Question

A solid sample containing some \(\mathrm{Fe}^{2+}\) ion weighs 1.062 g. It requires \(24.12 \mathrm{mL} 0.01562 \mathrm{M} \mathrm{KMnO}_{4}\) to titrate the \(\mathrm{Fe}^{2+}\) in the dissolved sample to a pink end point. a. How many moles \(\mathrm{MnO}_{4}\) - ion are required? _____________ moles b. How many moles \(\mathrm{Fe}^{2+}\) are there in the sample? _____________ moles c. How many grams of iron are there in the sample? ______________g d. What is the percentage of Fe in the sample? _______________%

Short answer

a. 0.0003765 moles of MnO₄⁻ ion are required. b. There are 0.0018825 moles of Fe²⁺ ion in the sample. c. There are 0.1052 g of iron in the sample. d. The percentage of Fe in the sample is approximately 9.91%.

Step-by-step solution

Write the balanced chemical equation

Before starting with the calculations, we need to write the balanced chemical equation for the reaction between Fe²⁺ ions and MnO₄⁻ ions. The balanced chemical equation is: \[\mathrm{5Fe^{2+}(aq) + MnO_4^-(aq) + 8H^+(aq)} \rightarrow \mathrm{5Fe^{3+}(aq) + Mn^{2+}(aq) + 4H_2O(l)}\]

Calculate the moles of MnO₄⁻ ion required

We are given the volume (24.12 mL) and molarity (0.01562 M) of KMnO₄ solution used in the titration. To calculate the moles of MnO₄⁻ ions, use the formula: \( moles = molarity * volume \) Note that the volume needs to be in liters (L). Convert the volume: 24.12 mL = 0.02412 L Now, plug in the numbers: moles of MnO₄⁻ = 0.01562 M * 0.02412 L moles of MnO₄⁻ = 0.0003765 mol #a.# The required moles of MnO₄⁻ ion are 0.0003765 moles.

Calculate the moles of Fe²⁺ ion in the sample

Using the stoichiometry of the balanced chemical equation, we can determine the moles of Fe²⁺ ions. From the equation, every 1 mole of MnO₄⁻ reacts with 5 moles of Fe²⁺. So, we can calculate the moles of Fe²⁺ using the ratio: \( moles\ Fe^{2+} = \dfrac{5\ moles\ Fe^{2+}}{1\ mole\ MnO_4^{-}} \times moles\ MnO_4^{-} \) Plug in the numbers: moles of Fe²⁺ = 5 * 0.0003765 mol moles of Fe²⁺ = 0.0018825 mol #b.# There are 0.0018825 moles of Fe²⁺ ion in the sample.

Calculate the grams of iron in the sample

To calculate the grams of iron, multiply the moles of Fe²⁺ by the molar mass of iron (Fe): grams of iron = moles of Fe²⁺ * molar mass of Fe The molar mass of iron is 55.85 g/mol, so: grams of iron = 0.0018825 mol * 55.85 g/mol grams of iron = 0.105185625 g #c.# There are 0.1052 g of iron in the sample (rounded to 4 significant figures).

Calculate the percentage of Fe in the sample

Finally, to find the percentage of iron in the sample, divide the grams of iron by the total weight of the sample and multiply by 100: percentage of Fe = (grams of iron / total weight of sample) * 100 We have the total weight of the sample, which is given as 1.062 g: percentage of Fe = (0.105185625 g / 1.062 g) * 100 percentage of Fe = 9.907 % #d.# The percentage of Fe in the sample is approximately 9.91 % (rounded to 3 significant figures).

Key concepts

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the different substances in a chemical reaction. It tells us the proportion in which reactants combine and products form. To utilize stoichiometry effectively, you need a balanced chemical equation that clearly represents what is happening in the reaction. For the redox titration exercise, stoichiometry helps us calculate the number of moles for each reactant and product involved in the reaction.

Balanced Equation: This specifies the ratio of moles of each substance involved. For example, in the titration reaction: \[5\,\mathrm{Fe^{2+}}(aq) + \mathrm{MnO_4^-}(aq) + 8\,\mathrm{H^+}(aq) \rightarrow 5\,\mathrm{Fe^{3+}}(aq) + \mathrm{Mn^{2+}}(aq) + 4\,\mathrm{H_2O}(l)\] The equation suggests that 5 moles of \(\mathrm{Fe^{2+}}\) react with 1 mole of \(\mathrm{MnO_4^-}\).

The Role of Molar Ratios: These ratios derived from the balanced equation are crucial for calculating the unknown quantities, such as the number of moles of \(\mathrm{Fe^{2+}}\) in the solution once the amount of \(\mathrm{MnO_4^-}\) used is known.

Stoichiometry simplifies complex chemical reactions into manageable calculations, allowing us to predict and quantify the amounts of products and reactants involved.

Oxidation-Reduction Reactions

Oxidation-reduction reactions, also known as redox reactions, involve the transfer of electrons between two substances. These reactions are characterized by changes in oxidation numbers of the involved elements, which reflects their loss or gain of electrons. In the titration example, \(\mathrm{Fe^{2+}}\) is oxidized, and \(\mathrm{MnO_4^-}\) is reduced, demonstrating a classic redox process.

Oxidation: This occurs when a substance loses electrons. In the given exercise, \(\mathrm{Fe^{2+}}\) (ferrous ion) is oxidized to \(\mathrm{Fe^{3+}}\) (ferric ion), indicating that electrons are lost.

Reduction: This happens when a substance gains electrons. Conversely, the \(\mathrm{MnO_4^-}\) ion is reduced to \(\mathrm{Mn^{2+}}\) as it gains electrons in the process.

Understanding redox reactions is essential in titration techniques because it helps track the electron transfer and helps quantify the reactants needed to reach the endpoint of a reaction.

Chemical Calculations

Chemical calculations are an integral aspect of chemistry that involves determining the amounts of substances involved in chemical reactions. With redox titrations, these calculations include conversions between concentrations, volumes, and masses of the reactants and products.

Calculating Moles: This involves using the formula: \[\text{moles} = \text{molarity} \times \text{volume (in liters)}\] For instance, the moles of \(\mathrm{MnO_4^-}\) used is calculated as \(0.01562 \,\mathrm{M} \times 0.02412 \,\mathrm{L} = 0.0003765\) moles.

Mass to Moles Conversion: Once the moles of iron are determined, they can be converted to grams using the element's molar mass. This connects stoichiometric coefficients with physical measurements, allowing us to determine the mass of the element in question.

Percentage Composition: Finally, the relative amount of a specific element within a compound or mixture is calculated as a percentage. This is done by comparing the mass of the element to the total mass of the sample, multiplied by 100 to express it as a percent.

In essence, chemical calculations provide a pathway from theoretical reactions to practical applications, allowing chemists to precisely analyze and interpret experimental data.