A \(1.100 \mathrm{~g}\) sample of copper ore is dissolved and the \(\mathrm{Cu}^{2+}(\mathrm{aq})\) is treated with excess \(\mathrm{KI}\). The liberated \(\mathrm{I}_{2}\) requires \(12.12 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) solution for titration. What is \(\%\) copper by mass in the ore?

Short Answer

Expert verified
The percentage of copper by mass in the ore is 3.50%

Step by step solution

01

Understand the Reaction

The reaction taking place is the reduction of iodine by sodium thiosulfate in the presence of iodide ions, liberating iodine. This iodine is then titrated with sodium thiosulfate solution. The balanced equation for the reaction is: \[2 \text{S}_2\text{O}_3^{2-}(aq) + \text{I}_2(s) -> \text{S}_4\text{O}_6^{2-}(aq) + 2 \text{I}^{-}(aq)\]
02

Calculate Moles of Na2S2O3

Calculate the moles of sodium thiosulfate (\text{Na}_2\text{S}_2\text{O}_3) used in the titration: \[ \text{Moles of Na}_2\text{S}_2\text{O}_3 = \text{Volume} \times \text{Molarity} = 12.12 \times 10^{-3} \text{ L} \times 0.10 \text{ M} = 1.212 \times 10^{-3} \text{ moles} \]
03

Calculate Moles of I2

From the stoichiometry of the reaction, the moles of iodine (\text{I}_2) will be half the moles of sodium thiosulfate, because two moles of sodium thiosulfate react with one mole of iodine: \[ \text{Moles of I}_2 = \frac{1}{2} \times \text{Moles of Na}_2\text{S}_2\text{O}_3 \] \[ \text{Moles of I}_2 = \frac{1}{2} \times 1.212 \times 10^{-3} = 6.06 \times 10^{-4} \text{ moles} \]
04

Calculate Moles of Cu2+

The moles of copper(\text{Cu}^{2+}) will be the same as the moles of iodine since the reaction that liberates iodine from copper is a 1:1 mole ratio: \[ \text{Moles of Cu}^{2+} = \text{Moles of I}_2 = 6.06 \times 10^{-4} \text{ moles} \]
05

Calculate Mass of Copper

We calculate the mass of copper by multiplying the moles of \text{Cu}^{2+} by the molar mass of copper (63.55 g/mol): \[ \text{Mass of Copper} = \text{Moles of Cu}^{2+} \times \text{Molar Mass of Copper} \] \[ \text{Mass of Copper} = 6.06 \times 10^{-4} \text{ moles} \times 63.55 \text{ g/mol} = 0.03852 \text{ g} \]
06

Calculate the Percentage of Copper in the Ore

Percentage of copper by mass in the ore is found by dividing the mass of copper by the mass of the ore sample and then multiplying by 100: \[ \% \text{Copper by mass} = \left( \frac{\text{Mass of Copper}}{\text{Mass of Ore Sample}} \right) \times 100 \]\[ \% \text{Copper by mass} = \left( \frac{0.03852 \text{ g}}{1.100 \text{ g}} \right) \times 100 = 3.5018\% \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stoichiometry
Stoichiometry is the mathematical relationship between the quantities of reactants and products in a chemical reaction. It is derived from the coefficients of a balanced chemical equation, which indicate the proportions in which chemicals react and are formed. Understanding stoichiometry is critical when conducting an iodometric titration to determine the concentration of a substance, such as copper in an ore sample.

In the given problem, stoichiometry plays a pivotal role in calculating the percentage of copper by mass. First, we define the balanced equation for the redox reaction, which guides us in determining the ratio of the participating compounds. In this case, the reaction shows that two moles of sodium thiosulfate react with one mole of iodine.Therefore, by measuring the amount of sodium thiosulfate required to titrate the liberated iodine, we can back-calculate the amount of iodine, and subsequently the amount of copper, present in the sample.
Mole Concept
The mole concept is a fundamental principle in chemistry that provides a bridge between the atomic scale and the macroscopic scale. A mole is defined as the amount of a substance that contains as many elementary entities (atoms, molecules, ions, etc.) as there are atoms in 12 grams of pure carbon-12. One mole corresponds to approximately \(6.022 \times 10^{23}\) entities, known as Avogadro's number.

