THE DETERMINATION OF PHYSICAL ORGANIC COMPOUND

ABSTRACT

A experiment has been done to determine the melting and boiling point of a substance .

On experiment to determine the big of boiling point a substance, while the determine of melting

point to identity a crystal compound and to get a purity of the compound.

A melting point apparatus was used to determine the melting point of which is the

sample X show the range of the substance start to it begin to melt until it is all melted is 121-122

C . A benzoic acid was suspected as sample X , because the melting point of them was same,the

benzoic acid have the range melting point is 121-122,4 C. The difference can cause because of the

determination technics.

The conclusion is a sample X has a big vanance was suspected as A compound have

been known, so in experiment should same, but in there is any different it meants the both

compound is not same , and if the different only a few degrees ( less than 1 C ) it meants the both

substance is same.

The resisitance in experiment is uncomplete of the appliance, so that difficult to get the

maximal result. The suggestion, laboratory worker to completed the material and appliance to get

maximal result.

APPLIANCE AND MATERIAL

A. Appliance

Melting Point Apparatus ( MPA )

Capiler (one the tip of it is close )

B. Material

Sample X ( representing a crystal )

PROCEDURE

The sample X should be in powder,than insert into capiler which the tip of it is

close.Than insert it into melting point apparatus and the temperature in celcius is noted, when it

begin to melt until it is all melt.

INTRODUCTION

Condensation [is] homogeneous mixture [among/between] two Iihat vitamin of ayau more.

Homogeneous burden that components in condensation cannot dibedalan one with is other. Ability

each;every dissolve Iihat vitamin type to [be] dissolve in a[n pelarut different each other. Solubility

a[n Iihat vitamin [is] also influenced by pressure. solid Solubility Iihat vitamin and hydrogen [do]

not bayak affect by pressure, and gas solubility will increase ever greaterly [his/its] [of] pressure.

This matter [of] ditunjukana by solid Iihat vitamin solubility will increase gone up it[him] gas

solubility and temperature will decrease gone up it[him] temperature.

Condensation a[n Iihat vitamin in pelarut [is] maximum concentration of the Iihat vitamin able to

be dissolve in its its[his]. Compound with incessant tonic bona generally better in pelarut of polar,

while Iihat vitamin which [do] not dissolve easier polar in pelarut [do] not polar. Ion compound

generally will be incessant better if/when its reaction [of] him of eksoterm. Function bunch of polar

will predominate the nature of condensation a[n small molecule, but for big molecule [of] bunch

polarity becoming not mean to be compared to part of other molecule which the non polar. Etanol

much more dissolve in water compared to heksanol a[n compound reacting with certain pereaksi

can change the nature of its condensation. Its it[him] insoluble amine alkyl in water will react with

acid solution of yielding salt of alkil dissolve ammonium in water.

Alcohol condensation in water caused by both earning each other hidorgen. However by increasing

long [him/ it] enchain its its[his], condensation in water also on the wane. Part of hydrocarbon a[n

alcohol have the character of hidrofob namely have the character of to refuse water molecules.

Long Mkin part of hydrocarbon will more and more to growing low [is] condensation of alcohol in

water. condensation of Alcohol also go up denagn increase of bunchs amount OH him.

Saturated condensation [is] a[n situation when a[n condensation have contained a[n Iihat vitamin

with maximum concentration. maximum Concentration value able to reach by a[n such this Iihat

vitamin with condensation and given [by] S lambing( Solubility). Condensation which enter can

dissolve referred [as] [by] dissolve Iihat vitamin [of] condensation less saturated. Condensation

which cannot again dissolve dissolve Iihat vitamin [is] so that formed [by] sediment referred [as]

[by] condensation pass saturatedly.

Result of condensation times;rill a[n Iihat vitamin ( Ksp) [is] result of ions concentration times;rill

in saturated condensation which [is] have rank [to] counted coefficient according to equation of

ionization. Price of Ksp [is] remain to [at] temperature which remain to, if/when temperature

boosted up [by] price of Ksp more and more big because condensation more and more big [at] high

temperature. Of price of Ksp and multiplication of ions can know [by] what is a[n condensation

have is saturated, not yet saturated or pass saturated pursuant to following :

a. If/When price of Ksp bigger than result of its ions concentration times;rill hence latutan not yet is

saturated ( [do] not happened sediment )

b. If/When price of Ksp smaller than result of its ions concentration times;rill hence condensation

pass saturatedly is ( happened [of] precipitation)

c. If/When price of Ksp [is] equal to result of its ion concentration times;rill hence condensation

earn saturatedly ( will tuang)

APPLIANCE AND MATERIALS.

A. Appliance - appliance.

Tube reaction of.,Drip pipette.,Beaker glass.

