Question

(a) Diethyl ether \(\left(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\right)\), the "ether" used medically as an anesthetic, is prepared commercially by treatment of ethyl alcohol \(\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}\right)\) with an acid. How many grams of diethyl ether would you obtain from \(40.0 \mathrm{~g}\) of ethyl alcohol if the percent yield of the reaction is \(87.0 \% ?\) $$ \begin{aligned} &2 \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}(l) \stackrel{\text { Acid }}{\longrightarrow} \mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l)+\mathrm{H}_{2} \mathrm{O}(l) \\ &\text { Ethyl alcohol } & \text { Diethyl ether } \end{aligned} $$ (b) How many grams of ethyl alcohol would be needed to produce \(100.0 \mathrm{~g}\) of diethyl ether if the percent yield of reaction is \(87.0 \% ?\)

Short answer

(a) 27.98 g of diethyl ether can be obtained. (b) 142.9 g of ethyl alcohol is needed.

Step-by-step solution

Calculate Molar Mass of Ethyl Alcohol

First, calculate the molar mass of ethyl alcohol (\( \mathrm{C}_2 \mathrm{H}_6 \mathrm{O} \)). - Carbon (C) has a molar mass of 12.01 g/mol, Hydrogen (H) is 1.01 g/mol, and Oxygen (O) is 16.00 g/mol. \[\text{Molar mass of } \mathrm{C}_2 \mathrm{H}_6 \mathrm{O} = 2 \times 12.01 + 6 \times 1.01 + 16.00 = 46.08 \text{ g/mol}. \]

Calculate Molar Mass of Diethyl Ether

Next, calculate the molar mass of diethyl ether (\(\mathrm{C}_4 \mathrm{H}_{10} \mathrm{O} \)). - Using the same atomic masses: \[\text{Molar mass of } \mathrm{C}_4 \mathrm{H}_{10} \mathrm{O} = 4 \times 12.01 + 10 \times 1.01 + 16.00 = 74.12 \text{ g/mol}. \]

Determine Moles of Ethyl Alcohol Used

Calculate the number of moles of ethyl alcohol available. Given the mass is 40.0 g: \[\text{Moles of } \mathrm{C}_2 \mathrm{H}_6 \mathrm{O} = \frac{40.0 \text{ g}}{46.08 \text{ g/mol}} = 0.8684 \text{ mol}\]

Calculate Ideal Grams of Diethyl Ether

According to the reaction, 2 moles of \(\mathrm{C}_2 \mathrm{H}_6 \mathrm{O}\) produces 1 mole of \(\mathrm{C}_4 \mathrm{H}_{10} \mathrm{O}\). Therefore: \[ 0.8684 \text{ mol } \mathrm{C}_2 \mathrm{H}_6 \mathrm{O} \times \frac{1 \text{ mol } \mathrm{C}_4 \mathrm{H}_{10} \mathrm{O}}{2 \text{ mol } \mathrm{C}_2 \mathrm{H}_6 \mathrm{O}} = 0.4342 \text{ mol } \mathrm{C}_4 \mathrm{H}_{10} \mathrm{O} \]Convert to grams: \[0.4342 \text{ mol} \times 74.12 \text{ g/mol} = 32.16 \text{ g}\]

Calculate Actual Yield of Diethyl Ether

Considering the percent yield of 87.0%, the actual grams obtained would be: \[32.16 \text{ g} \times \frac{87.0}{100} = 27.98 \text{ g} \]

Calculate Required Moles of Diethyl Ether for 100g

For part (b), determine moles of diethyl ether required: \[\text{Moles of } \mathrm{C}_4 \mathrm{H}_{10} \mathrm{O} = \frac{100.0 \text{ g}}{74.12 \text{ g/mol}} = 1.349 \text{ mol}\]

Calculate Required Moles of Ethyl Alcohol for Desired Ether

Using the reaction stoichiometry, find moles of \(\mathrm{C}_2 \mathrm{H}_6 \mathrm{O}\) : \[ 1.349 \text{ mol } \mathrm{C}_4 \mathrm{H}_{10} \mathrm{O} \times \frac{2 \text{ mol } \mathrm{C}_2 \mathrm{H}_6 \mathrm{O}}{1 \text{ mol } \mathrm{C}_4 \mathrm{H}_{10} \mathrm{O}} = 2.698 \text{ mol } \mathrm{C}_2 \mathrm{H}_6 \mathrm{O} \]

