Potassium Chromate Lab

Potassium Chromate Lab
The results obtained were consistent with the lab manual and the unknown sample aligned well with
the instructions that were given. The unknown sample was labeled "3" which was then combined
with HCl and centrifuged, decanted, combined with more HCl, centrifuged, and decanted again. The
precipitate formed from the two latter steps was then boiled and then combined with acetic acid and
Potassium Chromate to check for the presence of Pb2+. Upon mixing the precipitate with acetic acid
and Potassium Chromate, a bright yellow solution was formed with no apparent precipitate. This
bright yellow solution indicated that our unknown for part A contained Lead Chromate. In part, B
the liquid that was decanted in step 3 was to be tested for the presence of either the presence of
Barium or Calcium. The solution was mixed with Ammonium hydroxide and then it was tested to
make sure the solution was basic. Ammonium carbonate was then added and the solution was
heated, cooled, centrifuged, and the liquid was discarded. The precipitated was washed with water
and centrifuged. Acetic acid was mixed with the precipitated and dissolved until it was not cloudy.
Ammonium hydroxide was mixed with the unknown liquid. Potassium Chromate was then added
and a yellow precipitate was formed. The mixture was centrifuged and an orange liquid was formed
and then decanted. The remaining precipitate was then checked for the presence of barium by
adding HCl to dissolve it and was then ... Show more content on Helpwriting.net ...
The HCl itself helped to separate the Lead ions from the Calcium ions. This is why a precipitate was
formed for use in Part A and the liquid was saved for Part B. This worked due to differing KSP's as
one compound was more soluble rather than the other and thus the Lead, which was extremely
insoluble with a low KSP formed a precipitate with the addition of HCl since
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Precipitate Lab
The point of the experiment was to determine whether the chemical mixture will form a precipitate.
The first step was to combine compound 1 and compound 2. After mixing them together, if a solid
was created, a precipitate formed. Also for many of the chemical mixtures, the color changed. Silver
Nitrate + Sodium Chloride created a precipitate. The precipitate was AgCl (Silver Chloride). The
color of the precipitate was white. Copper Dichloride + Lead Dinitrate formed a precipitate. The
precipitate was PbCl2 (Lead Dichloride) and the color of this was blue. Ammonium Sulfate + Lead
Dinitrate also formed a precipitate. The precipitate was PbSO4 (Lead Sulfate) and the color of the
precipitate was white. Sodium Iodide + Barium Chloride did not form

Double Displacement Reaction Lab
Abstract
Introduction
This lab revolves around precipitate it reactions. The precipitate reactions happen when it had
cations (positive atoms) and anions (negative atoms) in aqueous solutions combined to form an
insoluble ionic solid, which is called a participate. In order for these reactions to occur they must be
Aqueous and this can be determined by using solubility rules. Predicting these reactions a sign test
to find out which ions are present in a solution. Double displacement reactions are the kinds used in
this lab. This means that the compound will switch partners and form a bond with a different part.
The reaction occurs in Aquarious solution and one of the products that is formed is in soluble.
Example: CdSO4(aq) + K2S(aq) >> CdS(s) + K2SO4(aq)
Both reactants are soluble and one of the products is a solid. A soluble substances dissolves with
water but an insoluble substance doesn't. In a solution, which is a homogenous mixture of two or
more substances, one will have a solvent and solute. The solvent ... Show more content on
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The solutions of the tests done on the software can be found in table 1. Choose the top pop up
experiment menu. The purpose of the software will be to practice identifying a noon's. From the
experiment menu, click the type of unknown that needs to start. Choose the one positive ion option.
When chosen, a picture of a solid will be seen: the unknown cation with the nitrate anion. This will
also show the results of putting the solid into water. To save the treated sample, click on one of the
seven saved samples slots in the upper right. To delete the treated simple click on the trashcan. Make
saved sample an active the sample by clicking on it. Only complete compounds can be added, for
example, if one wanted to add XS OHI– to the active sample, click on the NaI+, XSOHI–, and then
ADD. Record each reaction in the table as either a soluble or insoluble solution and record any

Importance Of Hydroxyapatite
CHAPTER 1
INTRODUCTION
1.1Importance of Hydroxyapatite
Hydroxyapatite is one of the calcium phosphatefamily.It containing water with a chemical formula
of Ca10(PO4)6(OH)2. (Yasushi Suetsugu)
Hydroxyapatite (HA)is one of the essentialbiominerals of interest in the calcium phosphate based
bioceramics and the main inorganic composition of teeth and bone.HA is utilized as material
fordental and bone substitution.This is because its similarity in chemical to nature bone matrix
.Hence,many research had been done to synthetic HA,so that it can use as a replacement to bone
,and as substitution in biomedical applications. In nowadays, HA has various of applications in
biomedical. For instance, it used as matrices for drug release control and as materials ... Show more
content on Helpwriting.net ...
This means that the solution has to contain more solute entities (molecules or ions) dissolved than it
would contain under the equilibrium (saturated solution).Crystallization is also a chemical solid–
liquid separation technique, in which mass transfer of a solute from the liquid solution to a pure
solid crystalline phase occurs.Usually,the crystalline solid have highly ordered three dimensional
arrangement of particles.
Crystallization can be divided into two processes, nucleation and crystal growth. Nucleation is the
initiation of a phase change in a small region, such as the formation of a solid crystal from a liquid
solution.It is a process in which a minimum number of atoms,ions or molecules join together to give
a stable nuclei.The rate of nucleation is believed to increase with increasing relative supersaturation.
The crystal growth is the subsequent growth of the nuclei that succeed in achieving the critical
cluster size.However,rate of the crystalgrowth is only moderate enhanced by high relative
supersaturation.Particles with smaller size is produced if crystal growth

Unknown Salt Synthesis Essay
Overall this experiment was successful, as it was possible to determine which unknown salt solution
the cation belonged too, and also the pairing anion. To figure out the cations, many tests and
observations were needed to be made. With the gathered flame test results, and the information
given in the procedure, conclusions could be made. Solution A is Sr2+ since it tested crimson red.
Solution C is Ba2+ because it tested green. Solution D is Li+ because it turned the flame carmine
red. Finally, solution G is Na+ because the flame had an intense colour of orange–yellow. The other
ions did not give any specific flame test colour. The next test used was the litmus test, and it
determined not only a cation, but also two anions. The acidity, and the basicity of each solution were
tested in the litmus test. When the solution is dropped on the paper and it turned the litmus paper
red, it means that the solution in acidic. In this case solution E turned the litmus paper red, proving
that H3O+ cations are present. Solution D and G turned the litmus paper blue which means they are
both basic solutions, and either OH– or CO32– anions are present. Left unknown still are solutions
B and F, however; solution B can be concluded to be Co2+ by qualitative observations. Solution B
was clear and light red/pink, and by conducting research it was determined that Co2+ solution
contained the same colour, (Brunning, October 3, 2015). Finally Ag+ was determined to be the
cation of unknown

Recrystallization Lab Report
The solid organic compounds that we use in reactions or that come from natural sources tend to have
impurities. One way for us to remove these impurities and obtain our desired product is by using a
technique called recrystallization. Our goal in this experiment is to prove that recrystallization is a
method to purify a solid organic compound with contaminants. In Part A we will purify acetanilide
by recrystallization and in Part B we will distinguish the best way to select the solvent required to
recrystallize the solute. Removing the solid impurities of an organic compound involves a series of
steps. The first step is to dissolve the solid in the minimum amount of the appropriate hot solvent.
Crystallization takes advantage of the fact that most solids are more soluble in a hot than in a cold
solvent. We need a solute that is soluble in ... Show more content on Helpwriting.net ...
Therefore, if a solid is first dissolved in an amount of hot solvent that does not dissolve it when cold,
crystals should form when the hot solution is allowed to cool. The second step is using hot vacuum
filtration, which removes the insoluble impurities from the hot mixture. The acetanilide given to us
had two impurities, one soluble and one insoluble in the hot water. The insoluble impurity can be
removed after the acetanilide is dissolved in the hot solvent and filtered. Most of the solvent may be
evaporated from the crystals as the vacuum pulls the air through the crystals on the funnel, but the
acetanilide will still contain the more soluble impurity. It is important to let the solution cool slowly
in order for the molecules to line up in an organized way and form the crystals. We also need to
make sure that the filter flask in clamped and that we use the appropriate tubes for the vacuum as
well as the appropriate sized filter paper. After letting the solution cool, we begin the process of
recrystallization in order to remove the soluble impurity.

