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

Answer 1

The equation is below:

2Cu + S --> Cu₂S

According to the equation, 2 moles of copper produce 1 mole of copper(I) sulfide. The molar mass of Cu₂S is 159.16 g/mol. Therefore,

Theoretical yield of Cu₂S = $0.0970\ mol(Cu)\times\frac{1\ mol(Cu_2S)}{2\ mol(Cu)}\times\frac{159.16\ g(Cu_2S)}{1\ mol(Cu_2S)}=7.72\ g$

$\%\ yield=\frac{actual\ yield}{theoretical\ yield}\times100\%=\frac{6.15\ g}{7.72\ g}\times100\%=79.7\%$

Answer: 79.7 %