Recrystallization of Benzoic Acid Outline Essay Sample

9 September 2017

Purification – physical separation of drosss and byproducts from an impure sample pure substance. *Distillation- for liquid compounds.
*Recrystallization- solid compounds

– precipitation of a solid cpd. from a concentrated solution in the signifier of crystals through chilling or vaporization -7 stairss:
1 ) choosing the appropriate dissolver
2 ) fade outing the solute
3 ) bleaching the solution
4 ) removing suspended solids
5 ) crystallising the solute
6 ) collection and rinsing the crystals
7 ) drying the merchandise
– 2 Kinds of Impurities: 1 ) more soluble
2 ) less soluble than the chief constituent
– impure solid foremost dissolved in an appropriate dissolver when heated – appropriate/ideal solvent = dissolves constituent to be purified when hot ( boiling point ) . = does NOT fade out constituent to be purified when cold= room temp. or 0oC = either dissolves drosss when cold OR does NOT fade out drosss when hot – constituents will so crystallise in its purer signifier as the solution is cooled. Less soluble constituent will crystallise first

As the crystals are formed. the right molecules that will suit in the crystal lattice are selected while the incorrect molecules are ignored. ensuing in a pure solid – Depends on: 1 ) solubility of the compound

2 ) differences in solubility of the desired solutes & A ; drosss dissolved in the dissolver – Common Solvents Used:
Solvent| Formula| Polarity| Boiling Pt. ( oC ) |
Water| H2O| Very Polar| 100|
Ethanol| CH3CH2OH| Polar| 78|
Methanol| CH3OH| Polar| 65|
dichloromethane| CH2Cl2| Slightly Polar| 40|
Diethyl ether| ( CH3CH2 ) 2O| Slightly Polar| 35|

IV. Experimental
0. 100 g of powdery impure benzoic acid. 1 little piece of boiling bit. and 2 milliliter of distilled H2O were added severally into a vial. While twirling the mixture invariably. it was so heated to boiling until the benzoic acid had dissolved wholly in the H2O.

The solution was so cooled by adding a few beads of distilled H2O. A pinch of activated C was added after. Then. the mixture was heated to boiling for a 2nd clip for a few proceedingss until the volume of the mixture had decreased by about 1 milliliters.

The hot mixture was so instantly poured into a 5mL disposable syringe ( without the acerate leaf ) with a cotton stopper indoors at the underside of the syringe tubing to move as a filter. The Piston of the syringe was so pressed to coerce the liquid through the filter. roll uping the filtrate into another phial.

The hot concentrated solution or filtrate was so cooled to room temperature before sealing the vial and slow-cooling it in an ice bath. The crystals yielded were collected in a pre-weighed filter paper on a 50 milliliter beaker ( as an jury-rigged filter paper support ) . Small increases of ice-cold H2O were added to rinse the phial to acquire the staying crystals.

The filter paper incorporating the gathered crystals was so carefully taken out from the filter system. so was folded with the crystals inside. and so pressed in between folded tissue paper to squash out extra H2O. After leting the filter paper and crystals to dry up wholly. the filter paper with the crystals was weighed. The weight of the filter paper entirely was subtracted from the obtained weight to acquire the weight of the yielded crystals. Percent recovery was computed for by spliting the weight of the yielded crystals by the weight of the impure benzoic acid.

V. Results
From the 0. 100 g of the impure sample of benzoic acid. 0. 030 g of pure benzoic crystals was yielded. holding a percent recovery of 30 % .

