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?
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\u00d763.5=6.1595g"
At 100% yield we expect
"\\frac{6.1595}{127}\u00d7159.16=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}\u00d7100=18.07\\%"
Comments
Leave a comment