experiment 1-laboratory techniques

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Feb 20, 2024

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Wednesday May 12, 2023 Saba Shakil CKCH 107 Shawn McFadden LABORATORY TECHNIQUES REPORT SHEET CALCULATIONS AND QUESTIONS A) The thermometer and its calculations 1) Atmospheric Pressure 99.25 KPa – Convert it to mm Hg mm Hg value = KPa (7.50062) mm Hg value = 99.25 (7.50062) mm Hg value = 744.40 2) True corrected temperature of boiling water b.p correction = (760 mm Hg - atmospheric pressure) x (0.037 °C/mm) The correction at 740.40 mm Hg is therefore: b.p correction = (760 mm Hg - 740.40 mm Hg) x (0.037 °C/mm) = 0.5772 °C The true corrected boiling point is thus 100.0 °C - 0.5772°C = 99.420 3) Percent relative error of thermometer (boiling water) Percent error= ( IValue observed-True value I)/ True value x 100% Percent error= (I98.50°C-99.420°CI)/99.420°C x 100% Percent error= (I-0.92°CI)/99.420°C x 100 Percent error =0.93 %

Wednesday May 12, 2023 Saba Shakil CKCH 107 Shawn McFadden 4) Calibration Curve B) Using the Balance to calibrate your 10 mL Pipette 1) Corrected Temperature from calibration curve: Temperature of water used in pipette (observed temperature)= 21.0 °C x= 21.0°C y=1.0249x - 1.5374 y=1.0249(21.0) - 1.5374 y=19.9855 °C Therefore the corrected temperature from the calibration curve is 20.0 °C

Wednesday May 12, 2023 Saba Shakil CKCH 107 Shawn McFadden 2) Volume delivered by pipette (Trial 1) Density= Mass/Volume The water in the pipette is 21 °C, according to table 1.1 the density is 0.997992 g/mL The mass of the water= Mass of Erlenmeyer Flask plus water - Mass of Erlenmeyer = 50.1701 g - 40.2362 g = 9.9339 g Volume= Mass/Density Volume= 9.9339 g /0.997992 g/mL Volume= 9.9539 mL 3) Volume delivered by pipette (Trial 2) Density= Mass/Volume The water in the pipette is 21 °C, according to table 1.1 the density is 0.997992 g/mL The mass of the water= Mass of Erlenmeyer Flask plus water - Mass of Erlenmeyer = 50.1685 g - 40.3050 g = 9.8635 g Volume= Mass/Density Volume= 9.8635 g /0.997992 g/mL Volume= 9.8833 mL

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Wednesday May 12, 2023 Saba Shakil CKCH 107 Shawn McFadden 4) Volume delivered by pipette (Trial 3) Density= Mass/Volume The water in the pipette is 21 °C, according to table 1.1 the density is 0.997992 g/mL The mass of the water= Mass of Erlenmeyer Flask plus water - Mass of Erlenmeyer = 50.2047 g - 40.2984 g = 9.9063 g Volume= Mass/Density Volume= 9.9063 g /0.997992 g/mL Volume= 9.9262 mL 5) Mean volume delivered by 10mL pipette Mean = (Trial 1 + Trial 2 + Trial 3 ) / 3 Mean = (9.9539 mL + 9.8833 mL + 9.9262 mL) /3 Mean = 9.9211 mL 6) Individual deviation from the mean (Trial 1, 2, 3) Individual deviation from mean = Value - Mean Deviation of 9.9539 from the mean = 9.9539 - 9.9211 = 0.0328 mL Deviation of 9.8833 from the mean = 9.8833 - 9.9211 = -0.0378 mL Deviation of 9.9262 from the mean = 9.9262 - 9.9211 = 0.0051 mL

