Answer:
Approximately [tex]28.6\%[/tex] (by mass.)
Explanation:
The empirical formula [tex]{\rm Li_{3} N}[/tex] gives the ratio between the atoms in this compound. This formula also gives the composition of each formula unit of this compound.
According to this formula, every formula unit of [tex]{\rm Li_{3} N}\![/tex] would consist of three [tex]{\rm Li}[/tex] atoms and one [tex]{\rm N}[/tex] atom.
The mass of one such formula unit would be:
[tex]\begin{aligned}& \text{mass of one ${\rm Li_{3} N}$ formula unit} \\ =\; & 3 \times (\text{mass of one ${\rm Li}$ atom}) \\ & + 1 \times (\text{mass of one ${\rm N}$ atom}) \\ =\; & 3 \times 6.94 \; {\rm amu} + 1 \times 14.01\; {\rm amu} \\ =\; & 48.97\; {\rm amu} \end{aligned}[/tex].
(The unit [tex]\text{amu}[/tex] is an abbreviation for atomic mass unit.)
Out of that [tex]48.97\; {\rm amu}[/tex], [tex]{\rm N}[/tex] would account for a mass of [tex]14.01\; {\rm amu}[/tex] (one atom.) The percentage of [tex]{\rm N}\![/tex] in this formula unit would be:
[tex]\displaystyle \frac{14.01\; {\rm amu}}{48.97\; {\rm amu}} \times 100\% \approx 28.6\%[/tex].
This percentage would approximately be the same as the mass percentage of [tex]{\rm N}\![/tex] in the [tex]{\rm Li_{3} N}[/tex] compound.
