Answer:
The empirical formula of the compound is [tex]Mg_3N_2[/tex].
Explanation:
Mass of magnesium in compound = 0.273 g
Mass of nitrogen in the compound = 0.105 g
Moles of magnesium =[tex]\frac{0.273 g}{24.305 g/mol}=0.0112 mol[/tex]
Moles of nitrogen = [tex]\frac{0.105 g}{14.0067 g/mol}=0.00750 mol[/tex]
Form empirical formula divides the lowest value of moles of an element present with all the moles of elements present.
Magnesium [tex]=\frac{0.0112 mol}{0.00750 mol}=1.5[/tex]
Nitrogen [tex]=\frac{0.00750mol}{0.00750 mol}=1[/tex]
The empirical formula of a compound:
[tex]Mg_{1.5}N_1=Mg_{\frac{15}{10}}N_1=Mg_{15}N_{10}=Mg_3N_2[/tex]
The empirical formula of the compound is [tex]Mg_3N_2[/tex].
