Lab Write-Up #1 – Name Emily Partners William and Tony Block 4A
SEPARATION OF A MIXTURE
Problem
How do you separate a mixture of iron filings, peppercorns, salt, and sand back to its original
state and mass?
Introduction
This experiment separates a mixture using the physical property of solubility. You will learn
magnetism, decantation, evaporation, and filtration, and will use a Bunsen burner to evaporate a
liquid. You will use the Law of Conservation of Mass to check your accuracy
Background Information
Mixtures are combinations of substances in which the components keep their individual
characteristics. They can be mixed in variable proportions and can be separated by simple
physical means. A mixture can be homogenous and heterogeneous. Homogenous mixtures have
the same composition and appearance throughout it (i.e. salt water), while heterogeneous
mixtures contain visibly different substances in it (i.e. peppercorns mixed with sand).
Compounds, on the other hand, are pure substances containing two or more elements that are
chemically bound together. Compounds cannot be separated by physical means.
Some of a mixture’s components have physical properties like melting point, boiling
point, or solubility that allow us to selectively remove one component from the mixture. Then
the percentage of each component in the original mixture can be calculated.
The methods of separation that will be used in this lab are filtration, magnetism, and
evaporation. The process of filtration is to pass a mixture through a filter (i.e. a funnel with filter
paper or a sieve) to isolate the substances in the mixture. The process of magnetism as a
separation method is to use magnetism to attract and isolate substances in a mixture that are
susceptible to magnetism. Be sure to put a plastic barrier around the magnet to easily extract the
substance. Evaporation as a separation method is the process of turning a liquid into vapor by
applying enough heat to reach the liquid’s boiling point.
The Law of Conservation of Mass states that mass is neither created nor destroyed in a
chemical reaction. This means that, in a closed system, the mass of a substance that undergoes a
chemical reaction will have the same mass after the chemical reaction. The Law of Conservation
of Mass can be utilized in this experiment to corroborate the fact that the after separating the
substances from the mixture, there should be the same amount as there was put into the mixture.
Calculating percent of error is a way to calculate how accurate an experimental value is.
Sources of error in measurement tools can cause a degree of error in a calculation, so you can
calculate this error with the following formula:
experimental value - theoretical value
Percent of Error =
 100
theoretical value
Works Cited
Curran, Gregory L. “Percent Error.” Science Help Online Chemistry. N.p., 7 Feb. 2008. Web. 16 Oct.
2011. <http://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson28.htm>.
The International Society for Complexity, Information, and Design. “Law of Conservation of Mass.”
ISCID. N.p., n.d. Web. 12 Oct. 2011. <http://www.iscid.org/encyclopedia/
Law_of_Conservation_of_Mass>.
MathMol. “Elements, Atoms, and Molecules.” MathMol. NYU/ACF Scientific Visualization
Laboratory, n.d. Web. 16 Oct. 2011. <http://www.nyu.edu/pages/mathmol/textbook/
compounds.html>.
Ophardt, Charles E. “Physical Properties.” Virtual Chembook. Elmhurst Coll., n.d. Web. 16 Oct. 2011.
<http://www.elmhurst.edu/~chm/vchembook/104Aphysprop.html>.
TutorVista.com. “Mixtures, Separation of Mixtures.” Tutor Vista. N.p., n.d. Web. 16 Oct. 2011.
<http://www.tutorvista.com/content/chemistry/chemistry-iii/chemistry-concepts/mixtures.php>.
Variables:
- Independent variable = separation technique
- Dependent variable = the mixture
- Control = original amounts of each substance in the mixture.
The control will be used to compare the experimental masses with and to calculate percent of
error. It is important to control the independent and dependent variables to minimize error.
Materials
1 mixture of sand and salt, iron and seeds
1 wash bottle with distilled water
0.1M AgNO3 solution in dropper bottle
1 magnet
1 Ziploc Bag
1 sieve
1 clay triangle
1 Bunsen burner
1 stirring rod
1 plastic funnel
1 filter paper
1 iron ring stand
1 500 mL beaker
3 100 mL beakers
1 electronic scale
2 400mL beakers
1 weighing boat
1 Petri dish
1 needle
1 test tube
1 pair of tweezers
1 graduated cylinder
1 hot plate
1 mitt
Safety:
Wear goggles when dealing with the evaporation of salt for the salt might splatter. Wear gloves
when handling chemicals and hot objects. Long hair should be tied back and no loose clothing
should be worn when near the Bunsen burner or other open flames.
Procedure:
1.
2.
3.
4.
Preparation
Clean off the weighing boat.
Place the weighing boat on the scale and zero it out.
