Question

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

Answer 1

`"5.00 mol NH"_4""^+"ions"` contain `3.01xx10^24"NH"_4""^+"ions"` and have a mass of `"90.2 g NH"_4""^+"ions"`.

Explanation:

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

The relationship between moles and ions is:

`1"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 `6.022xx10^23"ions/mol"`.

In the case of 5.00 mol `"NH"_4""^+"`:

`5.00cancel"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.

`"MM"_"NH4"=(1xx14.007"g/mol")+(4xx1.008"g/mol")="18.039 g/mol"`

`5.00cancel"mol NH"_4""^+xx(18.039"g NH"_4""^+)/(1cancel"mol NH"_4""^+)="90.2 g NH"_4""^+"` ions rounded to three significant figures