A sample of an organic compound with a mass of 0.420 g holds 16.67 % nitrogen by mass…

Step 1. $m(\text{N}) = 0.420\ \text{g} \times 0.1667 = 0.0700\ \text{g}$. The amount is $\dfrac{0.0700\ \text{g}}{14\ \text{g/mol}} = 5.00\ \text{mmol}$. Each nitrogen atom becomes one cyanide ion, so $n(\text{CN}^-) = 5.00\ \text{mmol}$.

Step 2. $\dfrac{5.00}{6} = 0.833\ \text{mmol}$ of $[\text{Fe}(\text{CN})_{6}]^{4-}$.

Step 3. Four iron(III) ions for every three complex ions: $0.833 \times \dfrac{4}{3} = 1.11\ \text{mmol}$ of $\text{Fe}^{3+}$.

Step 4. One formula unit uses three complex ions, so $n = \dfrac{0.833}{3} = 0.278\ \text{mmol}$. The mass is $0.278\ \text{mmol} \times 860\ \text{mg/mmol} = 239\ \text{mg} = 0.239\ \text{g}$.