Question

Ethanol (CH₃CH₂OH), the intoxicant in alcoholic beverages, is also used to make other organic compounds. In concentrated sulfuric acid, ethanol forms diethyl ether and water:

$\mathbf{2}\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{C}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathit{H}\mathbf{\left(}\mathit{I}\mathbf{\right)}\mathbf{\to }\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{C}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{C}{\mathit{H}}_{\mathbf{2}}\mathbf{\left(}\mathit{I}\mathbf{\right)}\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathit{g}\mathbf{\right)}$

In a side reaction, some ethanol forms ethylene and water:

$\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{C}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathit{H}\mathbf{\left(}\mathit{I}\mathbf{\right)}\mathbf{\to }\mathit{C}{\mathit{H}}_{\mathbf{2}}\mathbf{=}\mathit{C}{\mathit{H}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathit{g}\mathbf{\right)}$

(a) If 50.0 g of ethanol yields 35.9 g of diethyl ether, what is the percent yield of diethyl ether? (b) During the process, 45.0% of the ethanol that did not produce diethyl ether reacts by the side reaction. What mass of ethylene is produced?

Short answer

1. The percent yield of diethyl ether is89.3 %.
2. The mass of ethylene production is1.48 g.

Step-by-step solution

Calculating the percent yield of diethyl ether

The mass can be denoted as,

$\begin{array}{rcl}\mathrm{Mass}\mathrm{of}{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{OCH}}_2{\mathrm{CH}}_3\left(\mathrm{theoretical}\right)& =& 50\mathrm{gethanol}\times \frac{1\mathrm{molethanol}}{46.08\mathrm{gethanol}}\times \frac{1\mathrm{moldiethylether}}{2\mathrm{molethanol}}\times \frac{74.14\mathrm{g}\mathrm{diethylether}}{1\mathrm{moldiethylether}}\\ & =& 40.2\mathrm{gdiethylether}\end{array}$

The percent yield can be found as,

$$\begin{gathered} \%\mathrm{yield}=\frac{\mathrm{actual}\mathrm{yield}}{\mathrm{theoretical}\mathrm{yield}}\times 100 \\ \%\mathrm{yield}=\frac{35.9\mathrm{g}}{40.2\mathrm{g}}\times 100 \\ \%\mathrm{yield}=89.3\% \end{gathered}$$

Calculating the mass of ethylene production

The mass can be found as,

$\begin{array}{rcl}\mathrm{Mass}\mathrm{of}{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}(\mathrm{for}\mathrm{diethylether})& =& 35.9\mathrm{gdiethylether}\times \frac{1\mathrm{moldiethylether}}{4.14\mathrm{gdiethylether}}\times \frac{1\mathrm{molethanol}}{1\mathrm{moldiethylether}}\times \frac{46.08\mathrm{gethanol}}{1\mathrm{molethanol}}\\ & =& 44.6\mathrm{gethanol}\\ \mathrm{Mass}\mathrm{of}{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}(\mathrm{for}\mathrm{ethylene})& =& 0.45\times (50\mathrm{g}-44.6\mathrm{g})=2.43\mathrm{gethanol}\\ \mathrm{Mass}\mathrm{of}{\mathrm{CH}}_2& =& {\mathrm{CH}}_2=2.43\mathrm{gethanol}\times \frac{1\mathrm{molethanol}}{46.08\mathrm{gethanol}}\times \frac{1\mathrm{molethylene}}{1\mathrm{molethanol}}\times \frac{28.06\mathrm{gethylene}}{1\mathrm{molethylene}}\\ \mathrm{Mass}\mathrm{of}{\mathrm{CH}}_2& =& {\mathrm{CH}}_2=1.48\mathrm{g}\end{array}$