Measurements - High School Chemistry

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Question

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?

Answer

The formula for percent error is:

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

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

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

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Question

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.

Answer

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

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

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

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

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Question

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?

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

Answer

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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Question

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?

Answer

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

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

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:

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

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

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Question

After conducting an experiment that involved the reaction of solutions of and , 6.8 grams of was yielded. What is the percent error if the theoretical yield of for this experiment is 7.8 grams?

Answer

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.

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

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

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Question

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?

Answer

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.

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

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Question

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?

Answer

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.

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

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Question

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

Answer

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.

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

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Question

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

Answer

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.

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

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Question

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?

Answer

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.

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

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Question

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

Answer

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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Question

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?

Answer

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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Question

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?

Answer

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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Question

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?

Answer

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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Question

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?

Answer

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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Question

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?

Answer

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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Question

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

Answer

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:

accepted:

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

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

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Question

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?

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

Answer

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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Question

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?

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

Answer

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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Question

What is the concentration of Ca in a solution of 1 mol CaCl2 in 1 L of distilled water? (M = molarity, m= molality)

Answer

The definition of molality is moles of solute in 1 kg of the solvent, whereas molarity is the number of moles of solute per 1 L of solutioin. Since 1 mol of CaCl2 is added to 1 L of water, this means that the volume of the final solution is greater than 1 L. Thus, molality is the more accurate concentration determinant, since the solution is probably close to 1 L.

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