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AP Chemistry : Yield and Error

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Example Questions

Example Question #31 : Calculations

$A\rightleftharpoons B$ $\displaystyle A\rightleftharpoons B$

3 mols of $\displaystyle A$ are placed in a reactor and react to form $\displaystyle B$ as in the above reaction. At equilibrium, there are 1.75 mols of $\displaystyle B$ .

What is the percent conversion of $\displaystyle A$ ?

Possible Answers:

$50\%$ $\displaystyle 50\%$

$37\%$ $\displaystyle 37\%$

$58\%$ $\displaystyle 58\%$

$35\%$ $\displaystyle 35\%$

$45\%$ $\displaystyle 45\%$

Correct answer:

$58\%$ $\displaystyle 58\%$

Explanation:

To find the percent conversion, use the following equation:

$Percent \:Conversion =\frac{Initial-Final}{Initial}*100$ $\displaystyle Percent \:Conversion =\frac{Initial-Final}{Initial}*100$

Initially there are 3 mols of $\displaystyle A$ . We know that every 1 mol of $\displaystyle A$ that is consumed in the reaction yields 1 mol of $\displaystyle B$ . Therefore, the final amount of $\displaystyle A$ in mols is

$\displaystyle Final = 3-1.75=1.25$

Plugging this into the above equation gives a conversion rate of 58%.

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Example Question #32 : Calculations

You run an experiment in order to empirically find the gas constant, $\displaystyle R$ . You find

$R=8.7\:\frac{J}{mol\cdot K}$ $\displaystyle R=8.7\:\frac{J}{mol\cdot K}$

You know from the literature that

$R=8.314\:\frac{J}{mol\cdot K}$ $\displaystyle R=8.314\:\frac{J}{mol\cdot K}$

What is the percent error in your experimental value of $\displaystyle R$ ?

Possible Answers:

$3.8\%$ $\displaystyle 3.8\%$

$5.2\%$ $\displaystyle 5.2\%$

$4.4\%$ $\displaystyle 4.4\%$

$4.6\%$ $\displaystyle 4.6\%$

$4.2\%$ $\displaystyle 4.2\%$

Correct answer:

$4.6\%$ $\displaystyle 4.6\%$

Explanation:

To find the percent error in your $\displaystyle R$ value, you can use the following equation:

$Percent \:Error = \left | \frac{Experimental-Actual}{Actual} \right |*100$ $\displaystyle Percent \:Error = \left | \frac{Experimental-Actual}{Actual} \right |*100$

Doing this, we find the percent error to be 4.6%.

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Example Question #32 : Laboratory Techniques And Analysis

Suppose that a chemist working in a lab is trying to synthesize caffeine. In doing so, he predicts that he can produce $50\:mg$ $\displaystyle 50\:mg$ of caffeine. After the chemist finishes synthesizing the caffeine, he weighs it only to find out that $43\:mg$ $\displaystyle 43\:mg$ are produced. What is the percent yield of caffeine in this synthesis process?

Possible Answers:

$44\:\%$ $\displaystyle 44\:\%$

$36\:\%$ $\displaystyle 36\:\%$

$97\:\%$ $\displaystyle 97\:\%$

$86\:\%$ $\displaystyle 86\:\%$

Correct answer:

$86\:\%$ $\displaystyle 86\:\%$

Explanation:

When chemists synthesize compounds in the lab, a theoretical yield can be calculated by using knowledge of the full, balanced chemical equation, as well as the starting amounts of all reagents including the limiting reagent. However, due to a variety of factors, the actual yield obtained at the end of the experiment is nearly always going to be lower than the theoretical yield. Some of the things that can cause this are experimental error and side reactions, among others. Thus, chemists usually take the ratio of actual yield to theoretical yield to give the percent yield.

$\frac{actual\:yield}{theoretical\: yield}*100\%=percent\:yield$ $\displaystyle \frac{actual\:yield}{theoretical\: yield}*100\%=percent\:yield$

$\frac{43\: mg}{50\: mg}*100\: \%=86\: \%$ $\displaystyle \frac{43\: mg}{50\: mg}*100\: \%=86\: \%$

This means that the overall process was $86\:\%$ $\displaystyle 86\:\%$ efficient.

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AP Chemistry : Yield and Error

Study concepts, example questions & explanations for AP Chemistry

All AP Chemistry Resources

Example Questions

Example Question #31 : Calculations

Example Question #32 : Calculations

Example Question #32 : Laboratory Techniques And Analysis

All AP Chemistry Resources

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