# How many grams are there in 4.5 * 10^22 formula units of Ba(NO_2)_2?

Well, if there are $6.022 \times {10}^{23}$ $B a {\left(N {O}_{2}\right)}_{2}$ formula units there are $229.34 \cdot g$ by definition.
The molar mass of $B a {\left(N {O}_{2}\right)}_{2}$ is $229.34 \cdot g \cdot m o {l}^{-} 1$.
To get the mass of the given quantity, all we have do is divide this quantity by $\text{Avogadro's number}$, and then multiply by the molar mass.
$\frac{4.5 \times {10}^{22}}{6.022 \times {10}^{23} \cdot m o {l}^{-} 1}$ $\times 229.34 \cdot g \cdot m o {l}^{-} 1$ $=$ ??g