# What molar quantity of the metal salt are present in a 0.667*g mass of sodium dichromate?

$\text{5.10 mmol}$
$\text{Moles of sodium dichromate} = \frac{0.667 \cdot g}{261.97 \cdot g \cdot m o l} = 2.55 \times {10}^{-} 3 \cdot m o l$.
Given the composition of sodium dichromate, $N {a}_{2} C {r}_{2} {O}_{7}$, there are twice this molar quantity of sodium ions. How many oxygen atoms?