Problem 94 - General Chemistry

Question

A \(3.33-\mathrm{g}\) sample of iron ore is transformed to a solution of iron(II) sulfate, \(\mathrm{FeSO}_{4}\), and this solution is titrated with \(0.150 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) (potassium dichromate). If it requires \(43.7 \mathrm{~mL}\) of potassium dichromate solution to titrate the iron(II) sulfate solution, what is the percentage of iron in the ore? The reaction is $$ \begin{aligned} &6 \mathrm{FeSO}_{4}(a q)+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(a q)+7 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \\ &3 \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+7 \mathrm{H}_{2} \mathrm{O}(I)+\mathrm{K}_{2} \mathrm{SO}_{4}(a q) \end{aligned} $$

Step-by-Step Solution

Answer

The percentage of iron in the ore is 65.9%.

Step 1: Convert Volume to Liters

First, convert the volume of the potassium dichromate solution from milliliters to liters. Use the conversion factor that 1 L = 1000 mL. \[43.7 \, \text{mL} = \frac{43.7}{1000} \, \text{L} = 0.0437 \, \text{L}\]

Step 2: Calculate Moles of \( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \)

Use the molarity formula, \( \text{Molarity} = \frac{\text{moles}}{\text{volume (L)}} \), to find the moles of \( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \).\[\text{Moles of } \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} = 0.150 \, \text{mol/L} \times 0.0437 \, \text{L} = 0.006555 \, \text{mol}\]

Step 3: Determine Moles of \( \mathrm{FeSO}_{4} \)

According to the balanced reaction equation, 1 mole of \( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \) reacts with 6 moles of \( \mathrm{FeSO}_{4} \). Use this stoichiometry to find the moles of \( \mathrm{FeSO}_{4} \).\[\text{Moles of } \mathrm{FeSO}_{4} = 0.006555 \, \text{mol} \times 6 = 0.03933 \, \text{mol}\]

Step 4: Convert Moles of \( \mathrm{FeSO}_{4} \) to Grams of Iron

The molar mass of iron (Fe) is approximately 55.85 g/mol. Since each mole of \( \mathrm{FeSO}_{4} \) contains 1 mole of Fe, calculate the mass of iron.\[\text{Mass of iron} = 0.03933 \, \text{mol} \times 55.85 \, \text{g/mol} = 2.195 \, \text{g}\]

Step 5: Calculate Percentage of Iron in Ore

Determine the percentage of iron in the ore by using the formula:\[ \text{Percentage of Iron} = \left( \frac{\text{Mass of iron}}{\text{Mass of sample}} \right) \times 100\%\]Substitute in the known values:\[\text{Percentage of Iron} = \left( \frac{2.195 \, \text{g}}{3.33 \, \text{g}} \right) \times 100\% = 65.9\%\]

Key Concepts

Stoichiometry

Stoichiometry is the study of the quantitative relationships between the amounts of reactants and products in a chemical reaction. In the context of titration, it is crucial to determine how much of a substance is needed to fully react with another substance. This involves interpreting chemical reactions and using balanced chemical equations to find out the ratios in which chemicals react.

In our exercise, the balanced chemical equation is key. It shows that one mole of potassium dichromate ( \( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \) ) reacts with six moles of iron(II) sulfate ( \( \mathrm{FeSO}_{4} \) ). Thus, the stoichiometric coefficients tell us how to calculate the amount of each reactant needed based on the amount of other reactants or products involved. Recognizing these relationships allows us to precisely determine the moles of iron present and calculate the mass of iron in the sample.

Molarity

Molarity is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution and is expressed in moles per liter (mol/L). Molarity provides a way to express how much of a substance is dissolved in a specific volume of solvent.

In this titration problem, molarity helps us determine how much potassium dichromate is needed to complete the titration of the iron(II) sulfate solution. Knowing the molarity ( \( 0.150 \, \mathrm{M} \) ) and the volume of the solution used (converted into liters), we can calculate the moles of potassium dichromate using the formula:

• Molarity ( \( \text{M} \) ) = \( \text{Moles of solute} / \text{Volume of solution in liters} \)

Understanding molarity allows students to link the volume of a solution with its chemical potential, a vital concept in both analytical and synthetic chemistry.

Iron Ore Analysis

Iron ore analysis involves determining the composition of iron within ore samples, which is essential in assessing the ore's quality. This quantitative analysis is vital in industries such as steel manufacturing, where high iron content is desirable for efficiency and cost-effectiveness.

In the exercise, the goal is to find the percentage of iron in the 3.33-gram sample of iron ore. We did this by converting iron(II) sulfate into a soluble form and titrating it with potassium dichromate. This allowed us to measure the amount of iron present through the use of stoichiometry and molarity calculations. By calculating the moles of iron, we're able to measure its mass and then find its weight percentage in the sample, which informs us of the ore's iron quality.

Chemical Reactions

Chemical reactions describe the transformation of reactants into products, showcasing how atoms rearrange through bonds. In our problem, balancing the chemical reaction equation is crucial because it determines how reactants and products relate.

• Reactants: \( 6 \mathrm{FeSO}_{4}, \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}, 7 \mathrm{H}_{2} \mathrm{SO}_{4} \)
• Products: \( 3 \mathrm{Fe}_{2}(\mathrm{SO}_{4})_{3}, \mathrm{Cr}_{2}(\mathrm{SO}_{4})_{3}, 7 \mathrm{H}_{2} \mathrm{O}, \mathrm{K}_{2} \mathrm{SO}_{4} \)

Understanding the balanced equation helps ensure that the law of conservation of mass is respected. Each side of the equation has the same number of each type of atom. Knowing this is crucial for correctly predicting the outcomes of reactions and for the calculations made in stoichiometry to determine the amount of each chemical species involved. Emphasizing the role of chemical reactions not only explains the process but also the relationship between reactants and the transformation into products.