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

Md Ariful Islam

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Md Ariful Islam

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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.

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

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

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

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