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

Answer 1

Since 2 moles of Cu are contained in 1 mole of Cu₂S, we can calculate the expected amount of Cu₂S and use that as the reference for the 100%:

$0.0970\text{ mol Cu}\times \frac{1\text{ mol Cu}_2\text{S}}{2\text{ mol Cu}}\times \frac{159.158\text{ g Cu}_2\text{S}}{1\text{ mol Cu}_2\text{S}} \\ \text{ } \\ \implies 7.719163\text{ g Cu}_2\text{S} \approxeq 7.72\text{ g Cu}_2\text{S}$

Then we can calculate the yield percent as:

$\%(\text{yield})=\cfrac{m_{Cu_2S\text{ obtained}}}{m_{Cu_2S\text{ expected}}}\times 100 \\ \text{ } \\ \%(\text{yield})=\frac{2.98 \cancel{ \text{ g Cu}_2\text{S} } }{7.72 \cancel{\text{ g Cu}_2\text{S} } }\times 100=38.60 \%$

In conclusion, the yield percent for this reaction is 38.60%.

Reference

• Chang, R., & Goldsby, K. A. (2010). Chemistry. Chemistry, 10th ed.; McGraw-Hill Education: New York, NY, USA.