Many times errors are expressed in terms of percentage. The percent error is the absolute value of the difference of the true value and the experimental value, divided by the true value, and multiplied by 100 . Percent error \(=\frac{\mid \text { true value }-\text { experimental value } \mid}{\text { true value }} \times 100\) Calculate the percent error for the following measurements. a. The density of an aluminum block determined in an experiment was \(2.64 \mathrm{~g} / \mathrm{cm}^{3}\). (True value \(2.70 \mathrm{~g} / \mathrm{cm}^{3}\).) b. The experimental determination of iron in iron ore was \(16.48 \%\). (True value \(16.12 \% .)\) c. A balance measured the mass of a \(1.000-\mathrm{g}\) standard as \(0.9981 \mathrm{~g}\)

Short Answer

Expert verified
a. Percent error for aluminum block density: \(2.22\%\) b. Percent error for iron in iron ore: \(2.23\%\) c. Percent error for mass of a 1.000-g standard: \(0.19\%\)

Step by step solution

01

Identifying the true and experimental values

For this measurement, we are given the true value and experimental value for the aluminum block's density: True value = \(2.70 \frac{g}{cm^3}\) Experimental value = \(2.64 \frac{g}{cm^3}\)
02

Calculate the absolute difference

Calculate the absolute difference between the true value and the experimental value: Absolute difference = \(|2.70 - 2.64|\) Absolute difference = \(0.06\)
03

Calculate the percent error

Divide the absolute difference by the true value and multiply by 100 to get the percent error: Percent error = \(\frac{0.06}{2.70} \times 100\) Percent error = \(2.22\%\) #b. Iron in iron ore#
04

Identifying the true and experimental values

For this measurement, we are given the true value and experimental value for the iron content in iron ore: True value = \(16.12\%\) Experimental value = \(16.48\%\)
05

Calculate the absolute difference

Calculate the absolute difference between the true value and the experimental value: Absolute difference = \(|16.12 - 16.48|\) Absolute difference = \(0.36\)
06

Calculate the percent error

Divide the absolute difference by the true value and multiply by 100 to get the percent error: Percent error = \(\frac{0.36}{16.12} \times 100\) Percent error = \(2.23\%\) #c. Mass of a 1.000-g standard#
07

Identifying the true and experimental values

For this measurement, we are given the true value and experimental value for the mass of a 1.000-g standard: True value = \(1.0000~g\) Experimental value = \(0.9981~g\)
08

Calculate the absolute difference

Calculate the absolute difference between the true value and the experimental value: Absolute difference = \(|1.0000 - 0.9981|\) Absolute difference = \(0.0019\)
09

Calculate the percent error

Divide the absolute difference by the true value and multiply by 100 to get the percent error: Percent error = \(\frac{0.0019}{1.0000} \times 100\) Percent error = \(0.19\%\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Experimental Error Calculation
Understanding how to quantify uncertainty in experimental results is a key skill in chemistry. Experimental error calculation is a process that helps chemists evaluate the accuracy of their measurements compared to a known or accepted reference point, often referred to as the 'true value'. Calculating experimental error often involves finding the percent error, which is a ratio of the deviation of the experimental result from the true value relative to the true value itself, expressed as a percentage.

For instance, if a chemist is measuring the density of an aluminum block and determines it experimentally to be 2.64 g/cm3 while the true density is recognized to be 2.70 g/cm3, to calculate the experimental error, one would first calculate the difference between these two values. This difference is then divided by the true value to normalize it, providing a dimensionless figure that is then converted into a percent. This process is central in analyzing the quality and reliability of experimental data.
  • Identify true and experimental values.
  • Calculate the absolute difference.
  • Divide by the true value and multiply by 100 to find percent error.
True Value vs Experimental Value
In scientific experiments, especially in chemistry, distinguishing between true value and experimental value is crucial for interpreting results. The true value is a theoretical value based on accepted literature or standard values; it represents what would be obtained in an ideal scenario, free of error. On the other hand, the experimental value is the value actually measured during an experiment. It's susceptible to various sources of error, such as instrumental imprecision, human error, or environmental influences.

Importance of True and Experimental Values

Understanding the relationship between these two types of values is essential for error analysis. When measuring the iron content in iron ore, if the true value is 16.12% but the experimental determination yields 16.48%, this discrepancy must be addressed and understood. Acknowledging the difference helps in refining experimental techniques, calibrating instruments, and ultimately improves the accuracy of outcomes. It's the gap between the true and experimental values that guides chemists in evaluating their methods and procedures for potential sources of error.
Error Analysis in Chemistry Measurements
Error analysis in chemistry measurements is a critical process which involves assessing the accuracy and precision of experimental data. Accuracy pertains to how close a measurement is to the true value, while precision refers to the reproducibility of the measurements under unchanged conditions. Error analysis not only helps in documenting the reliability of results but also in improving experimental techniques.

Systematic vs Random Errors

Analyzing errors involves differentiating between systematic errors, which are consistent and repeatable deviations influenced by equipment or methodology, and random errors, which arise from unpredictable and fluctuating factors. For example, if a balance consistently measures a 1.000-g standard as 0.9981 g, there might be a systematic error in calibration to address. Understanding these concepts and learning to calculate the percent error from systematic and random sources are foundational skills for progressing in chemical experimentation and critical analysis.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The scientific method is a dynamic process. What does this mean?

Sterling silver is a solid solution of silver and copper. If a piece of a sterling silver necklace has a mass of \(105.0 \mathrm{~g}\) and a volume of \(10.12 \mathrm{~mL}\), calculate the mass percent of copper in the piece of necklace. Assume that the volume of silver present plus the volume of copper present equals the total volume. Refer to Table \(1.5\). Mass percent of copper \(=\frac{\text { mass of copper }}{\text { total mass }} \times 100\)

Match each description below with the following microscopic pictures. More than one picture may fit each description. A picture may be used more than once or not used at all. a. a gaseous compound b. a mixture of two gaseous elements c. a solid element d. a mixture of a gaseous element and a gaseo

In recent years, there has been a large push for an increase in the use of renewable resources to produce the energy we need to power our vehicles. One of the newer fuels that has become more widely available is a mixture of \(85 \%\) ethanol and \(15 \%\) gasoline, E85. Despite being more environmentally friendly, one of the potential drawbacks of E85 fuel is that it produces less energy than conventional gasoline. Assume a car gets \(28.0 \mathrm{mi} / \mathrm{gal}\) using gasoline at \(\$ 3.50 / \mathrm{gal}\) and \(22.5 \mathrm{mi} / \mathrm{gal}\) using \(\mathrm{E} 85\) at \(\$ 2.85 / \mathrm{gal}\). How much will it cost to drive 500 . miles using each fuel?

You have a \(1.0-\mathrm{cm}^{3}\) sample of lead and a \(1.0-\mathrm{cm}^{3}\) sample of glass. You drop each in separate beakers of water. How do the volumes of water displaced by each sample compare? Explain.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free