B. Materialss.

Water Refine.,Ether,Naoh 5%,Nahco3 5%,HCL 5%,Condensed H2SO4,H3Po4 95%,Paper

Litmus,Compound X.

C. Working procedure

This attempt [is] [done/conducted] to know faction from a compound X. First of all compound of X

enhanced [by] water, perceive what is compound of x the dissolve insoluble or. If compound of X

the dissolve in water hence taking again compound of X new and later;then enhance with ether,

dissolve otherwise hence compound of X that [is] including faction of S2 but if dissolve hence

compound of X that [is] tested with litmus paper, red litmus paper designate faction of SA, blue

litmus [of] faction of SB, unadjusted litmus [of] faction of S1. If compound of X the insoluble if

enhanced [by] water hence taking compound of X new and enhance Naoh 5%, if dissolve hence

enhancing with Nahco3 5%., dissolve result designate faction of A1 and do not be dissolve

designate faction of A2. Where at the (time) of addition of NaOH5% if the compound do not be

dissolve hence taking compound of X new and HCL 5%, dissolve result designate faction of B,

if/when do not be dissolve, compound of X which is just is enhanced [by] [of] H2So4 96%. Result

do not be dissolve designate faction of I while dissolve result hence compound of X which is just is

enhanced [by] [of] H3Po4 85 %, if dissolve designate faction of N1 and do not be dissolve

designate faction of N2.

RESULT OF PERCEPTION AND SOLUTION.

X1 ( white serbuk)

If X1 sample enhanced with water will be dissolve. While X1 sample added [by] insoluble ether

and there are white sediment. Hence condensation class of sample X1 [is] Glucose representing

faction condensation class of S2.

X2 ( white serbuk)

If X1 sample enhanced with insoluble water. While X2 sample if/when enhanced [by] Naoh 5%

will be dissolve, [is] later;then enhanced [by] Nahco3 5 is% dissolve. Hence condensation class of

sample X2 [is] Acid of benzoate representing condensation class of A1 X3 ( transparent

condensation)

If X3 sample enhanced with water will be dissolve. While X3 sample if/when enhanced with ether

will be dissolve, [is] later;then tested with litmus paper [there] no change. Hence condensation class

of sample X3 [is] Tertiary [of] Butanol representing faction condensation class of S1 X4 ( brown

condensation [of] kehitaman)

If X4 sample enhanced with insoluble water, While X4 sample enhanced with Naoh 5% nor is

dissolve, [is] later;then enhanced [by] HCL 5 is% dissolve. Hence condensation class of sample X4

[is] aniline representing faction condensation class B.

SOLUTION.

Sampel X1 which in form of white chromatic serbuk [is] Glucose ( C6H12O6). Dissolve Glucose

in water representing pelarut of polar but insoluble Glucose in ether and form white sediment.

Become glucose condensation class [is] S2. Which included in condensation class of S2 among

others [is] salt of acid of organic ( RCO2NA, RCO3NA), Amine of Hidroklorida ( RNH3CL), Sour

[of] Amino.

Sampel X2 which in form of white chromatic serbuk [is] acid of benzoate ( C6H5Cooh). Sour [of]

insoluble benzoate in water. This matter as according to literature that Acid of Benzoat have small

condensation in water and to heighten its condensation hence turned into sour form [of] its salt

form. In consequence when acid of benzoate enhanced [by] Naoh 5% causing dissolve benzoate

asan and form Natrium Benzoat ( its salt). acid of Benzoate dissolve also in Nahco3 5%. Becoming

class of[is acid solution of benzoate [is] A1.

Sampel X3 which in the form of transparent condensation [is] tert-butanol. Dissolve Tert-butanol in

water. This matter as according to literature that dissolve tert-butanol in water because less its

it[him] bunch of tert-butanol. Dissolve Tert-butanol also in ether and tested with litmus paper got

[by] result of that unadjusted litmus paper. This matter prove that tert-butanol [is] acid weak. This

matter as according to literature that acidity of alcohol progressively [is] downhill the than

sekunder primer,alcohol alcohol and of alcohol feeblest tertiary [of] its acidity. Condensation class

of tert-butanol [is] S1 Sampel X4 which in the form of tan condensation [of] kehitaman [is]

insoluble aniline.Anilin in water. dissolve nor Aniline in Naoh 5% But aniline earn dissolve in

HCL 5%. Becoming Aniline condensation class [is] B.

CONCLUSION.

Dissolve glucose in water but insoluble in ether. its Condensation class [is] S2.Acid of Benzote

dissolve tidal in water but dissolve in Naoh 5% and Nahco3 5%. its Condensation class [is] A1.

Dissolve Tert-butanol in ether and water. Tert-Butanol [is] acid weak. its Condensation class [is]

S1.Insoluble Aniline in and water of Naoh 5%. But dissolve in HCL 5%. its Condensation class [is]

B.