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High School Chemistry : Precision, Accuracy, and Error

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6 Diagnostic Tests 143 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Analytical Chemistry

Robert conducted an experiment in which he investigated how much water a paper towel could absorb. Initially, Robert found that one paper towel can absorb 12.8g of water. Later he found that his scale was not calibrated, so he had to repeat the experiment. After repeating the experiment with a new scale, Robert found that one paper towel can actually absorb 32.9g of water. What is the approximate percent error between the findings of the first and second experiments?

Possible Answers:

$39\%$ $\displaystyle 39\%$

$72\%$ $\displaystyle 72\%$

$157\%$ $\displaystyle 157\%$

$20\%$ $\displaystyle 20\%$

$61\%$ $\displaystyle 61\%$

Correct answer:

$61\%$ $\displaystyle 61\%$

Explanation:

The formula for percent error is:

$\frac{\left | \text{Accepted Value - Measured Value}\right |}{\text{Accepted Value}} \times100\%$ $\displaystyle \frac{\left | \text{Accepted Value - Measured Value}\right |}{\text{Accepted Value}} \times100\%$

In this case, the measured value is 12.8g and the accepted value is 32.9g.

$\frac{\left | 32.9g-12.8g\right |}{32.9g}\times 100\%=61\%$ $\displaystyle \frac{\left | 32.9g-12.8g\right |}{32.9g}\times 100\%=61\%$

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Example Question #2 : Precision, Accuracy, And Error

A reaction between one mole of sodium and one mole of chloride should yield 42 grams of sodium chloride. In your experiment, the actual yield is 32.73 grams. Calculate the percent error of your experiment.

Possible Answers:

$15.42\%\ \text{error}$ $\displaystyle 15.42\%\ \text{error}$

$25.13\%\ \text{error}$ $\displaystyle 25.13\%\ \text{error}$

$24.07\%\ \text{error}$ $\displaystyle 24.07\%\ \text{error}$

$22.07\%\ \text{error}$ $\displaystyle 22.07\%\ \text{error}$

Correct answer:

$22.07\%\ \text{error}$ $\displaystyle 22.07\%\ \text{error}$

Explanation:

To find percent error we need to use the following equation:

$\frac{\text{theoretical yield}-\text{actual yield}}{\text{theoretical yield}}*100\%$ $\displaystyle \frac{\text{theoretical yield}-\text{actual yield}}{\text{theoretical yield}}*100\%$

Plug in 42 for the theoretical yield and 32.73 for the actual yield and solve accordingly.

$\frac{42g-32.73g}{42g}*100\%=22.07\%$ $\displaystyle \frac{42g-32.73g}{42g}*100\%=22.07\%$

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Example Question #1 : Analytical Chemistry

In the following reaction, eight moles of sodium hydroxide is broken down into four moles of sodium oxide and four moles of water. What is the percent error if your experiment yields 195 grams of sodium oxide?

$2NaOH\rightarrow Na_{2}O + H_{2}O$ $\displaystyle 2NaOH\rightarrow Na_{2}O + H_{2}O$

Possible Answers:

$22.15\%\ \text{error}$ $\displaystyle 22.15\%\ \text{error}$

$34.17\%\ \text{error}$ $\displaystyle 34.17\%\ \text{error}$

$78.63\%\ \text{error}$ $\displaystyle 78.63\%\ \text{error}$

$21.37\%\ \text{error}$ $\displaystyle 21.37\%\ \text{error}$

Correct answer:

$21.37\%\ \text{error}$ $\displaystyle 21.37\%\ \text{error}$

Explanation:

To find the percent error we need to use the following equation:

$\frac{\text{theoretical yield}-\text{actual yield}}{\text{theoretical yield}}*100\%$ $\displaystyle \frac{\text{theoretical yield}-\text{actual yield}}{\text{theoretical yield}}*100\%$

But in order to do this, we first have to convert moles of sodium oxide into grams:

$4 mol\ Na_{2}O * \frac{62g}{1 mol}=248g\ NaO_2$ $\displaystyle 4 mol\ Na_{2}O * \frac{62g}{1 mol}=248g\ NaO_2$

This gives us a theoretical yield of 248g, which we plug in with our 195g actual yield.

$\frac{248g-195g}{248g}*100\%=21.37\%\ \text{error}$ $\displaystyle \frac{248g-195g}{248g}*100\%=21.37\%\ \text{error}$

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Example Question #61 : Measurements

If a given sample of silver and flourine ideally combine to form 0.6mol of AgF, what is the percent error if the actual yield is 43 grams?

