Question

When copper is heated with an excess of sulfur, copper(I) sulfide is formed. In a given experiment, 1.50 g of copper was heated with excess sulfur to yield 1.59 g copper(I) sulfide. What is the percent yield

Answer 1

Answer: The percent yield of the reaction is 83.2 %.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$      .....(1)

For copper:

Given mass of copper = 1.50 g

Molar mass of copper= 63.5 g/mol

Putting values in equation 1, we get:

$\text{Moles of copper}=\frac{1.50g}{63.5g/mol}=0.024mol$

The chemical equation for the reaction of copper and sulphur follows:

$2Cu+S\rightarrow Cu_2S$

Thus, copper is considered as a limiting reagent because it limits the formation of product and sulphur is in excess.

By Stoichiometry of the reaction:

2 moles of copper produces 1 mole of copper (I) sulphide

So,0.024 moles of copper will produce = $\frac{1}{2}\times 0.024=0.012moles$ of copper (I) sulphide

$\text{Mass of copper (I) sulphide}}=moles\times \text{Mass of copper (I) sulphide}=0.012\times 159=1.91g$

To calculate the percentage yield of copper (I) sulphide we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of copper (I) sulphide = 1.59 g

Theoretical yield of copper (I) sulphide= 1.91 g

Putting values in above equation, we get:

$\%\text{ yield of copper (I) sulphide}=\frac{1.59g}{1.91g}\times 100\\\\\% \text{yield of copper (I) sulphide}=83.2\%$

Hence, the percent yield of the reaction is 83.2 %.