Question

when copper is heated with an excess of sulfur, copper(i) sulfide is formed. in a given experiment, 0.0970 moles of copper were heated with excess sulfur to yield 1.68 g copper(i) sulfide. what is the percent yield?

Explanation:

Step1: Write the balanced chemical equation

$$2Cu + S ightarrow Cu_2S$$

From the equation, the mole ratio of \(Cu\) to \(Cu_2S\) is \(2:1\).

Step2: Calculate the theoretical moles of \(Cu_2S\)

Given \(n_{Cu}=0.0970\space mol\). Using the mole ratio \(\frac{n_{Cu_2S}}{n_{Cu}}=\frac{1}{2}\), so \(n_{Cu_2S}=\frac{0.0970\space mol}{2}=0.0485\space mol\)

Step3: Calculate the theoretical mass of \(Cu_2S\)

The molar mass of \(Cu_2S\) is \(M=(2\times63.55 + 32.07)\space g/mol=(127.1+32.07)\space g/mol = 159.17\space g/mol\)
Using \(m = n\times M\), the theoretical mass \(m_{theo}=0.0485\space mol\times159.17\space g/mol\approx7.72\space g\)

Step4: Calculate the percent yield

Percent yield formula is \(\%\text{yield}=\frac{m_{actual}}{m_{theo}}\times100\%\)
Given \(m_{actual} = 1.68\space g\) and \(m_{theo}\approx7.72\space g\)
\(\%\text{yield}=\frac{1.68\space g}{7.72\space g}\times 100\%\approx21.8\%\)

Answer:

The percent yield is approximately \(21.8\%\)