# Question 153f5

Feb 17, 2017

$6.44 \cdot {10}^{3} {\text{moles NiSO}}_{4}$

#### Explanation:

Technically, you should say that nickel(II) sulfate contains nickel(II) cations, ${\text{Ni}}^{2 +}$, not nickel atoms, $\text{Ni}$, but in this case, you can go for nickel atoms of nickel(II) cations.

Start by converting the number of atoms of nickel to moles. To do that, use the fact that $1$ mole of any element contains $6.022 \cdot {10}^{23}$ atoms of that element $\to$ this is known as Avogadro's constant.

Therefore, you know that

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{1 mole Ni" = 6.022 * 10^(23)color(white)(.)"atoms Ni}}}}$

You can thus say that your sample o nickel(II) sulfate contains

3.88 * 10^(27) color(red)(cancel(color(black)("atoms Ni"))) * "1 mole Ni"/(6.022 * 10^(23)color(red)(cancel(color(black)("atoms Ni")))) = 6.44 * 10^3color(white)(.)"moles Ni"#

Now, the chemical formula for nickel(II) sulfate looks like this

${\text{NiSO}}_{4}$

Notice that $1$ mole of nickel(II) sulfate formula units contains $1$ mole of nickel atoms, which means that your sample will contain

$6.44 \cdot {10}^{3} \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{moles Ni"))) * "1 mole NiSO"_4/(1color(red)(cancel(color(black)("moles Ni")))) = color(darkgreen)(ul(color(black)(6.44 * 10^(3)color(white)(.)"moles NiSO}}_{4}}}}$

The answer is rounded to three sig figs.