Question

Copper(I) sulfide reacts with \(\mathrm{O}_{2}\) upon heating to give copper metal and sulfur dioxide. (a) Write a balanced equation for the reaction. (b) What mass of copper metal can be obtained from \(500 . \mathrm{g}\) of copper \((\mathrm{I})\) sulfide?

Short answer

399.16 g of copper can be obtained from 500 g of copper(I) sulfide.

Step-by-step solution

Write the Unbalanced Reaction Equation

Copper(I) sulfide is represented as \( \text{Cu}_2\text{S} \), and it reacts with oxygen \( \text{O}_2 \) to form copper metal \( \text{Cu} \) and sulfur dioxide \( \text{SO}_2 \). The unbalanced chemical equation is:\[ \text{Cu}_2\text{S} + \text{O}_2 \rightarrow \text{Cu} + \text{SO}_2 \]

Balance the Chemical Equation

To balance the equation, ensure that the number of atoms of each element on the reactant side equals the number on the product side. The balanced equation is:\[ \text{Cu}_2\text{S} + \text{O}_2 \rightarrow 2\text{Cu} + \text{SO}_2 \]

Calculate the Molar Mass of Copper(I) Sulfide

Copper(I) sulfide \( \text{Cu}_2\text{S} \) has a molar mass calculated as follows: - Copper (Cu) has a molar mass of approximately 63.55 g/mol, so \( \text{Cu}_2 \) is \( 2 \times 63.55 = 127.1 \;\text{g/mol}\).- Sulfur (S) has a molar mass of approximately 32.07 g/mol.Thus, the molar mass of \( \text{Cu}_2\text{S} \) is \( 127.1 + 32.07 = 159.17 \;\text{g/mol}\).

Calculate the Moles of Copper(I) Sulfide

To find the moles of \( \text{Cu}_2\text{S} \), divide the mass by the molar mass:\[ \text{Moles of } \text{Cu}_2\text{S} = \frac{500 \;\text{g}}{159.17 \;\text{g/mol}} \approx 3.14 \;\text{mol} \]

Determine the Moles of Copper Produced

The balanced equation \( \text{Cu}_2\text{S} + \text{O}_2 \rightarrow 2\text{Cu} + \text{SO}_2 \) shows that 1 mole of \( \text{Cu}_2\text{S} \) produces 2 moles of \( \text{Cu} \). Therefore:\[ \text{Moles of Cu} = 2 \times 3.14 \;\text{mol} \approx 6.28 \;\text{mol} \]

Calculate the Mass of Copper Produced

To find the mass of copper, multiply the moles of copper by the molar mass of copper \( 63.55 \;\text{g/mol} \):\[ \text{Mass of Cu} = 6.28 \;\text{mol} \times 63.55 \;\text{g/mol} = 399.16 \;\text{g} \]

Final Answer Summary

The mass of copper metal that can be obtained from 500 g of copper(I) sulfide is approximately 399.16 g.

Key concepts

Balancing Chemical Equations

Balancing chemical equations is an essential skill in chemistry. It helps ensure the law of conservation of mass holds true. This law states that matter cannot be created or destroyed. In a chemical reaction, the total mass of the reactants must equal the total mass of the products.
To balance a chemical equation, you need to ensure the number of each type of atom is the same on both sides of the equation. Let's look at the example of copper(I) sulfide reacting with oxygen. The unbalanced equation is:

\[ \text{Cu}_2\text{S} + \text{O}_2 \rightarrow \text{Cu} + \text{SO}_2 \]

Our job is to balance it:

• Count the atoms of each element in the reactants and products.
• For copper (Cu), ensure there are 2 atoms on each side. You do this by putting a coefficient 2 in front of \( \text{Cu} \).
• Check sulfur (S): 1 atom in \( \text{Cu}_2\text{S} \) and 1 atom in \( \text{SO}_2 \) - already balanced.
• Finally, balance oxygen (O). There are 2 atoms in \( \text{O}_2 \) and 2 in \( \text{SO}_2 \) - also balanced.

So, the balanced equation is:
\[ \text{Cu}_2\text{S} + \text{O}_2 \rightarrow 2\text{Cu} + \text{SO}_2 \]
Balancing chemical equations is often a trial-and-error process. It involves adjusting the coefficients until all elements have the same number of atoms on both sides.

Stoichiometry

Stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. Once the equation is balanced, stoichiometry allows us to predict how much product will form from given amounts of reactants.
In our example, the balanced equation \( \text{Cu}_2\text{S} + \text{O}_2 \rightarrow 2\text{Cu} + \text{SO}_2 \) indicates that 1 mole of \( \text{Cu}_2\text{S} \) yields 2 moles of copper (Cu) metal. This stoichiometric relationship is key in further calculations.
To use stoichiometry effectively:

• Convert all given information into moles (using molar masses).
• Apply the mole ratio from the balanced equation to connect quantities of different substances.
• Finally, convert moles back into grams if needed (using molar masses again).

Understanding stoichiometry gives you the tools to predict amounts of substances used and produced in any chemical reaction. This makes it a foundational concept in chemistry.

Copper(I) Sulfide

Copper(I) sulfide, represented as \( \text{Cu}_2\text{S} \), is a compound of copper and sulfur. It's typically found in nature as an ore, called chalcocite. In chemical reactions, it is a useful starting material for producing copper metal.
When heated with oxygen, as in our exercise, copper(I) sulfide reacts to produce pure copper metal. This is summarized in the equation:

\[ \text{Cu}_2\text{S} + \text{O}_2 \rightarrow 2\text{Cu} + \text{SO}_2 \]

Copper sulfide’s role in metallurgy is vital since it provides a relatively simple and cost-effective way to purify copper. Understanding its reactions is important in industrial processes and helps chemists and engineers design efficient methods to extract metals.

Molar Mass Calculations

Molar mass calculations are essential for converting between the mass of a substance and the amount in moles. The molar mass is a compound’s mass per mole of its entities, with units of g/mol. This helps relate how many grams a mole of a substance weighs.
In our exercise, the molar mass of copper(I) sulfide \( \text{Cu}_2\text{S} \) is calculated as:

• Copper (Cu): Approximately 63.55 g/mol.
• Sulfur (S): Approximately 32.07 g/mol.

Since there are two copper atoms, the molar mass of copper contributes \( 2 \times 63.55 = 127.1 \;\text{g/mol} \) and for sulfur, it's \( 32.07 \;\text{g/mol} \). Thus, the total molar mass of \( \text{Cu}_2\text{S} \) is \( 159.17 \;\text{g/mol} \).
This allows us to convert from grams to moles using:
\[ \text{Moles} = \frac{\text{mass in grams}}{\text{molar mass}} \]
Molar mass calculations are central in stoichiometry and any quantitative chemical analysis, making it a crucial concept in both basic and advanced chemistry.