Comparison of Penny Densities Pre and Post 1982

Part A Comparison of Penny Densities Pre and Post 1982 The density of pennies made pre-1982 and post 1982 will be determined and compared. To obtain these densities masses will be obtained from balance measurements and volumes will be obtained from liquid displacement. To obtain experimental data you will observe the following video and make the measurements below. Sometimes he moves on a little fast so I would recommend you pause the video to make and record each measurement into the data table below. Procedure: 1. Record the mass of each set of 5 pennies dated pre-1982 and record in table 1. 2. Record the initial volume of water before adding pre-1982 pennies and record in table 1. 3. Record the final volume after adding 5 pre-1982 pennies and record in table 1. 4. Repeat steps 2 & 3 for each set of pre-1982 pennies and record in table 1. 5. Determine the penny volume for each set of pre-1982 pennies and record in table 1. 6. Determine the density for each set of pre-1982 pennies and record in table 1. *** be sure to include your units and appropriate significant figures Observations/Results Table 1: Mass and Volume Data for Density of Pre 1982 Pennies Set Mass 1 Initial Water Volume 15.402 g 21.8 mL 2 38.804 g 23.3 mL 3 46.101 g 25.0 mL 4 61.506 g 26.8 mL 5 76.857 g 28.4 mL Final Water Volume 23.3 mL 25.0 mL 26.8 mL 28.4 mL 30.1 mL Penny Set Volume 1.5 mL 3.2 mL 5.0 mL 6.6 mL 8.3 mL Penny Set Density 10 g/mL 9.6 g/mL 9.2 g/mL 9.3 g/mL 9.3 g/mL Deviation in Density 0.5 g/mL 0.1 g/mL 0.3 g/mL 0.2 g/mL 0.2 g/mL ****Show all Density and Deviation Calculations on the space provided on the next page**** Average Density:_9.5 g/mL_____________ 7. Record the mass of each set of 5 pennies dated post-1982 and record in table 2 . 8. Record the initial volume of water before adding post-1982 pennies and record in table 2. 9. Record the final volume after adding 5 post-1982 pennies and record in table 2. 10. Repeat steps 2 & 3 for each set of post-1982 pennies and record in table 2. 11. Determine the penny volume for each set of post-1982 pennies and record in table 2. 12. Determine the density for each set of post-1982 pennies and record in table 2. Table 2: Mass and Volume Data or Density of Post 1982 Pennies Set Mass Initial Final Penny Penny Deviation Water Water Set Set in Density Volume Volume Volume Density 1 12.456 g 21.2 mL 23.0 1.8 mL 6.9 0.3 g/mL mL g/mL 2 25.024 g 23.0 mL 24.8 3.6 mL 7.0 0.2 g/mL mL g/mL 3 37.538 g 24.8 mL 26.3 5.1 mL 7.4 0.2 g/mL mL g/mL 4 50.033 g 26.4 mL 28.0 6.8 mL 7.4 0.2 g/mL mL g/mL 5 62.510 g 28.0 mL 39.9 8.7 mL 7.2 0 g/mL mL g/mL 13. Determine the average density for the pre-1982 pennies and record in the space below table 1. 14. Determine the deviation in density for each set of pre-1982 pennies and record in the table 1. 15. Determine the average density for the post-1982 pennies and record in the space below table 2. 16. Determine the deviation in density for each set of post-1982 pennies and record in the space below table 2. ****Show all Density and Deviation Calculations on the space provided on the next page**** Average Density:__7.2 g/mL______________ Calculations: Show the calculations for each density and deviation calculation here. Th ey may be typed in using the word equation editor or they can be hand written and a picture taken and inserted here. Make sure if you take a picture that it is clear enough for your instructor to read Density table 1 g mL Set 1 Pre- 15.402 g ÷ 1.5 mL=10 Set 2 Pre- 38.804 g ÷3.2 mL=9.6 g mL Set 3 Pre- 46.101 g÷ 5.0 mL=9.2 g mL Set 4 Pre- 