Recrystallization Of Naphthalene Lab Report
Introduction In this experiment, the objective is to perform recrystallization properly to purify
multiple organic solids. Recrystallization has become a fundamental technique in organic chemistry
and the pharmaceutical industry, for separating and purifying organic compounds. Out of all the
techniques used for purifying organic compounds, recrystallization is the most vital one. This is due
to its effectiveness, convenience, and industrial applications. Recrystallization has to do with
dissolving organic compounds to purify them. The solid acts as the solute as it dissolves in an
appropriately selected hot solvent. The solvent is then cooled which allows for the solution to
become saturated with the solute. This leads to recrystallization, ... Show more content on
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Exposure to skin will lead to irritation as it is a permeator and a sensitizer. Exposure to eyes will
also lead to irritation and discomfort. Prolonged inhalation is dangerous as it may lead to nausea,
weakness, vomiting, and other harmful effects. Wearing gloves will counteract most of the potential
danger (NIOSH 2014).
Water
Molecular Formula: H2O / IUPAC Name: Water / Molar Mass: 18.02 g per mol / Melting Point: 0°C
/ Boiling Point: 100°C / No hazards
Ethanol
Molecular Formula: C2H6O / IUPAC Name: Ethanol / Molar Mass: 46.07 g per mol / Melting
Point: –114.1°C / Boiling Point: 78.5°C / is flammable and should be used with caution. Skin/eye
exposure may cause irritation. Inhalation should be avoided if possible.
Methanol
Molecular Formula: CH4O / IUPAC Name: Methanol / Molar Mass: 32.04 g per mol / Melting
Point: –97.8°C / Boiling Pont: 64.5°C / is flammable and should not be handled around flames.
Exposure to skin/eyes will cause irritation and discomfort. Is a permeator. Extended inhalation will
be harmful.

A Experiment On Separation And Purification Methods
This experiment introduces separation and purification methods used in research quite frequently.
These methods include hot vacuum filtration and recrystallization to purify a crude sample of
acetanilide containing two impurities. Recrystallization is a very common method that is used to
purify solids. This process dissolves a crude solid with impurities in a mildly boiling solvent, and
cooling down the mixture afterwards for crystals to reconstruct themselves in solution. This method
allows impurities to separate and remain in solution as precipitate or remain uniform in solution.
Solids are more soluble in hot solvents than in cold ones, allowing maximum dissolution for proper
separation and subsequent crystallization to occur. After purifying the compound, the melting point
is determined using the Mel–Temp technology to compare to the written melting point to see how
pure the sample truly is. The proper steps of recrystallization include selecting a solvent that fits the
characteristic of all the components, water in this experiment. Then, dissolving a crude sample in the
chosen solvent (at or close to the solvent boiling point), forming solid in the solution as cooling
takes place, using vacuum filtration to isolate the purified solid, ridding the crystals of all impurities
through drying. In this experiment, recrystallization allows purification of 150 mg of crude
acetanilide. Hot vacuum filtration is a process used. The process requires using a filter flask, Pasteur

Understanding Of Basic Solubility Principles And...
The purpose of this experiment was to apply the understanding of basic solubility principles and
precipitation reactions to identify two unknown cations in a given solution. The first two processes
involved adding various acids and basis to the unknown solution to observe different precipitation
reactions. The unknown solution contained either silver or lead and either barium or calcium. For
unknown solution one, adding a sample of hydrochloric acid to the unknown solution started the
experiment. This was imperative because it triggered a precipitation reaction, creating PbCl2 or
AgCl. HCl was an ideal acid that was used because it was soluble to Calcium or Barium but atleast
partially insoluble to Ag and Pb, allowing for the Calcium or Barium to have remained in solution,
become isolated in a new test tube, and set aside for later use.1 The chloride ions are much more
attractive to the silver and lead ions compared to the calcium or barium, which allowed some to
precipitate while others to have remained in solution.2
After the initial decanter, deionized water was added to the precipitate and the tube was placed in
boiling water. This is an essential step, because this allows for the dissolution of any PbCl that might
have precipitated. Since PbCl is slightly insoluble, if there was Pb in the unknown solution, the
heating process would have caused it to become mostly soluble in the water. The addition of heat as
a source of energy was enough to break the attractive forces

Calorimetry Lab Report
Experiment One
1. Turn on the electrically heated sand bath and allow for sand to begin heating
2. Place 50 milligrams (mg) of Benzoic Acid into a 10 X 100–millimeter (mm) reaction tube
3. Using a Pasteur pipette, add 1.0 milliliter (mL) of hot water drop–by–drop into the reaction tube
to reach complete solubility of the solute
4. Place the reaction tube into an electrically heated sand bath and observe the tube until the liquid
inside starts to boil and the benzoic acid has saturated fully
5. Remove the reaction tube from the heated sand bath and remove the wood applicator stick from
the liquid
6. If impurities are present, add a small amount of Pelletized Norit to the solution and allow to boil
for a few minutes, then add activated charcoal to absorb impurities.
7. ... Show more content on Helpwriting.net ...
Allow for the reaction tube to cool to room temperature (22°C), then cool further in ice
8. Using the Pasteur pipette, filter out water as much as possible from the tube
9. Add a few drops of cold ethanol into the tube to separate the crystals from water and allow the
water to be removed.
10. Connect a water aspirator onto the reaction tube and place the tube in a beaker of boiling water
for a few minutes to allow for water to evaporate out
11. Scrape crystals off the tube and onto filter paper using a spatula
12. Using the filter paper, dry off excess water from crystals and then allow for crystals to
completely dry off
13. Weigh the dry crystals and record the calculated percentage (%) of recovered purified solute.
Experiment Two
1. In reaction tube #1, Recrystallize 50 milligrams of Benzoic Acid from water by repeating
experiment one.
2. In reaction tube #2, Add 50 milligrams (mg) of Benzoic Acid into the tube and add the minimum
amount needed of hot Methanol to completely dissolve the solid substance
3. Add hot water to the solution in a drop–wise manner, shaking after each addition of water, until
solution becomes saturated and the recrystallization process has begun to

