# How can i Calculate the mass of oxygen in 350 g of Al(C2H3O2)3?

Well, what is your molar quantity of $\text{aluminum acetate}$?
You gots....(350*g)/(204.11 *g·mol^-1)=1.72*mol
And given ${\text{Al(O(O=)CCH"_3")}}_{3}$....CLEARLY there are SIX moles of oxygen per mole of aluminum salt.
And so we got $10.3 \cdot m o l \cdot \text{oxygen}$ in this quantity...