A compound contains only Fe and O. A \(0.2729 \mathrm{g}\) sample of the compound was dissolved in \(50 \mathrm{mL}\) of concentrated acid solution, reducing all the iron to \(\mathrm{Fe}^{2+}\) ions. The resulting solution was diluted to \(100 \mathrm{mL}\) and then titrated with a \(0.01621 \mathrm{M} \mathrm{KMnO}_{4}\) solution. The unbalanced chemical equation for reaction between \(\mathrm{Fe}^{2+}\) and \(\mathrm{MnO}_{4}^{-}\) is given below. \(\begin{aligned} \mathrm{MnO}_{4}^{-}(\mathrm{aq})+& \mathrm{Fe}^{2+}(\mathrm{aq}) \longrightarrow \mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{Fe}^{3+}(\mathrm{aq}) \quad(\text { not balanced }) \end{aligned}\) The titration required \(42.17 \mathrm{mL}\) of the \(\mathrm{KMnO}_{4}\) solution to reach the pink endpoint. What is the empirical formula of the compound?

Step by step solution

01

Balance the Reaction

First, balance the half-reactions in acidic solution. For \(\mathrm{MnO}_{4}^{-}\to \mathrm{Mn}^{2+}\), the balanced half-reaction is \(5\mathrm{e}^{-}+8\mathrm{H}^{+}+\mathrm{MnO}_{4}^{-}\) gives \(\mathrm{Mn}^{2+} + 4\mathrm{H}_{2}\mathrm{O}\). And for \(\mathrm{Fe}^{2+}\) to \(\mathrm{Fe}^{3+}\), the balanced half-reaction is \(\mathrm{Fe}^{2+}\to \mathrm{Fe}^{3+}+\mathrm{e}^{-}\). For the overall balanced reaction, multiply the second half-reaction by 5 and add them to get 5\(\mathrm{Fe}^{2+}+ \mathrm{MnO}_{4}^{-}+ 8\mathrm{H}^{+}\) gives \(5\mathrm{Fe}^{3+}+ \mathrm{Mn}^{2+}+ 4\mathrm{H}_{2}\mathrm{O}\).

02

Find the Moles of \(\mathrm{KMnO}_{4}\)

The moles of \(\mathrm{MnO}_{4}^{-}\) can be calculated using the formula \(\mathrm{Moles}\ =\ \mathrm{Molarity} \times \mathrm{Volume}\). Here, molarity of \(\mathrm{KMnO}_{4}\) is 0.01621 M and volume used is 42.17 mL or 0.04217 L. Therefore, moles of \(\mathrm{MnO}_{4}^{-}\) = \(0.01621 \mathrm{M} \times 0.04217 \mathrm{L}\) = \(684.0 \times 10^{-6} \mathrm{mol}\).

03

Find the Moles of \(\mathrm{Fe}^{2+}\)

From the balanced chemical equation, five moles of \(\mathrm{Fe}^{2+}\) react with one mole of \(\mathrm{MnO}_{4}^{-}\). Thus, the moles of \(\mathrm{Fe}^{2+}\) is five times the moles of \(\mathrm{MnO}_{4}^{-}\), which is \(5 \times 684.0 \times 10^{-6} \mathrm{mol}\) = \(3420 \times 10^{-6} \mathrm{mol}\) or \(3.42 \times 10^{-3} \mathrm{mol}\).

04

Calculate the Empirical Formula

Now, find the mass of Fe in the sample by using the moles of Fe and its molar mass (\(55.85 \mathrm{g/mol}\)). The mass of Fe = \(3.42 \times 10^{-3} \mathrm{mol} \times 55.85 \mathrm{g/mol}\) = \(0.191 \mathrm{g}\). Subtracting this mass from the total mass of the compound, one gets the mass of oxygen, \(0.2729 \mathrm{g} - 0.191 \mathrm{g}\) = \(0.0819 \mathrm{g}\). Dividing this mass by the molar mass of oxygen (\(16.00 \mathrm{g/mol}\)) gives its number of moles, which is \(0.00512 \mathrm{mol}\). So, the mole ratio of Fe to O is about \(3.42 / 0.00512 = 668\). As the ratio needs to be integer, this is approximately \(2:1\), resulting in an empirical formula of \(Fe_2O\).

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