# How many moles are in 4.81 * 10^14 molecules of NH_3?

There are $\frac{4.81 \times {10}^{14}}{6.022 \times {10}^{23} \cdot m o {l}^{-} 1}$ ammonia molecules.
The mole is simply a number (like a dozen, or 10, or 100); admittedly it's a very large number. If I have Avogadro's number (${N}_{A}$) or ammonia molecules then I have Avogadro's number of nitrogen atoms, and $3 \times {N}_{A}$ of hydrogen atoms. Are you with me? So all you have to do is divide the given quantity by Avogadro's number as set out in the intro. Can you tell me the mass of this quantity of ammonia?