Calorimetric Analysis Lab Report
Introduction The purpose of the lab was to use different solubilities of iron (III) and nickel (II) ions
to separate a mixture of the two. This was done by precipitating both of the solutions with hydroxide
and after a solution of ammonia was used to re–solvate the nickel as Ni(NH3)62+ (Rice 2017). The
separation of chemicals is a crucial step in many chemical procedures and experiments as it can
allow for the identification of the elements within an unknown solution (Altig 2009).
Methods
The experiment was conducted by pouring one mL of Ni2+ and Fe3+ each into a small test tube
along with dropwise amount of 6 M NaOH until no precipitate formed. After the solution was
placed in a centrifuge for two minutes to let the precipitate settle ... Show more content on
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Through the experiment the iron was precipitated by using 6M NaOH to create a reaction where
Fe3+ and NaOH formed Fe(OH)3(s). The remaining solution in which the precipitate was settled
became Ni(OH)2(aq). The nickel was precipitated by using dimethylglyoxime to form a solid with
nickel consisting of a compound Ni(dmg)2(s). A possible errors that could have occurred during the
experiment was the removal of the nickel ion precipitate/ while removing supernatant in the second
part of the experiment. This could have reduced the amount of precipitate and the addition of
ammonia to the precipitate could have been affected. The results of the lab were conclusive and a
nickel precipitate was formed. The goal of the lab was met and the solid precipitate of nickel and
iron were separated and

Melting Point Lab
Experiment #1: Purification of a Solid: Crystallization and Melting Point: Joseph Katz and Melissa
Colwell Binghamton University, Department of Chemistry, Binghamton, New York 13902
Introduction: One of the most important parts of this experiment was the concept of the melting
point. The melting point is the temperature at which a substance turns from a solid into a liquid.
Every substance has a specific temperature of melting point although any impurities in a substance
will lower the temperature required to melt it. This means that as more impurities are removed, the
closer the melting point of a substance will get closer to the real melting point. ... Show more
Although the amount recovered was high, the crystals still had a yellow color, showing that
impurities remained in the sample. However, the melting point was within the range of the acid
identified. Conclusion: The experiment determined that the unknown sample was phthalic acid due
to the melting point of 208.9 degrees Celsius. The correction of the machine was found to be 0.3
degrees Celsius, and it was machine #5. The percent recovered was determined to be 91.67%.
Acknowledgements: I would like to acknowledge my lab partner Melissa Colwell for assistance
during the lab. References: 1) Organic Chemistry Laboratory Manual, Spring 2015 Edition,
Department of Chemistry, Binghamton University, Binghamton New York, 2012 pp 3–4, 77–86. 2)
"Liquids." Melting Point, Freezing Point, Boiling Point.
http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch14/melting.php. 3) Lide, David R. CRC
Handbook of Chemistry and Physics: A Ready–reference Book of Chemical and Physical Data.
Boca Raton, FL: CRC, 1994.

Determination Of Solubility Of Fenoprofen
RESULTS
Determination of solubility of fenoprofen The solubility data is shown in Table I and it is presented
in Figs.1, 2 and 3. The solubility of fenoprofen amongst various oils investigated was found to be
highest in oleic acid (198.93±2.91mg/ml) followed by triacetin, labrafil M1944, capryol 90. The
solubility of drug was insignificantly different between labrafac lipophil, IPM, and labrafac PG.
Amongst surfactants, labrasol showed maximum solubility (129.17±1.4mg/ml) followed by tween
80, and span 20. Propylene glycol showed highest solubility among the cosurfactants
(210.5±1.57mg/ml), followed by glycerol, plurol oleique, and lauroglycol 90.
Preliminary screening of surfactants for emulsification ability The % transmittance values of
dispersions are given in Table II. These studies indicated that tween 80 had good ability to emulsify
oleic acid followed by labrasol, while span 20 failed to emulsify oleic acid.
Preliminary screening of cosurfactants The % transmittance values of surfactant/cosurfactant
combinations are given in Table III. These studies revealed that no significant difference between
propylene glycol and glycerol ability to improve nanoemulsification of tween 80. However,
propylene glycol was selected because it showed better drug solubility than glycerol.
Construction of pseudo–ternary phase diagrams The transparent to translucent microemulsion region
for each investigated system are presented in phase diagram with one axis representing

The Chemistry Of Material Chemistry
Material chemistry is important as it leads to synthesis of advanced materials with new and superior
properties and improved compositions. It is one of the integral parts of material science. It mainly
deals with four components. (i)Materials, (ii) Synthesis and processing by different possible routes,
(iii) Structure and Properties and finally (iv) Characterization and evaluation. Here the oxygen
carrier chosen for the experiment is a mixed oxide of both Cu and Mn. Cu and Mn both belong to
transition metal category with atomic numbers 29 and 25 respectively. Oxides of Cu and Mn are not
at all toxic in nature and are environmental benign. Also they are less costly compared to other metal
oxides. Copper oxides always show better reactivity and Manganese oxides have the ability to
change to various oxidation states. Cu has atomic mass of both Cu and Mn oxides separately have
showed high reactivity being an oxygen carrier, but literature shows that these oxides have
limitations when used individually like limited recyclability, agglomeration, sintering etc. This lead
to exploring the use of these single metal oxides in a mixed form. The three dimensional or
geometrical arrangement of molecules or atoms as points in a specific form in space is called a
crystal lattice. The smallest unit of a crystal lattice is called a unit cell. A unit cell is the smallest
repeating unit of a crystal. The unit cells have different structures. These arrangements are also
called Bravais

Lab Report: Predicting Precipitation
Predicting Precipitations Lab Report: explanation data of lab along hypothesis, and identifying
possible errors of lab for further investigations.
The reactions from 1A through 1D all have a precipitation formed because physical changes
appeared. AgNo3 is a clear solution; when it is mixed with a yellow FeCl3, a precipitate is formed in
a different color than the original colors. After a while, the precipitate became slightly solid. For 1B
and 1C, the solid was formed in similar ways, it might be because both reactants contains a similar
element. Similar data have also found in 2B and 2C. These two groups of reaction both contain a
similar element in the solution, so it might be the reason to cause the solid form in similar colors.
The reaction of 1D also become a precipitate, but appeared in a different color. For 3C, the result
was not clear enough to make sure if a precipitate was formed or not, but after a while, the solution
turned slightly foggy. Solution 2D, 3D, and 4D had no color change, therefore no precipitation was
formed. ... Show more content on Helpwriting.net ...
According to the predictions, solutions 1A through 1D all have a precipitation formed; referring
back to the data table, solutions 1A through 1D did have a solid formed during the lab. Another
example of why parts of the hypothesis is correct is because the 2B and 2C were predicted as the
same solid would form. This matched the data results because in the lab, 2B and 2C were identical
after the reactants were mixed together. However, one part of the hypothesis was wrong because the
prediction for 3C did not match the data for 3C. Solution 3C was predicted no precipitate will form,
but according to the data, a white foggy colored solid was

Dispersion strengthening usually design in nuclear power...
Dispersion strengthening usually design in nuclear power plants , hypersonic aircraft , and space
vehicles are continually seeking materials that have high strength at elevated temperature . These
requirements have been met in part by the precipitation–strengthened "supperalloys" that are
suitable for application up to around 1800°F . The refractory metals , tungsten , molybdenum ,
columbium , and tantalum , may sometimes be used when the service temperature exceeds the
useful temperature of the super alloys . However , the refractory metals are expensive , difficult to
fabricate , and have poor resistance to oxidation . There is a limit to which one can extend the
service temperature by precipitation strengthening . In ... Show more content on Helpwriting.net ...
Other potential application appear in the area of high–temperature heat exchangers and
regeneratively cooled rocket engines . Nucleation and growth in solid–state reaction , is the overall
rate or kinetics of the transformation process depends on both nucleation and growth . If more nuclei
are present at a larger number particular temperature , growth occurs from a larger number of sites
and the phase transformation is completed in a shorter period of time . At higher temperature , the
diffusion coefficient is higher , growth rates are more rapid and again expect the transformation to
be complete in a shorter time , assuming an equal number of nuclei .
Exceeding the Solubility Limit to alloy for strengthened
1. Widmanstatten structure – The precipitation of a second phase from the matrix when there is a
fixed crystallographic relationship between the precipitation and matrix crystal structures .
2. Interfacial energy – The energy associated with the boundary between two phases .
3. Dihedral angle – The angle that defines the shape of a precipitate particles in the matrix .
4. Coherent precipitation – A precipitation whose crystal structure and atomic arrangement have a
continuous relationship with the matrix from which the precipitation is formed .
A similar

