Question

Wk-14

Date
Page
Assignment
1. Calculate the percentage of iron in K3 Fe (CN)
[K= 39, Fe =56, C= 12, N =14]

Answer 1

The mass percentage of iron in
`\rm K_3[Fe(CN)_6]` (potassium ferricyanide) is approximately
`17\%`.

Step-by-step explanation:

Start by finding the formula mass of potassium ferricyanide,
`\rm K_3[Fe(CN)_6]`, using relative atomic mass data from the question:

`\begin{aligned} & M(\mathrm{K_3[Fe(CN)_6]}) \\ &\approx 3 * 39 + 56 + 6 * (12 + 14) \\ &= \rm 329 \; g \cdot mol^(-1)\end{aligned}`.

In other words, every one mole of
`\rm K_3[Fe(CN)_6]` formula units have a mass of approximately
`329\; \rm g`.

On the other hand, note that there is one iron
`\rm Fe` atom in every one mole of
`\rm K_3[Fe(CN)_6]` formula units. Hence, there would be exactly one mole of iron atoms in every one mole of
`\rm K_3[Fe(CN)_6]` formula units. The mass of that one mole of iron atoms is approximately
`56\; \rm g` (again, from the relative atomic mass data of the question.)

Therefore:

`\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}}* 100\% \\ &= \frac{M(\mathrm{Fe}) * (\text{Number of $\mathrm{Fe}$ atoms in each $\mathrm{K_3[Fe(CN)_6]}$ formula unit)}}{M(\mathrm{K_3[Fe(CN)_6]})} * 100\%\end{aligned}`

`\begin{aligned}&\approx (1 * 56\; \rm g \cdot mol^(-1))/(329\; \rm g \cdot mol^(-1))\end{aligned}`.