# What is the total mass of oxygen in 1.00 mole of Al_2(CrO_4)_3?

In $1$ $m o l$ $A {l}_{2} {\left(C r {O}_{4}\right)}_{3}$ there are $12$ $\text{mol oxygen atoms}$, or $192$ $g$.
Here, I speak of oxygen ATOMS (and not oxygen molecules). Clearly, there are 4 moles of oxygen atoms in one mole of chromate ion. We have a molar quantity of aluminum chromate, and thus there is 12-fold molar quantity of oxygen ATOMS: i.e. a mass of $12$ $m o l$ $\times$ $16.00 \cdot g \cdot m o {l}^{-} 1$ $=$ $192.0$ $g$.