In the context of an iodometric titration, knowing the mole concept allows us to calculate the number of moles of each reactant and product involved. For instance, after measuring the volume and molarity of sodium thiosulfate used in the titration, we calculate the moles of sodium thiosulfate, and through stoichiometric relationships, can determine the moles of iodine and ultimately the moles of copper in the ore. This quantitative relationship is crucial for determining the percent composition of an element within a compound, as demonstrated in the exercise.
Redox Reaction
Redox reactions are chemical reactions that involve the transfer of electrons between two species. The term 'redox' is a contraction of 'reduction-oxidation'. In these reactions, one substance is oxidized (loses electrons) while the other is reduced (gains electrons). An understanding of redox reactions is essential when working with iodometric titrations, which are often employed to determine the amount of a reducing agent in a sample.

In the exercise, the redox reaction occurs when the copper ions in the ore sample react with iodide ions to liberate iodine. This iodine is then titrated with sodium thiosulfate, which acts as the reducing agent. The balanced redox equation tells us not only the stoichiometry of the reaction but also the specific oxidation and reduction processes taking place. It is the foundation for calculating the mass and, therefore, the percentage of copper present in the ore sample, highlighting the interconnected nature of redox chemistry, stoichiometry, and the mole concept in quantitative chemical analysis.

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Most popular questions from this chapter

A particular acid-rain water has \(\mathrm{SO}_{3}{ }^{2-}\). If a \(25.00 \mathrm{~mL}\) sample of this water requires \(34.08 \mathrm{~mL}\) of \(0.01964 \mathrm{M} \mathrm{KMnO}_{4}\) for its titration, what is the molarity of \(\mathrm{SO}_{3}{ }^{2-}\) in acid-rain? \(2 \mathrm{MnO}_{4}^{-}+5 \mathrm{SO}_{3}^{2-}+6 \mathrm{H}^{+} \longrightarrow 5 \mathrm{SO}_{4}^{2-}+2 \mathrm{Mn}^{2+}+3 \mathrm{H}_{2} \mathrm{O}\)

\(25 \mathrm{~mL}\) of \(0.017 \mathrm{H}_{2} \mathrm{SO}_{3}\) in strongly acidic medium required \(16.9 \mathrm{~mL}\) of \(0.01 \mathrm{M} \mathrm{KMnO}_{4}\) and in neutral medium required \(28.6 \mathrm{~mL}\) of \(0.01 M\) \(\mathrm{KMnO}_{4}\) for complete conversion of \(\mathrm{SO}_{3}{ }^{2-}\) to \(\mathrm{SO}_{4}^{2-}\). Assign the oxidation no. of Mn in the product formed in each case.

\(\mathrm{Mn}^{2+}(\) aq \()\) can be determined by titration with \(\mathrm{MnO}_{4}^{-}(\) aq \()\) $$ 3 \mathrm{Mn}^{2+}+2 \mathrm{MnO}_{4}^{-} \longrightarrow 6 \mathrm{MnO}_{2}+2 \mathrm{H}_{2} \mathrm{O} $$ A \(25.00 \mathrm{~mL}\) sample of \(\mathrm{Mn}^{2+}\) (aq) requires \(34.77 \mathrm{~mL}\) of \(0.05876 M\) \(\mathrm{KMnO}_{4}\) (aq) for its titration. What is the molarity of the \(\mathrm{Mn}^{2+}\) (aq)?

Reaction, \(2 \mathrm{Br}^{-}{ }_{(\mathrm{aq})}+\mathrm{Cl}_{2}(\mathrm{aq}) \longrightarrow 2 \mathrm{Cl}^{-}_{(\mathrm{aq} .)}+\mathrm{Br}_{2(\mathrm{aq} .)}\), is used for commercial preparation of bromine from its salts. Suppose we have \(50 \mathrm{~mL}\) of a \(0.060 \mathrm{M}\) solution of NaBr. What volume of a \(0.050 \mathrm{M}\) solution of \(\mathrm{Cl}_{2}\) is needed to react completely with the Br?

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