# Question e25c7

Jul 2, 2015

9.12 moles and 164.5 g.

#### Explanation:

Ammonium sulfide, ${\left(N {H}_{4}\right)}_{2} S$, is an ionic compound formed when two ammonium cations, $N {H}_{4}^{+}$, form an ionic bond with one sulfide anion, ${S}^{2 -}$.

This means that every mole of ammonium sulfide will contain 2 moles of ammonium cations and 1 mole of sulfide ions.

Since you have 4.560 moles of ammonium sulfide, you'll also have

$4.560 \cancel{\text{moles"(NH_4)_2S) * ("2 moles "NH_4^(+))/(1cancel("mole"(NH_4)_2S)) = color(green)("9.120 moles } N {H}_{4}^{+}}$

To determine how many grams of ammonium sulfide would contain this many moles, use the ion's molar mass

$9.120 \cancel{\text{moles") * "18.04 g"/(1cancel("mole")) = color(green)("164.5 g } N {H}_{4}^{+}}$

ALTERNATIVE APPROACH

You can also determine how many grams of ammonium ions you'd get by using percent composition.

(2 * 18.04cancel("g/mol"))/(68.154cancel("g/mol")) * 100 = "52.94%" $N {H}_{4}^{+}$

This means that 100 g of ammonium sulfide will contain 52.94 g of ammonium ions.

The mass of ammonium sulfide would be

4.560cancel("moles") * "68.154 g"/(1cancel("mole")) = "310.78 g"#

This means that you get

$310.78 \cancel{\text{g"(NH_4)_2S) * ("52.94 g "NH_4^(+))/(100cancel("g"(NH_4)_2S)) = color(green)("164.5 g } N {H}_{4}^{+}}$