Answer:
The mass percentage of iron in [tex]\rm K_3[Fe(CN)_6][/tex] (potassium ferricyanide) is approximately [tex]17\%[/tex].
Explanation:
Start by finding the formula mass of potassium ferricyanide, [tex]\rm K_3[Fe(CN)_6][/tex], using relative atomic mass data from the question:
[tex]\begin{aligned} & M(\mathrm{K_3[Fe(CN)_6]}) \\ &\approx 3 \times 39 + 56 + 6 \times (12 + 14) \\ &= \rm 329 \; g \cdot mol^{-1}\end{aligned}[/tex].
In other words, every one mole of [tex]\rm K_3[Fe(CN)_6][/tex] formula units have a mass of approximately [tex]329\; \rm g[/tex].
On the other hand, note that there is one iron [tex]\rm Fe[/tex] atom in every one mole of [tex]\rm K_3[Fe(CN)_6][/tex] formula units. Hence, there would be exactly one mole of iron atoms in every one mole of [tex]\rm K_3[Fe(CN)_6][/tex] formula units. The mass of that one mole of iron atoms is approximately [tex]56\; \rm g[/tex] (again, from the relative atomic mass data of the question.)
Therefore:
[tex]\begin{aligned}& \text{Mass percentage of $\mathrm{Fe}$ in $\mathrm{K_3[Fe(CN)_6]}$} \\ &= \frac{\text{Mass of $\mathrm{Fe}$ in one mole of $\mathrm{K_3[Fe(CN)_6]}$ formula units}}{\text{Mass of one mole of $\mathrm{K_3[Fe(CN)_6]}$ formula units}}\times 100\% \\ &= \frac{M(\mathrm{Fe}) \times (\text{Number of $\mathrm{Fe}$ atoms in each $\mathrm{K_3[Fe(CN)_6]}$ formula unit)}}{M(\mathrm{K_3[Fe(CN)_6]})} \times 100\%\end{aligned}[/tex]
[tex]\begin{aligned}&\approx \frac{1 \times 56\; \rm g \cdot mol^{-1}}{329\; \rm g \cdot mol^{-1}}\end{aligned}[/tex].
