The percentage yield of the Copper(i) sulfide is 32.02%.
Let us first write the equation of the reaction of copper with sulfur to form Copper(i) sulfide,
2Cu + S → Cu₂S
According to this reaction,
2 moles of Cu gives 1 mole of Cu₂S
It is given, tat 0.0970 moles of copper is reacting,
So,
2 moles copper = 1 mole copper(i)sulfide
1 mole copper = 1/2 mole copper(i)sulfide
0.0970 moles copper = 0.0970/2 moles of copper(i)sulfide
We know,
Moles = formed mass/molar mass
Molar mass of copper sulfide is 160 grams/mol
Formed mass of copper sulfide = 160 x 0.097
Formed mass of copper sulfide = 15.52 grams.
Percentage yield = Experimental mass/theoretical mass x 100
Percentage yield = 4.97/15.52 x 100
Percentage yield = 32.02%
So, the percentage yield is 32.02%.
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