CHEM115 LAB 2 THICKNESS OF FOIL

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Chemistry

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Dec 6, 2023

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Brianna Krzyminski CHEM115 Lab 2 TITLE: Aluminum Foil Thickness Determination BACKGROUND INFORMATION: Much of what takes place in chemistry occurs on a scale so small that humans cannot directly observe the individual interactions. Additionally, within a normal chemistry laboratory, there frequently is not sufficient equipment to measure quantities as small as an individual atom. As a result, scientists must use measurements and observed properties to indirectly measure quantities. In this experiment, you will be examining aluminum atoms which have an atomic radius of 118.0 pm. This is much too small for the human eye to decipher and for any measuring device in an educational chemistry laboratory. You will need to use other indirect measurements to determine how many aluminum atoms thick a sheet of aluminum foil is. OBJECTIVE: In this experiment, you will determine how many atoms thick a sheet of aluminum foil is. EQUIPMENT: Balance Graduated Cylinder Ruler PROCEDURES: You will need to design a procedure to determine how many atoms thick a sheet of aluminum foil is. Your analysis must contain at least three significant figures. Your experiment should follow accepted scientific practices and should be conducted at least three times to help ensure precision. Use the average of your trials to get the atomic thickness of the aluminum foil. Assume that the atomic radius of aluminum is 118.0 pm and all of the atoms in the foil are connected end to end to each other for your calculations. Also assume that the density of aluminum foil is that of pure aluminum (2.6989 g/cm 3 ). ANALYSIS: Prepare a data table that will encompass and organize all of the appropriate data for the lab. METHOD #1: METHOD #2:
M Method I using D L W y solving for True Density of aluminum foil D 2.69899m em's mass of experimental aluminum foil m 5.9575g Length of exllerimental aluminum foil L Width of experimental aluminum foil w 38949m 5 95759 2.698991cm go.com x 30.4cm X Heem M 19 5.9575g L x w x D d Hk t go 8cm x 30.4cm x 2.698991cm 5 9575 50.8 30.4 2.6989 4167.96525 0.00142935cm 4 0.00 43cm calculate atomic km pm o Ham x Ign x Ym Pm significant figure Thickness 0.00143M IN 1012pm 0.00143 10 pm 100 0.00143 1010 I 100cm IN 08814388888 14300000 pm atomic thickness it aluminum atomic radius 118pm H 1pm 118pm d 14300000 pm 1100 pm 1211006.441 atomicthickness of aluminum foil 1.21 105
Method 2 using Displacement measure volume torch then using v Lxwx it solve for H step l calculatevolume of experimental aluminum volume cylinder without aluminum foil 67.0mL volume cylinder with aluminum 011 70.2mL 70.2mL 67.0mL 3.2mL ML glam volume of experimental aluminum 67 3.29 cm3 Length of experimental aluminum L1 50.8cm widthof experimental aluminum W 30.4cm L X W X H d 3 29km3 50.8cm x 30.4cm X H 3 29km3 1544.32 X H 4 3.29 Cm 1544.32 1544.32 1544.32 0.00207211 0.0021cm calculate 1h91mn1s Edm often x Ign x Ym Pm significant figure 0.0021 ax IN 1012pm 0.0021410 pm 100 0.0021 1010 I 100cm IN 088 24888888 21000000 pm atomic thickness it aluminum atomic radius 118pm H 1pm floopm d 2000000 Pm 177966.102 1100 pm atomicthickness of aluminum foil 1.78 105
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QUESTIONS: 1.) Calculate a percent error for the thickness of the Al foil that you determined. Use the true value for the thickness of the Al foil as being 1.93 x 10 5 atoms. % error = | Theoretical Value – Experimental Value | Theoretical Value METHOD #1: Theoretical Value: 1.93 x 10 5 Experimental Value: 1.21x 10 5 METHOD #2: Theoretical Value: 1.93 x 10 5 Experimental Value: 1.78 x 10 5 2.) How many atoms of Al are there in 34.2 g of aluminum chloride? 3.) How would your results change if all of your measurements only gave two significant figures in your measurement? Displayed below are the calculations for each method utilizing only 2 significant digits and how they differ from the original calculations made. 1.93 1.21 0,73 0.37305699 100 1 93 method It error 37.31 1.939 31.78 9 43 0.07772021 4100 method It error 7771 A 143 molecular Weight Aix 1 27 27 4 3 35 105 1329 34.29 All's lying 6 gg 02BmkfAm19tom 1.56x1023Alatoms 34.2gAld sum Fmx 2.7cm 0.0014524328 0.0015cm É I 0.0015cm gym 1,01pm 150,08 000pm 127118.64407 1.3 105 vs 1.21 105 Calculating volume 70mL67mL 3.0mL 3 bg em3 5kmx30imxH 1530 1530 tf H 0.0019607843 0.002cm 0.0020cm Jim 11pm 200,08 000pm 169491.52542 1.7 105 vs 1.78 105
4.) Using water displacement to find the volume of a piece of aluminum foil, would it make a difference if you used tap water or deionized water? Explain your answer. For measuring volume, I do not believe results will be effected whether utilizing tap or deionized water, as I believe the aluminum foil would displace each the same way. Using deionized water vs. tap water would make more of a difference if being used in a chemical solution as the purified state of the deionized water would likely yield more accurate results in that setting. 5.) If the aluminum foil is not pure aluminum (it has several other components such as carbon), how would this affect your determination? If the aluminum foil is not pure and has other components this would effect accuracy of the experiment as the aluminum would no longer be pure aluminum and would then be a compound which would change the mass of the material. 6.) Aluminum costs about $0.85 (85 cents) per pound. One pound = 453.592 g. How much is one atom of aluminum worth? 85 Cents LB ORF 85 Cents 453.92g of Al Al molecular weight 27g Mol 4532.591mg 6.411,1023 1.011678157 1025 101.17 1023 101.17 1023 Al atoms 85 Cents 101.4 1023 8.401700108 10 24 0.84 10 cents