Place the mixture on the weighing boat.
Measure and record the mass of the mixture.
Filtration
5. Clean off a 500 mL beaker.
6. Filter the mixture through a sieve and into the 500 mL
beaker.
Image 2a Filtration Step 6
7. Pour the peppercorns caught in the sieve into a 100 mL
beaker.
8. Use a needle to filter any stuck granules.
9. Pour the mixture out of the sieve and into the weighing
boat and use tweezers to pick out any remaining salt
granules.
10. Swipe the surface of the sieve with a magnet in a Ziploc
bag to pick up any excess iron fillings.
11. Zero the electronic scale with a weighing boat on it.
Image 2b Filtration Step 8
12. Pour the peppercorns onto the weighing boat and measure and record its mass.
13. Pour the peppercorns back into the 100 mL beaker.
Magnetism
14. Wash and dry a Petri dish.
15. Pour the mixture into a Petri dish.
16. Swipe the magnet in a Ziploc bag over the mixture to
separate the iron filings.
17. Place the iron filings in a 100 mL beaker by placing the bag
over the beaker and taking out the magnet.
18. Repeat steps 16-17 until you pick up all the iron filings.
19. Clean off the weighing boat.
20. Zero the electronic scale with a weighing boat on it.
21. Pour the iron filings onto the weighing boat.
22. Weigh and record the mass of the iron filings.
23. Pour the iron filings back into the 100 mL beaker.
Filtration
24. Measure the mass of a clean 400 mL beaker and record it.
25. Measure the mass of a clean filter paper and record it.
26. Place the salt and sand mixture in a 100mL beaker.
27. Measure out 30 mL of distilled water with a graduated
cylinder into another 100 mL beaker.
28. Place the water on a hot plate and warm up the water.
29. Pour the warm water into the salt and sand mixture.
30. Stir the mixture with a stirring stick so the salt it dissolved.
31. Set up an iron ring stand over the 400 mL beaker.
32. Place a funnel in the iron ring.
33. Fold a piece of filter paper.
34. Wet the filter paper with distilled water in a wash bottle.
35. Filter the mixture once by pouring it while stirring it through
the funnel.
36. Wash out the sand 3 more times with distilled water from a
wash bottle
37. Catch two drops of the mixture dripping out of the funnel in
small test tube.
38. Add 2 drops of 0.1M AgNO3 to the filtrate in the test tube
and swish it around.
39. Rinse the sand with distilled water one more time and repeat
steps 37-39 if the filtrate comes out a cloudy white.
40. Pour your sample in the test tube down the sink.
Image 3a Magnetism Step 16
Image 4a Filtration Step 33
Image 4b Filtration Step 35
41. Wait for the water to stop dripping out of the filter paper.
42. Lift the filter paper off the filter.
43. Place the filter paper on a watch glass on a hot plate and
warm it up to dry the sand.
44. Allow the filter paper and dried sand to cool to around
room temperature.
45. Measure the mass of the sand on the filter paper and record
it.
Image 4c Filtration Step 38
Evaporation
46. Measure the mass of a clean watch glass.
47. Place the 400 mL beaker with the solution on a clay
triangle on the iron ring stand.
48. Place the watch glass on the beaker.
49. Place a Bunsen burner under the ring stand.
50. Heat the mixture with a very gentle flame.
51. Lift the beaker and watch glass off the ring stand with mitts
once the salt has dried.
52. Allow the beaker to cool for 5 minutes.
53. Measure the mass of the dried salt with the beaker and
record it.
54. Measure the mass of the mass of the watch glass with the
Image 5a Evaporation Step 51
salt and record it.
Data Collection:
The Percent of Error in Salt, Sand, Peppercorn, and Iron Filings Samples
Theoretical Mass
Percent of Error
Experimental Mass (g) (g)
(%)
2.0 ±0.1
2.15
7.0
Salt
1.9 ±0.1
1.93
1.6
Sand
2.0 ±0.1
2.09
4.3
Peppercorn
2.6 ±0.1
2.09
24
Iron Filings
Table 1: The experimental mass, theoretical mass, and percent of error in the salt, sand,
peppercorn and iron filings mixture.
Information Table
Measurement (grams)
Mixture unknown code number:
Mass of mixture sample
Mass of 400 mL beaker
Mass of clean filter paper
Mass of watch glass
Mass of watch glass with salt
Mass of evaporating dish
Mass of evaporating dish with salt
Mass of beaker with dried salt
Mass of filter paper with dried sand
#6
8.19±0.01
164.1±0.1
0.8±0.1
54.7±0.1
54.8±0.1
45.0±0.1
45.1±0.1
166.0±0.1
2.7±0.1
Table 2: Measurements collected throughout the lab.
Mass of Sand = 2.7g  0.8g  1.9g
Mass of Salt  (166.0g  45.1g)  (164.1g  45.0g)  2.0g
Data Processing:
Mass of your sample of mixture
8.19 g
Original amount
8.26 g
Mass of iron filings in the mixture sample
2.6 g
Original amount
2.09 g
Mass of peppercorns in the mixture sample 2.0 g
Original amount
2.09 g
Mass of salt in the mixture sample
2.0 g
Original amount
2.15 g
Mass of sand in the mixture sample
1.9 g
Original amount
1.93 g
Percentage of iron in the original mixture 25.30%
2.09 g
x

