1.50 g of copper was heated with excess sulfur to yield 1.76 g copper(I) sulfide. What is the percent yield?

Answer 1

Note: I updated my answer after Alexis G. kindly noted a math error (noted in bold and underline).

Hello, Emi,

We need to have a balanced equation to start. The chemical formula for copper(1) sulfide is Cu₂S. So we can write a balanced equation:

2Cu + S = Cu₂S

This tells us we should get 1/2 mole Cu₂S for every 1 mole Cu.

Find moles Cu by dividing the mass by the molar mass of Cu: (1.50g/(63.55 g/mole) = 0.0236 moles Cu. We have excess sulfur, so the equation preducts we should get 0.0118 moles of Cu₂S.

(0.0118 moles Cu₂S)*(159.2 grams Cu₂S/mole Cu₂S) = 1.88 grams Cu₂S.

We actually got 1.76 grams Cu₂S. That means the yield was (1.76g/1.88g) = 0.937 or 93.7%

Not bad.

Bob