# How many moles of magnesium, chlorine, and oxygen are in 7.90 mol of "Mg"("ClO"_4)_2?

Jun 15, 2017

There are 7.90 mol of $\text{Mg}$, 15.8 mol of $\text{Cl}$, and 63.2 mol of $\text{O}$.

#### Explanation:

The calculation might be easier if we multiply all the atoms inside parentheses by the subscript 2.

Then the formula becomes ${\text{MgCl"_2"O}}_{8}$.

We see that one formula unit of "Mg"("ClO"_4)_2 contains 1 $\text{Mg}$ atom, 2 $\text{Cl}$ atoms, and 8 $\text{O}$ atoms.

Then 1 mol of "Mg"("ClO"_4)_2 contains 1 mol of $\text{Mg}$ atoms, 2 mol of $\text{Cl}$ atoms, and 8 mol of $\text{O}$ atoms.

$\text{Moles of Mg atoms" = 7.90 color(red)(cancel(color(black)("mol Mg"("ClO"_4)_2))) × "1 mol Mg atoms"/(1 color(red)(cancel(color(black)("mol Mg"("ClO"_4)_2)))) = "7.90 mol Mg atoms}$

$\text{Moles of Cl atoms" = 7.90 color(red)(cancel(color(black)("mol Mg"("ClO"_4)_2))) × "2 mol Cl atoms"/(1 color(red)(cancel(color(black)("mol Mg"("ClO"_4)_2)))) = "15.8 mol Cl atoms}$

$\text{Moles of O atoms" = 7.90 color(red)(cancel(color(black)("mol Mg"("ClO"_4)_2))) × "8 mol O atoms"/(1 color(red)(cancel(color(black)("mol Mg"("ClO"_4)_2)))) = "63.2 mol O atoms}$