Question

4. A student started with a 0.032 g sample of copper which he took through the series of reactions described in this experiment. At the end of the experiment he obtained 0.038 g of a black product. What was his percent yield

Answer 1

Answer:

$Y=48.6\%$

Explanation:

Hello,

In this case, we can consider the following chemical reaction for the oxidation of copper which only occurs at high temperatures:

$2Cu+O_2\rightarrow 2CuO$

In such a way, for 0.032 grams of copper, the following grams of copper (II) oxide (black product) are yielded:

$m_{CuO}=0.032gCu*\frac{1molCu}{63.546gCu} *\frac{2molCuO}{2molCu}*\frac{79.546gCuO}{1molCuO} =0.078gCuO$

Therefore, the percent yield is:

$Y=\frac{0.038g}{0.078g}*100\%\\ \\Y=48.6\%$

Best regards.