# A 26.9-g sample of ammonium carbonate contains how many mol of ammonium ions?

Jun 18, 2017

The sample of ammonium carbonate contains ${\text{0.560 mol NH"_4}}^{+}$.

#### Explanation:

The chemical formula for ammonium carbonate is ("NH"_4)_2"CO"_3". The formula indicates that in one mole of ammonium carbonate, there are two moles of ammonium ions, ${\text{NH"_4}}^{+}$.

To calculate moles of ammonium ions, requires two steps.

First determine moles ("NH"_4")"_2"CO"_3 in ${\text{26.9 g (NH"_4")"_2"CO}}_{3}$ by multiplying by the inverse of its molar mass, $\text{96.086 g/mol}$.

26.9color(red)cancel(color(black)("g (NH"_4")"_2"CO"_3))xx(1"mol (NH"_4")"_2"CO"_3)/(96.086color(red)cancel(color(black)("g (NH"_4")"_2"CO"_3)))="0.280 mol (NH"_4")"_2"CO"_3"

Next determine the mol ${\text{NH"_4}}^{+}$ in the sample, by multiplying mol ("NH"_4")"_2"CO"_3 by the mole ratio of ${\text{NH"_4}}^{+}$ and ("NH"_4")"_2"CO"_3.

0.280color(red)cancel(color(black)("mol (NH"_4")"_2"CO"_3))xx(2"mol NH"_4^+)/(1color(red)cancel(color(black)("mol (NH"_4")"_2"CO"_3)))="0.560 NH"_4"^+

rounded to three sig figs