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Jan 9, 2024
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Uploaded by CaptainWildcatMaster787
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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