42.36 %
What's the formula of ammonium sulfide?
Consider the two ions:
The charge on each sulfide ion is twice that on each ammonium ion. You shall expect two ammonium ions for each sulfide ion:
What's the mass of nitrogen in one mole formula unit of ammonium sulfide?
There are two moles of ammonium ions in one mole formula unit of [tex](\text{NH}_4})_2\text{S}[/tex]. There's one nitrogen atom in each ammonium ion. Therefore, there are two moles of nitrogen atoms in each mole formula unit of [tex](\text{NH}_4})_2\text{S}[/tex].
Relative atomic mass from a modern periodic table:
The mass of that two moles of nitrogen atoms will be [tex]14.007 \times 2 = 28.014 \; \text{g}[/tex].
What's the %mass of nitrogen in [tex](\text{NH}_4})_2\text{S}[/tex]?
[tex](\text{NH}_4})_2\text{S}[/tex] is an ionic compound. The composition of this compound shall be uniform throughout its ionic lattice. As a result, the percentage mass of nitrogen in 6.289 moles formula units of [tex](\text{NH}_4})_2\text{S}[/tex] shall be the same as that in one mole formula units of [tex](\text{NH}_4})_2\text{S}[/tex]. (However, finding the value in the case with one mole of the substance is likely easier.)
Now, what's the mass of one mole formula units of [tex](\text{NH}_4})_2\text{S}[/tex]?
[tex]M((\text{NH}_4})_2\text{S}) = 2 \times \underbrace{(14.007 + 3 \times 1.008)}_{\text{NH}_4} + \underbrace{32.06}_{\text{S}} = 66.122 \; \text{g} \cdot \text{mol}^{-1}[/tex].
[tex]m((\text{NH}_4})_2\text{S}) = n((\text{NH}_4})_2\text{S}) \cdot M((\text{NH}_4})_2\text{S}) = 66.126 \; \text{g}[/tex].
What the percentage mass of nitrogen in ammonium sulfide?
[tex]\% m(\text{N}) = \dfrac{m(\text{N})}{m((\text{NH}_4)_2\text{S})} \times 100\% = \dfrac{28.014}{66.122} \times 100\% = 42.37 \%[/tex].