Wednesday May 12, 2023 Saba Shakil CKCH 107 Shawn McFadden 7) Average deviation from the mean: Absolute deviation of 0.0328 = |0.0328| = 0.0328 Absolute deviation of -0.0378 = |-0.0378| = 0.0378 Absolute deviation of 0.0051 = |0.0051| = 0.0051 Calculate the sum of the absolute deviations. Sum of absolute deviations = 0.0328 + 0.0378 + 0.0051 = 0.0757 Average deviation = Sum of absolute deviations / Number of values Average deviation = 0.0757 / 3 = 0.0252 mL C)Determination of Density 1)Density of objects Density = Mass/Volume Water = 9.80/9.75 = 1.00 g/mL Ethanol = 10.55/19.50 = 0.540 g/mL Tall Shiny Metal = 80.25/8.00 = 10.0 g/mL Short Golden Metal = 90.96/9.10 = 10.0 g/mL Short Dull Metal = 30.26/10.00 = 3.10 g/mL Rectangular Prism = 20.88/2.00 = 10.4 g/mL

Wednesday May 12, 2023 Saba Shakil CKCH 107 Shawn McFadden QUESTIONS 1. Using table 1.1 creates a graph relating temperature to density of water. (see Appendix A for proper graphing techniques) 2. Determine the density of water at 34 °C using your graph. Place this point on the graph

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Wednesday May 12, 2023 Saba Shakil CKCH 107 Shawn McFadden 3.The relationship between temperature and density is linear for the region above 4°C up to close to the boiling point. Consult Appendix A regarding linear relationships and using the method of least squares determine the equation that relates temperature to density. Make sure to show all your work. m = Σ(Δx * Δy) / Σ(Δx^2) where Σ denotes the sum of the values. Σ(Δx * Δy) = -0.008644 + (-0.006579) + (-0.004642) + (-0.002982) + (-0.001659) + (-0.000693) + (-0.000131) + 0.000006 + (-0.000321) + (-0.000927) + (-0.002465) + (-0.004565) + (-0.006786) + (-0.009892) = -0.054227 Σ(Δx^2) = 43.086641 + 31.005441 + 21.004641 + 12.750441 + 6.634641 + 2.460241 + 0.327121 + 0.184041 + 0.511441 + 1.188441 + 4.771441 + 20.191441 + 29.445841 + 42.263641 = 214.526617 m = -0.054227 / 214.526617 ≈ -0.000252 b = ȳ - m * x̄ where x̄ is the mean of x and ȳ is the mean of y. x̄ = 21.571 ȳ = 0.997762 b = 0.997762 - (-0.000252) * 21.571 ≈ 0.997813 Therefore, the equation of the line that relates temperature (x) to density (y) is: y = -0.000252x + 0.997813 4. Calculate the density of water at 34 °C using the equation of the line from question #3.

Wednesday May 12, 2023 Saba Shakil CKCH 107 Shawn McFadden y = -0.000252x + 0.997813 y = -0.000252(34) + 0.997813 y= 0.989245 g/mL 5. Calculate % relative error for density of water and ethanol. Density of water % Relative error: Percent Relative Error = |(Your Value - Literature Value) / Literature Value| * 100% Percent Relative Error = |(1.00 - 0.997992) / 0.997992| * 100% Percent Relative Error = |0.002008 / 0.997992| * 100% Percent Relative Error = 0.002011 * 100% Percent Relative Error = 0.201% Density of ethanol percent Relative error Percent Relative Error = |(Your Value - Literature Value) / Literature Value| * 100% Percent Relative Error = |(0.540 - 0.7892) / 0.7892| * 100% Percent Relative Error = |-0.2492 / 0.7892| * 100% Percent Relative Error = 0.3154 * 100% Percent Relative Error = 31.54%

Wednesday May 12, 2023 Saba Shakil CKCH 107 Shawn McFadden 6. Compare the volumes of the metal cylinders determined by the Archimedes’ method with the calculated volume based upon length and diameter. Object Length and diameter Archimedes Method Tall Shiny Metal 34.5 cm 3 8.00 mL Short Golden Metal 48.7 cm 3 9.10 mL Short Dull Metal 50.2 cm 3 10.00 mL Rectangular prism 3.6 cm 3 2.00 mL 7. Being mindful of propagation of error, to what certainty is your calculated volume for the rectangular metal object? (Appendix B for propagation of error) (ie; what is the ± in cm 3 ? ΔV = V * sqrt((Δr/r)^2 + (Δh/h)^2) a) Short dull V = π * r^2 * h = π * (2.2 cm)^2 * 3.3 cm ≈ 50.486 cm³ Δr = 0.1 cm (assuming a reasonable uncertainty) Δh = 0.1 cm (assuming a reasonable uncertainty) ΔV = 50.486 cm³ * sqrt((0.1 cm / 2.2 cm)^2 + (0.1 cm / 3.3 cm)^2) ≈ 1.702 cm³ b) Short gold V = π * r^2 * h = π * (2.2 cm)^2 * 3.2 cm ≈ 43.648 cm³

