# How many atoms in a mass of ONE MOLE of "magnesium nitrate"?

You have the formula $M g {\left(N {O}_{3}\right)}_{2}$....
And so there are $1 \times M g + 2 \times N + 6 \times O \cdot \text{atoms}$.......
...which is $148.30 \cdot g \cdot m o {l}^{-} 1$. And thus in $\text{Avogadro's number of magnesium nitrate formula units,}$ $\text{i.e. a mass of 148.3 g}$, ${N}_{A} \equiv 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$, there are $1 \times {N}_{A} \left(M g\right) + 2 \times {N}_{A} \left(N\right) + 6 \times {N}_{A} \left(O\right) \equiv 9 {N}_{A}$ atoms in toto....