# How many moles are in 9.461 * 10^24 formula units of manganese(III) dichromate?

##### 1 Answer
Dec 18, 2015

$\text{15.71 moles}$

#### Explanation:

As you know, ionic compounds form giant lattice structures in which the positively charge ions, or cations, and the negatively charged ions, or anions, are held together by electrostatic attraction.

This of course implies that you cannot speak of molecules of an ionic compound. Instead, the formula unit is used to express the smallest ratio that exists between the cations and anions in the ionic compound.

Simply put, the formula unit is the empirical formula of an ionic compound.

This means that one mole of manganese(II) dichromate, ${\text{MnCr"_2"O}}_{7}$, will contain $6.022 \cdot {10}^{23}$ formula units of manganese(II) dichromate - this is known as Avogadro's number.

So, you know that you're dealing with $9.461 \cdot {10}^{24}$ formula units of manganese(II) dichromate. Since you have more than $6.022 \cdot {10}^{23}$ formula units, you can predict that you will have more than one mole of the compound.

In fact, looking at the exponents, you can say that you'll have more than ten moles of the compound.

Confirm this by doing

9.461 * 10^(24)color(red)(cancel(color(black)("form. units"))) * ("1 mole MnCr"_2"O"_7)/(6.022 * 10^(23)color(red)(cancel(color(black)("form. units")))) = color(green)("15.71 moles MnCr"_2"O"_7)

The answer is rounded to four sig figs, the number of sig figs you have for the number of formula units.