The Thermodynamics Of Solidification Process
THERMODYNAMICS
PROJECT
Prof: Srinivasan G. Srivilliputhur
Thermodynamics of Solidification.
– Finally Draft –
Chun–Yu Lin and Yiyang Wan
University of North Texas
Material Science and Engineering department
Denton, TX 76207
Oct 30 2014
Outline
1. Introduction of Solidification
2. Principle of Solidification Process
a. Nucleation
b. Growth of solid
c. Alloy's phase diagram
d. Alloy Solidification
3. Real Application of Solidification Process
a. Eutectic Solidification Process
b. Peritectic Solidification Process
4. Conclusion
Introduction
Solidification is process through which crystalline materials, such as metals and alloys, transform
from non–crystallized state into crystallized state. This process is a basic technique used in alloy
casting, growth of single–phase semiconductors, welding and etc. We need to understand what's
happening during solidification and how it affect the structure of final materials, which directly
determines the properties of products. Besides, a proper set of solidification parameters also helps to
improve energy efficiency.
Principle of solidification process
1. Nucleation

The whole process begins with the creation of a combination of atoms randomly, followed by
stabilization of these tiny cores (homogeneous nucleation). Thermodynamics plays an important role
because it determines if the liquid continues to solidify or remains in equilibrium.
When the environment temperature is below the melting

Separation of a Carboxylic Acid, a Phenol, and a Neutral...
Objective: The objective of this experiment is to use acid–base extraction techniques to separate a
mixture of organic compounds based on acidity and/or basicity. After the three compounds are
separated we will recover them into their salt forms and then purify them by recrystallization and
identify them by their melting points.
Procedure:
Extraction of Carboxylic Acid
A pre–weighed (0.315g) mixture of Carboxylic acid, a phenol, and neutral substance was placed into
a reaction tube (tube 1). tert–Butyl methyl ether (2ml) was added to the tube and the solid mixture
was dissolved. Next, 1 ml of saturated NaHCO3 solution was added to the tube and the contents
were mixed separating the contents into three layers. Once this was completed ... Show more
Next, add concentrated HCl drop wise until the litmus paper indicates that it is acidic. During this
procedure, I added to much HCl because many CO2 gas bubbles evolved. This error in the
experiment will cause the recovery % to be lower.
Recovery of Phenol by Acidification
To tube 3 a piece of litmus paper was placed into the tube. Then as is tube 2 concentrated HCl was
added drop wise until the litmus paper indicated that it is acidic. No CO2 gas will evolve.
Purification of neutral substance
Tube 4 now should only have crude solid in the tube and it is then weighed. The tube is placed into a
50℃ water bath and then approximately 0.5 –1 ml of methanol is added, as well as H2O until the
solution gets cloudy, once the solution is dissolved it is cooled to room temperature and then iced.
The crystals are then collected using a Hirsh funnel. Next a small amount (~ 0.1g) of the crystals are
placed into a melting point tube and placed into the melting point machine to record the unknown
neutral substances melting point.
Purification of Carboxylic Acid and Phenol
To tube 2 and tube 3 a boiling chip is added. The two tubes are boiled to remove any residual ether.
Next, the tubes are cooled to room temperature and placed into an ice bath to allow for
crystallization. The solution is then removed from the solid in each tube and discarded. To tube 2
and 3 ~0.5 ml of H2O is added for recrystallization, the tubes

Recrystallization Of Naphthalene Lab Report
Experiment 4: Recrystallization of Pure Phthalic Acid, Benzoic Acid, and Naphthalene
Charles Nyberg
Gilbert Immanuel and Rachel Zigelsky
Mussie Gide Introduction Chemical experiments are often described, when they are finished, by the
yield or recovery of the target product or solid. Also melting point can be tested to determine purity
in a compound. Most times these chemicals are not completely pure so they can be recrystallized to
remove the impurities. The technique of recrystallization learned and utilized in this lab works best
to remove contaminates from nonvolatile organic solids.1 A solvent is used that dissolves the solid
at room temperature and then the impurities can be removed via pipette or other simplified removal
process. When all impurities are dissolved and removed, a lattice or an ordered structure in the form
of crystals is able to be collected and observed. Some substances cannot be recrystallized with only
one solvent. Therefore, two solvents, or solvent pairs, can help dissolve the substance at boiling
point only and not room temperature.2 Solvent pairs are able to be dissolved into each other but not
anything else effectively. One part of the pair is used to prevent dissolving at room temperature and
the other half is used to help dissolve at boiling point. Lastly, activated charcoal is utilized to
remove brightly colored impurities from the dissolved solid solutions. The charcoal attracts the
impure substances and then is able to be

How Does Temp Affect The Growth Rate Of Crystals?
A crystal is a solid whose atoms, or other components of matter, are structured in a uniform pattern
as seen in Figure 1. This repetitive pattern forms an extending crystal lattice that gives crystals their
unique geometric shape, known as the crystal structure. The crystal structure can be seen in Figure 2
below. The primary macroscopic characteristic of a crystal is its shape, typically consisting of flat
faces, also known as facets, meeting at sharp angles. As a crystal grows, the new atoms attach far
more easily to the rougher parts of the surface than the flatter, smoother parts of the surface. It is
through this that facets are formed, typically as large and smooth surfaces, with these planes thus
forming the entire crystal. ("How Do ... Show more content on Helpwriting.net ...
The larger the amount of substance dissolved in the solvent and the more pressure there is on that
substance, the larger crystals will be. ("How Does Temp Affect the Growth Rate of Crystals?"
2018).
Crystals grown in higher temperatures are not necessarily better than those grown in cooler
temperatures; different types of crystals are simply produced. Crystals grown in cooler solutions
force the minerals together, thus creating bonds and impurities in their structure that interrupt the
crystal pattern. This results in forming a large number of smaller crystals. In contrast, in warmer
temperatures, the distance between the atoms (or other components of matter) is larger. This thus
allows the crystals to form as larger, purer shapes. ("How Does Temp Affect the Growth Rate of
Crystals?" 2018).
In addition to this, the temperature of the solution also affects the rate at which the water evaporates.
When the water evaporates slowly from the solution, only a few crystals begin their growth – this
also means the crystals have more time to grow large before all the water has dissolved. In
comparison, when the water evaporates quickly from the solution, more crystals begin their growth
– however, these crystals do not have as much time to grow as large. ("How Does Temp Affect the
Growth Rate of Crystals?"