Calculate Adjusted Required Moles Considering Yield

Because the yield is only 87.0%, more ethyl alcohol is needed: \[\text{Adjusted moles of } \mathrm{C}_2 \mathrm{H}_6 \mathrm{O} = \frac{2.698}{0.87} = 3.101 \text{ mol}\]

Convert Moles of Ethyl Alcohol to Grams

Finally, convert moles to grams for the required ethyl alcohol: \[3.101 \text{ mol} \times 46.08 \text{ g/mol} = 142.9 \text{ g}\]

Key concepts

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, are transformed into different substances called products. In the case of this reaction, ethyl alcohol (\(\mathrm{C}_2 \mathrm{H}_6 \mathrm{O}\)) reacts in the presence of an acid to form diethyl ether (\(\mathrm{C}_4 \mathrm{H}_{10} \mathrm{O}\)) and water. This reaction can be represented by the equation: \[ 2 \mathrm{C}_2 \mathrm{H}_6 \mathrm{O}(l) \rightarrow \mathrm{C}_4 \mathrm{H}_{10} \mathrm{O}(l) + \mathrm{H}_2 \mathrm{O}(l) \]The coefficients in the balanced chemical equation, such as the "2" in front of ethyl alcohol, show the molar ratio of the reactants and products. In this example, it takes 2 moles of ethyl alcohol to produce 1 mole of diethyl ether. These coefficients are essential for calculating the amounts of reactants needed or products formed in a reaction.

• Reactants: Substances that go through chemical changes.
• Products: New substances formed as a result of the reaction.
• Molar Ratio: Corresponds to the amounts of substances as indicated by the balanced equation.

Understanding chemical reactions allows us to predict the outcomes and quantities in reactions, crucial for applications like manufacturing or pharmaceuticals.

Percent Yield

Percent yield is a measure used in chemistry to determine how efficient a chemical reaction has been. It compares the actual yield (amount of product actually produced) to the theoretical yield (amount predicted by stoichiometric calculations based on perfect conditions). The percent yield is calculated using the formula:\[\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\%\]In practical situations, the actual yield is often less than the theoretical yield due to factors like incomplete reactions, side reactions, or experimental errors. In the given exercise, a percent yield of 87.0% suggests that only 87% of the diethyl ether that could have been produced was actually obtained. This is a common aspect to consider in experimental chemistry, as it reflects the limitations of real-world conditions.

• Theoretical Yield: Calculated amount of product expected based on initial reactants.
• Actual Yield: Amount of product actually obtained from the reaction.
• Factors Affecting Yield: Incomplete reactions, impurities, measurement errors.

Molar Mass Calculations

Calculating molar mass is essential for stoichiometry, as it provides a link between the mass of a substance and the amount in moles. The molar mass is determined by summing the atomic masses of all atoms in a molecule. For example, the molar mass of ethyl alcohol (\(\mathrm{C}_2 \mathrm{H}_6 \mathrm{O}\)) can be calculated by adding the masses of carbon, hydrogen, and oxygen:\[\text{Molar mass of } \mathrm{C}_2 \mathrm{H}_6 \mathrm{O} = 2 \times 12.01 + 6 \times 1.01 + 16.00 = 46.08 \text{ g/mol}\]For diethyl ether (\(\mathrm{C}_4 \mathrm{H}_{10} \mathrm{O}\)), the calculation is similar:\[\text{Molar mass of } \mathrm{C}_4 \mathrm{H}_{10} \mathrm{O} = 4 \times 12.01 + 10 \times 1.01 + 16.00 = 74.12 \text{ g/mol}\]These calculations are crucial for understanding reactions, as they help in converting between mass and moles, allowing chemists to predict how much product can be made from given reactants.

• Molar Mass: Mass of one mole of a substance expressed in grams per mole (g/mol).
• Importance: Helps convert between gram quantity and mole quantity, which is pivotal for calculations in chemical reactions.