61.506 g ÷ 6.6 mL=9.3 g mL Set 5 Pre- 76.857 g ÷ 8.3 mL=9.3 g mL Deviation Density Table 1 g g g −9.5 ∨¿ 0.5 mL mL mL Set 1 Pre- ¿ 10.0 Set 2 Pre-| 9.6 g g g −9.5 ∨¿ 0.1 mL ml mL Set 3 Pre- ¿ 9.2 g g g −9.5 ∨¿ 0.3 mL mL mL Set 4 Pre-| 9.3 g g g −9.5 ∨¿ 0.2 mL mL mL ¿ 9.3 g g g −9.5 ∨¿ 0.2 mL mL mL Set 5 Pre- Density Table 2 Set 1 Post- 12.456 g ÷ 1.8 mL=6.9 g mL Set 2 Post- 25.024 g ÷3.6 mL=7.0 g mL Set 3 Post- 37.538 g ÷ 5.1 mL=7.4 g mL Set 4 Post- 50.033 g ÷ 6.8 mL=7.4 g mL Set 5 Post- 62.510 g ÷ 8.7 mL=7.2 g mL Deviation Density Table 2 Set 1 Post- ¿ 6.9 g g g −7.2 ∨¿ 0.3 mL mL mL Set 2 Post- ¿ 7.0 g g g −7.2 ∨¿ 0.2 mL mL mL Set 3 Post- ¿ 7.4 g g g −7.2 ∨¿ 0.2 mL mL mL Set 4 Post- ¿ 7.4 g g g −7.2 ∨¿ 0.2 mL mL mL Set 5 Post- ¿ 7.2 g g g −7.2 ∨¿ 0 mL mL mL Questions: 1. How did the densities of the two time periods differ? The density of the Pre 1982 pennies was greater than the density of the Post 1982 pennies. 2. Do a google search to find the difference between these time periods. How do these time period for pennies differ in terms of metals used? Using another google search compare the densities of those metals. Based on what you see do your results make sense? Provide an explanation. Before 1982, pennies were made of all copper. Since then, pennies have been made with and outside coating of copper and an inner core of a different metal called Zinc. Copper has a density of 8.94 g/mL which can compare to the densities found in table one, that are all close to 9 and average about 9.5 g/mL. Zinc has a density of 7.140 g/mL which can compare to the densities found in table two that are all around or near 7 with an average of 7.2 g/mL. 3. If the true value for the density of a pre-1982 penny is 8.8 g/mL, calculate the absolute error and relative percent error based on your average density for a pre-1982 penny. Show all work to receive credit. 9.5−8.8 =.080 8.8 9.5−8.8 Relative percent error×100 %=8.0 % error 8.8 Absolute Error - 4. If the true value for the density of a post-1982 penny is 7.2 g/mL, calculate the absolute error and relative percent error based on your average density for a post-1982 penny. Show all work to receive credit. Absolute error - 7.2 −7.2 =0 7.2 Relative Percent Error- 7.2 −7.2 ×100 %=0 % error 7.2 5. For each of the following samples, describe in detail how one would determine its density. a. a ball bearing First measure the mass by a mass balance and volume by using the formula radius. Then density = 4 3 V = π r , r is the 3 mass volume b. granular sodium chloride First find the mass of the substance using a mass balance, then find the volume using displacement of a liquid in a graduated cylinder. The liquid will not be water though you will need to use a non-polar liquid like hexane. You can do this by taking a 100mL graduated cylinder and fill it up with water to the 50mL line then very slowly add the sodium chloride using a funnel, being sure not to spill any. Now read where the water is now on the graduated cylinder. Find the volume with the equation FinalInitial= Volume. Then to find the density divide mass by volume. c. a lump of sodium metal First find the mass of the metal using a mass balance, then same as “b” find the volume using displacement of water in a graduated cylinder following the same steps. In this case you can use water as the liquid. Record the initial volume of the water and the final value once the metal is dropped in. subtract initial from final and that will be the volume then use the formula density = calculate the density. d. an irregular piece of sealed wood mass volume to First find the mass of the