Question

\(32 \mathrm{~g}\) of a sample of \(\mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\) were dissolved in dilute sulphuric acid and water and it's volume was made up to 1 litre, \(25 \mathrm{~mL}\) of this solution required \(20 \mathrm{~mL}\) of \(0.02 \mathrm{M} \mathrm{KMnO}_{4}\) solution for complete oxidation. Calculate the weight \(\%\) of \(\mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\) in the sample. (a) \(34.75\) (b) \(69.5\) (c) \(89.5\) (d) None of these

Short answer

The weight percent of \(\mathrm{FeSO}_4 \cdot 7\mathrm{H}_2\mathrm{O}\) in the sample is approximately 89.5%.

Step-by-step solution

Determine the molar mass of \(\mathrm{FeSO}_4 \cdot 7\mathrm{H}_2\mathrm{O}\)

Calculate the molar mass by summing the atomic masses of each element in one mole of \(\mathrm{FeSO}_4 \cdot 7\mathrm{H}_2\mathrm{O}\): \(\mathrm{Fe} = 55.85, \mathrm{S} = 32.07, \mathrm{O} = 16.00, \mathrm{H}_2\mathrm{O} = 18.015\). The molar mass is equal to \(1 \times 55.85 + 1 \times 32.07 + 4 \times 16.00 + 7 \times 18.015\).

Calculate the molarity of the \(\mathrm{FeSO}_4\) solution

Determine the moles of \(\mathrm{FeSO}_4 \cdot 7\mathrm{H}_2\mathrm{O}\) in 32 g of the sample, then divide by the total volume of the solution in liters to find the molarity.

Write down the balanced redox reaction

For the reaction between \(\mathrm{FeSO}_4\) and \(\mathrm{KMnO}_4\) in acidic medium, the balanced equation is: \(10 \mathrm{Fe}^{2+} + 2 \mathrm{MnO}_4^{-} + 16 \mathrm{H}^+ \rightarrow 5 \mathrm{Fe}_2^{3+} + 2 \mathrm{Mn}^{2+} + 8 \mathrm{H}_2\mathrm{O}\).

Determine the moles of \(\mathrm{KMnO}_4\) used

Use the volume and molarity of \(\mathrm{KMnO}_4\) to calculate the moles. The formula is \(\text{Moles} = \text{Molarity} \times \text{Volume in liters}\).

Use stoichiometry to find the moles of \(\mathrm{FeSO}_4 \cdot 7\mathrm{H}_2\mathrm{O}\) oxidized

Using the balanced equation from Step 3, calculate the moles of \(\mathrm{FeSO}_4 \cdot 7\mathrm{H}_2\mathrm{O}\) that reacted based on the stoichiometric ratios.

Calculate the mass of \(\mathrm{FeSO}_4 \cdot 7\mathrm{H}_2\mathrm{O}\) that reacted

Multiply the moles of \(\mathrm{FeSO}_4 \cdot 7\mathrm{H}_2\mathrm{O}\) by its molar mass to find the mass.

Calculate the weight percent of \(\mathrm{FeSO}_4 \cdot 7\mathrm{H}_2\mathrm{O}\) in the sample

The weight percent is calculated by the formula \(\text{Weight percent} = \frac{\text{Mass of solute in the sample}}{\text{Total mass of the sample}} \times 100\)%.

Key concepts

Molar Mass Calculation

Understanding molar mass is crucial for solving redox titration problems, as it helps to convert between grams of a substance and moles, a fundamental step in stoichiometry. This is especially true on exams like the JEE, where accuracy is key. The molar mass is the mass of one mole of a substance (the mass of Avogadro's number of molecules), and it is expressed in grams per mole (g/mol).

To calculate the molar mass, you sum up the atomic masses of each element present in the compound, as found in the periodic table, multiplied by the number of atoms of that element in the formula. For instance, in the exercise provided, we calculate the molar mass of hydrated ferrous sulfate, \(\mathrm{FeSO}_4 \cdot 7\mathrm{H}_2\mathrm{O}\), by adding the atomic masses of iron (Fe), sulfur (S), oxygen (O), and seven times the mass of water (H2O).

Stoichiometry

Stoichiometry is the aspect of chemistry that involves the calculation of reactants and products in chemical reactions. It is based on the conservation of mass where the quantity of each element must remain the same before and after a chemical reaction. In JEE redox titration problems, stoichiometry enables you to determine the relationship between the amounts of reactants used and products formed.

Using stoichiometry, you can calculate how much of a chemical is needed to react completely with a given amount of another chemical. This is done through mole-to-mole conversions obtained from a balanced chemical equation. For example, knowing that it takes 5 moles of \(\mathrm{Fe}^{2+}\) to react with 1 mole of \(\mathrm{MnO}_4^{-}\), as in the balanced equation provided in the exercise, is essential for determining the amount of \(\mathrm{FeSO}_4 \cdot 7\mathrm{H}_2\mathrm{O}\) that reacts.

Molarity Calculation

Molarity is a measure of concentration, expressed as moles of solute per liter of solution (M or mol/L). In the context of JEE problems, calculating the molarity is a fundamental skill that allows you to understand the quantitative aspects of chemical solutions. When you're given the mass of a solute and the volume of the solution, you can calculate the molarity by dividing the number of moles of the solute by the volume of the solution in liters.

Here is the formula for molarity:
\[\text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}}\].
In the provided exercise, initially the molarity of the \(\mathrm{FeSO}_4 \cdot 7\mathrm{H}_2\mathrm{O}\) solution must be calculated, which involves finding the number of moles of \(\mathrm{FeSO}_4 \cdot 7\mathrm{H}_2\mathrm{O}\) and dividing it by the volume of the solution prepared.

Balanced Chemical Equations

A balanced chemical equation ensures that the law of conservation of mass is upheld by having the same number of atoms of each element on both sides of the reaction. JEE chemistry problems often require balancing redox reactions, where transfer of electrons between reactants must be accounted for. In redox titrations, understanding this balance is pivotal for doing stoichiometric calculations.

The balanced equation enables the correct molar ratios between reactants and products to be used in stoichiometric calculations. For example, with the balanced redox reaction in the exercise, where \(10 \mathrm{Fe}^{2+}\) ions react with \(2 \mathrm{MnO}_4^{-}\) ions and \(16 \mathrm{H}^+\) ions to produce \(5 \mathrm{Fe}_2^{3+}\), \(2 \mathrm{Mn}^{2+}\), and \(8 \mathrm{H}_2\mathrm{O}\), we can determine the exact molar amounts involved in the reaction. Correctly balancing the equation is the foundation for accurate stoichiometric and molarity calculations.