What is the empirical formula for a compound that consists of 72.2% magnesium and 27.8% nitrogen by mass?

It wouldn't be a bad idea to start your guessing with #"Mg"_3"N"_2#, as that is the expected ionic compound when #"Mg"^(2+)# forms a compound with #"N"^(3-)#. Ironically, that is the answer.

The molar masses of each atom are:

#"M"_"Mg" ~~ "24.305 g/mol"#

#"M"_"N" ~~ "14.007 g/mol"#

Using the knowledge that #72.2%# magnesium by mass, we can do this:

#%"Mg" = ("M"_"Mg,total")/("M"_"N,total" + "M"_"Mg,total")#

#0.722 = ("M"_"Mg,total")/("M"_"N,total" + "M"_"Mg,total")#

#0.722"M"_"N,total" + 0.722"M"_"Mg,total" = "M"_"Mg,total"#

#0.722"M"_"N,total" = 0.278"M"_"Mg,total"#

We should get:

#color(green)("M"_"Mg,total"/"M"_"N,total") = 0.722/0.278#

#~~ color(green)(2.5971)#

So we have the approximate mass ratio of magnesium to nitrogen in the compound, but not the #"mol"# ratio. Note that if we have a 1:1 ratio of #"mol"#s of magnesium to nitrogen, we have:

#color(green)(("M"_"Mg")/("M"_"N")) = 24.305/14.007 ~~ color(green)(1.735)#

But the ratio we have is:

#("M"_"Mg,total")/("M"_"N,total") ~~ 2.5971#

So, we have more magnesium (or less nitrogen) than we would have in a 1:1 ratio. As a result, if we divide these two numbers like so, we get the #\mathbf"mol"# ratio of magnesium to nitrogen:

#2.5971/1.735 ~~ 1.5#

Since #1.5 = 3/2#, we can scale this up to determine the empirical formula. It's evidently:

#color(blue)("Mg"_3"N"_2)#

In 100 g of compound there are 72.2 g magnesium, and 27.8 g nitrogen.

We divide thru by the molar masses of each element:

For the metal: #(72.2*cancelg)/(24.31*cancel(g)*mol^-1)# #~=# #3# #mol#;

And for nitrogen: #(27.8*cancelg)/(14.01*cancel(g)*mol^-1)# #~=# #2# #mol#.

So the empirical formula is #Mg_3N_2#. Because this is clearly NOT a molecular species, the empirical formula is all that is needed to identify it.