chemistry labsperation techqniue

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Harvard University *

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PHYSICAL C

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Chemistry

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Jan 9, 2024

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Fernando Vargas, Hector Gomez, Maximo Chavarro Period 1 8/24/23 Separation Techniques Laboratory Filtration and Chromatography Data Part 2 – Chromatography Sharpie Color Pigment Color Distance Pigment Traveled Distance Solvent Traveled (mm) RF Pink Pink 38mm 90mm 0.4222 Magenta 15mm 0.1666 Light purple 76mm 0.8444 Black 88mm 0.9777 Blue 84mm 0.9333 Purple 8mm 0.0888 Light pink 18mm 0.2000 Orange Orange 72mm 90mm 0.8000 Yellow 81mm 0.9000 Black Black 1mm 90mm 0.0111 Purple 10mm 0.1111 Light gray 16mm 0.1777 Blue 78mm 0.8666 Brown 86mm 0.9555 Part 3 Pre-Filtration Measurements Mass of 50 mL volumetric cylinder 89.92 g Mass of 50 mL volumetric cylinder with sea water 115.07 g Mass of 25 mL of sea water 25 mL Mass of sea water 25.15 g Density of sea water 1.006 g/mL Temperature of sea water 23 C ° Part 5 Post Filtration Measurements Mass of 50 mL volumetric cylinder 89.96 g Mass with filtered sea water 114.089 g Mass of filtered sea water 24.123 g Volume of filtered water 25mL Density of sea water 0.965 g/mL
Temperature of sea water 23 C ° Calculations Mass of sea water: 15.07g – 89.92g = 25.15g (mass with sea water) – (mass without sea water) = answer Density of sea water: 25.15g / 25mL = 1.005 g/mL (volume/mass=density) Mass of filtered sea water: 114.089g - 89.96g = 24.123g (mass with filtered sea water) – (mass without filtered sea water) = answer Density of filtered sea water: 24.123g / 25mL = 0.965 g/mL Post lab questions 1. Why would you never use a beaker to measure the volume of the sea water? The reason we do not use beakers for measuring volume is mainly because beakers are not intended for mixing. Beakers are made for mixing, reacting, and filtering solutions and substances. In addition to this beaker are known to have around a 5% error range when attempting to measure any substances, therefore making any measurements in a beaker volume wise inaccurate. 2. Why did the pigments in the chromatography separate? The reason for this is due to the chemical reaction the pigment had with the sea water. This is because certain pigment colors attach better to the water than other substances. This allows for variation of distance each pigment traveled as pigment reactions vary. 3. What does the retardation factor (Rf) number tell you about the pigment during chromatography? The retardation factor acts as the ratio of the solvent's travel distance to that of the solute. It tells me what how far the pigment traveled in comparison to the entire solution. 4. Which water sample had the largest density (before or after filtration)? Why do you think this is the case? I concluded that the water before it was filtered was much denser than after it was filtered. This was because when the water was filtered, I could observe particles of sand, salt and other substances on the filter paper that would still be in the water had it not been filtered (extra mass in unfiltered water). Based on this observation and the logic of how density is calculated ( d = M/V, where d is density, M is mass, and V is volume) I concluded that unfiltered water had a greater density than filtered water. 5. Why is it important to measure the temperature of the water when we are trying to calculate its density? It is important to measure the water temperature before calculating any density because temperature influences volume of a substance. If the temperature of the sea water after it is filtered is higher than the original sample, your comparisons won’t be to scale. This is because water expands at higher temperatures due to the atoms of the water spreading further apart. The more heat you add to a substance the further it expands. This directly affects the volume of the substance in question which then affects your final calculation for density as the equation is Mass/Volume relying on mass and volume.
6. Calculate the Error and Percent Error for your value of Density for the filtered sea water (The accepted value for the density of sea water is 1.03 g/mL). Given that the accepted value for density of sea water is 1.03 g/mL I calculated it to be -6.31% rounded to the nearest hundred. My work is listed below for this problem. Assuming the formula ((Estimated Number – Actual Number)/ Actual number) x 100 then, ((0.965 g/mL – 1.03 g/mL)/1.03 g/mL) x 100 = (-0.065 g/mL/1.03 g/mL) x 100= -0.0631068 x 100 = -6.31067961% (ALL WORDS FOR POSTLAB QUESTIONS AND LAYOUT SHOULD NOT BE THE SAME DELETE WHEN U READ THIS)
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