# What is the mass of 19.42 atoms of F?

Well $6.022 \times {10}^{23}$ atoms of fluorine (as the atom, not as ${F}_{2}$) have a mass of $19.00 \cdot g$.
And thus $\frac{19}{6.022 \times {10}^{23} \cdot m o {l}^{-} 1} \times 19.00 \cdot g \cdot m o {l}^{-} 1$
$=$ $\text{not too many grams}$.
We cannot really speak of $0.42 \cdot \text{atoms}$ of fluorine. We can have the one atomic particle; we cannot really have 1/2 of an atom.