# How many grams of O are contained in 8.52 g K_2CO_3?

Nov 12, 2015

Moles of ${K}_{2} C {O}_{3}$ $=$ $\frac{8.52 \cdot g}{131.21 \cdot g \cdot m o {l}^{-} 1}$ $=$ ?? $m o l$.
But there are 3 mol of oxygen in each mole of ${K}_{2} C {O}_{3}$.
So moles of oxygen $=$ 3xx(8.52*g)/(131.21*g*mol^-1) = ?? mol.