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Percent yields are generally calculated from masses. Would the result be the same if amounts (mol) were used instead? Why?

Short Answer

Expert verified
The result would be the same because the molar mass cancels out in the percent yield calculation when using moles.

Step by step solution

01

- Understand Percent Yield

Percent yield is calculated by comparing the actual yield (the amount of product obtained from a reaction) to the theoretical yield (the amount of product expected based on stoichiometric calculations). It is given by the formula: Percent Yield = \( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100 \)
02

- Analyze Units of Measurement

The actual and theoretical yields can be measured either in masses (grams) or in amounts (moles). When converting between these units, the molar mass of the substance acts as the conversion factor: \( \text{Mass (grams)} = \text{Amount (moles)} \times \text{Molar Mass (g/mol)} \).
03

- Examine Impact on Percent Yield Calculation

When calculating percent yield using masses, both actual and theoretical yields are in grams. If instead amounts (moles) are used, the molar mass cancels out in the fraction \( \frac{\text{Actual Yield (moles)} \times \text{Molar Mass}}{\text{Theoretical Yield (moles)}} \times \frac{1}{\text{Molar Mass}} \), resulting in: Percent Yield = \( \frac{\text{Actual Yield (moles)}}{\text{Theoretical Yield (moles)}} \times 100 \). Thus, the result is the same.
04

- Summarize the Explanation

Since both grams and moles can be converted using the molar mass, the fraction for percent yield remains the same whether masses or amounts are used in the calculation.

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Key Concepts

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

Actual Yield
The actual yield is the measured amount of a product obtained from a chemical reaction.
This value is typically determined through experimental data.
  • It is often less than the theoretical yield due to losses or inefficiencies in the reaction process.
  • For example, if you were synthesizing water from hydrogen and oxygen, the actual yield would be the water you can physically measure after the reaction.
Understanding actual yield is important, as it represents the real-world efficiency of a chemical reaction, reflecting practical constraints like incomplete reactions, side reactions, and measurement errors.
Theoretical Yield
The theoretical yield is the maximum amount of product that could be formed from a chemical reaction based on stoichiometric calculations.
This is an ideal value assuming perfect conversion of reactants to products with no losses.
  • It is calculated using the balanced chemical equation and the initial amounts of the reactants.
  • For instance, if a reaction predicts 10 grams of water should form from a given amount of oxygen and hydrogen, that 10 grams is the theoretical yield.
Knowing the theoretical yield is crucial for determining the percent yield and evaluating the efficiency of a chemical reaction by comparing it to the actual yield.
Molar Mass
Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol).
It serves as a conversion factor between the mass of a substance and the amount in moles.
  • To calculate molar mass, simply sum the atomic masses of all atoms in a molecule. For example, the molar mass of water (H2O) is calculated as the sum of the masses of 2 hydrogen atoms and 1 oxygen atom, resulting in approximately 18 g/mol.
  • It’s essential in converting between the mass of a substance and the amount in moles, thus bridging the gap between theoretical calculations and experimental measurements.
Understanding molar mass is key to accurately performing stoichiometric calculations, vital for determining theoretical yields and converting mass data to moles and vice versa.

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Most popular questions from this chapter

Nitrogen (N), phosphorus (P), and potassium (K) are the main nutrients in plant fertilizers. By industry convention, the numbers on a label refer to the mass percents of \(\mathrm{N}, \mathrm{P}_{2} \mathrm{O}_{5},\) and \(\mathrm{K}_{2} \mathrm{O},\) in that order. Calculate the N/P/K ratio of a \(30 / 10 / 10\) fertilizer in terms of moles of each element, and express it as \(x / y / 1.0\).

Calculate each of the following: (a) Mass \% of I in strontium periodate (b) Mass \% of Mn in potassium permanganate

What is the difference between an empirical formula and a molecular formula? Can they ever be the same?

Aspirin (acetylsalicylic acid, \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\) ) is made by reacting salicylic acid \(\left(\mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}\right)\) with acetic anhydride \(\left[\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O}\right]:\) $$ \mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}(s)+\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}(s)+\mathrm{CH}_{3} \mathrm{COOH}(l) $$ In one preparation, \(3.077 \mathrm{~g}\) of salicylic acid and \(5.50 \mathrm{~mL}\) of acetic anhydride react to form \(3.281 \mathrm{~g}\) of aspirin. (a) Which is the limiting reactant (the density of acetic anhydride is \(1.080 \mathrm{~g} / \mathrm{mL}) ?\) (b) What is the percent yield of this reaction? (c) What is the percent atom economy of this reaction?

List three ways compositional data may be given in a problem that involves finding an empirical formula.

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