The calculation might be easier if we multiply all the atoms inside parentheses by the subscript 2.
Then the formula becomes #"MgCl"_2"O"_8#.
We see that one formula unit of #"Mg"("ClO"_4)_2# contains 1 #"Mg"# atom, 2 #"Cl"# atoms, and 8 #"O"# atoms.
Then 1 mol of #"Mg"("ClO"_4)_2# contains 1 mol of #"Mg"# atoms, 2 mol of #"Cl"# atoms, and 8 mol of #"O"# atoms.
#"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"#
#"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"#
#"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"#
