# What is the mass and number of moles of 2.408*10^21 molecules of ammonia?

$6.022 \times {10}^{23}$ ammonia molecules have a mass of $17.03 \cdot g$. And so.........
$\frac{2.408 \times {10}^{21}}{6.022 \times {10}^{23} \cdot m o {l}^{-} 1} \times 17.03 \cdot g \cdot m o {l}^{-} 1$ $\cong$ $7.0 \times {10}^{-} 2 \cdot g$ of ammonia.
As you are no doubt aware, the number $6.022 \times {10}^{23}$ is $\text{Avogadro's number, } {N}_{A}$, which is the link between the micro world of atoms and molecules to the macro world of grams and kilograms. ${N}_{A}$ ammonia molecules have a mass of $17.03 \cdot g$.
To belabour the point, $\text{Avogadro's number, } {N}_{A}$ ammonia molecules, which is equivalent to $3 \times {N}_{A}$ hydrogen atoms, and $1 \times {N}_{A}$ nitrogen atoms, has a mass of $17.03 \cdot g$. Capisce?