Possible Answers:

$51.98\%\ \text{error}$ $\displaystyle 51.98\%\ \text{error}$

$47.81\%\ \text{error}$ $\displaystyle 47.81\%\ \text{error}$

$56.43\%\ \text{error}$ $\displaystyle 56.43\%\ \text{error}$

$43.57\%\ \text{error}$ $\displaystyle 43.57\%\ \text{error}$

Correct answer:

$43.57\%\ \text{error}$ $\displaystyle 43.57\%\ \text{error}$

Explanation:

Our first step to complete this problem is to convert moles AgF into grams:

$0.6 mol\ AgF*\frac{127g}{1 mol}=76.2g\ AgF$ $\displaystyle 0.6 mol\ AgF*\frac{127g}{1 mol}=76.2g\ AgF$

This gives us a theoretical yield of 76.2 grams. We can find the percent error by plugging in 76.2g for our theoretical yield and 43g for the actual yield in the following equation:

$\frac{\text{theoretical yield}-\text{actual yield}}{\text{theoretical yield}}*100\%$ $\displaystyle \frac{\text{theoretical yield}-\text{actual yield}}{\text{theoretical yield}}*100\%$

$\frac{76.2g-43g}{76.2g}*100\%=43.57\%\ \text{error}$ $\displaystyle \frac{76.2g-43g}{76.2g}*100\%=43.57\%\ \text{error}$

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Example Question #3 : Precision, Accuracy, And Error

After conducting an experiment that involved the reaction of solutions of $\displaystyle Na_2SO_4$ and $\displaystyle Ba(NO_3)_2$, 6.8 grams of BaSO_4 $\displaystyle BaSO_4$ was yielded. What is the percent error if the theoretical yield of BaSO_4 $\displaystyle BaSO_4$ for this experiment is 7.8 grams?

Possible Answers:

$12.8\%$ $\displaystyle 12.8\%$

$13.2\%$ $\displaystyle 13.2\%$

$9.2\%$ $\displaystyle 9.2\%$

$14.7\%$ $\displaystyle 14.7\%$

Correct answer:

$12.8\%$ $\displaystyle 12.8\%$

Explanation:

Use the following formula to find the percent error:

$\frac{|\text{accepted value-experimental value}|}{\text{accepted value}}\cdot100\%$ $\displaystyle \frac{|\text{accepted value-experimental value}|}{\text{accepted value}}\cdot100\%$

For this experiment, our accepted value is the same as the theoretical value.

$\text{Percent error}=\frac{|7.8-6.8|}{7.8}\cdot100\%$ $\displaystyle \text{Percent error}=\frac{|7.8-6.8|}{7.8}\cdot100\%$

$\textup{Percent error}=0.128\cdot100\%=12.8\%$ $\displaystyle \textup{Percent error}=0.128\cdot100\%=12.8\%$

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Example Question #71 : Measurements

A student measures that a piece of string is 2.6 cm, but the actual length of the string is 2.9 cm. What is the student's percent error?

Possible Answers:

$19.7\%$ $\displaystyle 19.7\%$

$38.5\%$ $\displaystyle 38.5\%$

$10.3\%$ $\displaystyle 10.3\%$

$20.6\%$ $\displaystyle 20.6\%$

Correct answer:

$10.3\%$ $\displaystyle 10.3\%$

Explanation:

Use the following formula to find the percent error:

$\frac{|\text{accepted value-experimental value}|}{\text{accepted value}}\cdot100\%$ $\displaystyle \frac{|\text{accepted value-experimental value}|}{\text{accepted value}}\cdot100\%$

For this experiment, our accepted value is the same as the theoretical value.

$\text{Percent error}=\frac{|2.9-2.6|}{2.9}\cdot100\%=10.3\%$ $\displaystyle \text{Percent error}=\frac{|2.9-2.6|}{2.9}\cdot100\%=10.3\%$

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Example Question #1 : Calculating Error

A standard mass of 100.0 grams was placed on a balance. The balance indicated that the mass was 101.6 grams. What is the percent error for the balance?

Possible Answers:

$4.7\%$ $\displaystyle 4.7\%$

$3.2\%$ $\displaystyle 3.2\%$

$1.6\%$ $\displaystyle 1.6\%$

$2.9\%$ $\displaystyle 2.9\%$

Correct answer:

$1.6\%$ $\displaystyle 1.6\%$

Explanation:

Use the following formula to find the percent error:

$\frac{|\text{accepted value-experimental value}|}{\text{accepted value}}\cdot100\%$ $\displaystyle \frac{|\text{accepted value-experimental value}|}{\text{accepted value}}\cdot100\%$

For this experiment, our accepted value is the same as the theoretical value.

