Question

ATOM ECONOMY: Ethylene oxide, \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O},\) is an important industrial chemical las it is the starting place to make such important chemicals as ethylene glycol (antifreeze) and various polymers \(1 .\) One way to make the compound is called the "chlorohydrin route." $$ \mathrm{C}_{2} \mathrm{H}_{4}+\mathrm{Cl}_{2}+\mathrm{Ca}(\mathrm{OH})_{2} \rightarrow \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}+\mathrm{CaCl}_{2}+\mathrm{H}_{2} \mathrm{O} $$ Another route is the modern catalytic reaction. $$ \mathrm{C}_{2} \mathrm{H}_{4}+1 / 2 \mathrm{O}_{2} \rightarrow \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O} $$ (a) Calculate the \(\%\) atom economy for the production of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\) in each of these reactions. Which is the more efficient method? (b) What is the percent yield of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\) if \(867 \mathrm{g}\) of \(\mathrm{C}_{2} \mathrm{H}_{4}\) is used to synthesize \(762 \mathrm{g}\) of the product by the catalytic reaction?

Short answer

(a) Catalytic route is more efficient (Atom economy: 100% vs. 25.45%). (b) Percent yield is 87.9%.

Step-by-step solution

Calculating Molar Masses for Chlorohydrin Route

For the chlorohydrin route reaction, first calculate the molar masses of all the reactants and products. \(\mathrm{C}_{2}\mathrm{H}_{4} = 28.05 \, \mathrm{g/mol}, \mathrm{Cl}_{2} = 70.90 \, \mathrm{g/mol}, \mathrm{Ca(OH)}_{2} = 74.10 \, \mathrm{g/mol}, \mathrm{C}_{2}\mathrm{H}_{4}\mathrm{O} = 44.05 \, \mathrm{g/mol}, \mathrm{CaCl}_{2} = 110.98 \, \mathrm{g/mol}, \mathrm{H}_{2}\mathrm{O} = 18.02 \, \mathrm{g/mol}. \) Calculate the molar mass of the reactants: \(28.05 + 70.90 + 74.10 = 173.05 \, \mathrm{g/mol}.\) So the total molar mass of reactants is \(173.05 \, \mathrm{g/mol}.\)

Calculating Atom Economy for Chlorohydrin Route

Using the formula for atom economy, \(\text{Atom Economy} = \frac{\text{Molar mass of desired product}}{\text{Total molar mass of reactants}} \times 100\%%.\) For the chlorohydrin route, it is \(\frac{44.05}{173.05} \times 100\% = 25.45\%\) atom economy.

Calculating Molar Masses for Catalytic Route

For the catalytic route, we have fewer reactants: \(\mathrm{C}_{2}\mathrm{H}_{4} = 28.05 \, \mathrm{g/mol}, \mathrm{O}_{2} = 32.00 \, \mathrm{g/mol}.\) The total molar mass of reactants is \(28.05 + \frac{32.00}{2} = 44.05 \, \mathrm{g/mol}.\)

Calculating Atom Economy for Catalytic Route

Using the same atom economy formula, \(\text{Atom Economy} = \frac{44.05}{44.05} \times 100\% = 100\%\). The catalytic route has an atom economy of \(100\%\), making it more efficient than the chlorohydrin route.

Calculating Percent Yield

Determine the theoretical yield of \(\mathrm{C}_{2}\mathrm{H}_{4}\mathrm{O}\) from \(867 \, \mathrm{g}\) of \(\mathrm{C}_{2}\mathrm{H}_{4}\): since the ratio is 1:1, theoretical yield is \(867 \, \mathrm{g}\) of \(\mathrm{C}_{2}\mathrm{H}_{4}\mathrm{O}\). The percent yield formula is \(\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%.\) Thus, \(\frac{762}{867} \times 100\% = 87.9\%\) percent yield.

Key concepts

Molar Mass Calculation

Molar mass calculation is an essential step in understanding chemical reactions as it helps to determine the amount of reactants and products involved. The molar mass is the mass of one mole of a substance, typically measured in grams per mole (g/mol). To find the molar mass of a compound, simply add up the atomic masses of all the atoms in its chemical formula.

For example, let's look at ethylene oxide (C₂H₄O). The atomic masses for carbon (C), hydrogen (H), and oxygen (O) are approximately 12.01, 1.008, and 16.00 respectively. Therefore, the molar mass of C₂H₄O is calculated as follows:

• 2 Carbons: 2 × 12.01 g/mol = 24.02 g/mol
• 4 Hydrogens: 4 × 1.008 g/mol = 4.032 g/mol
• 1 Oxygen: 1 × 16.00 g/mol = 16.00 g/mol

Adding these up, the molar mass of C₂H₄O is 44.052 g/mol.

Having accurate molar masses is crucial for reaction calculations such as those in atom economy and yield efficiency.

Chlorohydrin Route

The chlorohydrin route is a traditional method for producing ethylene oxide, involving several reactants. In this reaction, ethylene (C₂H₄) is combined with chlorine (Cl₂) and calcium hydroxide (Ca(OH)₂) to yield ethylene oxide (C₂H₄O), along with side products such as calcium chloride (CaCl₂) and water (H₂O).

To understand the efficiency of this route, one can use the concept of atom economy. Atom economy is a measure of how much of the reactants' mass is converted into the desired product. The formula for atom economy is:

• Atom Economy = (Molar mass of desired product / Total molar mass of reactants) × 100%

In this route, only 25.45% of the reactant mass becomes the desired product, ethylene oxide.

While the chlorohydrin route is functional, it is not the most efficient due to the production of multiple byproducts. This makes it less appealing compared to more modern methods.

Catalytic Reaction

The catalytic route is a more modern and efficient method for producing ethylene oxide. In this process, ethylene (C₂H₄) reacts directly with oxygen (O₂) in the presence of a catalyst to produce ethylene oxide (C₂H₄O).

The beauty of this method lies in its simplicity and efficiency. With fewer reactants involved, this reaction often achieves an atom economy of 100%. This means that all the mass of the reactants is converted directly into the desired product, ethylene oxide. Such high atom economy is favorable because it indicates minimal waste production, making the process more sustainable and cost-effective.

Catalysts are substances that speed up chemical reactions without being consumed in the process. They are crucial in the catalytic route as they lower the activation energy, allowing the reaction to occur under less extreme conditions. This contributes to the industrial preference for the catalytic method over the chlorohydrin route.

Percent Yield Calculation

Percent yield is a concept used to compare the efficiency of a chemical reaction in practical terms. It provides insight into how much product is actually obtained from a reaction compared to what was theoretically possible.

To calculate percent yield, you use the equation:

• Percent Yield = (Actual Yield / Theoretical Yield) × 100%

The actual yield is the amount of product collected from the reaction, while the theoretical yield is the maximum amount of product that could be formed based on the reactants used.

In the given problem, if 867 g of ethylene (C₂H₄) is used to synthesize 762 g of ethylene oxide (C₂H₄O), the theoretical yield (based on the 1:1 molar ratio) is assumed to be 867 g. Thus, the percent yield is calculated as:

• (762 g / 867 g) × 100% = 87.9%

A percent yield of 87.9% indicates a highly efficient reaction, although some loss is expected due to practical limitations like incomplete reactions or side reactions.