Question

If the mass of isolated ${\mathbf{Fe}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}$were 0.300 g, what would be the average mass of iron per tablet?

Short answer

The average mass of iron per tablet will be 17.5 mg

Step-by-step solution

Define average mass

An average atomic mass, also known as an atomic weight, is the weighted average mass of the atoms in a naturally occurring sample of an element.

Determine the average mass of iron  

Evaluate the number of moles of isolated ${\mathrm{Fe}}_2{\mathrm{O}}_2$corresponding to 0.300 g of $F{e}_2{O}_2$.

The number of moles of aferric oxide is calculated as:

$$\begin{gathered} {\mathrm{Fe}}_2{\mathrm{O}}_2=\frac{0.300\mathrm{g}}{159.69\mathrm{g}/\mathrm{mol}} \\ =0.0019\mathrm{mol} \end{gathered}$$

Each mol of ${\mathrm{Fe}}_2{\mathrm{O}}_2$has two mol of iron and so, 0.019mol of ${\mathrm{Fe}}_2{\mathrm{O}}_3$has $2\times 0.0019=0.0038$mol iron.

The mass of the iron in iron (i.e., 0.300g of ${\mathrm{Fe}}_2{\mathrm{O}}_3)$ is:

$$\begin{gathered} 0.0038\mathrm{mol}\mathrm{Fe}=\frac{\mathrm{mass}\mathrm{of}\mathrm{Fe}}{\mathrm{atomic}\mathrm{mass}\mathrm{of}\mathrm{Fe}} \\ \mathrm{mass}\mathrm{of}\mathrm{Fe}=0.0038\mathrm{mol}\times 55.845\mathrm{gFe} \\ =0.21\mathrm{gFe} \end{gathered}$$

Therefore, the average mass of the iron per iron tablet in each of the twelve tablets is:

$$\begin{gathered} 0.21\mathrm{g}/12=0.0175\mathrm{g} \\ =17.5\mathrm{mg} \end{gathered}$$