Question

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

Answer 1

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.

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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)`

Answer 2

`Mg_3N_2`

Explanation:

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.