Answer : The empirical formula of the compound is, (c) [tex]NaNH_2[/tex]
Solution : Given,
Moles of sodium, Na = 0.5 moles
Moles of nitrogen, N = 0.5 moles
Moles of hydrogen, H = 1.0 moles
Divide the each moles value by the smallest number of moles of given element.
[tex]\text{ Moles of Na}=\frac{0.5moles}{0.5moles}=1[/tex]
[tex]\text{ Moles of N}=\frac{0.5moles}{0.5moles}=1[/tex]
[tex]\text{ Moles of H}=\frac{1moles}{0.5moles}=2[/tex]
The mole ratio of the elements are,
Na : N : H = 1 : 1 : 2
The mole ratio of the element is represented by the subscripts in the empirical formula.
The empirical formula is written as, [tex]Na_1N_1H_2[/tex] or [tex]NaNH_2[/tex]
Therefore, the empirical formula of the compound is, (c) [tex]NaNH_2[/tex]
