Analysis of an Aluminum - Zinc Alloy

! ! ! ! ! Analysis of an Aluminum - Zinc Alloy ! Procedure: In this Þnal laboratory experiment, the procedure written down in my pre lab, was somewhat done the way it was stated. Some things that were changed were that the mass of alloy we used was about 0.06g, rather than about 0.120g, as I said in the pre lab. Additionally, we did not dilute the 6M HCl solution to a 3M solution for part C, we still use the 6M HCl. Lastly, instead of just measuring the Þnal volume in the buret, we also measured the initial volume of the buret, and I will take the difference between the two values. ! Experimental Data (Part A) Room Temperature (¡C) Mass of Aluminum (g) 22.5 Initial Buret Volume (mL) Final Buret Volume (mL) Height of Water Column (cm) 0.030 10.1 50.8 27.00 Calculated Results (Part A) Initial Final Volume Partial Pressure of Atmospheric H2 gas Volume Volume of H2 Pressure H2O from Pressure actual yield (mL) (mL) (L) (ΔV) of H2 room (atm) (Mols H2) (atm) temperature (atm) 10.1 50.8 0.0407 0.934 0.0269 0.987 0.00157 Theoretical Percent Yield ! yield (Mols H2) (%) 0.00167 94.0 Sample Calculations (Part A) Volume of H2 (mL) (ΔV):$ $ $ $ $ $ $ $ 50.8 mL! -10.1 mL! 40.7 mL / 1000 = 0.0407 L Pressure of H2O from room temperature (atm) - From table 9.1: 0.0269 atm Partial Pressure of H2 (atm): PH2 = Patm - ( 1 atm x height ) - PH2O! $ $ $ $ $ 1026 cm ! $ $ $ =>$ PH2 = 0.987 - ( 1 atm x 27.00 ) - 0.0269 = 0.934 atm $ $ $ $ $ 1026 cm ! H2 gas actual yield (nH2): $ $ $ $ nH2 = PH2VH2 RT$ Where R = 0.0821atm (Universal Gas Constant)! T = Temperature in degrees kelvin (¡C+273.15)! ! ! ! ! $ $ $ $ $ $ nH2 = $ (0.934 atm)(0.0407 L) (0.0821 atm)(295.65 K)! = 0.00157 mol ! Theoretical Yield of H2:! Balanced EQ = 2 Al (s) + 6HCl (aq) Ñ> 2 AlCl3 (aq) + 3 H2 (g)! Mass of Al * 1 Mol of Al * 3 Mols of H2 Molar Mass of Al * 2 Mols of Al => 0.030g Al * 1 Mol Al * 3 Mol H2 26.98 g Al * 2 Mol Al $ $ Percent Yield: $ $ $ $ Actual Yield *100$ Theoretical Yield $ $ = Theoretic Amount of H2 (g)! k! $ = 0.00167 Mol H2 (g) $ ! => $ $ $ 0.00157 mol * 100 = 94.0% 0.00167 mol ! Experimental Data (Part B) Room Temperature (¡C) Mass of Zinc (g) 22.5 Initial Buret Volume (mL) Final Buret Volume (mL) Height of Water Column (cm) 0.123 5.2 49.8 28.72 Calculated Results (Part B) Initial Final Volume Partial Pressure of Atmospheric H2 gas Volume Volume of H2 Pressure H2O from Pressure actual yield (mL) (mL) (L) (ΔV) of H2 room (atm) (Mols H2) (atm) temperature (atm) 5.2 49.8 0.0446 0.932 0.0269 0.987 0.00171 Theoretical Percent Yield ! yield (Mols H2) (%) 0.00188 Sample Calculations (Part B) Volume of H2 (mL) (ΔV):$ $ $ $ $ $ $ $ 49.8 mL! $ -5.2 mL! 44.6 mL / 1000 = 0.0446 L Pressure of H2O from room temperature (atm) - From table 9.1: 0.0269 atm Partial Pressure of H2 (atm): PH2 = Patm - ( 1 atm x height ) - PH2O! $ $ $ $ $ 1026 cm ! $ $ $ =>$ PH2 = 0.987 - ( 1 atm x 28.72 ) - 0.0269 = 0.932 atm $ $ $ $ $ 1026 cm ! 90.9 ! ! ! ! H2 gas actual yield (nH2): $ $ $ $ $ $ $ $ nH2 = PH2VH2 RT$ nH2 = $ $ Where R = 0.0821atm (Universal Gas Constant)! T = Temperature in degrees kelvin (¡C+273.15)! (0.932 atm)(0.0446 L) (0.0821 atm)(295.65 K)! $ = 0.00171 mol! Theoretical Yield of H2:! Balanced EQ = Zn (s) + 2HCl (aq) Ñ> ZnCl2 (aq) + H2 (g)! Mass of Zn * 1 Mol of Zn * 1 Mol of H2 Molar Mass of Zn * 1 Mol of Zn = => 0.123g Zn * 1 Mol Zn * 1 Mol H2 65.38 g Zn * 1 Mol Zn $ $ Percent Yield: $ $ $ $ $ $ $ Actual Yield *100$ Theoretical Yield $ $ ! Theoretic Amount of H2 (g)! k! = 0.00188 Mol H2 (g) $ $ ! => $ $ $ 0.00171 mol * 100 = 90.9% 0.00188 mol ! Experimental Data (Part C) The unknown Alloy number was #16 Room Temperature (¡C) Trial Number 22.5 Mass of Unknown (g) Initial Buret Volume (mL) Final Buret Volume (mL) Height of Water Column (cm) 1 0.060 4.6 35.25 35.11 2 0.060 3.2 40.4 32.73 Calculated Results (Part C) The unknown Alloy number was #16 Trial # Pressure of Initial Final Volume Partial Volume Volume of H2 Pressure H2O from room (mL) (mL) (L) (ΔV) of H2 temperature (atm) (atm) Atmospheric H2 gas Pressure actual (atm) yield Mass Al Percent in Alloy Al in (g) Alloy (%) Percent Zinc in Alloy (%) (Mols H2) 1 4.6 35.25 0.0306 0.926 0.0269 0.987 0.00117 0.0063 10.3 89.7 2 9.2 40.4 0.0312 0.928 0.0269 0.987 0.00119 0.0067 11.2 88.8 Sample Calculations (Part C) ! ! ! ! Volume of H2 (mL) (ΔV):$ $ $ $ $ $ $ $ 35.25 mL$ $ $ $ -4.6 mL$ $ $ 30.65 mL / 1000 = 0.0306 L 40.4 mL! -9.2 mL ! 31.2 mL / 1000 = 0.0312 L Pressure of H2O from room temperature (atm) - From table 9.1: Partial Pressure of H2 (atm): PH2 = Patm - ( $ $ $ $ $ $ $ Trial 1$ =>$ PH2 = 0.987 $ $ $ $ $ $ $ Trial 2 $=>$ PH2 = 0.987 $ $ $ $ $ H2 gas actual yield (nH2): $ $ $ $ $ $ $ $ $ $ $ $ Trial 1 =>$ $ $ Trial 2 =>$ $ $ nH2 = PH2VH2 RT$ nH2 = $ nH2 = $ $ $ $ $ $ $ $ $ Trial 1 =>$ $ $ Trial 2 =>$ $ $ $ $ 1 atm x height ) - PH2O! 1026 cm ! ( 1 atm x 35.11 ) - 0.0269 = 0.926 atm 1026 cm ! ( 1 atm x 32.73 ) - 0.0269 = 0.928 atm 1026 cm ! Where R = 0.0821atm (Universal Gas Constant)! T = Temperature in degrees kelvin (¡C+273.15)! (0.926 atm)(0.0306 L) (0.0821 atm)(295.65 K)! (0.928 atm)(0.0312 L) (0.0821 atm)(295.65 K)! Mass of Aluminum in Alloy (g): MassAl = $ $ $ $ $ $ $ $ $ $ 0.0269 atm = 0.00117 mol! = 0.00119 mol! nH2 - (0.0153)(MassAlloy)! $ 0.0403! 0.00117 - (0.0153)(0.060)$ $ 0.0403! 0.00119 - (0.0153)(0.060)$ $ 0.0403! = 0.0062 g! = 0.0067 g! Percent of Aluminum in Alloy (%): MassAl% = MassAl * 100 ! $ $ $ $ $ $ MassAlloy$ ! $ $ $ $ ! $ $ $ $ Trial 1 =>$ = 0.0062g * 100 = 10.3% $ $ $ $ $ $ 0.060g$ ! $ $ $ $ Trial 2 =>$ = 0.0067g * 100 = 11.2% $ $ $ $ $ $ 0.060g$ ! Percent of Zinc in Alloy (%): $$ $ $ $ ! $ $ $ Trial 1 => $ $ $ Trial 2 => MassZinc% = 100% - MassAl% ! MassZinc% = 100% - 10.3% = 89.7 % MassZinc% = 100% - 11.2% = 88.8 % ! Discussion Questions: 1. As always, error is always introduced in experiments. First of all, we have accuracy to talk about. I feel that our accuracy was fairly good. I seemed to get similar mols and percents for the alloy part of the lab. Secondly, I feel that our precision was very good. For both parts A and B we got over 90 precent for percent yield. Im not sure how precise we were for part ! ! C, however. ! This is because I'm not sure what percent aluminum or percent zinc my alloy was. Even though I feel pretty! conÞdent in my results. Next, is systematic error. As always there is some systematic error, however because we were not transferring solutions from many different containers. I think we did a good job at keeping our systematic error at a minimum. Lastly, as always there was some random error. Some of this random error could have been from gas escaping while putting on the stopper to the test tube. Or even that the test tube wasnÕt sealed enough and let out some of the H2 pressure to the atmosphere. However, from our results, we did a good job at keeping it to a minimum. 2. Some important assumptions from the composition of the alloy were that we assumed that the alloy was 100 percent aluminum and zinc combined, and that there were absolutely no other metals in the alloy. Also the percent of aluminum in the alloy was about 10 percent, so ! assumed that zinc was 90 percent. 3. The percent yield for Al and Zn that I obtained from parts A and B effect the results for our alloy in the experiment by having similar percent yields for the alloy. Because both of our percent yields for part A and B were around 90 percent, we can estimate that the yield for ! our alloy was around that number too.