Question

What is the % nitrogen by weight in 6.389 mol of ammonium sulfide? Answer in units of %.

Answer 1

42.36 %

Explanation

What's the formula of ammonium sulfide?

Consider the two ions:

• Ammonium: ${\text{NH}_4}^{+}$.
• Sulfide: $\text{S}^{2-}$.

The charge on each sulfide ion is twice that on each ammonium ion. You shall expect two ammonium ions for each sulfide ion:

• Ammonium sulfide: $(\text{NH}_4})_{\bf 2}\text{S}$.

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 $(\text{NH}_4})_2\text{S}$. There's one nitrogen atom in each ammonium ion. Therefore, there are two moles of nitrogen atoms in each mole formula unit of $(\text{NH}_4})_2\text{S}$.

Relative atomic mass from a modern periodic table:

• Nitrogen: 14.007;
• Hydrogen: 1.008;
• Sulfur: 32.06.

The mass of that two moles of nitrogen atoms will be $14.007 \times 2 = 28.014 \; \text{g}$.

What's the %mass of nitrogen in $(\text{NH}_4})_2\text{S}$?

$(\text{NH}_4})_2\text{S}$ 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 $(\text{NH}_4})_2\text{S}$ shall be the same as that in one mole formula units of $(\text{NH}_4})_2\text{S}$. (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 $(\text{NH}_4})_2\text{S}$?

$M((\text{NH}_4})_2\text{S}) = 2 \times \underbrace{(14.007 + 3 \times 1.008)}_{\text{NH}_4} + \underbrace{32.06}_{\text{S}} = 66.122 \; \text{g} \cdot \text{mol}^{-1}$.

$m((\text{NH}_4})_2\text{S}) = n((\text{NH}_4})_2\text{S}) \cdot M((\text{NH}_4})_2\text{S}) = 66.126 \; \text{g}$.

What the percentage mass of nitrogen in ammonium sulfide?

$\% m(\text{N}) = \dfrac{m(\text{N})}{m((\text{NH}_4)_2\text{S})} \times 100\% = \dfrac{28.014}{66.122} \times 100\% = 42.37 \%$.