8.26 g 100
8.26x  209
x  25.30%
Percentage of peppercorns in the original mixture 25.30%
2.09 g
x

8.26 g 100
8.26x  209
x  25.30%
Percentage of sand in the original mixture 23.37%
1.93 g
x

8.26 g 100
8.26x  193
x  23.37%
Percentage of salt in the original mixture
26.03%
2.15 g
x

8.26 g 100
8.26x  215
x  26.03%
Experimental percentage of iron in original mixture
31.48%
2.6 g
x

8.26 g 100
8.26x  260
x  31.48%
Experimental percentage of peppercorns in original mixture 24.21%
2.0 g
x

8.26 g 100
8.26x  200
x  24.21%
Experimental percentage of sand in original mixture
23.00%
1.9 g
x

8.26 g 100
8.26x  190
x  23.00%
Experimental percentage of salt in original mixture
24.21%
2.0 g
x

8.26 g 100
8.26x  200
x  24.21%
Total percent recovery
102.9%
31.48% + 24.21% + 23.00% + 24.21% = 102.9%
Percent of Error:
Salt: 7.0% error
2.0  2.15
Salt % Error =
• 100  7.0%
2.15
Sand: 1.6% error
1.9  1.93
Sand % Error =
• 100  1.6%
1.93
Peppercorn: 4.3% error
2.0  2.09
Peppercorn % Error =
• 100  4.3%
2.09
Iron Filings: 24% error
2.6  2.09
Iron Filings % Error =
• 100  24%
2.09
Graph1: This graph compares the experimental to the theoretical mass of the salt, sand,
peppercorn, and iron filings in mixture #6.
Graph 2: Graph 2 shows the percent of error found in between the experimental and theoretical
mass of the salt, sand, peppercorn, and iron filings in mixture #6.
Conclusion
1. Many students do not recover 100% of the original mixture. Describe at least two
possible problems that could cause less than 100% recovery of the mixture.
There are many errors that could cause one to recover less than 100% of the original mixture.
One such error could be not extracting all of the iron filings from the mixture. Since it requires
more than one round of swiping a magnet in a Ziploc bag over the mixture to extract all the iron
filings, if one didn’t do this enough times then there could still be some iron filings in the
mixture. This would cause the mass of the separated iron filings to be less than what was
originally put in. Another possible error could be that the mixture was not filtered enough times
to extract all the salt in the mixture. Filtration of the mixture to isolate the salt also required one
to wash the mixture thoroughly with distilled water several times. To test if there was any salt
leftover in the filter paper, two drops of the water dripping from the filter was added into a test
tube. Silver nitrate was then added to the test tube and one would observe if the mixture would
come out cloudy or clear. If the mixture was a little cloudy and the mixture was not filtered
more, then some of the salt would have been trapped in the filter paper. Failure to thoroughly
extract a substance from the mixture would have resulted in less than 100% recovery of the
mixture.
2. Pretend that you received a mixture that actually contained 50.0% salt and 50.0% sand.
At the end of the experiment, this student calculated that his sample was composed of
45.2% salt and 54.8% sand. What might he have done (or not done correctly) during
the experiment to cause this error in his results?
If the student calculated that the sample was composed of 45.2% salt and 54.8% sand, then some
of the salt must have not been extracted from the sand. When using a filter and filter paper to
filter the salt from the sand, the student must have not have rinsed the sand enough times so that
all of the salt ended up in the beaker below. Some of this salt might have been trapped in the
filter paper. When the filter paper was dried with the sand, the salt would have crystallized again.
Thus, the mass of the sand would have included the mass of the sand along with some of the salt.
This would cause the experimental mass of the sand to be greater than the experimental mass of
the salt. Another possible error that might have occurred in the student’s experiment is that the
student did not add enough water to dissolve all of the salt. If there were not an adequate amount
of water added to the mixture, then the solubility would not have allowed some of the salt to be
