Percent Yield Question

Question

When copper is heated with an excess of sulfur, copper(I) sulfide is formed. In a given experiment, 0.0970 moles of copper was heated with excess sulfur to yield 3.77 g copper(I) sulfide. What is the percent yield?

Answer 1

Explanation:

When calculating percent yield, it is helpful to approach the problem in a calculated manner. While first learning, always write out the given information and think about what pieces are missing.

First, determine the balanced chemical equation of the items that you are working with. In this case, we know that sulfur and copper are heated to form copper (I) sulfide. From that balanced chemical equation, we learn the ratio between products and reactants. In this case, the ratio of copper and copper (I) sulfide is 2:1.

To calculate percent yield, we need both theoretical yield and the actual yield. We are only given the actual yield in this problem, thus we need to calculate the theoretical yield. To do such, we start with the known amount of copper and use basic stoichiometry to arrive to grams of Cu₂S. Recall that theoretical yield basically represents the maximum product formation possible based on the limiting reactant. So percent yield is a measure of how successful a given reaction was performed. In this case, our answer was 48.8% (Ensure answer is rounded to 3 sig-figs).

Work:

Balanced Chemical Equation:

2 Cu + S ---> Cu₂S

Theoretical Yield:

(0.0970 moles Cu) (1 mole Cu₂S) (159.16 grams Cu₂S)

------------------------ x -------------------- x --------------------------------- = 7.72 g Cu₂S

(2 mole Cu) (1 mole Cu₂S)

Actual Yield:

3.77 g Cu₂S

Percent Yield:

Actual Yield 3.77 g Cu₂S

Percent yield = ------------------------ x 100 = -------------------- x 100 = 48.8%

Theoretical Yield 7.72 g Cu₂S

Final Answer = 48.8% Yield