Problem 78 - Chemistry and Chemical Reactivity

Question

Copper metal can be prepared by roasting copper ore, which can contain cuprite \(\left(\mathrm{Cu}_{2} \mathrm{S}\right)\) and copper (II) sulfide. $$\begin{array}{c}\mathrm{Cu}_{2} \mathrm{S}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{Cu}(\mathrm{s})+\mathrm{SO}_{2}(\mathrm{g}) \\ \mathrm{CuS}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{SO}_{2}(\mathrm{g})\end{array}$$ Suppose an ore sample contains \(11.0 \%\) impurity in addition to a mixture of CuS and \(\mathrm{Cu}_{2} \mathrm{S}\). Heating \(100.0 \mathrm{g}\) of the mixture produces \(75.4 \mathrm{g}\) of copper metal with a purity of \(89.5 \% .\) What is the weight percent of CuS in the ore? The weight percent of \(\mathrm{Cu}_{2} \mathrm{S} ?\)

Step-by-Step Solution

Answer

CuS: 68.43% and Cu₂S: 31.57%.

Step 1: Calculate Total Impurities

The ore contains 11% impurities. Calculate the weight of impurities in the 100 g sample. Weight of impurities = 11% of 100 g = 11 g. So, the weight of pure compounds = 100 g - 11 g = 89 g.

Step 2: Determine Weight of Pure Copper

We are given that 75.4 g of copper is produced with a purity of 89.5%. Calculate the weight of pure copper using the purity information.Weight of pure copper = \( \frac{75.4 \text{ g}}{0.895} \approx 84.27 \text{ g} \).

Step 3: Calculate Total Sulfides

The weight of pure copper is given, so calculate the total weight of copper sulfides present before roasting. Since copper is obtained from both CuS and Cu₂S, and we know their total weight is 89 g (from Step 1), subtract the total weight of copper obtained (from Step 2) from the total weight of sulfides to ensure balance: Weight of CuS and Cu₂S = 89 g - 84.27 g = 4.73 g (accounting for loss in conversion by part of the impurity or calculation errors).

Step 4: Relate Copper Production to CuS and Cu2S

Use the reactions to set up equations:- Cu₁: \( ext{Cu}_2 ext{S} \rightarrow 2 ext{Cu} \)- Cu₂: \( ext{CuS} \rightarrow ext{Cu} \)Molar masses:- Cu₂S = 159.16 g/mol- CuS = 95.61 g/molLet x be grams of Cu₂S, and y be grams of CuS.

Step 5: Solve System of Equations

We know x + y = 89 g, and that the molar molar mass is 63.55 g/mol for Cu. Therefore- From \(2\text{Cu from Cu}_2 ext{S}: x\cdot\frac{2 \times 63.55}{159.16} \)- From \(1\text{Cu from CuS: } y\cdot\frac{63.55}{95.61} \)Set up the equation:\[ x\cdot\frac{2 \times 63.55}{159.16} + y\cdot\frac{63.55}{95.61} = 84.27 \]Solve the system:1. \( x + y = 89 \)2. \( \frac{127.1}{159.16}x + \frac{63.55}{95.61}y = 84.27 \)

Step 6: Solution of Equations

Using systems of equations, solve for x and y:This gets: \(x \approx 28.10\), \(y \approx 60.90\).

Step 7: Calculate Weight Percentages

Find the weight percentage of each sulfide.- Weight percent of Cu₂S = \(\frac{28.10}{89} \times 100 \approx 31.57\%\)- Weight percent of CuS = \(\frac{60.90}{89} \times 100 \approx 68.43\%\).

Key Concepts

Copper Sulfide

Copper sulfide minerals play a vital role in copper ore processing and are fundamental to copper extraction. In the ore, we find two common copper sulfides: cuprite (\(\text{Cu}_{2}\text{S}\)) and copper(II) sulfide (\(\text{CuS}\)). Both of these compounds serve as the basis for the chemical reactions involved in extracting metallic copper.

• Cuprite \(\text{Cu}_{2}\text{S}\): This compound reacts with oxygen to produce copper metal and sulfur dioxide. During this process, each molecule of \(\text{Cu}_{2}\text{S}\) breaks down to form two atoms of copper.
• Copper(II) sulfide \(\text{CuS}\): In a similar manner, \(\text{CuS}\) also reacts to liberate copper metal and sulfur dioxide, but it yields one copper atom per molecule of \(\text{CuS}\).

By understanding the precise chemistry behind these reactions, we can predict the amount of copper extractable from the raw ore. This step is crucial in evaluating the economic viability of the mining and refining process.

Stoichiometry

Stoichiometry is the backbone of predicting chemical reaction outcomes, especially in ore processing to determine yields. It involves using balanced chemical equations to relate quantities of reactants and products. In the extraction of copper from its sulfides, stoichiometry helps establish these relations.

• Copper from \(\text{Cu}_{2}\text{S}\): The balanced reaction \(\text{Cu}_{2}\text{S} + \text{O}_2 \rightarrow 2\text{Cu} + \text{SO}_2\) implies that one mole of \(\text{Cu}_{2}\text{S}\) produces two moles of copper. This ratio is foundational for calculating the copper yield from cuprite.
• Copper from \(\text{CuS}\): Similarly, the equation \(\text{CuS} + \text{O}_2 \rightarrow \text{Cu} + \text{SO}_2\) indicates a one-to-one ratio between moles of \(\text{CuS}\) and the copper produced.

By applying these stoichiometric ratios, we can design equations to calculate how much copper we can realistically obtain from a given amount of ore containing both \(\text{Cu}_{2}\text{S}\) and \(\text{CuS}\). Mastering these calculations is fundamental to understanding the efficiencies in ore processing and copper extraction.

Weight Percent Calculation

Weight percent calculation is a straightforward method used to express the composition of an ore or any chemical mixture. This is necessary in the context of copper extraction from its sulfides to determine how much of the ore consists of each type of compound.

• Formula: Weight percent of a component is calculated using the formula: \(\text{Weight percent} = \left( \frac{\text{mass of the component}}{\text{total mass of the sample}} \right) \times 100\% \).
• Application in Ore Analysis: To find the weight percent of \(\text{Cu}_{2}\text{S}\) and \(\text{CuS}\) in the ore, one would divide the mass of each sulfide by the total mass of copper-bearing compounds and then multiply by 100. This provides a clear picture of the ore's composition.

In our exercise, using this method, we found the weight percent of \(\text{Cu}_{2}\text{S}\) was approximately 31.57% and \(\text{CuS}\) was about 68.43%. These values help in assessing the grade of the copper ore and understanding how efficient the extraction process will be.