Smart Solution For Solid Waste Management Essay
SMART SOLUTION FOR SOLID WASTE MANAGEMENT ABSTRACT:: when we go arround
we notice one annoying thing that is overflowing of garbage bins in public places due to excess
disposal of waste which degrades the quality of environment in india.This even leads to bad smell
unhygienic conditions for people and also spoils the beauty of environement at tourist places
.Eventhough government take measures to overcome this problem but still it is continuing due to
irresponsibility of people and improper management of waste. keeping all these in mind we are
proposing an "iot based waste management system for smart cities".This proposed system includes
IR sensor which sends message to concerned person when the threshold limit is reached.Threshold
limit is nothing but the maximum limit of garbage the dustbin holds.Each dustbin is given a unique
ID.this system is implemented with low cost embedded system which helps to track the dustbin
using these unique ID 'S. One happy advantage is that the details can be accessed by authorities
from their place using internet and can make decisions or actions to clean the concerned
dustbin.When a message is reached dustbin gets locked and will not allow people to throw anymore
waste and immediate action is taken by authority by sending waste collector vehicle to concerned
place and later dustbin is unlocked KEYWORDS: IOT,IR sensor,RFID INTRODUCTION:: "IOT–
internet of things " is one of the most trending

Digi Melt
Using a Digi Melt to find the melting point, then determining an appropriate solvent, the identity of
an unknown solid was found, dissolved, gravity filtrated, and recrystallized to produce a pure solid.
By preforming a boiling point test with a thermometer, capillary tube, and water bath, the identity of
an unknown liquid was discovered
Procedure:
MACROSCALE UNKNOWN A
Obtain macroscale unknown and weight out 1 gram *Mass = 1.0518 grams
*Appearance: The compound was white, grainy, and appeared to have sand in it
To take a melting point:
1. A small amount of the unknown was placed on a piece of dry, new filter paper
2. The open end of a capillary tube was thrusted into the unknown to allow a small amount of solid
to be trapped in the tube ... Show more content on Helpwriting.net ...
A piece of filter paper was placed on the top of the funnel, placed on top of a filter flask with a
stopper, and clamped to a ring stand
2. A piece of heavy–walled tubing was then clamped to the side arm of the flask and connected to
the water trap and then connected to a vacuum source
3. The vacuum adapter was turned on and the filter paper was sucked down, drying out the crystals
4. To wash them, cold recrystallization solvent was used
5. To remove the crystals, a small amount of the cold recrystallization solvent and a spatula was
used to get the crystals out of the flask
6. The vacuum was left turned on for 5 min to make sure the crystals were dry
7. After the crystals were dried by filtration, the crystals were taken out of the funnel and set aside to
dry *During vacuum filtration, I was able to get most of the crystals out of the E. flask, but some
were left in there, which could result in a lower percent yield
When the crystals were completely dry, the were weighed by:
– Taking a vial and weighing it empty
– Putting the product in the vial and weighing total
– Subtract the vial weight from the total weight *Final weight of pure compound: 0.3894 grams
*Final melting point: 114.1 – 114.6 degrees

Iron carbon phase diagrams Essay
The Iron–Iron Carbide (Fe–Fe3C) Phase Diagram
In their simplest form, steels are alloys of Iron (Fe) and Carbon (C).
The Fe–C phase diagram is a fairly complex one, but we will only consider the steel part of the
diagram, up to around 7% C b d Carbon.
Phases present α–ferrite, γ–ferrite, δ–ferrite, Fe3C (iron carbide or cementite)
Fe–C liquid solution
School of Mechanical and Building Sciences, VIT University, Vellore
1
Phases in Fe–Fe3C Phase Diagram α–ferrite – solid solution of C in BCC Fe
Stable form of iron at room temperature.
The maximum solubility of C is 0.022 wt%
T
Transforms t FCC γ–austenite at 912 °C f to t it t γ–austenite – solid solution of C in FCC Fe
Fe3C (iron ... Show more content on Helpwriting.net ...
PM
L(0.53% C) + δ (
(
)
(BCC Ferrite of 0.1% C)
)
B

→ γ (FCC Austenite of 0.18% C)
(
)
N
The maximum solubility of carbon in BCC δ–iron is 0.1% (point M) whereas in FCC γ–iron, it is
greater. The presence of carbon influences the allotropic changes. As carbon is increased or added to
the iron, the temperature increases from 1400ºC to 1493ºC at
0.1%
0 1% carbon. b School of Mechanical and Building Sciences, VIT University, Vellore
7
Peritectic reaction – Fe–Fe3C System
Consider the portion NMPB in Peritectic Reaction
On cooling, the portion NM represents the beginning of the crystal structure change from BCC δ
iron to FCC γ iron for alloys
PM
B
containing less than 0 1% carbon
0.1% carbon.
Line MP represents the beginning of crystal structure change by
N
means of peritectic reaction for the alloys between 0.1 & 0.18%
Carbon.
Line NP represents the end of crystal structure change for alloys containing less than 0.18% C.
Portion PB represents the end of crystal structure by means of peritectic reaction for the alloys
between 0.18– 0.5% carbon. Here the reaction takes place isothermally (i.e.) at constant
temperature.

At the peritectic reaction point, liquid of 0.53% C combines with δ ferrite of 0.1% C to form FCC γ
austenite of 0.18% C
School of Mechanical and Building Sciences, VIT University, Vellore
8
4
Eutectic reaction –

Double Displacement Reaction Lab Report
Problem/Background Information:
Creating a solid from two liquid is an exceptional type of double displacement reaction that
produces a precipitate, it happen when anions and cations is merge together in an aqueous solution
to form an insoluble ionic solid called precipitate. Reactions are decide to be resolved, depends on
the solubility rules for solids in order to know the product and know how to write a net ionic
equation, because some aqueous reaction can not form precipitates. It permits scientist and the
people to foretell which ions are exist in a solution. The problem is choosing the right chemicals to
mix to produce a precipitate. Choosing the factors to make a precipitate can be different, because it
depend on the temperature ... Show more content on Helpwriting.net ...
The information cannot be unremembered. Recording data is very easy but helpful in arrange
everything.
Use a notebook, not a paper because it is very easy to lost.
Write it bullet points or full sentence to record data.
Draw when you need to.
Write the date every time you record or collect information.
Put a title on every time you write.
When you finished collecting and recording every data and information your from observation,
answer the question of the problem from your experiment.
For conclusion, write down you ideas and perspective about your experiment.
Safe Procedure:
1. You must wear a goggles or safety glasses when experimenting to protect your eyes from
chemical substances. Make a routine of wearing it before experimenting and keeping it on until you
have completed cleaning up everything.
2. In the laboratory, do not eat, drink, taste or smoke.
3. You should wear long sleeved shirt, gloves and leather topped shoes all the time when doing
experiments.
4. Your hair should be tied back to prevent hair falling into flames or

Copper Reaction Lab Report
AP Chemistry
12/13/11
Round–Trip Copper Reactions Lab
The purpose of this lab was to evaluate our skills of decanting a supernatant liquid without losing
the solid and successful completion of a series of reactions. This was done through five chemical
reactions involving copper. In this lab, elemental copper was put through five different chemical
reactions in order to convert it into different compounds. By the end of the fifth reaction, the copper
was back to its elemental state. In the first reaction, 0.95 g of pure copper was reacted with 4.0 mL
of concentrated nitric acid under the fume hood. The solution was swirled until all of the copper had
dissolved. The balanced equation for this reaction is as follows:
Cu (s) + ... Show more content on Helpwriting.net ...
The black precipitate was allowed to settle and then the supernatant, the clear liquid that lies above a
precipitate, was decanted, or poured carefully off. Then, 200 mL of hot distilled water was added
and the precipitate was allowed to settle to repeat the decanting process again. In the fourth reaction,
15 mL of 6.0 M sulfuric acid was added to the copper (II) oxide while stirring. The balanced
equation for this reaction is as follows:
CuO (s) + H2SO4 (aq) à CuSO4 (aq) + H2O (l) + SO2 (g)
When the sulfuric acid was added to copper (II) oxide, the solution turned blue. This was due to the
formation of aqueous copper (II) sulfate, which produced the copper ions to change the color of the
solution. In the fifth reaction, 2.29 g of zinc was added to the copper (II) sulfate solution under the
fume hood. The balanced equation for this reaction is written as followed:
Zn (s) + CuSO4 (aq) à ZnSO4 (aq) + Cu (s)
When the zinc was added to the copper (II) sulfate solution, the solution started to bubble. As the
solution was stirred, it turned a cloudy blue. Small flecks of a brown solid were visible. As the
solution became colorless, the brown solid settled to the bottom of the beaker. The solid formed was
copper in its elemental state. The color faded from the solution as the copper ions slowly formed
into solid copper. The copper was poured into a funnel with filter paper and washed three times with
25 mL