dissolved. Thus, when filtering the mixture some of the salt would not have passed through the
filter paper. The mass of the salt that was still in the sand mixture would then be added to the
mass of the sand, causing the mass of the sand to be greater than the mass of the salt.
3. A student obtains the following data:
Mass of evaporating dish
Mass of dish with mixture sample
Mass of beaker
Mass of beaker with dried salt
Mass of evaporating dish with dried sand
25.87 g
28.12 g
146.36 g
147.10 g
???
However, this student spills her sand sample out of the evaporating dish before weighing it.
If the student believes in the Law of Conservation of Mass, what should have been the
weight of the evaporating dish with the sand in it? Show your work.
In order to find the mass of the evaporating dish with the sand in it, you have to first subtract the
mass of the evaporating dish from the mass of the dish with mixture sample.
Mass of mixture = 28.12 g - 25.87 g
 2.25 g
Next, you have to subtract the mass of the beaker from the mass of beaker with dried salt to find
the mass of the salt.
Mass of salt = 147.10 g - 146.36 g
 0.74 g
After that, you have to subtract the mass of the salt from the mass of the mixture to find the mass
of the dried sand.
Mass of Sand = 2.25 g - 0.74 g
=1.51 g
Lastly, to find the mass of the evaporating dish with dried sand you have to add the mass of the
evaporating dish to the mass of the dried sand.
Mass of evaporating dish with dried sand = 25.87 g + 1.51 g
= 27.38 g
Thus, the mass of the evaporating dish with dried sand is 27.38 grams.
4.
A student receives a sample of a mixture with three components, solid iodine that is
removed first from the mixture by evaporation, solid salt that is dissolved to separate it
from the third component, and solid sand. The salt and sand are dried and weighed but
the iodine escapes as a gas and is not recovered. The student starts with 8.35 g of
mixture and recovers 2.37 g of salt and 4.11 g of sand. What is the percent of each
component in the original mixture? Show your work. (Attach a separate sheet of paper, if
necessary.)
In order to find the percent of each component in the original mixture, you have to first use
proportions to calculate the percent of salt and sand in the mixture.
Percent of Salt in Mixture
Percent of Sand in Mixture
2.37 g
x
4.11 g
x


8.35 g 100
8.35 g 100
8.35x  237
8.35x  411
x  28.4%
x  49.2%
The mixture was composed of 28.4% salt and 49.2% sand. To find the percentage of the mixture
that the solid iodine composed of, you have to subtract 28.4% and 49.2% from 100%.
Percent of Solid Iodine = 100%  28.4%  49.2%
 22.4%
Thus, the mixture was composed of 22.4% solid iodine, 28.4% solid salt, and 49.2% solid sand.
5. What would two things could you have improved on or would do differently next time?
If I were to have the opportunity to repeat this lab, I would do some things differently. One
thing that I would change about my procedure was to mass the 400 mL beaker beforehand.
Although the procedure says to mass the 400 mL beaker beforehand, that instruction was
added to the procedure after I realized that I recorded the mass of the 100 mL beaker used not
the 400 mL beaker. However, I couldn’t record the mass of another 400 mL beaker to replace
the one that was already used to evaporate the salt, for the mass of each beaker is different.
Due to this error, the mass of the salt was inaccurate. Another thing I would do differently is
to make sure that the magnets used were properly cleaned before using it. Since, while using
a magnet to clean off the iron filings stuck in the sieve, the plastic barrier broke, some of the
iron filings were stuck onto the magnet. However, there was already some iron filings stuck
on the magnet from another experiment. Thus, when we tried to recover the iron filings lost
by the broken Ziploc bag, we recovered some iron filings that weren’t ours. This is why the
percent recovery for our iron filings was over 100%. In the future, I would be sure to record
the mass of all the appropriate materials and to make sure that they were properly cleaned
before proceeding with my experiment.