VI. Discussion
The chemical equation for the reaction of benzoic acid with H2O is as follows: C6H5COOH ( aq ) + H2O ( cubic decimeter ) C6H5COO- ( aq ) + H3O+ ( aq ) .
It is a reversible reaction allows the dissolved benzoic acid to recrystallize or travel back to its solid signifier as a purer compound. Water= appropriate or ideal dissolver for benzoic acid due to immense solubility difference in H2O. * indissoluble @ low temp 2. 1g of benzoic acid dissolves in 1L of H2O at 10oC * soluble @ high temp 68g of benzoic acid dissolves in 1L of H2O at 95oC The mixture invariably swirled as the mixture was being heated * additions rate of disintegration and information of the system * increasing the interaction between H2O and benzoic acid molecules. *The complete disintegration of benzoic acid led to a clear solution. Boiling chip- induces boiling

– added to the mixture while @ ROOM temp– NOT boiling temp– to forestall spilling or boiling over Activated charcoal- used for decolorization
– made up of finely separated C atoms w/ great surface country – adsorb the drosss from sol’n ( aka it causes drosss to attach & amp ; be trapped onto the wood coal atom’s surface ) . – C atoms excessively BIG to go through thru filter paper minimizes drosss! *NOTE: Excessively much activated C loss of the pure substance. Filter syringe- used to increase pureness to filtrate out activated wood coal & A ; other drosss on the cotton stopper Sol’n was filtered quickly to hold every bit small residue of benzoic acid crystals as possible yielded from crystallisation already taking topographic point as the temp. & A ; solubility of the benzoic acid was diminishing. Slow chilling in an ice bath- to assist crystals set up in a finer undistorted molecular geometry. excepting drosss from the crystal formation. In instance crystallisation doesn’t take topographic point while chilling usage SEED CRYSTAL * Its similar construction to the pure crystals will function as “source code” merely molecules that will suit absolutely in the lattice will crystallise. go forthing out drosss with a different construction. Complete drying of the filter paper and crystals minimize inaccuracy in consequences because the added mass of H2O while weighing affects or increases the % recovery. Formula for per centum recovery was used:

% recovery= weight recovered benzoic acid x 100 weight impure benzoic acid

VII. Answers to Questions
1. List the belongingss of an ideal dissolver to execute the purification of an organic compound by the crystallisation technique. * fade out the solute @ high temperatures or boiling platinum
* does NOT fade out solute @ depression or room temp.
* either dissolves drosss ( @ room temp ) or NOT fade out them at all ( @ high temp ) so that drosss WON’T crystallize with solute & A ; can be therefore removed by filtration. * Inert= should NOT respond with solute or there will be a lesser output of pure crystals due to molecular interaction * volatile so it can be easy removed from the solute crystals.

2. In sublimating by crystallisation from a dissolver. is it advisable to chill the solution easy or quickly? Explain. COOL SLOWLY so that crystals will organize in its most right & A ; undistorted molecular geometry. go forthing drosss out from the crystal formation. *Rapid chilling will ensue a greater output in the terminal ( because drosss will hold been included ) . therefore taking to inaccurate consequences.

3. What advantages does H2O hold as a crystallisation dissolver? * non-flammable
* safe and non-toxic
* used for a broad scope of temperatures
* high heat capacity and can fade out different substances * polar. therefore. it can fade out substances. particularly polar 1s. easy.

4. Solubility of benzoic acid in H2O: 0. 21g/ 100mL H2O at 10oC ; 0. 27g/
100mL at 18oC ; 2. 75g/ 100mL at 80oC ; 6. 80g/ 100mL at 95oC. Two pupils crystallized 10 g samples of benzoic acid from H2O. the first dissolution benzoic acid at 80oC and filtrating at 10oC. the 2nd fade outing the acid at 95oC and filtrating at 18oC. Calculate the measure of H2O each pupil was required to utilize and the maximal recovery of benzoic acid possible in each instance. STUDENT A

@ 80oC 10 g =2. 75 g s
ten milliliter 100mL
x= 363. 636 milliliter H2O
@ 10oC y gms = 0. 21 g vitamin D
363. 636 milliliter 100mL
y= 0. 7636g benzoic acid
9. 24g benzoic acid
9. 24g benzoic acid

