Question

When copper is heated with excess of sulfur, copper 1 sulfide is formed. In a given experiment, 0.0970 moles of copper was heated with excess sulfur to yield 1.43 g copper 1 sulfide. What is the percent yield?

Answer 1

First, write the chemical equation,

2Cu+S$\longrightarrow$ Cu₂S

From the equation, two moles of copper react to produce one mole of Cu₂S. The relation is therefore

127g Cu $\equiv$ 159.16g Cu₂S

Mass of copper available is

Mass=moles×mollar mass

$0.097×63.5=6.1595g$

At 100% yield we expect

$\frac{6.1595}{127}×159.16=7.91g$

Percentage yield is the quotient between the expected mass at 100% production and the mass produced then multiply by 100.

$\frac{1.43}{7.91}×100=18.07\%$