Nerolin Lab Report
Nerolin was prepared with a mixture of methanol, 2–naphthol, potassium hydroxide (KOH), and
iodoethane using a reflux apparatus. A reaction mixture was obtained, and allowed to freeze. Nerolin
crystals were obtained by way of gravity filtration and vacuum filtration in a successive manner. The
nerolin crystals were further analyzed to show purity of the experimental product by examining the
shape and size of the crystals, along with the melting point of the crystals. From this experiment,
6.44% of nerolin crystals were recovered, with a melting point measured as 33–35C. The
preparation of nerolin was performed due to its role in maintaining long–lasting fragrance release
properties in fabrics.
Nerolin, the product of this experiment, has ... Show more content on Helpwriting.net ...
The decanted liquid had to be placed back into the freezer in order to refreeze the crystals to allow
them to be decanted off. This could have caused a portion of the product to have been lost, thus
affecting the percent yield and purity of the final product. Furthermore, there was an oiled substance
thought to be a contaminant in the product after the addition of hot methanol and hot water to the
Erlenmeyer flask. This contaminant was left in the Erlenmeyer flask to see whether or not it had an
effect on the final product, but once the activated charcoal was added into the Erlenmeyer flask, the
contaminant seemed to have latched onto the charcoal. Since it latched onto the charcoal and the
charcoal was filtered out by way of gravity filtration, the contaminant is thought to have been
filtered out as well. While the contaminant seemed to have been filtered out by the gravity filtration,
there could be a small portion of the contaminant that remained in the filtered–out product, thus
affecting the percent yield and purity of the final product as

Geometric Isomers Post Lab Report
Marwah Alabbad
Analysis and Identification of two Geometric Isomers post Lab
9/2/15
1) As conducted in pre lab question, melting point for Maleic acid is 139– 140◦C and Furmaic acid
is 287◦C, and in the lab experiment the sample of A and B was identified first by test the melting
point for each sample, and the result came as test tube B melted faster than test tube A, which leads
to test tube A is Furamic acid and test tube B is Maleic acid, because of the low melting point for
Maleic acid, it melted faster than Furamic. Furamic acid requires more thermal energy in order to
break the hydrogen bonds in the molecular.
2) In the second experiment, the solubility of sample A and B was observed, and the result was that
test tube B has higher ... Show more content on Helpwriting.net ...
For example the results of melting point and solubility experiments if test tube B came as test tube B
has higher melting point and more solubility in water than test tube A, therefore, test tube B is a
Maleic acid and test tube A is Furamic acid. Based on the structure of each acid, maleic has a cis
configuration of carboxylic group in the same side which makes it polar and the hydrogen bonds are
weak which take less thermal energy to break them apart therefore more soluble in water. Whereas
in Furmaic (test tube A) structure has a Trans (opposite) carboxylic group which make it less soluble
and higher melting point than Maleic

Crystals : What Are They
Crystals: What Are They The study of crystals is called crystallography. A crystal is a solid that is
made up of the numerous atoms or molecules being arranged in a specific repeating pattern. This
ends up in the material having a specific shape and color, and having other characteristic. Crystals
may be big or little, but in a way they are the same shape. If you take a look at the display of crystals
in a lab. Salt and sugar are examples of crystals. Table salt has a cube–shaped structure. Snow has a
form of six–sided structure. Diamonds are an example of a crystal, they are made of pure carbon.
How are crystals grown?
Crystals grow in water or another liquid because it can only hold a certain amount of solute. When
the temperature of the solution is increased, hot water can dissolve more solid substances than cold
water. This can happen because warm water makes molecules move apart. This makes room for
more solid substance to dissolve. When a solid can't be dissolved anymore, the solution is said to be
saturated. As this solution cools, the water molecules tighten up again and there 's less room for the
solution to hold onto the dissolved solid.
Crystals begin to form and build when water starts to let go of extra solute. This process is called
recrystallization.
How do crystals form? If the compound is a solid when it is pure, there will be some force between
these solute molecules. Most of the time when these solute molecules meet. They will stay together
for a

Freezing Point Depression Lab
Freezing Pt Depression Lab
Purpose– To predict compare the actual and theoretical van't Hoff Factors for three compounds via
freezing point depression
Day 1
Procedure
Fill a beaker ¾ full with ice from the freezer and add ¼ to ½ inch of table salt. Stir until mixed and
check that the temperature is ~≤ –8 ºC.
Fill a test tube ½ full with tap water
Put the temp probe in the test tube and immerse the test tube in the ice–salt bath.
Stir the water in the test tube gently with a thermometer while keeping track of the temperature.
When the first ice crystals appear on the inside wall of the test tube, record the temperature. This
should be the freezing point of the liquid. (In this step water is the pure solvent).
Repeat as many times as necessary ... Show more content on Helpwriting.net ...
It is necessary to find the freezing point of pure water because our pure water does not necessarily
freeze at 0°C; it freezes around 0°C. Due to the temperature in our room and the pressure of the
atmosphere, our water will not freeze at exactly 0°C. Yes it would freeze near that point but we need
to be accurate in order to produce the best results for our lab. Even though pure water is said to boil
at 0°C, that is determined upon the pressure and temperature
Is the van't Hoff factor higher or lower than expected? Explain why this happens. In our lab the
calculated Van't Hoff Factor came out to be higher than the theory twice, and it came out lower than
the expected once. For a majority of this lab, the Van't Hoff Factor is higher than the expected
amount. The Van't Hoff factor shows the experimenter the average amount of ions that form in a
solution containing one solute. When the number that researchers calculate turns out to be lower
than the average Van't Hoff factor, researchers say that 0%–99% was dissociated. When the number
turns out to be equal to the Van't Hoff factor, 100% dissociates. When the Van't Hoff calculated is
higher than than the expected, there is most likely a high chance of error somewhere throughout the
lab. However, it is possible that the calculated Van't Hoff factor was higher than expected due to

