Problem 51 - General Chemistry: Principles and Modern Applications

Question

A \(1.100 \mathrm{g}\) sample of copper ore is dissolved, and the \(\mathrm{Cu}^{2+}(\mathrm{aq})\) is treated with excess KI. The liberated \(\mathrm{I}_{3}^{-}\) requires \(12.12 \mathrm{mL}\) of \(0.1000 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\) for its titration. What is the mass percent copper in the ore?

Step-by-Step Solution

Answer

The mass percent of copper in the ore is 3.50%.

Step 1: Write balanced chemical equations

But before we proceed, the balanced equations for the reactions are: \(\mathrm{Cu}^{2+} + 2\mathrm{I}^{-} \rightarrow \mathrm{CuI}_{2}\) and \(\mathrm{I}_{3}^{-} + 2\mathrm{S}_{2}\mathrm{O}_{3}^{2-} \rightarrow 3\mathrm{I}^{-} + 2\mathrm{S}_{4}\mathrm{O}_{6}^{2-}\). Now, from the second equation, we see that one mole of I3- consumes 2 mol of \(\mathrm{S}_{2}\mathrm{O}_{3}^{2-}\).

Step 2: Calculate moles of Na2S2O3

We first need to calculate the moles of the titrating agent, \(\mathrm{Na}_{2}\mathrm{S}_{2}\mathrm{O}_{3}\). Moles can be calculated using the formula \(Moles = Molarity \times Volume\) where volume is in liters. Converting the given volume from mL to L, we get \(12.12 \mathrm{mL} = 0.01212 \mathrm{L}\). Thus, the moles of \(\mathrm{Na}_{2}\mathrm{S}_{2}\mathrm{O}_{3}\) is \(0.1000 M \times 0.01212 L = 0.001212 mol\).

Step 3: Calculate moles of I3-

From the stoichiometry of the equation in Step 1, we know 1 mol of I3- reacts with 2 mol of \(\mathrm{S}_{2}\mathrm{O}_{3}^{2-}\). Therefore, the moles of I3-, which was produced from the reaction with copper is \(0.001212 mol \div 2 = 0.000606 mol\).

Step 4: Calculate moles of Cu2+

Now, again based on the stoichiometry of the reaction, 1 mol of I3- was produced for each mole of Cu2+ that reacted. Thus, the moles of \(\mathrm{Cu}^{2+}\) is also \(0.000606 mol\).

Step 5: Calculate mass of Copper (Cu)

The molar mass of copper (Cu) is 63.55g/mol. So, the mass of copper can be calculated as \(Mass = moles \times Molar mass\). Therefore, the mass of copper in the ore is \(0.000606 mol \times 63.55 g/mol = 0.0385 g\).

Step 6: Calculate mass percent of Copper

The mass percent of copper in the ore can be calculated as \((Mass of Cu / Total mass of the ore) \times 100\)%. Substituting the values we get \((0.0385 g / 1.100 g) \times 100 = 3.50 \% \)

Key Concepts

Chemical Reactions

In chemical reactions, substances known as reactants transform into different substances called products. Balancing these reactions is crucial in chemistry to ensure the law of conservation of mass is followed. This principle states that matter is neither created nor destroyed in a chemical reaction.
To balance a chemical equation, we count the number of atoms of each element on both sides of the equation and adjust the coefficients to make them equal.

• For the reaction between copper ions and iodide ions, the equation is: \[ \text{Cu}^{2+} + 2\text{I}^{-} \rightarrow \text{CuI}_2 \]
• In this reaction, one copper ion reacts with two iodide ions to form copper iodide.

Similarly, we balance the iodide reaction with the titrant as: \[ \text{I}_3^- + 2\text{S}_2\text{O}_3^{2-} \rightarrow 3\text{I}^- + 2\text{S}_4\text{O}_6^{2-} \]Understanding the details of these reactions helps in tackling complex problems involving multiple steps and reagents.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It connects the macroscopic properties of amounts of substances with the atomic-scale representations. In our example, stoichiometry was essential to understand how the amount of consumed titrant was related to the initial copper content.
We used stoichiometry to determine:

• The relationship between moles of sodium thiosulfate, \(\text{Na}_2\text{S}_2\text{O}_3\), and the tri-iodide ion, \(\text{I}_3^-\).
• The connection between \(\text{I}_3^-\) moles and the initial copper ions, \(\text{Cu}^{2+}\).

This step-by-step conversion of units helps us figure out how much product forms from a given set of reactants, and vice versa. It's like a recipe that tells us how to make the exact amount of brownies from a specific amount of ingredients.

Copper Titration

Copper titration is a method used to analyze the concentration of copper ions in a solution. One popular technique involves the oxidation of iodide ions by copper ions. As the iodide ions are oxidized, they form tri-iodide ions \(\text{I}_3^-\), which can then be titrated using a solution of sodium thiosulfate \(\text{Na}_2\text{S}_2\text{O}_3\).
This process of titration is carried out until the tri-iodide ions are completely reacted with the sodium thiosulfate, as indicated by a sudden change in color. Here's what happens during the titration process:

• The tri-iodide ions produced from copper react with sodium thiosulfate, leading to the transformation back into iodide ions.
• The amount of sodium thiosulfate used provides information about the initial amount of copper in the sample.

By carefully measuring how much standard solution is needed to reach the endpoint, we can calculate the mass percent of copper in the sample, giving us insights into its purity and composition.