# How do you calculate the percentage composition of Oxygen in Zn(NO_3)_2?

Feb 25, 2016

$\frac{6 \times 15.999 \cdot g \cdot m o {l}^{-} 1}{189.36 \cdot g \cdot m o {l}^{-} 1}$ $\times$ 100% $=$ ??
$15.999 \cdot g \cdot m o {l}^{-} 1$ is of course the molar mass of the oxygen atom. $189.36 \cdot g \cdot m o {l}^{-} 1$ is the molar mass of zinc nitrate. So it is a bit under 50% by mass.
$1 \times Z n + 2 \times N + 6 \times O$