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

##### 2 Answers

It wouldn't be a bad idea to start your guessing with

The molar masses of each atom are:

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

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

Using the knowledge that **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

#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 **ratio of magnesium to nitrogen**:

#2.5971/1.735 ~~ 1.5#

Since *scale this up* to determine the **empirical formula**. It's evidently:

#color(blue)("Mg"_3"N"_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:

And for nitrogen:

So the empirical formula is