# How many oxygen ATOMS are present in a 75*g mass of water?

Approx. $70 \cdot g$ oxygen.
There are $\frac{75 \cdot g}{18.0 \cdot g \cdot m o {l}^{-} 1}$ $=$ $4.17 \text{ moles of water}$.
So, we simply have to mulitply this molar quantity by the mass of atomic oxygen: $4.17 \cdot m o l \times 15.999 \cdot g \cdot m o {l}^{-} 1$ $\cong 70 \cdot g \text{ oxygen}$