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

Dec 28, 2015

It wouldn't be a bad idea to start your guessing with ${\text{Mg"_3"N}}_{2}$, as that is the expected ionic compound when ${\text{Mg}}^{2 +}$ forms a compound with ${\text{N}}^{3 -}$. Ironically, that is the answer.

The molar masses of each atom are:

$\text{M"_"Mg" ~~ "24.305 g/mol}$

$\text{M"_"N" ~~ "14.007 g/mol}$

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

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

$0.722 = \left(\text{M"_"Mg,total")/("M"_"N,total" + "M"_"Mg,total}\right)$

$0.722 \text{M"_"N,total" + 0.722"M"_"Mg,total" = "M"_"Mg,total}$

$0.722 \text{M"_"N,total" = 0.278"M"_"Mg,total}$

We should get:

$\textcolor{g r e e n}{\text{M"_"Mg,total"/"M"_"N,total}} = \frac{0.722}{0.278}$

$\approx \textcolor{g r e e n}{2.5971}$

So we have the approximate mass ratio of magnesium to nitrogen in the compound, but not the $\text{mol}$ ratio. Note that if we have a 1:1 ratio of $\text{mol}$s of magnesium to nitrogen, we have:

$\textcolor{g r e e n}{\left(\text{M"_"Mg")/("M"_"N}\right)} = \frac{24.305}{14.007} \approx \textcolor{g r e e n}{1.735}$

But the ratio we have is:

$\left(\text{M"_"Mg,total")/("M"_"N,total}\right) \approx 2.5971$

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 $\setminus m a t h b f \text{mol}$ ratio of magnesium to nitrogen:

$\frac{2.5971}{1.735} \approx 1.5$

Since $1.5 = \frac{3}{2}$, we can scale this up to determine the empirical formula. It's evidently:

$\textcolor{b l u e}{{\text{Mg"_3"N}}_{2}}$

Dec 28, 2015

$M {g}_{3} {N}_{2}$

#### 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: $\frac{72.2 \cdot \cancel{g}}{24.31 \cdot \cancel{g} \cdot m o {l}^{-} 1}$ $\cong$ $3$ $m o l$;

And for nitrogen: $\frac{27.8 \cdot \cancel{g}}{14.01 \cdot \cancel{g} \cdot m o {l}^{-} 1}$ $\cong$ $2$ $m o l$.

So the empirical formula is $M {g}_{3} {N}_{2}$. Because this is clearly NOT a molecular species, the empirical formula is all that is needed to identify it.