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 1.76 g copper(I) sulfide. What is the percent yield?

Answer 1

Answer:

Percent yield = 22.8 %

Explanation:

Step 1: Data given

Numbers of moles  copper = 0.0970 moles

Mass of copper(I) sulfide = 1.76 grams

Step 2: The balanced equation

2Cu + S ⇒ Cu2S

Step 3:  Calculate moles of Cu2S

For 2 moles Cu we need 1 mol S to produce 1 mol Cu2S

For 0.0970 moles Cu we'll hace 0.0970 / 2 = 0.0485 moles

Step 4: Calculate mass of Cu2S

Mass Cu2s = moles Cu2S * molar mass Cu2S

Mass Cu2S = 0.0485 moles * 159.16 g/mol

Mass Cu2S = 7.72 grams

Step 5: Calculate percent yield

Percent yield = (actual yield/ theoretical mass) * 100%

Percent yield = (1.76 grams / 7.72 grams)*100%

Percent yield = 22.8 %