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Chapter 3: Problem 66

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

Fluorine is so reactive that it forms compounds with several of the noble gases. (a) When \(0.327 \mathrm{~g}\) of platinum is heated in fluorine, \(0.519 \mathrm{~g}\) of a dark red, volatile solid forms. What is its empirical formula? (b) When \(0.265 \mathrm{~g}\) of this red solid reacts with excess xenon gas, \(0.378 \mathrm{~g}\) of an orange-yellow solid forms. What is the empirical formula of this compound, the first to contain a noble gas? (c) Fluorides of xenon can be formed by direct reaction of the elements at high pressure and temperature. Under conditions that produce only the tetra- and hexafluorides, \(1.85 \times 10^{-4} \mathrm{~mol}\) of xenon reacted with \(5.00 \times 10^{-4} \mathrm{~mol}\) of fluorine, and \(9.00 \times 10^{-6} \mathrm{~mol}\) of xenon was found in excess. What are the mass percents of each xenon fluoride in the product mixture?

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

Calculate each of the following: (a) Mass fraction of \(\mathrm{Cl}\) in calcium chlorate (b) Mass fraction of \(\mathrm{N}\) in dinitrogen trioxide

Narceine is a narcotic in opium that crystallizes from solution as a hydrate that contains 10.8 mass \(\%\) water and has a molar mass of \(499.52 \mathrm{~g} / \mathrm{mol} .\) Determine \(x\) in narceine \(x \mathrm{H}_{2} \mathrm{O}\).

Chlorine gas can be made in the laboratory by the reaction of hydrochloric acid and manganese(IV) oxide: $$ 4 \mathrm{HCl}(a q)+\mathrm{MnO}_{2}(s) \longrightarrow \mathrm{MnCl}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2}(g) $$ When 1.82 mol of \(\mathrm{HCl}\) reacts with excess \(\mathrm{MnO}_{2}\), how many (a) moles of \(\mathrm{Cl}_{2}\) and (b) grams of \(\mathrm{Cl}_{2}\) form?
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