Exp. 1 Density of Water Slides

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Pennsylvania State University, Altoona *

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Chemistry

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Feb 20, 2024

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7/27/2023 1 CHEM 111: Density of Water Density Learning Objectives: 1. Measure mass using an analytical balance. 2. Measure volume using water displacement method. 3. Calculate density with the correct number of significant figures. 4. Calculate and define average of a set of values. 5. Perform error analysis: standard deviation and percent error 6. Graph data in Excel and insert a linear trendline and equation. 7. Describe what the trendline equation and R-squared value represents. 1 2

7/27/2023 2 Purpose of Experiment Part A Purpose: Determine the average density of water. Calculate percent error and standard deviation for the four samples. Part B Purpose: Determine the degrees brix or % sugar and density of sugar in two different fruit juices. Part A. Measuring Mass • Make sure balance plate is clean and close balance doors. • Tare balance: 0.0000 g • Place plastic 25-mL graduated cylinder inside. • Close balance doors and record exact mass of empty graduated cylinder. Mass of empty graduated cylinder: 12.3637 g 3 4

7/27/2023 3 5 Part A. Measuring Volume of Water 25.0 mL Graduated Cylinder • Fill a 150-mL beaker with deionized water and measure the temperature of the water using a thermometer. Ex. 21.3ºC • Fill a disposable small plastic pipette twice with water and transfer the water to the 25.0 mL graduated cylinder. • Record this volume in the column labeled as the “Volume of Water” for sample #1 in Table 1 2 x 6 Measuring from a Graduated Cylinder 25.0 mL Graduated Cylinder • There are 10 lines or calibration marks between the 20 and 25. • 25.0 mL – 20.0 mL = 5.0 mL • Therefore, each line represents 0.5 mL . • 5.0/10 = 0.5 mL . 21.5 mL 5 6

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7/27/2023 4 Table 1. Sample Volume of Water (mL) Total Mass of GC + Water (g) Mass of Water Sample (g) Density of the Sample (g mL -1 ) 1 5.0 2 3 4 Average Density = Percent Error (of average density) = Table 1. The masses and volumes of samples of water in determining density . Part A. Total Mass • Using the electronic balance, determine the total mass of the 25.0 mL graduated cylinder and the water you added. • Record this mass in the column labeled “Total mass of GC + water ” in Table 1. Total mass: 17.4085 g 7 8

7/27/2023 5 Table 1. Sample Volume of Water (mL) Total Mass of GC + Water (g) Mass of Water Sample (g) Density of the Sample (g mL -1 ) 1 5.0 17.4085 2 3 4 Average Density = Percent Error (of average density) = Table 1. The masses and volumes of samples of water in determining density . Part A. Measuring Mass of Water 17.4085 g (total mass) – 12.3637 g (mass of graduated cylinder) = 5.0448 g Sample Volume of Water (mL) Total Mass of GC + Water (g) Mass of Water Sample (g) Density of the Sample (g mL -1 ) 1 5.0 17.4085 5.0448 2 3 4 Average Density = Percent Error (of average density) = 9 10

7/27/2023 6 Part A. Calculate Density Calculate density. Round off the final answer to the correct number of significant figures. = = -1 -1 5.0 mL Mass 5.0448 g Density= = 1.00896 g mL 1.0 g mL Volume Sample Volume of Water (mL) Total Mass of GC + Water (g) Mass of Water Sample (g) Density of the Sample (g mL -1 ) 1 5.0 17.4085 5.0448 1.0 2 3 4 Average Density = Percent Error (of average density) = Part A. Calculate Average Density To compute the average, ҧ𝑥 , (mean), add up all the individual values ( x 1 , x 2 , x 3 , …) and divide by the total number of values, N , as follows 1 2 3 N x x x x x N + + + = -1 -1 -1 1 -1 - 2.73 g mL + 2.72 g mL + 2.74 g x = =2.73 g mL 4 mL + 2.73 g mL 11 12

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7/27/2023 7 13 Part A. Calculate Percent Error Percent error represents how close your experimental value is to the accepted or true value (0% error perfect!, 100% error completely wrong!) Experimental Value - Accepted Value Percent Error = x 100 Accepted Value For example, the accepted value for the density of aluminum is 2.70 g mL -1 and the experimental value is the average of 2.73 g mL -1 ) -1 -1 -1 (2.73 g mL - 2.70 g mL Percent Error = x 100 = 1.11 % 2.70 g mL Creating a Graph Linear regression equation: y = mx + b y = y-axis (vertical values) is mass (g) values x = x-axis (horizontal values) is volume (mL) values m= slope ( rise over run) y/x is density (g/mL) b= y-intercept should be near zero y = 0.9967x + 0.2057 R² = 0.9994 0.00000 5.00000 10.00000 15.00000 20.00000 25.00000 0.0 5.0 10.0 15.0 20.0 25.0 Mass of Water (g) Volume of water (mL) Plotted Correctly 13 14

7/27/2023 8 R-squared Value The R-squared (R 2 ) value is an assessment of how well the data fits the trend-line (sometimes referred to as the scatter in the data). R 2 values range from 0 (poor fit) to 1 (perfect fit). y = 23829x - 0.0021 R² = 1 0 0.2 0.4 0.6 0.8 1 1.2 1.4 0.00E+00 1.00E-05 2.00E-05 3.00E-05 4.00E-05 5.00E-05 6.00E-05 Absorbance at 427nm Tartarzine (yellow #5) concentration (M) R 2 = 1 (perfect fit) R 2 = 0.651 (poor fit ) 16 Part A. Standard Deviation Standard deviation ( ) measures the central value without outliers. The smaller the standard deviation, the more precise your data is. 𝜎 This Photo by Unknown Author is licensed under CC BY-NC-ND • 68% of the values within one standard deviation ( ± σ ) of the average (mean) • 95% of the values within two standard deviations ( ± 2 σ ) of the average (mean) • 99.7% of the values in the sample within three standard deviations ( ± 3 σ ) of the average (mean) 15 16

7/27/2023 9 17 Part A. Standard Deviation ( ) ( ) ( ) ( ) 2 2 2 1 2 N x - x + x - x +… + x - x σ = N-1 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 1 2 2 2 2 2 3 2 2 4 x -x = 2.73- = 0.0000 x -x = 2.72- = 0.0001 x -x = 2.74- = 0.0001 x -x = 2.73- = 0.0000 2.73 2.73 2.73 2.73 For example, 4 density values are calculated: 2.73 g mL -1 , 2.72 g mL -1 , 2.74 g mL -1 , and 2.73 g mL -1 . The average is 2.73 g mL -1 . Calculate the standard deviation. First, calculate the square of the difference between each value and the average . 18 Part A. Standard Deviation Next, add all the values to calculate the sum: Sum of values = 0.0000 + 0.0001 + 0.0001 + 0.0000 = 0.0002 g mL -1 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 1 2 2 2 2 2 3 2 2 4 x -x = 2.73- = 0.0000 x -x = 2.72- = 0.0001 x -x = 2.74- = 0.0001 x -x = 2.73- = 0.0000 2.73 2.73 2.73 2.73 17 18

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7/27/2023 10 19 Part A. Standard Deviation Finally take the square root of the sum divided by (total # of values -1) ( ) ( ) ( ) ( ) 2 2 2 1 2 N x - x + x - x +… + x - x σ = N-1 ( )         -1 0.0002 σ = = ±0.00816 g mL 4 samples -1 Report value with average: 2.73 ± 0.00 816 g mL -1 = 2.73 ± 0.01 g mL -1 68% of the values lie within the range of 2.72 g mL -1 to 2.74 g mL -1 . 19