Question

The mineral claudetite contains the element arsenic in the form of arsenic(III) oxide, \(A s_{2} O_{3}\). The \(A s_{2} O_{3}\) in a \(0.562-g\) sample of the impure mineral was converted first to \(\mathrm{H}_{3} \mathrm{AsO}_{3}\) and then titrated with a \(0.0480 \mathrm{M}\) solution of \(\mathrm{I}_{3}^{-},\) which reacts with \(\mathrm{H}_{3} \mathrm{AsO}_{3}\) according to the following balanced net ionic equation $$\begin{aligned}\mathrm{H}_{3} \mathrm{AsO}_{3}(\mathrm{aq})+3 \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{I}_{3}^{-}(\mathrm{aq}) & \rightarrow \\\\\mathrm{H}_{3} \mathrm{AsO}_{4}(\mathrm{aq}) &+2 \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+3 \mathrm{I}^{-}(\mathrm{aq})\end{aligned}$$ If the titration required \(45.7 \mathrm{mL}\) of the \(\mathrm{I}_{3}^{-}\) solution, what is the percentage of \(\mathrm{As}_{2} \mathrm{O}_{3}\) in the mineral sample?

Short answer

The percentage of \( \mathrm{As}_{2} \mathrm{O}_{3} \) in the mineral sample is approximately 38.6\%.

Step-by-step solution

Calculate moles of I3-

First, find the moles of \ \( \mathrm{I}_{3}^{-} \ \) used in the titration. The concentration is given as \ \( 0.0480 \, \text{M} \ \) and the volume used is \ \( 45.7 \, \text{mL} = 0.0457 \, \text{L} \ \). Thus, the moles are calculated as: \ \[ \text{Moles of } \mathrm{I}_{3}^{-} = 0.0480 \, \text{M} \times 0.0457 \, \text{L} = 0.0021936 \, \text{mol}.\]

Determine moles of H3AsO3

According to the balanced chemical equation, one mole of \ \( \mathrm{I}_{3}^{-} \ \) reacts with one mole of \ \( \mathrm{H}_{3} \mathrm{AsO}_{3} \ \). Therefore, the moles of \ \( \mathrm{H}_{3} \mathrm{AsO}_{3} \ \) is the same as the moles of \ \( \mathrm{I}_{3}^{-} \ \): \[ 0.0021936 \, \text{mol}.\]

Calculate moles of As2O3

The conversion from \ \( \mathrm{As}_{2} \mathrm{O}_{3} \ \) to \ \( \mathrm{H}_{3} \mathrm{AsO}_{3} \ \) is one-to-one for arsenic atoms. Since one \ \( \mathrm{As}_{2} \mathrm{O}_{3} \ \) gives two \ \( \mathrm{H}_{3} \mathrm{AsO}_{3} \ \), calculate the moles of \ \( \mathrm{As}_{2} \mathrm{O}_{3} \ \) using: \[ \text{Moles of } \mathrm{As}_{2} \mathrm{O}_{3} = \frac{0.0021936}{2} = 0.0010968 \, \text{mol}.\]

Find mass of As2O3

Calculate the mass of \ \( \mathrm{As}_{2} \mathrm{O}_{3} \ \) using its molar mass (197.84 g/mol): \[ \text{Mass of } \mathrm{As}_{2} \mathrm{O}_{3} = 0.0010968 \, \text{mol} \times 197.84 \, \text{g/mol} = 0.2169 \, \text{g}.\]

Calculate percentage of As2O3 in sample

The percentage of \ \( \mathrm{As}_{2} \mathrm{O}_{3} \ \) in the sample can be calculated as: \[ \text{Percentage} = \left( \frac{0.2169 \, \text{g}}{0.562 \, \text{g}} \right) \times 100\% \approx 38.6\%.\]

Key concepts

Titration Calculation

Titration is a fundamental analytical technique used to determine the concentration of an unknown solution. It involves the gradual addition of a reagent, known as the titrant, to a solution with a substance of interest until the reaction reaches completion. In the context of the exercise, Arsenic (III) Oxide (\(\mathrm{As}_2\mathrm{O}_3\)) is first transformed into \(\mathrm{H}_3\mathrm{AsO}_3\), and then it is titrated with \(\mathrm{I}_3^-\) in solution.
Let's break down the steps involved in calculating titration results:

