Cameron KahnMuditha DiasExperiment 5-Postlab26 March 2014Experiment 5-Determination of Total Fe in Sediment Samples1) Show all data from steps 2, 13, 14AmountSample 1.0055gSite 3 1.0108gSite 10 1.0133gSite 10AbsobanceWavelength (λ)0.8471 5620.8464 5620.8433 562Standard deviation0.0020224Average0.8456Sample AbsorbanceWavelength (λ)0.2296 5620.2307 5620.2299 562Standard deviation0.00056862Average0.23006667Site 3 AbsorbanceWavelength (λ)0.4831 5620.4829 5620.4826 562Standard deviation0.00025166Average0.482866671mL of Site 10 with 3mL of FerrozineAbsorbance Wavelength (λ)0.8861 5620.8875 5620.888 562Average 0.8872Standard deviation 0.00098489 2) Calculate the success of your recovery standard analysis. Report Fe recovered (in mg Fe/g sample). Solve for the standard deviation and 95% confidence interval of your answer. Did you recover the same amount to the 95% level of confidence?***Dilution factor of 2.5Concentration of Iron 2mL Sample: 2mL Ferrozine (M)*trial 1 trial 2 trial 3 AverageSample 1.45E-05 1.46E-05 1.45E-05 1.45E-05Site 3 3.72E-05 3.72E-05 3.72E-05 3.72E-05Site 10 6.98E-05 6.98E-05 6.95E-05 6.97E-05theoretical yield=.191×1.0055=0.191mg1.45E-05 * 3500/1.0055= 0.0505mg/g1.46E-05 * 3500/1.0055= 0.0508 mg/g1.45E-05 * 3500/1.0055=0.0505 mg/gAverage= 0.0506 mg/gStandard Deviation= 0.000173.0506.191×100=26.49 %recoveryConfidence interval: 0.0506 ± z×σ mg/g0.0506±4.303×.0001730.0506±7.44e-4We did not recover the same amount to the 95% level of confidenceBlankAbsorbance Wavelength (λ)0.0677 5620.0678 5620.0677 562Standard deviation5.7735E-05Average0.067733333) Calculate the concentration of iron (in mg Fe/g sample) in both of your samples, taking into account the recovery standard and absorbance blank. Show your calculations and commute error from the recovery standard calculation.*the commute error was just the standard deviation for both samples (site 3 and site 10)A=ε ×b × cWhere A=absorbance (average of absorbance – average of blank), ε= molar absorbance coefficient (27900m-1cm-1), b=path length (1cm), and c= concentration of Fe.Sample=(0.23006667−067733)=27900 ×1× Cc=5.80e-6 M ×2.5=1.45e-5 M^^^This calculation was done for each “trial” for site 3 and site 10. This process would yield the concentration of Fe in each trial for both samples which would also take into account the blank.0.19mg1 g× 1.0055= 0.191 mg Theoretical yieldConcentration of Iron 2mL Sample: 2mL Ferrozine (M)*trial 1 trial 2 trial 3 AverageSample 1.45E-05 1.46E-05 1.45E-05 1.45E-05Site 3 3.72E-05 3.72E-05 3.72E-05 3.72E-05Site 10 6.98E-05 6.98E-05 6.95E-05 6.97E-05Site 3: 3.72E-05 * 3500/1.0108= 0.129mg/gGiven standard: 0.19mg per 1 gramCalculated Standard: 0.129 mg per 1.0108 grams 0.129/1.0108 =0.128 mg per 1 gram0.128/0.19=0.6717=67.17% ± 0Commute error (the standard deviation)Site 10: 6.97E-05 * 3500/1.0133= 0.241mg/g Given standard: 0.19mg per 1 gramCalculated Standard: 0.241 mg per 1.0133 grams 0.241/1.0133 =0.238 mg per 1 gram0.238/0.19=1.25=125.18%± 5.77e-4Commute error (the standard