10 g- 0. 7636g=

Student B
@ 95oC 10 g = 6. 80 g vitamin D
ten milliliter 100mL
x= 147. 058 milliliter H2O
@ 18oC y gms = 0. 27 g vitamin D
147. 058mL 100mL
9. 60g benzoic acid acid
9. 60g benzoic acid acid
y= 0. 397 g benzoic acid

10 g- 0. 397g=

5. A Solid ( X ) is soluble in H2O to the extent of 1g per 100g of H2O at room temperature and 10g per 100g of H2O at the boiling point. a. How would you sublimate X from a mixture of 10g of Ten with 0. 1g of dross Y. which is wholly indissoluble in H2O. and 1g of dross Z. holding the same solubility features in H2O as X? Separating Yttrium

1. Dissolve the mixture in 100g H2O and furuncle to 100?C. Everything should fade out except for Y. 2. Cool mixture to room temperature
3. Filter. This will divide dross Y from X and dross Z. Separating Z
1. Add H2O up to 100mL to the filtrate.
2. Heat to boiling ( 100oC )
3. Slowly cool to room temp
4. Filter the mixture. The filter paper residue= pure X crystals. *Formation of Z is undistinguished because it is merely 1g and has a different crystal construction than X. excepting Z from X’s crystallisation.

90 % or 9g of Ten will be obtained
90 % or 9g of Ten will be obtained
B. How much pure Ten could be obtained after one recrystallization from H2O? 10g of X – 1g X dissolved ten 100 % =
10 g of X *Solubilities of different solutes are independent of each other. therefore. 1 g of solutes X and Z would fade out at room temperature.

c. How much pure Ten could be obtained after one recrystallization from a mixture of 10g of Ten with 9g of Z? * A = 1000g H2O to fade out 10g X @ room temp.
A = 1000g H2O to fade out 10g X @ room temp.
Ten and Z must be dissolved foremost. Solubilities of X and Z are independent of each other. Solid X. being the larger sum. is used as footing. 10g of X = 1g of Ten
A g H2O 100g H2O
* B= 900g H2O to fade out 9g Z @ room temp.
B= 900g H2O to fade out 9g Z @ room temp.
Water must so be boiled until 900g of H2O is left to guarantee dross Z won’t dissolve and crystallise with X under room temperature. 9g of Z = 1g of Omega
B g H2O 100g H2O
* C= 9g of X dissolved in
900g H2O @ room temp.

C= 9g of X dissolved in
900g H2O @ room temp.

Slowly cool the solution to organize X crystals.
C g of X = 1g of Ten
900g H2O 100g H2O

1 g of X crystallized
1 g of X crystallized

10g X- 9g Ten dissolved=

d. Based on the consequence obtained. what is suggested about the usage of crystallisation as a purification technique? Using crystallisation to divide and sublimate solid substances is non so efficient because of its heavy dependance on the substance’s solubility. doing it hard to divide substances of the same solubility and of important sums. VIII. Decision

Recrystallization. although a common purification technique for its simpleness and utility. is non such an efficient method because it CAN’T: * sublimate non-solids
* separate compounds whose constituents ( the chief constituent and dross ) have the same solubility due to recrystallization’s heavy dependance on solubility. particularly when both constituents exist in important sums * wholly sublimate the end-product

IX. Mentions

“Crystallization. ” University of Colorado. University of Colorado. n. d. Web. 4 Feb. 2013 & lt ; hypertext transfer protocol: //orgchem. Colorado. edu/CCCE/frame/images/handbook. pdf & gt ; . “Recrystallization. ” Hartnell College. Hartnell College. n. d. Web. 17 Dec.
2012. & lt ; hypertext transfer protocol: //www. hartnell. edu/faculty/shovde/chem12a/lablecturehandouts/recrystallizationlecture. htm & gt ; . “Recrystallization: a purification technique for solids. ” German Jordanian University. German Jordanian University. n. d. Web. 17 Dec. 2012. & lt ; World Wide Web. gju. edu. jo/admin/s32files/3-RECR. physician & gt ; .

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