Experiment 2
Measurements and Graphing
During the first day of this lab, you will measure the diameter of a set of steel balls (ball bearings) and a set of wooden balls. The intent of this part of the experiment is to be able to obtain a mathematical relationship between your measured diameter (or derived radius) and the mass of the solid sphere. After you determine the diameter of a particular ball, you must also measure the corresponding mass of that ball (you must keepthe mass and diameter of each ball together for subsequent calculations). These correlated values will be graphed during the second day of this lab using the computer and LabWorks software. From these computer-generated graphs, you will be able to determine the correct mathematical relationship between length (diameter) and mass (volume). Mass and volume are directly related, based on the definition of density (D=m/V).
After you have obtained the correct mathematical relationship, and the equation describing this relationship, you will measure the diameter and mass of another steel ball (this will be a large steel ball, using the procedure described below as a Learning Outcome). You will then use the mathematical equation derived above, and calculate its mass, based on the measured diameter. After you determine the mass based on the mathematical relationship (using the equation for the straight line), you will do a percent error analysis of the actual measured mass to this mathematically calculated mass.
For the second part of this experiment, you will use several different measuring containers to measure out 40 mL of water. For this part of the experiment, you will use a 100-mL (or 150-mL) beaker, a 100-mL graduated cylinder, a 50-mL graduated cylinder, and a 10-mL graduated cylinder. From these measurements, you will determine which measuring device gives you the most accurate results. You will determine an error analysis for this part of the experiment.
Day 1: Mass and Diameter Measurement of the Balls and Volume Measurements of Water
Diameter and Mass of Steel and Wooden Balls
Obtain a set of steel balls from the cart. Using a venier caliper (an explanation on its use will be given by your instructor) to measure the diameter of each ball in the set. After you have determined the diameter for each of the balls in the set, you must obtain their masses and record these data in your notebook (remember that you must correlate the mass and radius for each ball). Both you, and your partner, will determine, independently values for this part of the experiment.
Optional, based on Instructor recommendations: Obtain a set of wooden balls and make diameter and mass measurements as before (both students collect and record independent measuring data). It is easiest to weigh the ball by placing each ball on a pre-zeroed balance using the lid of the container placed upside down on the balance pan to prevent the balls from rolling around.
There will be no calculations or manipulation of your data during the first day. All computer-based analyses will be performed on the second day of the lab.
Measurements of 40 mL of Water
The purpose of this part of the experiment is two-fold: (1) determining which measuring device gives you the most accurate measurement of 40 mL of water, and (2) performing a percent error analysis of your measurements based on the theoretical amount of water you should obtain (theoretically, you should have 40.00 mL of water, and using the density at the temperature you used, you can determine the theoretical mass of water that you should have) during each measurement. Be sure to measure the temperature of the water you are using, since the density of water varies with temperature. The density of water, at the temperature you measured, can be obtained from the CRC handbook (your instructor will show you how to read this table). Use D.I. water for all measurements.
To begin this part of the experiment, obtain a clean and dry 100- or 150-mL beaker from your drawer and determine is mass. You will use this beaker for all four volume measurements, but it only needs to be absolutely dry for the first measurement.
These are the only measurements you need to perform during the first day of the lab.
Day 2: Determination of the relationship between mass and diameter, and percent error calculations for the water measurements.
Diameter vs Mass Analysis
You will follow the step-by-step procedures outlined in the published experimental protocols for the computer analysis of the measurements you made on the set of steel balls and wooden balls. There really is no alteration of this explanation, since the use of the computer and of LabWorks is very well described.
