Problem 12
Question
A \(1.000 \mathrm{~mL}\) sample of acetone, a common solvent used as a paint remover, was placed in a small vial whose mass was known to be \(4.002 \mathrm{~g}\). The following values were obtained when the acetone-filled vial was weighed: \(4.531 \mathrm{~g}, 4.525 \mathrm{~g}\), and, \(4.537 \mathrm{~g}\). How would you characterize the precision and accuracy of these measurements if the true mass of the acetone was \(0.7795 \mathrm{~g}\) ?
Step-by-Step Solution
Verified Answer
The measurements are precise (small range) but not accurate (large deviation from true mass).
1Step 1: Calculate the Measured Mass of Acetone
First, calculate the mean mass of acetone from the given measurements. Subtract the mass of the empty vial from each measurement to find three acetone masses: \[ 4.531 - 4.002 = 0.529 \, \text{g} \]\[ 4.525 - 4.002 = 0.523 \, \text{g} \]\[ 4.537 - 4.002 = 0.535 \, \text{g} \] Calculate the average mass: \[ \text{Average} = \frac{0.529 + 0.523 + 0.535}{3} = 0.529 \, \text{g} \]
2Step 2: Assess Precision
To assess precision, analyze how close the measurements are to each other. Calculate the range of measurements as: \[ \text{Range} = 0.535 - 0.523 = 0.012 \, \text{g} \] A small range indicates good precision among the measurements.
3Step 3: Assess Accuracy
To assess accuracy, compare the average measured mass to the true mass of the acetone (0.7795 g). The difference between the average mass and the true mass is: \[ |0.529 - 0.7795| = 0.2505 \, \text{g} \]This large difference indicates poor accuracy.
Key Concepts
mass measurementerror analysisprecision vs. accuracy
mass measurement
Mass measurement is a fundamental practice in scientific experiments as it helps to determine the amount of substance being used or analyzed. In the exercise with acetone, mass measurements are made by first weighing the vial alone and then with acetone. This allows for calculating the mass of the acetone by subtraction, i.e., the mass of the filled vial minus the mass of the empty vial. Using three different mass measurements, students compute the acetone's mass each time:
- 4.531 g - 4.002 g = 0.529 g
- 4.525 g - 4.002 g = 0.523 g
- 4.537 g - 4.002 g = 0.535 g
error analysis
Error analysis is a technique used to understand the uncertainties and potential errors in measurements. In our example of acetone's mass, it's clear that repeated measurements give a distribution of results around a central value. The key aspects of error analysis involve calculating:
- The average value: This is a central tendency measure determined here by \[ \frac{0.529 + 0.523 + 0.535}{3} = 0.529 \, \text{g} \]
- The range: Here, calculated as the difference between the highest and lowest values, \[ 0.535 - 0.523 = 0.012 \, \text{g} \]. A smaller range indicates closely grouped measurements, suggesting lower random error.
- The deviation from true value: This analyses systematic error by comparing the average measured value to the true value, \[ |0.529 - 0.7795| = 0.2505 \, \text{g} \].
precision vs. accuracy
Understanding the difference between precision and accuracy is crucial in scientific measurements. Precision refers to how close multiple measurements are to each other, while accuracy indicates how close a measurement is to the true value.In the acetone example:
- Precision: We observed a small range of 0.012 g among the measurements \( \text{(0.535 g, 0.523 g, 0.529 g)} \), indicating high precision, as the values are closely spaced.
- Accuracy: The accuracy is poor since the average value 0.529 g is far from the true value of 0.7795 g. The difference of 0.2505 g exemplifies significant inaccuracy.
Other exercises in this chapter
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