Chem Lab Project 2 Essay
Investigating the Determining Characteristics of Cations and Anions
Chem 111 Sec 560
Introduction: The purpose of this lab was to study the specific characteristics of cations and anions,
and ultimately to be able to identify an unknown substance based on our studies and tests using the
logic trees developed through the experiment. A logic tree is a graphical display of the findings from
this lab which, through a series of yes/no questions, elimination tests, will help us to identify and
unknown sample. This lab will result in 2 logic trees, one for identifying cations and the other,
anions.
The lab was broken into three parts: Part 1 dealt with identifying a cation, Part 2 an ... Show more
The resulting color flashes were recorded. The cation solutions were then disposed of in the
appropriate waste containers.
From the information in each of these tests, a Cation logic tree was formed for easy identification of
unknown cations.
Part 2: Anion Analysis: The experiment was performed on 4 anions: chloride (Cl
), sulfate (SO42
),
nitrate (NO3
), and carbonate (CO32
). A sample of each anion was prepared by placing 10 drops of
each anion solution into a centrifuge tube. The tubes were labeled accordingly. The original color of
the solution was recorded. The first test performed was an Anion Elimination Test, the Silver Nitrate
Test. 0.1 M AgNO3 was added dropwise to each tube until a precipitate was formed or 5 drops of
0.1
M AgNO3 were administered. First, 2 drops of 0.1 M AgNO3 was added to the tube containing Cl
‐
and the results recorded, noting the appearance (texture), color, and shape. This process was
repeated for all the anions as follows: SO42: 20 drops, NO3
: 20 drops, CO32
: 1 drop of 0.1 M
AgNO3 was added.
For each solution which formed a precipitate (Cl
,CO32
) the solution was centrifuged for 5 minutes
and the supernatant was decanted; 5 drops of distilled water were added to both

Of The Melting Point Of Benzoic Acid And Erlenmeyer Fask
Introduction Solid substances that contain impurities may be purified in a series of steps. The
process known as recrystallization allowed for purification to occur. The solubility and melting point
of the substances are accounted for certain substances to be dissolved in certain solvents. The
melting point is a certain temperature at which a given substance, in this case a solid, is melted. The
melting point of benzoic acid and naphthalene in this experiment are needed to understand at which
temperature the substances are purified. The melting points were denoted in ranges that explained at
which points the substance began to melt, and were completely melted. In relation, the melting
points of certain substances noted, aided in the selection of which solvents were used. The solvents
used were selected in terms that the substance completely will dissolve; however, the impurity is
particularly insoluble to that same solvent, for the pure crystals to be collected. Benzoic acid was
dissolved in water, being the solvent, because the solubility is higher at higher temperatures. The
same goes for naphthalene being dissolved in ethanol. These concepts allowed the process of
recrystallization to occur if followed, resulting in the products of purified crystals.
Experimental
Into a 25 ml Erlenmeyer flask was placed 1.0 grams of benzoic acid into 13.32 ml of boiling water
at 250̊ C. Small portions of water were added until all the benzoic acid dissolved. After all the
benzoic

Micro-Reactions: Predicting the Products of Double...
Micro–Reactions: Predicting the Products of Double Replacement Reactions
Introduction – A double replacement reaction is a chemical reaction between two compounds where
the positive ion of one compound is exchanged with the positive ion of another compound. If you
have the reactants of two reaction solution that you can determine the products. All you need to do is
pair the positive parts of the compounds with the other compounds negative part. Once you find the
products you can determine their phase of matter by using Table H. You can also use Table F to
determine the solubility guidelines for aqueous solutions. If the product falls under soluble or
exceptions to insoluble it is in the aqueous stare. If the product falls under ... Show more content on
Helpwriting.net ...
If the product/reactant falls under insoluble or the soluble exception column it is a precipitate. You
can identify a precipitate in a reaction if the reactant or product is in solid form.
6. Balanced equation and state:
a. No Reaction
b. NaOH + CuSO4 ––> Na2SO4 + Cu(OH)2 (s)
c. 2Na3PO4 + 3CuSO4 ––> 3Na2SO4 + Cu3(PO4)2 (s)
d. NaCl + AgNO3 ––> NaNO3 + AgCl (s)
e. NaOH + AgNO3 ––> NaNO3 + AgOH (s)
f. Na3PO4 + AgNO3 ––> NaNO3 + Ag3PO4 (s)
Conclusion – The main idea of this experiment is that if you have the reactants of a double
replacement reaction that you should be able to find out several things. First, you can figure out the
products from switching the positive ion with the other positive ion. Once you have found the
products you can determine the phase of matter they are in from Table F and H. Finally, you can also
determine whether the reaction went to completion or not. The observations and data above leads
me to believe that our experiment is valid. We were very careful and observant of the directions that
needed to be taken. Once source of error could have been that certain substances were placed in the
wrong wells, causing your data to stray from the correct information. One way this experiment
could be applied to a real life situation is he residue in the bottom of a glass in your bathroom. There
is stuff dissolved in water, and any trace of water left in a glass will eventually evaporate and leave
the residue.

Recrystallization Of Pure Phthalic Acid
Experiment 4: Recrystallization of Pure Phthalic Acid, Benzoic Acid and Naphthalene
Introduction
Recrystallization is a technique frequently used in organic chemistry to purify solid organic
compounds. The goal of this technique is to allow organic compounds to form crystal lattice
structures, and to remove any of the impurities that do not align within this crystal structure [3]. The
theory behind recrystallization revolves around entropy; as heat will cause a organic compound to
dissolve (increase in entropy), a decrease in heat will then allow that organic compound to reform
(decrease in entropy) and become more pure [2].
The recrystallization technique utilizes the ability of a compound to dissolve within a hot solvent
and produce a ... Show more content on Helpwriting.net ...
The solution is then cooled and recrystallization of the solute occurs. For a solvent pair to work
correctly, the solvents used must be miscible with one another [1].
Although most organic compounds are not colored, it might be necessary to use activated charcoal
as a decolorization agent within the recrystallization process if a solution is colored [3].
Activated charcoal works well for decolorizing a solution because it is composed of carbon, and
thus is non–polar and will attract non–polar impurities [2]. Although charcoal will not work for all
solutions that are colored, as some solutes might be non–polar and thus product will be lost, it can
be a useful in many circumstances. Charcoal can be activated by heating it within an oven, thus
opening pores to add additional surface area for more impurities to be attracted to [3].
Experimental Section
Recrystallization of Phthalic Acid
Recrystallization of Benzoic Acid
Recrystallization of Naphthalene
Test tubes were filled with
200mg of solid phthalic acid, water was added dropwise then solution was heated in sand bath. As
acid boiled, water was dropwise until solid dissolved.
Test tube was corked.
After test tube reached 22*C, it was cool in ice bath and stirred using Pasteur pipette.
Air was expelled an Pipette tip was moved to the bottom of test tube.
Water was withdrawn with pipette from bottom of the tube. Test tube was tapped on wood surface to
compress

Describe How Several Ways Mineral Crystals Can Form
1. What is a mineral, as geologists understand the term? How is this definition different from the
everyday usage of the word?
To a geologist, a mineral is naturally occurring solid, formed by geologic processes that have a
crystalline structure and a definable chemical composition. Its internal structure characterized by an
orderly arrangement of atoms, ions, or molecules in a crystalline lattice. Almost all minerals are
inorganic, for instance, sugar is an organic chemical in which is made by carbon, hydrogen, and
oxygen that it is not a mineral.
2. Why is glass not a mineral?
A glass isn't a mineral because crystalline structure in which the atoms make up a mineral aren't
distributed randomly and cannot move around easily, and the atoms in a glass aren't arranged in an
orderly pattern. Moreover, both minerals and glasses are solids, and they can retain their shape, but a
mineral is crystalline ,and glass is not because the atoms, ions, or molecules are the elements in a
mineral are ordered into a crystal lattice, and glass doesn't has those elements and disorderly
arrange.
3. Describe the several ways that mineral crystals can form.
The mineral crystals can form in five ways. First, they can form by the solidification of a melt which
the freezing of a liquid, such as ice crystals are a type of mineral made by freezing water. Second,
they ... Show more content on Helpwriting.net ...
The formation of LIPs associated with superplumes might have affected sea level, climate, and may
have caused the extinction of some species. Because when volcanic eruptions begin, the huge
quantities of basaltic lava spew out of the ground, the hot basaltic lava has low viscosity that can
erupts the localities, and it can flow tens to hundreds of kilometers across the

