper(II) nitrate:}$

$= \frac{3.50 \cdot g}{187.56 \cdot g \cdot m o {l}^{-} 1} = 1.87 \times {10}^{-} 2 \cdot m o l$

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

In one formula unit of $C u {\left(N {O}_{3}\right)}_{2}$, clearly, there are $9$ actual atoms:$1 \times C u , 2 \times N , 6 \times O$. Agreed?

So we work out the product, $1.87 \times {10}^{-} 2 \cdot m o l \times 9 \cdot \text{atoms} \cdot m o {l}^{-} 1 = 0.168 \cdot m o l \times {N}_{A}$.

Now by definition, in one mole of stuff there are ${N}_{A} \equiv 6.022 \times {10}^{23}$ individual items of stuff.

And thus moles of atoms, $= 0.168 \cdot m o l \times 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$

$= 1.01 \times {10}^{23}$ individual atoms.