What is the mass in grams of 4.6*10^21 molecules of carbon tetrabromide?

Apr 15, 2016

$\left(4.6 \times {10}^{21} {\text{ molecules")/(6.022xx10^23" molecules mol}}^{-} 1\right) \times 331.61 \cdot g \cdot m o {l}^{-} 1$ $=$ ?? $g$
$6.022 \times {10}^{23}$ molecules, i.e. a mole, of $C B {r}_{4}$, have a mass of $331.61 \cdot g$.
(4.6xx10^21*"molecules")/(6.022xx10^23*"molecules"*"mol"^-1)xx331.61*g*mol^-1=?*g.