# If I have 5.00 moles of "NH"_4""^+", how many ions do I have? Grams?

Mar 10, 2016

$\text{5.00 mol NH"_4""^+"ions}$ contain $3.01 \times {10}^{24} \text{NH"_4""^+"ions}$ and have a mass of $\text{90.2 g NH"_4""^+"ions}$.

#### Explanation:

I edited the question to reflect the fact that $\text{NH"_4""^+}$ is the ammonium cation, not a molecule. It is a polyatomic ion.

The relationship between moles and ions is:

$1 \text{mol ions"=6.022xx10^23"ions}$

In order to determine the number of ions in a given number of moles, multiply the given moles by $6.022 \times {10}^{23} \text{ions/mol}$.

In the case of 5.00 mol $\text{NH"_4""^+}$:

$5.00 {\cancel{\text{mol NH"_4""^+xx(6.022xx10^23"NH"_4""^+)/(1cancel"mol NH"_4""^+)=3.01xx10^24"NH"_4}}}^{+}$ ions rounded to three significant figures

In order to determine the mass of a given number of moles, multiply the moles by the molar mass (MM), which is determined by multiplying the subscript of each element by its atomic weight in g/mol from the periodic table.

$\text{MM"_"NH4"=(1xx14.007"g/mol")+(4xx1.008"g/mol")="18.039 g/mol}$

$5.00 \cancel{\text{mol NH"_4""^+xx(18.039"g NH"_4""^+)/(1cancel"mol NH"_4""^+)="90.2 g NH"_4""^+}}$ ions rounded to three significant figures