• First, determine the amount or volume of titrant used in the reaction. This is measured in mL or L.
• Next, using the concentration of the titrant, calculate the moles of titrant added. In this case, the titrant is a \(0.0480\, \text{M}\) \(\mathrm{I}_3^-\) solution, and the volume used is \(45.7\, \text{mL} = 0.0457\, \text{L}\). \[\text{Moles of } \mathrm{I}_3^- = 0.0480 \times 0.0457 = 0.0021936 \text{ mol}\].

In titration, the stoichiometry of the reaction provides a direct relationship between the amounts of reactants and products. This is crucial for calculating unknown concentrations.
In summary, titration calculations provide an accurate method for determining the concentration of an analyte when the stoichiometry of the chemical reaction is well understood.

Stoichiometry

Stoichiometry is an essential concept in chemistry dealing with the calculation of reactants and products in a chemical reaction. It's pronounced like stoy-kee-aw-met-tree, which, much like the Swiss Army knife of chemistry, helps you balance equations and find amounts in reactions.
In the exercise example, stoichiometry helps us understand the mole relationships in the reaction between \(\mathrm{H}_3\mathrm{AsO}_3\) and \(\mathrm{I}_3^-\). The balanced equation: \[\mathrm{H}_3\mathrm{AsO}_3(\mathrm{aq}) + 3 \mathrm{H}_2\mathrm{O}(\ell) + \mathrm{I}_3^-(\mathrm{aq}) \rightarrow \mathrm{H}_3\mathrm{AsO}_4(\mathrm{aq}) + 2 \mathrm{H}_3\mathrm{O}^+(\mathrm{aq}) + 3 \mathrm{I}^-(\mathrm{aq})\]shows that one mole of \(\mathrm{H}_3\mathrm{AsO}_3\) reacts with one mole of \(\mathrm{I}_3^-\), meaning the reaction is direct and can be equimolar.
When calculating the moles \(\mathrm{H}_3\mathrm{AsO}_3\) in the sample:

• Stoichiometry confirmed that the moles of \(\mathrm{H}_3\mathrm{AsO}_3\) was equal to the moles of \(\mathrm{I}_3^-\) used, i.e., \(0.0021936 \text{ mol}\).

Further stoichiometric calculations also relate \(\mathrm{As}_2\mathrm{O}_3\) and \(\mathrm{H}_3\mathrm{AsO}_3\). Specifically, \(1 \text{ mol}\) of \(\mathrm{As}_2\mathrm{O}_3\) yields \(2 \text{ mol}\) of \(\mathrm{H}_3\mathrm{AsO}_3\), highlighting the transformation factor needed in such conversions. By these principles, stoichiometry aids significantly in deducing the amount of substance in the sample.

Percentage Composition

Percentage composition in chemistry refers to the percentage by mass of each element in a compound. It's a way to quantify how much of a specific substance, like \(\mathrm{As}_2\mathrm{O}_3\) in claudetite, is in a sample.
Here's how you can calculate it step-by-step:

• First, determine the mass of the substance in question in the sample. For \(\mathrm{As}_2\mathrm{O}_3\), it's \[0.0010968 \, \text{mol} \times 197.84 \, \text{g/mol} = 0.2169 \, \text{g}.\]
• Next, use this mass relative to the total sample mass to calculate the percentage composition:\[\text{Percentage} = \left( \frac{0.2169 \, \text{g}}{0.562 \, \text{g}} \right) \times 100\% = 38.57\%\].

This calculation tells us that \(38.57\%\) of the total sample mass consists of \(\mathrm{As}_2\mathrm{O}_3\). Percentage composition provides insight into the purity of a sample and can inform adjustments or conclusions in scientific or industrial processes.
Remember, each ingredient's contribution to a compound's total mass reveals the richness or paucity of constituent elements, making percentage composition a useful tool for comparison and assessment in various chemical analyses.