When copper is heated with an excess of sulfur, copper(I) sulfide is formed. In a given experiment, 1.50 g of copper was heated with excess sulfur to yield 1.56 g of copper(I) sulfide. What is the percent yield?

• A. 92.3%
• B. 78.9%
• C. 85.7%
• D. 100.0%

Answer: (C)

Verified step by step guidance

1. Determine the balanced chemical equation for the reaction: Copper (Cu) reacts with sulfur (S) to form copper(I) sulfide (Cu2S). The balanced equation is: 2Cu + S -> Cu2S.
2. Calculate the molar mass of copper (Cu) and copper(I) sulfide (Cu2S). The molar mass of Cu is approximately 63.55 g/mol, and the molar mass of Cu2S is approximately 159.16 g/mol.
3. Convert the mass of copper used (1.50 g) to moles using its molar mass: moles of Cu = mass of Cu / molar mass of Cu.
4. Use the stoichiometry of the balanced equation to find the theoretical yield of Cu2S. Since 2 moles of Cu produce 1 mole of Cu2S, calculate the moles of Cu2S that should form from the moles of Cu calculated.
5. Convert the theoretical moles of Cu2S to grams using its molar mass, then calculate the percent yield using the formula: Percent Yield = (actual yield / theoretical yield) * 100, where the actual yield is 1.56 g.