# How many moles of oxygen are in 1 mole of Ca(NO_3)_2?

Dec 23, 2015

The only thing this is really asking you to do is to understand subscripts and parentheses.

Examining ${\left({\text{NO}}_{3}\right)}_{2}$ only, we see that the subscript $3$ indicates $\setminus m a t h b f \left(3\right)$ $\setminus m a t h b f \left(\text{mol}\right)$s of oxygen atoms within each $\text{mol}$ of ${\text{NO}}_{3}^{-}$ (nitrate) polyatomic ions.

The subscript $2$ indicates $\setminus m a t h b f \left(2\right)$ $\setminus m a t h b f \left(\text{mol}\right)$s of ${\text{NO}}_{3}^{-}$ polyatomic ions. As a result, we double the number of oxygens as well as the number of nitrogens (but we only care about the oxygens here) that we already have.

Thus, $\left(\text{O} \times 3\right) \times 2 = \setminus m a t h b f 6$ moles of oxygen atoms.