object using a mass balance. Then using the displacement of a liquid theory on how to find the volume. But you will need to use a low-density liquid such as Alcohol (ethyl, methyl, propyl) because if not then the piece of wood would float. You find the initial volume of the water before the object is inserted then you find the final volume of water after the object is inserted. To find the volume you will subtract the initial volume from the final volume to find the volume of the piece of sealed wood. Then using the formula density= 6. mass to find the density. volume Density can be used a conversion factor. A density of 2.50 g/ml means, a 2.50 g sample has a volume of 1 ml., or 2.50 g = 1ml. As discussed in lab 1, equalizes can be used to convert between units. Density is used to convert between mass and volume. To answer these question look on-line for the density of each metal at room temperature (25 C). Show all work. a. The mass of a sample of aluminum whose volume is 5.90 cm3 D=10.000 g/cm^3 Mass=Density * Volume Mass= 10.000 g × 5.90 cm3 = 59 g 1 cm3 b. The volume of a sample of lead whose mass is 3.970 g D= 1.930 g/ cm3 Volume= mass density 3.970 g = .3501 g/ cm3 1 1.34 g/cm3 Part B: Density of Water by different measuring device To complete part B answer questions 1 and 2. 1. Based on the video you watched in part A, and the videos on measuring the volume of liquids, describe in detail how you could design an experiment to determine not only the density of water, but compare finding the volume using a graduated cylinder versus a volumetric pipet. In your descriptions indicate how mass, volume and density would be determined, how many trials should be run, etc. First you need to find the mass of an empty graduated cylinder, then record the mass in grams in a table. Pour 100 mL of water into the graduated cylinder. Next, weigh the graduated cylinder with the water in it on an electric scale. Then record the mass in grams. To find the mass of only the water, subtract the mass of the empty graduated cylinder from the mass of the cylinder with water in it. Record the mass of 100 mL of water in the table. Next use the mass and volume of the water to calculate density. Density is equal to the mass divided by the volume in g/ cm3 . To gather more trials, por out the water until you have 50 mL of water in the graduated cylinder. Complete the same steps as you did for 100mL of water and record the data. Lastly you can pour out the water until you have 25 mL of water and repeat the same steps. At the end you can compare the results of the densities of the 3 trials. In order to record a more accurate and precise density of the water, a volumetric pipet can be used by repeating the same steps taken to find the density of water in a graduated cylinder. However, the volumetric pipet will only allow you to record up to 20 mL of water and is readable up to 0.01mL.The volumetric pipet should first be measured by itself in grams, then measured with the 20mL of water in it. The mass of the water should be the mass of the empty pipet subtracted from the pipet with 20mL of water. To find the density you would divide the mass over the volume of the water in the pipet. For the next trials, each volume of the pipet should be poured out to 15mL of water, and then to 10mL for the final trial. After gathering all the data, you an compare the densities of water from the three trials and compare the densities between the volumetric pipet and graduated cylinder. 