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

1 Answer
Mar 10, 2016

Answer:

#"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