Problem 99 - Chemistry

Question

Ethylene reacts with hydrogen chloride to form ethyl chloride: $$\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{HCl}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}(g) $$Calculate the mass of ethyl chloride formed if \(4.66 \mathrm{~g}\) of ethylene reacts with an 89.4 percent yield.

Step-by-Step Solution

Answer

To find the mass of ethyl chloride formed, convert the given mass of ethylene to moles using its molar mass. Use the stoichiometry of the balanced chemical equation to determine the theoretical moles of ethyl chloride it would produce under 100% yield, then apply the given percent yield to find the actual produced moles. Convert these moles back to grams using the molar mass of ethyl chloride to find the mass produced.

Step 1: Determine Moles of Ethylene

Use the molar mass of ethylene \( (\mathrm{C}_{2} \mathrm{H}_{4}) \), which is roughly 28.05 g/mol to convert grams of ethylene to moles. Thus the moles of ethylene can be computed as follows: \( moles = \frac{mass}{molar ~ mass} = \frac{4.66 ~ g}{28.05 ~ g/mol} \)

Step 2: Use Stoichiometry of Reaction

The balanced chemical equation shows that 1 mole of ethylene reacts to produce 1 mole of ethyl chloride. Therefore, under 100% yield, the moles of ethylene would be equal to the moles of ethyl chloride.

Step 3: Apply Percent Yield

Apply the provided percent yield of 89.4% to account for the actual amount of ethyl chloride produced. Multiply the moles of ethyl chloride obtained in step 2 by the percent yield: \(actual ~ moles = theoretical ~ moles * (yield / 100) \)

Step 4: Calculate Actual Mass

Lastly, convert the actual moles of ethyl chloride to grams using the molar mass of ethyl chloride (\( \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}) \), which is roughly 64.5 g/mol. The mass of ethyl chloride is given by \( mass = actual ~ moles * molar ~ mass = actual ~ moles * 64.5 ~ g/mol \)

Key Concepts

Chemical Equation Balancing

Balancing a chemical equation is a fundamental aspect of stoichiometry and a vital first step in solving any stoichiometry problem. A balanced equation respects the law of conservation of mass by ensuring that the number of atoms for each element is the same on both the reactant and product sides. In essence, what you start with must be equal to what you end up with - no atoms are lost or created during a chemical reaction.

For example, the chemical reaction in the original exercise is balanced as such: \( \mathrm{C}_2\mathrm{H}_4(g) + \mathrm{HCl}(g) \longrightarrow \mathrm{C}_2\mathrm{H}_5\mathrm{Cl}(g) \)Here, there are two carbon atoms, four hydrogen atoms, and one chlorine atom before and after the reaction occurs. In some cases, coefficients may need to be adjusted to achieve a balanced equation; that was not necessary here. Understanding this balance is crucial when using stoichiometry to find how much product is formed from given reactants, as it helps us determine the mole ratio between reactants and products.

Percent Yield Calculation

The percent yield is a measure of the efficiency of a chemical reaction and indicates how much product is obtained compared to the maximum amount possible, known as the theoretical yield. It is calculated using the formula:\( \text{Percent Yield} = \frac{\text{Actual Yield (g)}}{\text{Theoretical Yield (g)}} \times 100\% \)In practice, reactions rarely proceed to completion, and this can be due to several factors including incomplete reactions, side reactions, or loss of product during collection. Thus, percent yield provides a realistic expectation of the experiment's outcome. As in the exercise provided, applying the given percent yield (89.4%) to the theoretical yield allows for the calculation of the actual amount of product formed, and this step is essential in bridging the gap between theoretical predictions and experimental reality.

Molar Mass

The molar mass is a fundamental concept in stoichiometry that corresponds to the mass of one mole of a substance and is expressed in grams per mole (g/mol). It can be calculated by summing the atomic masses (from the periodic table) of all the atoms in the molecule. Molar mass acts as a conversion factor between the mass of a substance and the number of moles, which are key to solving stoichiometry problems.

For example, ethylene (\( \mathrm{C}_2\mathrm{H}_4 \)) has a molar mass of 28.05 g/mol, meaning 28.05 grams of ethylene is equivalent to one mole of ethylene molecules. This conversion is crucial in the exercise when translating the given mass of ethylene to moles, which then can be used in stoichiometric calculations to relate reactants to products. Knowing the molar masses of both reactant and product allows for the entire stoichiometric process, from reacting moles to predicting the mass of product formed.