Chemistry 321L – 02MPost-LabExperiment 1: Dilution, Calibration and Error AnalysisBy: Cameron Kahn12 February 2014TA: Muditha DiasKahn 1 | P a g eSingle investigator tasks:1. Flow chart depicting dilution strategiesParallel Dilution-Strategy 1(30% sucrose written on working bottle)Kahn 2 | P a g eMass of solution=10.0519gDiluted to 10mLMass of solution=10.0826gDiluted to 10mLMass of solution=10.1137gDiluted to 10mLMass of solution=10.202gDiluted to 10mLMass of solution=10.236gDiluted to 10mLParallel dilutions use multiple dilutions to determine the relationship between % sucrose and densityStrategy 1 involved 5 steps for each dilution: Tare volumetric flask, add known volume via Eppendorf, dilute to the mark (10mL), determine themass of the solution, and calculate densityCalculated density= 1.011g/mLCalculated density= 1.008g/mLCalculated density= 1.005g/mLCalculated density= 1.0236g/mLCalculated density= 1.020g/mLAdded 2,500μL of known sucroseAdded 2,000μL of known sucroseAdded 1,500μL of known sucroseAdded 1,000μL of known sucroseAdded 500μL of known sucroseThe temperatureduring dilutions was 24oCSerial Dilution(30% sucrose written on working bottle)Kahn 3 | P a g eThe temperatureduring dilutions was 24oCSD3 density=0.9939g/mLSD5Density=0.9825g/mLSD4Density=0.9875g/mLThis process continues by taking the newly labeled calibrator previously made, add 2,000μLfrom previous calibrator into volumetric flask, dilute to 10mL, and then re-label new calibrator concentration (SD3, SD4, and SD5)This is now labeled SD2and has a density=0.9946g/mLDiluted to 10mLAdded 2000μL from calibrator to volumetric flaskWe took the most concentrated calibrator and label it SD1.From parallel dilutions, SD1 has a density= 1.0236g2. Tabulated data setsCalibration of Eppendorf 1 (10 to 100 μL)Mass (g) Average Mass (g)10 0.0089 0.00880.00880.0087standard deviation 0.000125 0.0228 0.02260.02260.0225standard deviation 0.000250 0.0479 0.04750.04740.0473standard deviation 0.000375 0.0719 0.07180.07180.0718standard deviation 5.7735E-05100 0.0963 0.09670.09700.0967standard deviation 0.0004Calibration of Eppendorf 2 (100 to 1,000 μL)Mass (g)Average Mass(g)100 0.094 0.09440.0940.095standard deviation 0.001250 0.242 0.2420.2420.243standard deviation 0.001500 0.493 0.48970.4920.484standard deviation 0.005750 0.747 0.74230.7410.740Kahn 4 | P a g estandard deviation 0.00401000 0.994 0.99030.9880.989standard deviation 0.003Parallel DilutionSet Volumes (μL)Volumes after pipette calibrations (μL)% sucroseDensity (g/mL)500 494.74 1.48 1.0051000 996.09 2.99 1.008261500 1497.44 4.49 1.011372000 1998.79 6.0 1.020212500 2500.15 7.50 1.023633. Calibration Curves0 20 40 60 80 100 12000.020.040.060.080.10.12f(x) = 0x − 0R² = 1Calibration of Eppendorf 1 (10- 100μL) Linear ()Set Volume μL Average Mass (g)Kahn 5 | P a g eSerial DilutionSet Volumes (μL) Volumes after pipette calibrations (μL)% sucroseDensity (g/mL)2000 1998.79 7.50 1.02362000 1998.79 1.50 0.99462000 1998.79 0.30 0.99392000 1998.79 0.06 0.98732000 1998.79 0.01 0.98250 200 400 600 800 1000 120000.20.40.60.811.2f(x) = 0x − 0.01R² = 1Calibration of Eppendorf 2 (100-1000μL) Linear ()Set Volume (μL)Average Mass (g)1 1.01 1.01 1.02 1.02 1.0302468f(x) = 294.07x − 293.6R² = 0.96Parallel Dilution(% sucrose versus density) Linear ()Density g/mL% Sucrose0.98 0.99 0.99 1 1 1.01 1.01 1.021.02 1.03 1.03012345678f(x) = 195.11x − 192.53R² = 0.95Serial Dilution (% sucrose versus density) Linear ()Density (g/ml)% sucrose4. % Sucrose in group’s unknown using regression calculationsKahn 6 | P a g eFor Parallel Dilution, y=294.07x-293.6 x=0.99863g/mL (density calculated for 10-fold dilution of unknown) y(unknown %sucrose)=0.067x10= 0.67%For Serial Dilution, y=195.11x-192.53x=0.99863g/mLy=(unknown %sucrose)=2.31x10=23.13%5. Solve for the 95% confidence interval around the unknown accounting for the pipettor calibrations.**we only took one measurement for the density of the unknown therefore to use thet-test properly; we made up two other density measurements.Density= 0.99863g/mL, 0.99899g/mL, 0.99888g/mLParallel dilutionx=0.99863g/mL unknown % sucrose (y-value) was 0.67%x=0.99899g/mL unknown % sucrose (y-value) was 1.73%x=0.99888g/mL unknown % sucrose (y-value) was 1.41%The 95% confidence interval of unknown sucrose 1.27±2.34Serial dilutionx=0.99863g/mL unknown % sucrose (y-value) was 23.13%x=0.99899g/mL unknown % sucrose (y-value) was 23.83%x=0.99888g/mL unknown % sucrose (y-value) was 23.61%The 95% confidence interval of unknown sucrose 23.53±1.546. Show the hypothesis was correct by using the t-test at the 95% confidence levelCompare the 95% confidence interval of unknown sucrose from Parallel dilution and serial dilution.Parallel dilution 1.27±2.34Serial dilution 23.53±1.54Texpected value=59.21Texp>Ttable , then the two results are different59.21>4.303 , therefore the two results are differentKahn 7 | P a g eCollective tasks:1. Using the eight values in the data set provided, solve for the class average and 95% confidence interval of the slope and y-intercept from both dilution techniques.Collective TasksParallelSerial22.55 20.4827.35 23.123.48 19.8626.36 27.4729.09 27.2923.4 238.2 21.76.1 21.923.9 22.323.2 28.9Ttable=2.36Parallel dilution:Average= 21.3695% confidence interval= 21.36±18.43Serial dilution:Average=23.695% confidence interval=23.6±7.44Texp=0.25Texp<Ttable , then the results are the same (this was not asked for but I was curious)2. Show that the hypothesis, the values for slope and y-intercepts from the given data set, was correct using the pooled standard deviation method for both dilution techniques.Using Spool equation I get 5.94. To find the Texp substitute Spool value along with the other information already calculated into the Texp calculation (that’s associated with Spool).The Texp value is 0.75If Texp<Ttable, then the results are statistically the same0.75<2.36, therefore our null hypothesis is correct (holds true) that the class data from the two methods yielded the same results. Kahn 8 | P a g