# Question #4dd5f

Jan 7, 2018

$1.3 \cdot {10}^{23}$ $\text{formula units}$

#### Explanation:

For starters, it's worth pointing out that you're not dealing with molecules of iron(III) fluoride here because this compound is ionic.

So a more accurate statement here would be that the problem wants you to find the number of formula units of iron(III) fluoride present in that sample.

Now, the key here is the molar mass of iron(III) fluoride, which you will find listed as ${\text{112.84 g mol}}^{- 1}$.

Since you know that the molar mass of the compound tells you the mass of exactly $1$ mole of iron(III) fluoride, you can say that you will get $\text{112.84 g}$ for every $1$ mole of iron(III) fluoride present in your sample.

You should also know that Avogadro's constant tells you that you need $6.022 \cdot {10}^{23}$ formula units of iron(III) fluoride in order to have $1$ mole of this compound.

So if you know that $1$ mole contains $6.022 \cdot {10}^{23}$ formula units and has a mass of $\text{112.84 g}$, you can say that your sample contains

$24 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{g"))) * (6.022 * 10^(23) quad "formula unit FeF"_3)/(112.84color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)(1.3 * 10^(23) quad "formula units FeF}}_{3}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the mass of iron(III) fluoride.