Question

Ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\), an important industrial organic chemical, can be prepared by heating hexane \(\left(\mathrm{C}_{6} \mathrm{H}_{14}\right)\) at \(800^{\circ} \mathrm{C}\) : $$ \mathrm{C}_{6} \mathrm{H}_{14} \stackrel{\Delta}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{4}+\text { other products } $$ If the yield of ethylene production is 42.5 percent, what mass of hexane must be used to produce \(481 \mathrm{~g}\) of ethylene?

Short answer

Approximately 1158.52 g of hexane is required.

Step-by-step solution

Determine the Molar Mass of Ethylene

First, find the molar mass of ethylene (\(\mathrm{C}_2 \mathrm{H}_4\)). Carbon has a molar mass of 12.01 g/mol and hydrogen has a molar mass of 1.01 g/mol. Therefore, the molar mass of ethylene is calculated as follows:\[\text{Molar mass of } \mathrm{C}_2 \mathrm{H}_4 = 2(12.01) + 4(1.01) = 28.05 \text{ g/mol}\]

Calculate Moles of Ethylene Needed

Next, calculate the number of moles of ethylene that corresponds to 481 g using the molar mass:\[\text{Moles of } \mathrm{C}_2 \mathrm{H}_4 = \frac{481 \text{ g}}{28.05 \text{ g/mol}} \approx 17.15 \text{ moles}\]

Adjust for Yield Percentage

Given the yield is 42.5%, calculate the moles of ethylene that need to be theoretically produced to actually get the desired amount:\[\text{Theoretical moles needed} = \frac{17.15}{0.425} \approx 40.35 \text{ moles}\]

Mole Ratio from Hexane to Ethylene

The balanced chemical equation for the reaction isn't provided, but we assume hexane completely decomposes to produce multiple moles of ethylene per mole of hexane. Typically, each molecule of hexane can produce up to three molecules of ethylene. This implies a 1:3 ratio of hexane to ethylene. Calculate the moles of hexane needed:\[\text{Moles of } \mathrm{C}_6 \mathrm{H}_{14} = \frac{40.35}{3} \approx 13.45 \text{ moles}\]

Calculate Mass of Hexane Required

Finally, calculate the mass of hexane required using its molar mass. The molar mass of hexane \((\mathrm{C}_6 \mathrm{H}_{14})\) is 86.18 g/mol:\[\text{Mass of } \mathrm{C}_6 \mathrm{H}_{14} = 13.45 \times 86.18 \approx 1158.52 \text{ g}\]

Conclusion

Therefore, to produce 481 g of ethylene with a 42.5% yield, approximately 1158.52 g of hexane is required.

Key concepts

Molar Mass Calculation

Understanding molar mass is a vital aspect of chemistry, particularly in reactions involving gases and large molecules. Molar mass, the mass of a given substance divided by its amount of substance measured in moles, allows chemists to convert between the mass of a substance and the number of moles. In the context of our exercise, we calculated the molar mass of ethylene (\( \mathrm{C}_2 \mathrm{H}_4 \)). This molecule contains two carbon atoms and four hydrogen atoms. Carbon has an atomic mass of 12.01 g/mol while hydrogen's atomic mass is 1.01 g/mol. By adding these up based on the number of each atom in the molecule, we arrive at a molar mass of 28.05 g/mol for ethylene. This serves as a starting point for determining how much of a given reagent or product is needed or produced in a reaction.
Accurate calculation of molar masses is crucial for understanding the quantitative aspects of chemical reactions. This is because it directly influences the conversion of mass to moles and vice versa, a step essential for stoichiometric calculations.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. These transformations are governed by the principles of chemical change and stoichiometry. When we discuss the preparation of ethylene from hexane, we're exploring a specific type of reaction that occurs under certain conditions (e.g., 800°C). Although the reaction in the exercise isn't fully balanced, the production of ethylene from hexane involves breaking and forming new chemical bonds, leading to the formation of ethylene and other byproducts.
The hexane molecule is decomposed, which can be visualized as the breaking apart of its chemical structure to form smaller molecules like ethylene. Understanding the basics of chemical reactions includes knowing how reactants are transformed into products and predicting the types of products formed based on the nature of the reactants and reaction conditions. In practice, knowing the chemical behavior helps in designing processes to maximize the efficiency and yield of desired products.

Stoichiometry

Stoichiometry is the quantitative heart of chemistry. It involves the calculation of reactants and products in chemical reactions. Here, it allows us to relate the amounts of hexane required to produce ethylene. In stoichiometry, each chemical equation is a recipe that indicates the proportionate amounts of reactants needed to form products. However, in this exercise, the exact reaction isn't fully detailed, yet we can deduce the primary stoichiometric relationship between hexane and ethylene.
The assumed 1:3 mole ratio implies that one mole of hexane can produce up to three moles of ethylene. Using stoichiometry, we translate the number of moles into the mass of the substances involved, leveraging the calculated molar masses. Correct stoichiometric calculations enable chemists to predict product yields, adjust experimental conditions, and optimize industrial chemical processes.

Yield Calculation

Yield calculation is crucial in both laboratory and industrial chemical processes. It involves determining the efficiency of a chemical reaction. The yield is expressed as a percentage of the theoretical maximum amount of product that a reaction could produce. In this exercise, only 42.5% of ethylene's theoretical yield is realized. Understanding this concept allows chemists to evaluate the actual performance of chemical processes.

• Theoretical Yield: The maximum product amount calculated based on stoichiometry and complete conversion of reactants.
• Actual Yield: The amount of product obtained from an experiment or industrial process.
• Percent Yield: A comparison of the actual yield to the theoretical yield, expressed as a percentage.

To adjust for yield inefficiencies, we divide the desired mass of ethylene by the yield percentage, resulting in the theoretical moles required. From this, we can calculate the actual amount of hexane needed to achieve the target ethylene production. Accurate yield calculations contribute to reducing waste, saving costs, and improving the sustainability of chemical processes.