Question

A A compound has been isolated that can have either of two possible formulas: (a) \(\mathrm{K}\left[\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]\) or (b) \(\mathrm{K}_{3}\left[\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right] .\) To find which is correct, you dissolve a weighed sample of the compound in acid, forming oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\). You then titrate this acid with potassium permanganate, \(\mathrm{KMnO}_{4}\) (the source of the \(\left.\mathrm{MnO}_{4}-\text { ion }\right) .\) The balanced, net ionic equation for the titration is $$ \begin{aligned} 5 \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq})+& 2 \mathrm{MnO}_{4}-(\mathrm{aq})+6 \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq}) \rightarrow \\ & 2 \mathrm{Mn}^{2+}(\mathrm{aq})+10 \mathrm{CO}_{2}(\mathrm{g})+14 \mathrm{H}_{2} \mathrm{O}(\ell) \end{aligned} $$ Titration of \(1.356 \mathrm{g}\) of the compound requires \(34.50 \mathrm{mL}\) of \(0.108 \mathrm{M} \mathrm{KMnO}_{4} .\) Which is the correct formula of the iron-containing compound: (a) or (b)?

Short answer

The formula is (a) \( \mathrm{K} \left[\mathrm{Fe}\left(\mathrm{C}_2\mathrm{O}_4\right)_2\left(\mathrm{H}_2\mathrm{O}\right)_2\right] \).

Step-by-step solution

Determine Moles of Permanganate Used

Calculate the moles of \( \mathrm{KMnO}_4 \) used during the titration. Use the molarity formula:\[\text{Moles of } \mathrm{MnO}_4^- = 0.03450 \, \text{L} \times 0.108 \, \text{mol/L} = 0.003726 \, \text{mol}\]

Utilize Stoichiometry of Reaction

From the balanced net ionic equation, observe that 2 moles of \( \mathrm{MnO}_4^- \) react with 5 moles of \( \mathrm{H}_2\mathrm{C}_2\mathrm{O}_4 \). Use this stoichiometry ratio to find moles of oxalic acid:\[\text{Moles of } \mathrm{H}_2\mathrm{C}_2\mathrm{O}_4 = \frac{5}{2} \times 0.003726 = 0.009315 \, \text{mol}\]

Find Moles of Compound Corresponding to Molar Mass of Options

For compound (a) \( \mathrm{K} \left[\mathrm{Fe}\left(\mathrm{C}_2\mathrm{O}_4\right)_2\left(\mathrm{H}_2\mathrm{O}\right)_2\right] \), each formula unit contains 2 moles of \( \mathrm{C}_2\mathrm{O}_4^{2-} \). Thus:\[\text{Moles of compound (a)} = \frac{0.009315}{2} = 0.0046575 \, \text{mol}\]Calculate the molar mass of compound (a):\[\text{Molar mass} = 39.10 + 55.85 + 2 \times (2 \times 12.01 + 4 \times 16.00) + 2 \times (2 \times 1.08 + 16.00) = 304.1 \, \text{g/mol}\]

Consider Second Formula Option

For compound (b) \( \mathrm{K}_3\left[\mathrm{Fe}\left(\mathrm{C}_2\mathrm{O}_4\right)_3\right] \), each formula unit contains 3 moles of \( \mathrm{C}_2\mathrm{O}_4^{2-} \). Thus:\[\text{Moles of compound (b)} = \frac{0.009315}{3} = 0.003105 \, \text{mol}\]Calculate the molar mass of compound (b):\[\text{Molar mass} = 3 \times 39.10 + 55.85 + 3 \times (2 \times 12.01 + 4 \times 16.00) = 491.1 \, \text{g/mol}\]

Compare Experimental and Theoretical Masses

For compound (a), multiply moles by its molar mass:\[\text{Mass of (a)} = 0.0046575 \, \text{mol} \times 304.1 \, \text{g/mol} = 1.4167 \, \text{g} \]For compound (b):\[\text{Mass of (b)} = 0.003105 \, \text{mol} \times 491.1 \, \text{g/mol} = 1.525 \, \text{g} \]Compare these to the actual mass used in the experiment (1.356 g).

Draw Conclusion

The calculated mass for compound (a) (1.4167 g) is closer to the experimental mass (1.356 g) than compound (b) (1.525 g). Therefore, the correct formula is:\[\mathrm{K} \left[\mathrm{Fe}\left(\mathrm{C}_2\mathrm{O}_4\right)_2\left(\mathrm{H}_2\mathrm{O}\right)_2\right]\]

Key concepts

Stoichiometry

The concept of stoichiometry revolves around the quantitative relationships between reactants and products in a chemical reaction. In this exercise, stoichiometry helps us determine the amount of one substance required or produced when a certain amount of another substance is involved. An important part of stoichiometry is using the balanced chemical equation to find the mole ratio of the involved substances.
This mole ratio is essential to convert moles of a known substance to moles of another substance. In our exercise, we used the balanced net ionic equation:

• 5 moles of oxalic acid react with 2 moles of the permanganate ion.

By knowing the moles of potassium permanganate used, we can employ stoichiometry to find the moles of oxalic acid produced, crucial for identifying the compound's formula.

Titration

Titration is an analytical technique that involves the gradual addition of a solution (the titrant) to a known volume of a second solution (the analyte) until the reaction reaches the endpoint, which is often indicated by a color change.
In the given exercise, the titration process helps determine the amount of oxalic acid formed from the unknown compound during dissolution. Using a known concentration and volume of permanganate (\( \mathrm{KMnO}_4 \)), we can find out the exact amount of oxalic acid in the solution.
This information is used to backtrack and identify the correct formula of the iron-containing compound, by measuring the amount of oxalic acid titrated in the reaction.

Molar Mass Calculation

Calculating the molar mass involves adding up the atomic masses of all the atoms present in a compound's formula. This is a crucial step in converting between moles and grams, a core component of many chemistry calculations.
For both possible compounds in our exercise, their molar masses were calculated by summing the atomic masses:

• Compound (a): \( \mathrm{K} \left[\mathrm{Fe}\left(\mathrm{C}_2\mathrm{O}_4\right)_2\left(\mathrm{H}_2\mathrm{O}\right)_2\right] = 304.1 \, \text{g/mol} \)
• Compound (b): \( \mathrm{K}_3\left[\mathrm{Fe}\left(\mathrm{C}_2\mathrm{O}_4\right)_3\right] = 491.1 \, \text{g/mol} \)

This calculation helped in determining which compound was closer to the experimentally found mass, hinting at the correct formula.

Chemical Formula Determination

Determining the chemical formula is a process guided by experimental data and chemical laws.
In our exercise, both possible formulas resulted in different masses, which we calculated theoretically. By comparing these theoretical masses to the experimental mass recorded (1.356 grams), we are able to narrow down the correct chemical formula for the compound.
This comparison showed that the experimental mass was closer to the calculated mass for compound (a) than for compound (b). Therefore, chemical formula determination in this case relied on comparing experimental results to theoretical predictions, highlighting which structure suits the observed data best.