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Chapter 3: Problem 88

Why is the actual yield of a reaction almost always smaller than the theoretical yield?

Short Answer

Expert verified
The actual yield of a reaction is usually lower than the theoretical yield due to several factors, including incomplete reactions or equilibrium being reached before all reactants are converted, side reactions leading to different products, and practical losses during separation and purification processes.

Step by step solution

01

Understanding Theoretical Yield

Theoretical yield is the maximum amount of product that could be formed from the reactants. This value is obtained by calculations from the stoichiometric coefficients in the balanced chemical equation.
02

Understanding Actual Yield

Actual yield is the amount of product that is actually formed when the reaction is carried out in the laboratory. It is usually determined by experiments in a laboratory setting.
03

Reason 1: Reaction Kinetics

Different reactions proceed at different rates. Some reactions do not proceed to completion because they reach equilibrium before all the reactants are converted to products.
04

Reason 2: Side Reactions

The reactants may react in ways different from those required, leading to formation of different products. This is often the case if reaction conditions are not strictly controlled.
05

Reason 3: Practical Losses

Some amount of product might be lost during the separation or purification steps in the process. Instrumental errors can also contribute to the reduction in actual yield.

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Most popular questions from this chapter

A certain metal M forms a bromide containing 53.79 percent Br by mass. What is the chemical formula of the compound?

Consider the reaction of hydrogen gas with oxygen gas: $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) $$

The empirical formula of a compound is CH. If the molar mass of this compound is about \(78 \mathrm{~g},\) what is its molecular formula?

(a) For molecules having small molecular masses, mass spectrometry can be used to identify their formulas. To illustrate this point, identify the molecule that most likely accounts for the observation of a peak in a mass spectrum at 16 amu, 17 amu, 18 amu, and 64 amu. (b) Note that there are (among others) two likely molecules that would give rise to a peak at 44 amu, namely, \(\mathrm{C}_{3} \mathrm{H}_{8}\) and \(\mathrm{CO}_{2} .\) In such cases, a chemist might try to look for other peaks generated when some of the molecules break apart in the spectrometer. For example, if a chemist sees a peak at 44 amu and also one at 15 amu, which molecule is producing the 44 -amu peak? Why? (c) Using the following precise atomic masses \(-\mathrm{H}\) ( 1.00797 amu), \({ }^{12} \mathrm{C}(12.00000 \mathrm{amu}),\) and \({ }^{16} \mathrm{O}(15.99491 \mathrm{amu})-\) how precisely must the masses of \(\mathrm{C}_{3} \mathrm{H}_{8}\) and \(\mathrm{CO}_{2}\), be measured to distinguish between them?

Because of its detrimental effect on the environment, the lead compound described in Problem 3.148 has been replaced by methyl tert-butyl ether (a compound of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O}\) ) to enhance the performance of gasoline. (This compound is also being phased out because of its contamination of drinking water.) When \(12.1 \mathrm{~g}\) of the compound are burned in an apparatus like the one shown in Figure \(3.6,30.2 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(14.8 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) are formed. What is the empirical formula of the compound?
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