# What is the molar mass of magnesium nitride, ##”Mg”_3″N”_2##, if the mass of one mole of magnesium is 24.31 g and of one mole of nitrogen is 14.01 g?

##”100.95 g mol”^(-1)##

The idea here is that you need to use the chemical formula for magnesium nitride, ##”Mg”_3″N”_2##, to calculate the mass of one mole of the compound, i.e. its molar mass.

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What is the molar mass of magnesium nitride, ##”Mg”_3″N”_2##, if the mass of one mole of magnesium is 24.31 g and of one mole of nitrogen is 14.01 g?

So, magnesium nitride is an ionic compound made up of magnesium cations, ##”Mg”^(2+)##, and nitride anions, ##”N”^(3-)##. The subscripts used in the compound’s chemical formula tell you how many ions you get in one formula unit of magnesium nitride.

In this case, you have

##”Mg”_ color(red)(3) “N”_ color(blue)(2) -> color(red)(3)”Mg”^color(blue)(2+) + color(blue)(2)”N”^color(red)(3-)##

So, one formula unit of magnesium nitride contains ##color(red)(3)## magnesium cations and ##color(blue)(2)## nitride anions.

This means that one mole of magnesium nitride will contain ##color(red)(3)## moles of magnesium cations and ##color(blue)(2)## moles of nitride anions.

Now, you know that one mole of magnesium has a mass of ##”24.31 g”##. Since you get three moles of magnesium cations per mole of the compound, magnesium’s contribution to the molar mass will be

##color(red)(3) color(red)(cancel(color(black)(“moles Mg”^(2+)))) * “24.31 g”/(1color(red)(cancel(color(black)(“mole Mg”^(2+))))) = “72.93 g”##

Likewise, you know that one mole of nitrogen has a mass of ##”14.01 g”##. Since you get two moles of nitride anions per mole of the compound, nitrogen’s contribution will be

##color(blue)(2) color(red)(cancel(color(black)(“moles N”^(3-)))) * “14.01 g”/(1color(red)(cancel(color(black)(“mole N”^(3-))))) = “28.02 g”##

Therefore, the total mass of one mole of magnesium nitride will be

##”72.93 g” + “28.02 g” = “100.95 g”##

Since this is how many grams you get per mole, you can say that the molar mass will be

##”molar mass” = color(green)(|bar(ul(color(white)(a/a)color(black)(“100.95 g mol”^(-1))color(white)(a/a)|)))##

I’ll leave the answer rounded to five , despite the fact that you only have four sig figs for the molar masses of the two .

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