# Question 1dca1

Sep 22, 2017

$\text{4.81 g}$

#### Explanation:

Start by converting the number of formula units of nickel(II) oxalate, ${\text{NiC"_2"O}}_{4}$, to moles by using Avogadro's constant.

Since you know that in order to have $1$ mole of nickel(II) oxalate you need to have $6.022 \cdot {10}^{23}$ formula units of this ionic compound, you can say that your sample will be equivalent to

4.93 * 10^(22) color(red)(cancel(color(black)("formula units NiC"_2"O"_4))) * overbrace(("1 mole NiC"_2"O"_4)/(6.022 * 10^(23)color(red)(cancel(color(black)("formula units NiC"_2"O"_4)))))^(color(blue)("Avogadro's constant"))#

$= {\text{0.08187 moles NiC"_2"O}}_{4}$

Now, the chemical formula of nickel(II) oxalate tells you that $1$ mole of this compound contains

• one mole of nickel(II) cations, $1 \times {\text{Ni}}^{2 +}$
• one mole of oxalate anions*, $1 \times {\text{C"_2"O}}_{4}^{2 -}$

This means that your sample will contain $0.08187$ moles of nickel(II) cations.

Finally, to convert this to grams, use the fact that the mass of an ion is approximately equal to the mass of the neutral atom. This will allow you to use the molar mass of nickel to find the mass of the sample.

$0.08187 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles Ni"^(2+)))) * "58.6934 g"/(1color(red)(cancel(color(black)("mole Ni"^(2+))))) = color(darkgreen)(ul(color(black)("4.81 g}}}}$

The answer is rounded to three sig figs, the number of sig figs you have for the number of formula units of nickel(II) oxalate.