Question

Ethanol \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right),\) the intoxicant in alcoholic beverages, is also used to make other organic compounds. In concentrated sulfuric acid, ethanol forms diethyl ether and water: $$ 2 \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l) \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OCH}_{2} \mathrm{CH}_{3}(l)+\mathrm{H}_{2} \mathrm{O}(g) $$ In a side reaction, some ethanol forms ethylene and water: $$ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l) \longrightarrow \mathrm{CH}_{2} \mathrm{CH}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ (a) If \(50.0 \mathrm{~g}\) of ethanol yields \(35.9 \mathrm{~g}\) of diethyl ether, what is the percent yield of diethyl ether? (b) If \(45.0 \%\) of the ethanol that did not produce the ether reacts by the side reaction, what mass (g) of ethylene is produced?

Short answer

The percent yield of diethyl ether is approximately 89.3%. The mass of ethylene produced is approximately 6.84 g.

Step-by-step solution

Calculate moles of ethanol

Start with the amount of ethanol given: 50.0 g. Use the molar mass of ethanol (\text{C}_2\text{H}_5\text{OH}) to convert grams to moles. The molar mass of ethanol \text{C}_2\text{H}_5\text{OH} is 46.07 g/mol.\[ \text{Moles of ethanol} = \frac{50.0 \text{ g}}{46.07 \text{ g/mol}} \approx 1.085 \text{ moles} \]

Calculate theoretical yield of diethyl ether

Use the balanced chemical equation for the formation of diethyl ether to find the stoichiometric relationship. According to the equation, 2 moles of ethanol produce 1 mole of diethyl ether: \[ \text{Theoretical moles of diethyl ether} = \frac{1.085 \text{ moles}}{2} = 0.543 \text{ moles} \] Use the molar mass of diethyl ether (\text{C}_2\text{H}_5\text{OC}_2\text{H}_5}): 74.12 g/mol to find the theoretical mass: \[ \text{Theoretical mass of diethyl ether} = 0.543 \text{ moles} \times 74.12 \text{ g/mol} \approx 40.23 \text{ g} \]

Calculate percent yield of diethyl ether

Percent yield is calculated using the formula: \[ \text{Percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100 \] Substituting the given values: \[ \text{Percent yield} = \frac{35.9 \text{ g}}{40.23 \text{ g}} \times 100 \approx 89.3\text{ \text{%}} \]

Determine ethanol used in side reaction

To find the ethanol that did not produce the ether, subtract the amount used to form diethyl ether: \[ \text{Moles of ethanol not producing ether} = 1.085 \text{ moles} - 0.543 \text{ moles} = 0.542 \text{ moles} \] Given that 45.0% of this ethanol undergoes the side reaction: \[ \text{Moles of ethanol in side reaction} = 0.542 \text{ moles} \times 0.45 = 0.244 \text{ moles} \]

Calculate mass of ethylene produced

Use the molar mass of ethylene (\text{C}_2\text{H}_4): 28.05 g/mol.\text{ Using the stoichiometry of the side reaction: 1 mole of ethanol produces 1 mole of ethylene.\[ \text{Mass of ethylene produced} = 0.244 \text{ moles} \times 28.05 \text{ g/mol} \approx 6.84 \text{ g} \]

Key concepts

percent yield

In chemistry, percent yield is a crucial concept. It helps us understand how effective a reaction is. Percent yield compares the actual amount of product obtained to the theoretical amount. To calculate it, use this formula:

• \[ \text{Percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100 \]

The theoretical yield is what you expect if everything reacts perfectly. However, in real life, reactions are not always perfect. This leads to a different, usually lower actual yield.

In the given problem, we had ethanol forming diethyl ether. We used the formula to find a percent yield of 89.3%. This shows we obtained 89.3% of the diethyl ether we could have made in a perfect scenario.

side reactions

Side reactions are another important concept in chemistry. They occur when a reaction that follows a different pathway happens along with the main reaction. Side reactions can reduce the yield of the desired product.

In this exercise, along with forming diethyl ether, ethanol also forms ethylene and water as a side reaction. This alternative pathway utilizes some of our ethanol, which means less is available to form the desired diethyl ether. Let's break it down:

• In the main reaction, 2 moles of ethanol produce 1 mole of diethyl ether and water.
• In the side reaction, ethanol forms ethylene and water instead.

Understanding side reactions helps chemists find ways to minimize unwanted products and increase the efficiency of the desired reaction.

stoichiometry

Stoichiometry is the section of chemistry that evaluates the quantities of reactants and products in chemical reactions. It uses the balanced chemical equation to understand the relationship between different substances in a reaction.

Here are simple steps to follow in stoichiometry calculations:

• First, write a balanced chemical equation.
• Then, identify the molar ratios from the equation.
• Use molar masses to convert between grams and moles.
• Finally, apply these ratios to find the desired unknown quantity.

In our problem, we used stoichiometry to figure out how many moles of diethyl ether we could theoretically form from our given ethanol. Then, we considered the side reaction and calculated the remaining ethanol that forms ethylene. This process shows how stoichiometry helps predict product amounts and keep track of all reactants and products in a reaction.