2. Based on what you learned in the assigned videos, which measuring device (the graduated cylinder or the volumetric pipet) would you expect to give the more precise volume measurements? Which would you expect to give the more accurate measurements? The volumetric pipet will give you more precise measurements when compared to the graduated cylinder when measuring volume. The volumetric pipet can record up to 20mL and can be readable up to 0.01mL while the graduated cylinder can record up to 100mL (depending on size) and can be readable up to 0.1mL. When looking at accuracy, the volumetric pipet is also more accurate because its long size decreases error or misreading the meniscus when compared to the graduated cylinder. Accuracy and precision Questions 1. An experiment is performed to determine the % sulfate in a sample. The experiment is repeated four times. The true concentration of sulfate is 44.01%. Trial % Sulfate 1 44.02 2 44.11 3 43.98 4 44.09 Calculate the mean, deviation, % relative deviation, absolute error, and percent relative error. Show all work. Mean- 176 .2 4 = 44.05% Deviation – Trial 1- |44.02−44.0 5| = 0.03% Trial 2- |44.11 − 44.05| =0.06% Trial 3- |43.98−44.0 5| = 0.07% Trial 4- |44.09−44.0 5| = 0.04% Relative Deviation – Trial 1- ¿ 44.02−44.0 5∨ ¿ × 100 % = 6.8% 44.0 5 ¿ Trial 2- ¿ 44.11−44.0 5∨ ¿ ×100 % 44.0 5 = 13.6% ¿ ×100 % 44.05 = 15.8% ¿ ×100 % 44.05 = 9% ¿ Trial 3- ¿ 43.98−44.0 5∨ ¿ Trial 4- ¿ 44.09−44.0 5∨ ¿ Absolute ErrorTrial 1- |44.02-44.01|= 0.01 Trial 2- |44.11-44.01|= 0.1 Trial 3- |43.98-44.01|= 0.03 Trial 4- |44.09-44.01|= 0.08 Percent ErrorTrial 1- ¿ 44.02− 44.01∨ ¿ × 100 % = 2.2% 44.01 ¿ Trial 2- ¿ 44.11 −44.01 ∨ ¿ ×100 % 44.01 = 2.2% ¿ Trial 3- ¿ 43.98 −44.01∨ ¿ ×100 % 44.01 = 6.8% ¿ Trial 4- ¿ 44.09 −44.01∨ ¿ ×100 % = 18.1% 44.01 ¿ 2. The mass of a 45.750 g piece of copper is measured four times on two different balances with the following results: Trial Balance 1 (g) Balance 2 (g) 1 45.747 45.76 2 45.748 45.74 3 45.744 45.73 4 45.745 45.77 a) Calculate the mean, deviation, % relative deviation, absolute error, and % relative error. Show all Work. Balance 1 Mean- Balance 2 182 .984 4 = 45.746 g Mean- DeviationTrial 1- |45.747-45.746|=0.001 g Trial 2- |45.748-45.746|=0.002 g Trial 3- |45.744-45.746|=0.002 g Trial 4- |45.745-45.746|=0.001 g % relative deviationTrial 1Trial 2Trial 3Trial 4- 0.001 × 100 45.746 0.002 × 100 45.746 0.002 × 100 45.746 0.001 × 100 45.746 183 4 = 45.75 g DeviationTrial 1- |45.76-45.75|= 0.01g Trial 2- |45.74-45.75|= 0.01g Trial 3- |45.73-45.75|= 0.02g Trial 4- |45.77-45.75|= 0.02g % relative deviation- = 0.2% Trial 1- = 0.4% Trial 2- = 0.4% Trial 3- = 0.2% Trial 4- 0.01 ×100 45.75 0.01 ×100 45.75 0.02 ×100 45.75 0.02 ×100 45.75 = 2% = 2% =4% = 4% Absolute errorTrial 1- |45.747-45.750|= 0.003 g Trial 2- |45.748-45.750|= 0.002 g Trial 3- |45.744-45.750|= 0.006 g Trial 4- |45.745-45.750|= 0.005 g Absolute ErrorTrial 1- |45.76-45.750|=0.01g Trial 2- |45.74-45.750|=0.01 g Trial 3- |45.73-45.750|=0.02 g Trial 4- |45.77-45.750|=0.02 g % Error- % Error- Trial 1Trial 2Trial 3Trial 4- 0.003 ×100 45.750 0.002 ×100 45.750 0.006 ×100 45.750 0.005 ×100 45.750 b) = 0.7% Trial 1- =0.4% Trial 2- = 1% Trial 3- =1% Trial 4- 0.01 ×100 45.750 0.01 ×100 45.750 0.02 ×100 45.750 0.02 ×100 45.750 = 2% = 2% = 4% = 4% Which balance is more precise? Explain. Balance 1 is more precise. Balance as its average deviation is 0.001 which is less then balance 2 (0.01). A balance is more precise if deviation is lesser c) Which balance is more accurate? Explain. Balance 2 is more accurate. Balance as its observation are much closer to the actual mass of the copper metal than that of balance 1. The mean of balance 2 is 45.75 is closest to 45.750