If the theoretical yield of a reaction is \(0.118 \mathrm{~g}\) and the actual yield is \(0.104 \mathrm{~g}\), what is the percent yield?

Short Answer

Expert verified
The percent yield is 88.14%.

Step by step solution

01

Understand Percent Yield

The percent yield of a reaction is calculated by dividing the actual yield by the theoretical yield and then multiplying by 100 to convert it to a percentage. The percent yield shows the efficiency of a chemical reaction.
02

Calculate Percent Yield

To find the percent yield, use the formula: Percent Yield = (Actual Yield / Theoretical Yield) * 100. In this case, we will divide the actual yield, 0.104 g, by the theoretical yield, 0.118 g, and multiply the result by 100.
03

Perform the Calculation

Substitute the given yields into the formula: Percent Yield = (0.104 g / 0.118 g) * 100 = 88.14%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Theoretical Yield
Understanding the theoretical yield is essential for studying chemical reactions. It represents the maximum amount of product that could be generated from a given amount of reactants, based on the balanced chemical equation and stoichiometry. To calculate this, chemists use the mole ratio between the reactants and products from the balanced equation to determine what the yield would be if everything went perfectly and no product was lost.

For students, visualizing the process helps. Imagine you have enough material to build 10 toy cars, but due to some parts being defective, you end up with fewer. The theoretical yield is those 10 cars you expected to build before starting.
Actual Yield
On the flip side, we have the actual yield, which is the amount of product actually obtained from a chemical reaction. It's often less than the theoretical yield due to a multitude of real-world factors like incomplete reactions, impurities, or loss of material during the process. Consider the example of baking cookies; your recipe might indicate that you can make 24 cookies, but if some break you might only end up with 20 eatable cookies - this is your actual yield.

An important thing for students to note is how to accurately measure actual yield, which is typically done through weighing the product after the reaction has been completed. Also, documenting this correctly and understanding the reasons for any discrepancies is crucial for laboratory work.
Chemical Reaction Efficiency
Chemical reaction efficiency is quantified by percent yield, which compares the actual yield to the theoretical yield and expresses it as a percentage. It's a measure of how 'successful' a chemical reaction has been. In our toy car analogy, if you were able to build 8 cars out of the 10 possible, your efficiency would be 80%.

When solving problems related to percent yield, it's not enough to just do the math; it's also key to understand what might have led to the loss of yield. Reasons could be anything from evaporation to side reactions or even measurement errors. Discussing these factors adds depth to understanding efficiency in chemistry.
Stoichiometry
Stoichiometry is the calculation of reactants and products in chemical reactions. It is based on the law of conservation of mass; the mass of the products in a chemical reaction must equal the mass of the reactants. Understanding stoichiometry is like learning to follow a complex recipe with exact proportions—you need the right amount of each ingredient to achieve your desired product.

To become proficient in stoichiometry, students should practice converting between grams, moles, and molecules, using the mole ratio from the balanced equation, and applying these skills to determine the theoretical yield. Just as in cooking, precise measurements and ratios ensure the best outcome, and the same goes for chemical reactions.

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Most popular questions from this chapter

If the theoretical yield of a reaction is \(24.8 \mathrm{~g}\) and the actual yield is \(18.5 \mathrm{~g}\), what is the percent yield?

Consider the reaction between sulfur trioxide and water: $$ \mathrm{SO}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4}(a q) $$ A chemist allows \(61.5 \mathrm{~g}\) of \(\mathrm{SO}_{3}\) and \(11.2 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) to react. When the reaction is finished, the chemist collects \(54.9 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\). Determine the limiting reactant, theoretical yield, and percent yield for the reaction.

For the reaction shown, calculate how many grams of oxygen form when each quantity of reactant completely reacts. $$ 2 \mathrm{HgO}(\mathrm{s}) \longrightarrow 2 \mathrm{Hg}(l)+\mathrm{O}_{2}(g) $$ (a) \(2.13 \mathrm{~g} \mathrm{Hg} \mathrm{O}\) (b) \(6.77 \mathrm{~g} \mathrm{Hg} \mathrm{O}\) (c) \(1.55 \mathrm{~kg} \mathrm{Hg} \mathrm{O}\) (d) \(3.87 \mathrm{mg} \mathrm{HgO}\)

For the reaction shown, calculate the theoretical yield of the product in moles for each of the initial quantities of reactants. $$ \mathrm{Ti}(s)+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{TiCl}_{4}(s) $$ (a) \(2 \mathrm{~mol} \mathrm{Ti} ; 2 \mathrm{~mol} \mathrm{Cl}_{2}\) (b) \(5 \mathrm{~mol} \mathrm{Ti} ; 9 \mathrm{~mol} \mathrm{Cl}_{2}\) (c) \(0.483 \mathrm{~mol} \mathrm{Ti} ; 0.911 \mathrm{~mol} \mathrm{Cl}_{2}\) (d) \(12.4 \mathrm{~mol} \mathrm{Ti} ; 15.8 \mathrm{~mol} \mathrm{Cl}_{2}\)

Consider the generic chemical reaction: $$ 2 \mathrm{~A}+3 \mathrm{~B} \longrightarrow 3 \mathrm{C} $$ How many moles of B are required to completely react with: (a) \(6 \mathrm{~mol}\) of \(\mathrm{A}\) (b) \(2 \mathrm{~mol}\) of \(\mathrm{A}\) (c) \(7 \mathrm{~mol}\) of \(\mathrm{A}\) (d) 11 mol of \(\mathrm{A}\)

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