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Subtopics of Class 11 Chemistry Chapter 12 – Organic Chemistry – Some Basic Principles and Techniques
- General Introduction
- Tetravalence Of Carbon: Shapes Of Organic Compounds
- Structural Representations Of Organic Compounds
- Complete, Condensed And Bond-line Structural Formulas
- Three-dimensional Representation Of Organic Molecules
- Classification Of Organic Compounds
- Nomenclature Of Organic Compounds
- The IUPAC System Of Nomenclature
- Iupac Nomenclature Of Alkanes
- Nomenclature Of Organic Compounds Having Functional Group(S)
- Nomenclature Of Substituted Benzene Compounds
- Structural Isomerism
- Fundamental Concepts In Organic Reaction Mechanism
- Fission Of A Covalent Bond
- Nucleophiles And Electrophiles
- Electron Movement In Organic Reactions
- Electron Displacement Effects In Covalent Bonds
- Inductive Effect
- Resonance Structure
- Resonance Effect
- Electromeric Effect (E Effect)
- Types Of Organic Reactions And Mechanisms
- Methods Of Purification Of Organic Compounds
- Differential Extraction
- Qualitative Analysis Of Organic Compounds
- Detection Of Carbon And Hydrogen
- Detection Of Other Elements
- Quantitative Analysis
- Carbon And Hydrogen
Organic Chemistry Some Basic Principles and Techniques Chemistry Chapter 12
• Organic Chemistry
Organic chemistry is the branch of chemistry that deals with the study of hydrocarbons and their derivatives.
The Shapes of Carbon Compounds:
In organic or carbon compounds, s and p orbitals are involved in hybridisation. This leads to y three types of hybridisation which are sp3(in alkanes) – Tetrahedral in shape sp2(in alkenes) – Planar structure sp(in alkynes) – Linear molecule
Functional Group: The functional group are atom or group of atoms joined in a specific manner which determines the chemical properties of the organic compound. The examples are hydroxyl group (—OH), aldehyde group (—CHO) and carboxylic acid group (—COOH) etc.
• Homologous Series
A homologous series may be defined as a family of organic compounds having the same functional group, similar chemical properties and the successive members differ from each other in molecular formula by —CH2 units.
The members of a homologous series can be represented by same general molecular formula.
• Nomenclature of Organic Compounds
Common name (Common system): Before the IUPAC system of nomenclature, organic compounds were named after the sources of origin, for example, urea was so named because it was obtained from the urine of mammals. Formic acid was so named since it was extracted from red ants called formica.
• I UP AC (International Union of Pure and Applied Chemistry) System
According to IUPAC system, the name of an organic compound contains three parts: (i) word root, (ii) suffix, (iii) prefix.
(i) Word root: Word root represents the number of carbon atoms present in the principal chain, which is the longest possible chain of carbon atoms.
(ii) Suffix: Suffix are of two types, primary suffix, secondary suffix.
(a) Primary Suffix: It indicates the type of bond in the carbon atoms.
(b) Secondary Suffix: Secondary suffix is used to represent the functional group.
(iii) Prefix: Prefix is a part of IUPAC name which appears before the word root. Prefix
are of two types:
(a) Primary prefix: For example, primary prefix cyclo is used to differentiate cyclic compounds.
(b) Secondary prefix: Some functional groups are considered as substituents and denoted by secondary prefixes.
Substituted Group Secondary prefix.
— F Flupro
— Cl Chloro
— Br Bromo
— NO Nitroso
— NO2 Nitro
— CH3 Methyl
— OCH3 Methoxy
Naming of Compounds Containing Functional Groups: The longest chain of carbon atoms containing the functional group is numbered in such a manner that the functional group is attached at the carbon atoms possessing lowest possible number in the chain.
In case of polyfunctional compounds, one of the functional group is selected as principal functional group and the compound is named on that basis. The choice of principal functional group is made on the basis of order of preference.
The order of decreasing priority for the functional group is
When there are two or more compounds possessing the same molecular formula but different structural formula and different physical and chemical properties, the phenomenon is called isomerism. Such compounds are called isomers.
