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High School Chemistry : Precision, Accuracy, and Error

Study concepts, example questions & explanations for High School Chemistry

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Example Question #11 : Precision, Accuracy, And Error

A student found out through experimentation that it requires 23.6 mL of a NaOH solution to completely neutralize 25 mL of a HCl solution. However, the actual amount of NaOH solution needed to neutralize 25 mL of the HCl solution is 22.8 mL. What is the percent error?

Possible Answers:

$3.5\%$

$4.9\%$

$1.6\%$

$8.2\%$

Correct answer:

$3.5\%$

Explanation:

Use the following formula to find the percent error:

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

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

$\text{Percent error}=\frac{|22.8-23.6|}{22.8}\cdot100\%=3.5\%$

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Example Question #71 : High School Chemistry

Joseph calculated that the concentration of an acid solution was 13.2M. The actual concentration of the solution was 16.3M. What is Joseph's percent error?

Possible Answers:

$19.0\%$

$22.9\%$

$16.3\%$

$3.3\%$

Correct answer:

$19.0\%$

Explanation:

Use the following formula to find the percent error:

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

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

$\text{Percent error}=\frac{|16.3-13.2|}{16.3}\cdot100\%=19.0\%$

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Example Question #13 : Calculating Error

Patricia measured that the density of a given block of iron was $7.1\frac{g}{cm^3}$ . The actual density of iron is $7.9\frac{g}{cm^3}$ . What is Patricia's percent error?

Possible Answers:

$16.3\%$

$10.1\%$

$8.9\%$

$6.6\%$

Correct answer:

$10.1\%$

Explanation:

Use the following formula to find the percent error:

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

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

$\text{Percent error}=\frac{|7.9-7.1|}{7.9}\cdot100\%=10.1\%$

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Example Question #72 : High School Chemistry

Steven calculated that the melting point of tin is 462 degrees Fahrenheit. The actual melting point of tin is 450 degrees Fahrenheit. What is Steven's percent error?

Possible Answers:

$2.9\%$

$8.4\%$

$2.2\%$

$1.7\%$

Correct answer:

$2.2\%$

Explanation:

Use the following formula to find the percent error:

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

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

$\text{Percent error}=\frac{|450-460|}{450}\cdot100\%=2.2\%$

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Example Question #73 : High School Chemistry

Through experimentation, Luke found the boiling point of ethanol to be 143 degrees Fahrenheit. The actual boiling point of ethanol is 173 degrees Fahrenheit. What is Luke's percent error?

Possible Answers:

$6.7\%$

$13.6\%$

$19.2\%$

$17.3\%$

Correct answer:

$17.3\%$

Explanation:

Use the following formula to find the percent error:

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

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

$\text{Percent error}=\frac{|173-143|}{173}\cdot100\%=17.3\%$

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Example Question #81 : Measurements

After conducting an experiment, Kevin produced 56.2 kg of sulfur dioxide gas. However, he was supposed to have been able to produce 84.0 kg of sulfur dioxide gas. What was his percent error?

Possible Answers:

$19.4\%$

$33.1\%$

$13.2\%$

$42.1\%$

Correct answer:

$33.1\%$

Explanation:

Use the following formula to find the percent error:

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

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

$\text{Percent error}=\frac{|84.0-56.2|}{84.0}\cdot100\%=33.1\%$

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Example Question #11 : Calculating Error

At the beginning of the experiment, Jimmy calculated an expected yield of 1.75g of NaCl . At the end of the experiment, he only recovered 0.64g . Calculate his percent error.

Possible Answers:

$63.4\%$

$36.6\%$

$27.5\%$

$188\%$

-1.11

Correct answer:

$63.4\%$

Explanation:

Percent error is a calculation usually completed at the end of an experiment for data analysis. It is in essence a tool that can be used to calculate the precision of your calculations. This can be done by following the equation:

$\large percent_(error_)= \left |\frac{experimental- accepted}{accepted} \right | \cdot 100$

Experimental represents that value that was experimentally obtained. Accepted represents that value that would have been accepted.

This quickly turns into a "plug and chug" type of problem.

For this problem:

experimental: 0.64g

accepted: 1.75g

$= \left|\frac{0.64-1.75}{1.75}=\frac{-1.11}{1.75}\right|$

$=\left|-0.634 \cdot 100\right|=63.4\%$

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Example Question #18 : Calculating Error

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?

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

Possible Answers:

$78.63\%\ \text{error}$

$22.15\%\ \text{error}$

$34.17\%\ \text{error}$

$21.37\%\ \text{error}$

Correct answer:

$21.37\%\ \text{error}$

Explanation:

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

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

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

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

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

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

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Example Question #19 : Calculating Error

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

Possible Answers:

$22.15\%\ \text{error}$

$78.63\%\ \text{error}$

$21.37\%\ \text{error}$

$34.17\%\ \text{error}$

Correct answer:

$21.37\%\ \text{error}$

Explanation:

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

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

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

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

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

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

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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 #11 : Precision, Accuracy, And Error

Example Question #71 : High School Chemistry

Example Question #13 : Calculating Error

Example Question #72 : High School Chemistry

Example Question #73 : High School Chemistry

Example Question #81 : Measurements

Example Question #11 : Calculating Error

Example Question #18 : Calculating Error

Example Question #19 : Calculating Error

All High School Chemistry Resources

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