Multi Step Synthesis Of Aspirin
This experiment explored the multi–step synthesis of synthesis acetylsalicylic acid commonly
known as aspirin. A common procedure used in organic chemistry labs is multi step synthesis. this
procedure involves a cascade of reactions that ultimately lead to a desire product. The series of
reactions begins with commonly found or readily available reagents which are synthesized into
products which can be used for a secondary reaction, and so on and so forth until the final desired
product is synthesized (Ryerson Department of Chemistry, 2017).
In this experiment the synthesis of aspirin was used to find the importance of the efficiency of
multistep synthesis as well as a means to explore a number or purity tests. The purity tests used in
this experiment were a melting point range test, a thin layer chromatography test, a ferric chloride
test, and infrared spectroscopy.
After the completion of the experiment it was found that the percent yield was 24% which is very
low. This shows the importance of efficiency as the process of making aspirin, is a two step
synthesis (Atienza et al., 2010) and 74% of theoretical product was converted into waste or lost.
Also it was found that the end product was not pure aspirin as the sample of aspirin made failed two
purity test. This revealed that not merely one purity test is not sufficient to denote a substance as
pure.
Introduction
The first step of the synthesis is to take pure methyl salicylate and hydrolyzed into salicylic acid,

Crystals Growth Lab
Introduction
The crystal growth lab was done to display how crystals expand and grow through a changing
environment. A solution is a homogeneous mixture of two or more components, one liquid, and one
dissolved solid or solids. The amount of the solid introduced in the solution is known as the
concentration. Solubility is the property of a concentration to dissolve into a solution, describing
how well it dissolves in different temperatures and masses. Depending on the solubility a solution
can be saturated, undersaturated, or supersaturated.
When a solution has the maximum amount of solid concentration based on the temperature, crystals
will not form or dissolve into the solution being equally saturated. In an undersaturated solution,
there ... Show more content on Helpwriting.net ...
The solution became lighter in color as the seed crystalized, attaching to the bottom of the beaker.
The seed crystal was gradient in color started as dark blue and turning lighter blue as it grew upward
seen in Figure 2. Conclusion
In conclusion, from this experiment, we can say our hypothesis was not valid. We reject the
hypothesis because in the recording of immediate results, our solution reached temperatures much
less than 70° Celsius and yet no crystallization began. The solution showed no signs of crystal
growth at the predicted temperature, however, did ultimately crystalize under a much lower
temperature. Our results may have been impacted by the faulty thermometer and the insulation using
the Styrofoam cup. As our cooling rate is slightly slower, our seed crystal formed more individual
crystals, similar in size, rather than hundreds of very fine crystals. Our crystal grew from the seed
crystal and also from the bottom of the beaker. The seed crystal allowed the Copper Sulfate ions to
attach to it, growing outward as the solution stabilized. Crystals also formed on the bottom of the
beaker as the concentration was not fully dissolved, settling to the bottom of the beaker. To improve
the experiment, we could have used a better thermometer to record the appropriate data. We also
should have heated the solution further to ensure the concentration of Copper Sulfate was

Experimental Techniques For The Synthesis And...
CHAPTER 2
EXPERIMENTAL TECHNIQUES FOR THE SYNTHESIS AND CHARACTERIZATION OF
NANOMATERIALS.
In order to explore novel physical properties and phenomena and realize potential applications of
nanostructures and nanomaterials, the ability to fabricate and process nanomaterials and
nanostructures is the first corner stone in nanotechnology. There exist a number of methods to
synthesize the nanomaterials, which are categorized in two techniques "top down and bottom up".
Solid state route, ball milling comes in the category of top down approach, while wet chemical
routes like sol–gel, co–precipitation, etc. come in the category of bottom up approach. Secondly,
characterization of nanomaterials is necessary to analyze their various properties. Therefore, this
chapter describes the various methods of synthesis and characterization of nanomaterials.
Characterization techniques include XRD, SEM, TEM, EDAX, UV–Visible spectroscopy, FTIR
spectroscopy, etc.
2.2. Synthesis of Nanomaterials:
Fabrication of nanomaterials with strict control over size, shape, and crystalline structure has
become very important for the applications of nanotechnology in numerous fields including
catalysis, medicine, and electronics. Synthesis methods for nanoparticles are typically grouped into
two categories: "top–down" and "bottom–up" approach. The first involves the division of a massive
solid into smaller and smaller portions, successively reaching to nanometer size. This approach may
involve milling or

Recrystallization Lab
Purpose:
The purpose of recrystallization lab is to recrystallize cream of tartar from water.
Procedure:
1. Obtain and weigh a sample of potassium hydrogen tartrate; use between 200 and 300 mg.
2. In a boiling water bath, heat a sample of distilled water.
3. Dissolve your sample in a minimum of water. To do this, add the water drop–wise to your solid,
stirring well between each drop, until the solid just dissolves. You might have to heat the test tube
while you are adding the water to keep the solution hot.
4. After getting the solid to dissolve, allow the solution to cool slowly and look for crystals. After
the solution has cooled to close to room temperature, cool it in an ice bath. Collect the solid using
vacuum filtration and your Hirsh ... Show more content on Helpwriting.net ...
Weigh the solid and after drying.
Data:
Cream of tartar = 0.218g
Empty plastic dish = 2.161g
Cream of tartar recovered with plastic dish = 2.212g
Calculations:
%recovery= amount recovered / initial amount x 100%
2.212g – 2.161g = 0.051g
0.051/ 0.218 = 0.2339
0.2339 x 100 = 23.4%
Discussion:
My data is bad because in the calculations, it showed that it is 23.4 percent. However, the test tube
did show little bit crystals while using the spatula to scratch the tube.
Error Analysis:
There are few errors in the lab. The grams of cream of tartar might become less because it is hard to
scrape the crystals out of the test tube so it will be stuck inside the test tube and some of them is still
in the plastic boat dish. During the experiment, the water may have added too much that causing
precipitating out. Also, it is may not heat enough the test tube while adding hot water.
Modification:

We would spend more time scraping the test tube to create crystals and be more careful with adding
water so it won't precipitating out.
Conclusion:
We recrystallized cream of tartar from water with 23.4 percent recovery.
Answers to the post laboratory questions:
1) Define the following terms:
a) saturated solution – is a solution that can't add more solute into

The Product Being Used For An Organic Compound
Introduction When it comes to creating or performing a reaction, the solubility of the product being
used is very important. Knowing the biological and physical processes of your product is the start of
any experiment. Obtaining this information plays a key role in choosing solvents for processes such
a recrystallization. Recrystallization is a procedure for purifying an impure compound in a solvent.
It is used only to purify solids in large or small quantities. That is the purpose of this experiment.
The product being used for this experiment is acetanilide (C6H5NH). The goal is to purify and
recrystallize the organic compound in the appropriate solvent. Acetanilide was first introduced to the
world in 1886 as a pharmaceutical fever–reducing drug. When proven to be ineffective, it was
chosen as an alternative pain reliever since it is converted to acetaminophen when inside the body.
In the first part of this experiment we will learn to crystalize an organic compound then later on
recrystallize it with the use of an appropriate solvent. The molecules of this compound are able to
return to its previous solid form due to the solvents inability to hold on to its molecules and they
start to move freely out in the solution and begin to form solid crystals. The reaction of the product
being studied is shown below. The product of this reaction can be analyzed using NMR and IR
spectrums.
Experimental Acetanilide was synthesized from a starting material of aniline (2mL)

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