$\text{Percent error}=\frac{|100-101.6|}{100}\cdot100\%=1.6\%$ $\displaystyle \text{Percent error}=\frac{|100-101.6|}{100}\cdot100\%=1.6\%$

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Example Question #1 : Calculating Error

From an experiment, a student found the density of a rubber eraser was $3.3\frac{g}{cm^3}$ $\displaystyle 3.3\frac{g}{cm^3}$. However, the actual density of the eraser is $2.9\frac{g}{cm^3}$ $\displaystyle 2.9\frac{g}{cm^3}$. What is the student's percent error?

Possible Answers:

$13.8\%$ $\displaystyle 13.8\%$

$12.2\%$ $\displaystyle 12.2\%$

$11.7\%$ $\displaystyle 11.7\%$

$15.9\%$ $\displaystyle 15.9\%$

Correct answer:

$13.8\%$ $\displaystyle 13.8\%$

Explanation:

Use the following formula to find the percent error:

$\frac{|\text{accepted value-experimental value}|}{\text{accepted value}}\cdot100\%$ $\displaystyle \frac{|\text{accepted value-experimental value}|}{\text{accepted value}}\cdot100\%$

For this experiment, our accepted value is the same as the theoretical value.

$\text{Percent error}=\frac{|2.9-3.3|}{2.9}\cdot100\%=13.8\%$ $\displaystyle \text{Percent error}=\frac{|2.9-3.3|}{2.9}\cdot100\%=13.8\%$

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Example Question #1 : Calculating Error

The experimental value obtained for the specific heat of gold was $0.33\frac{J}{g^{\text o}C}$ $\displaystyle 0.33\frac{J}{g^{\text o}C}$. The known specific heat of gold is $0.219\frac{J}{g^{\text o}C}$ $\displaystyle 0.219\frac{J}{g^{\text o}C}$. What is the percent error?

Possible Answers:

$33.6\%$ $\displaystyle 33.6\%$

$50.7\%$ $\displaystyle 50.7\%$

$47.3\%$ $\displaystyle 47.3\%$

$40.2\%$ $\displaystyle 40.2\%$

Correct answer:

$50.7\%$ $\displaystyle 50.7\%$

Explanation:

Use the following formula to find the percent error:

$\frac{|\text{accepted value-experimental value}|}{\text{accepted value}}\cdot100\%$ $\displaystyle \frac{|\text{accepted value-experimental value}|}{\text{accepted value}}\cdot100\%$

For this experiment, our accepted value is the same as the theoretical value.

$\text{Percent error}=\frac{|0.219-0.33|}{0.219}\cdot100\%=50.7\%$ $\displaystyle \text{Percent error}=\frac{|0.219-0.33|}{0.219}\cdot100\%=50.7\%$

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Example Question #2 : Calculating Error

A reaction produced 98.5 grams of nitrogen dioxide gas. However, the calculated theoretical value indicates that 115.2 grams of nitrogen dioxide gas should have been produced. What is the percent error?

Possible Answers:

$24.9\%$ $\displaystyle 24.9\%$

$10.1\%$ $\displaystyle 10.1\%$

$16.3\%$ $\displaystyle 16.3\%$

$14.5\%$ $\displaystyle 14.5\%$

Correct answer:

$14.5\%$ $\displaystyle 14.5\%$

Explanation:

Use the following formula to find the percent error:

$\frac{|\text{accepted value-experimental value}|}{\text{accepted value}}\cdot100\%$ $\displaystyle \frac{|\text{accepted value-experimental value}|}{\text{accepted value}}\cdot100\%$

For this experiment, our accepted value is the same as the theoretical value.

$\text{Percent error}=\frac{|115.5-98.5|}{115.2}\cdot100\%=14.5\%$ $\displaystyle \text{Percent error}=\frac{|115.5-98.5|}{115.2}\cdot100\%=14.5\%$

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High School Chemistry : Precision, Accuracy, and Error

Study concepts, example questions & explanations for High School Chemistry

All High School Chemistry Resources

Example Questions

Example Question #1 : Analytical Chemistry

Example Question #2 : Precision, Accuracy, And Error

Example Question #1 : Analytical Chemistry

Example Question #61 : Measurements

Example Question #3 : Precision, Accuracy, And Error

Example Question #71 : Measurements

Example Question #1 : Calculating Error

Example Question #1 : Calculating Error

Example Question #1 : Calculating Error

Example Question #2 : Calculating Error

All High School Chemistry Resources

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