Basically, scientists and mathematicians manipulate data to obtain graphs. Usually, if you can produce a straight line from your experimental data, you can draw conclusions about the mathematical relationships between two variables. For example, the area of a circle has the formula of πr2, where r is the radius (diameter/2). If you plotted the area (y-axis) verses the radius (x-axis) you see a parabola or parabolic curve, since the area increases by the square of the radius, as shown below in the table. However, if you squared the radius values and plotted them against the area, you would now see a direct, linear correlation. This correlation implies that the area is function of the square of the radius, which corresponds to the formula for a circle, which includes the square of the radius ( πr2 ).
radius (cm) |
Area (cm2) |
radius squared (cm2) |
2.00 |
12.57 |
4.00 |
4.00 |
50.27 |
16.00 |
6.00 |
113.1 |
36.00 |
8.00 |
201.1 |
64.00 |
10.00 |
314.2 |
100.0 |
You can see that the relationship between radius and area is not just a linear relationship (e.g., for the values shown, the radius increases by a factor of 5, but the area increases by a factor of 25). The actual relationship turns out to be a square relationship between radius and area, and is easily observed by plotting the area verses the square of the radius as shown in the table. For your experiment, you will plot the measured diameter verses the corresponding mass for that ball. If your calculated "best fit" line does not fit your data points, then you should plot the mass verses the square of the diameter, and, if necessary, verses the cube of the diameter. One of these three mathematical relationships should fit your data best. In your notebook, explain why the relationship you chose (direct linear relationship, a squared relationship, or a cubed relationship) is most likely correct, based on other mathematical reasonings.
After you plot your data, and have an acceptable graph, print a single copy (make additional xerox copies for lab partners). Before you print your copy, be sure to add your names to the graph. (You will be graded on how accurately your follow these instructions.)
Learning Outcome: After you have determined the best mathematical relationship to fit your measurements, you will use the formula obtained for the straight-line plot to calculate an experimental mass for a large steel ball. Obtain one of the large steel balls from the instructor. Using a caliper, measure its diameter. Use this diameter as the variable in the equation for the straight line you just determined. This is the calculated (extrapolated) mass. Compare this calculated mass to its actually mass (using the balance) using the percent error formula shown below:
This percent error reflects how close your measured mass (from the balance) is to the calculated mass (using the mathematical equation). Percent error (%error) is calculated using the following formula (answer is an absolute value, always positive)
Percent Error in Water Measurements
Based on the theoretical density of water (from CRC handbook), obtain the experimental volume of water in your measurements (you measured mass, so divide your mass values by the density). Using the example below, set up a table to determine the volume of water actually present in each of your samples. Then, calculate the percent error (formaul shown above) for your measurements. For example, pure water at 20oC has a density of 0.9982063 g/mL.
The table below is an example that was made up by your instructor. It is not necessarily representative of your own data.
Measuring vessel |
Mass of water |
Volume of water |
Theoretical volume of water |
% Error |
150-mL beaker |
36.558 g |
36.622 mL |
40.00 mL |
8.445% |
100-mL cylinder |
40.125 g |
40.195 mL |
40.00 mL |
0.4875% |
50-mL cylinder |
39.581 g |
39.650 mL |
40.00 mL |
0.8750 |
10-mL cylinder |
33.256 g |
33.314 mL |
40.00 mL |
16.72% |
Temp (oC) |
Density (g/mL) |
Temp (oC) |
Density (g/mL) |
Temp (oC) |
Density (g/mL) |
12.0 |
0.99950 |
17.0 |
0.99877 |
22.0 |
0.99777 |
13.0 |
0.99938 |
18.0 |
0.99860 |
23.0 |
0.99754 |
14.0 |
0.99924 |
19.0 |
0.99841 |
24.0 |
0.99730 |
15.0 |
0.99910 |
20.0 |
0.99820 |
25.0 |
0.99704 |
16.0 |
0.99894 |
21.0 |
0.99799 |
26.0 |
0.99678 |
1If your temperature is not one of the temperatures shown above, you can extrapolate to get the approximate density for the temperature that you measured. For example, if the temperature you measured was 20.5oC, the density difference between 20.0oC and 21.0oC is 0.00021 g/mL, meaning that the density will vary by about 0.000021 g/mL for each 0.1oC temperature change. Therefore, the extrapolated density for water at 20.5oC would be about 0.99810 g/mL (actual density is 0.9981019 g/mL to two more significant figures).
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Copyright © Donald L. Robertson (Modified: 09/20/2006)