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?

1 Answer
Jun 10, 2016

Answer:

#"100.95 g mol"^(-1)#

Explanation:

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.

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 sig figs, despite the fact that you only have four sig figs for the molar masses of the two elements.