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Wednesday May 12, 2023 Saba Shakil CKCH 107 Shawn McFadden Δr = 0.1 cm (assuming a reasonable uncertainty) Δh = 0.1 cm (assuming a reasonable uncertainty) ΔV = 43.648 cm³ * sqrt((0.1 cm / 2.2 cm)^2 + (0.1 cm / 3.2 cm)^2) ≈ 1.449 cm³ c) Tall shiny metal V = π * r^2 * h = π * (1.7 cm)^2 * 3.8 cm ≈ 38.097 cm³ Δr = 0.1 cm (assuming a reasonable uncertainty) Δh = 0.1 cm (assuming a reasonable uncertainty) ΔV = 38.097 cm³ * sqrt((0.1 cm / 1.7 cm)^2 + (0.1 cm / 3.8 cm)^2) ≈ 2.233 cm³ d)Rectangular prism V = L * W * H = 1 cm * 1 cm * 3.6 cm = 3.6 cm³ ΔL = 0.1 cm (assuming a reasonable uncertainty) ΔW = 0.1 cm (assuming a reasonable uncertainty) ΔH = 0.1 cm (assuming a reasonable uncertainty) ΔV = 3.6 cm³ * sqrt((0.1 cm / 1 cm)^2 + (0.1 cm / 1 cm)^2 + (0.1 cm / 3.6 cm)^2) ≈ 0.211 cm³ Short dull- ±1.702 cm³

Wednesday May 12, 2023 Saba Shakil CKCH 107 Shawn McFadden Short gold- ±1.449 cm³ Tall shiny - ±2.233 cm³ Rectangular prism- ±0.211cm³ REFERENCES Ethyl Alcohol, Reagent Anhydrous . Ethyl alcohol. (n.d.). https://macro.lsu.edu/howto/solvents/ethanol.htm ΔV = V * sqrt((Δr/r)^2 + (Δh/h)^2) Where: ● V is the calculated volume ● Δr is the uncertainty in the radius ● Δh is the uncertainty in the height Let's calculate the uncertainties for each cylindrical object: a) Cylinder A: V = π * r^2 * h = π * (2.2 cm)^2 * 3.3 cm ≈ 50.486 cm³ Δr = 0.1 cm (assuming a reasonable uncertainty) Δh = 0.1 cm (assuming a reasonable uncertainty) ΔV = 50.486 cm³ * sqrt((0.1 cm / 2.2 cm)^2 + (0.1 cm / 3.3 cm)^2) ≈ 1.702 cm³ b) Cylinder B:

Wednesday May 12, 2023 Saba Shakil CKCH 107 Shawn McFadden V = π * r^2 * h = π * (2.2 cm)^2 * 3.2 cm ≈ 43.648 cm³ Δr = 0.1 cm (assuming a reasonable uncertainty) Δh = 0.1 cm (assuming a reasonable uncertainty) ΔV = 43.648 cm³ * sqrt((0.1 cm / 2.2 cm)^2 + (0.1 cm / 3.2 cm)^2) ≈ 1.449 cm³ c) Cylinder C: V = π * r^2 * h = π * (1.7 cm)^2 * 3.8 cm ≈ 38.097 cm³ Δr = 0.1 cm (assuming a reasonable uncertainty) Δh = 0.1 cm (assuming a reasonable uncertainty) ΔV = 38.097 cm³ * sqrt((0.1 cm / 1.7 cm)^2 + (0.1 cm / 3.8 cm)^2) ≈ 2.233 cm³ Therefore, the uncertainties in the volumes are approximately: a) ±1.702 cm³ b) ±1.449 cm³ c) ±2.233 cm³

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