It is of two types:
(1) Structural Isomerism
(1) Structural Isomerism: Structural isomerism is shown by compounds having the same molecular formula but different structural formulae differing in the arrangement of atoms.
(2) Stereoisomerism: When isomerism is caused by the different arrangements of atoms or groups in space, the phenomenon is called stereoisomerism. The steroeoisomers have same structural formula but differ in arrangement of atoms in space. Stereoisomerism is of two types:
(i) Geometrical or Cis-Trans Isomerism
(ii) Optical Isomerism
• Fundamental Concepts in Organic Reaction Mechanism
Fission of a covalent bond: A covalent bond can undergo Fission in two ways:
(i) By Homolytic Fission or Homolysis
(ii) By Heterolytic Fission or Heterolysis
Homolytic Fission: In this process each of the atoms acquires one of the bonding electrons.
Heterolytic Fission: In this process one of atoms aquires both of the bonding electrons when the bond is broken.
If B is more electronegative than A which thereby aquires both the bonding electrons and becomes negatively charged.
The products of heterolytic fission are ions.
Reaction Intermediates: Heterolytic and homolytic bond fission results in the formation of short-lived fragments called reaction intermediates. Among the important reaction intermediates are carbonium ions, carbanions, carbon free radicals and carbenes. Carbonium Ions (carbocations): Organic ions which contain a positively charged carbon atom are called carbonium ions or carbocations. They are formed by heterolytic bond fission.
where Z is more electronegative than carbon.
Tertiary carbonium ion is more stable than a secondary, which in turn is more stable than a primary because of +1 effect associated with alkyl group.
Carbanion: Organic ion which contains a negatively charged carbon atom are called carbanions. They are also formed by heterolytic bond fission.
Where Z is less electronegative than carbon. A primary carbanion is more stable than a secondary, which in turn is more stable than a tertiary, because of +1 effect associated with alkyl group.
Electrophile: It is positively charged or neutral species which is electron deficient, e.g.,
He–, H20+, CH3 , NH4+, AICl3 , S03 , CHCl2 , CCI3.
Nucleophile: It is negatively charged or neutral species with lone pair of electrons e.g., (HO–), Cyanide (C = N), H20: R3N, R2NH etc.
Electron Displacement Effects in Covalent Bonds: Electronic displacements in covalent bonds occurs due to the presence of an atom or group of different electronegativity or under the influence of some outside attaching group.
These lead to a number of effects which are as follows:
(i) Inductive effect (ii) Elecromeric effect
(iii) Resonance or Mesomeric effect (iv) Hyperconjugation effect.
Inductive Effect: It involues c electron. The a electrons which form a covalent bond are seldom shared equally between the two atoms. Due to different electronegatively electrons are displaced towards the more electronegative atom.
This introduces a certain degree of polarity in the bond.
The more electronegative atom acquires a small negative charge (δ–). The less electronegative atom acquires a small positive charge (δ+).
Consider the carbon-chlorine bond
As chlorine is more electronegative, it will become negatively charged with respect to the carbon atom.
Structure I- indicates the relative charges on the two atoms.
Structure II- indicates the direction in which the electrons are drawn.
Atoms or groups which lose electrons towards a carbon atom are said to have a +1 effect. Those atoms or groups which draw electrons away from a carbon atom are said to have a -I Effect.
Some common atoms or groups which cause +I or -I effects are shown below:
The inductive effect of C3 upon C2 is significantly less than the effect of the chlorine atom on C.
Resonance Structure: A number of organic compounds cannot be accurately represented by one structure.
For example, benzene is ordinarily represented as
Carbon-carbon double bond length = 1.34 A
Carbon-carbon single bond length = 1.54A. But it has been determined experimentally that all carbon-carbon bonds in benzene are identical and have same bond length (1.39A).
Thus the structure of benzene cannot be represented by single structure. It can be represented equally well by ‘he energetically similar structures I and II. The two structures are called resonance structures.
Actual structure of benzene is resonance hybrid of structures I and II.
Another example of resonance is provided by nitromethane (CH3N02) which can be represented by two Lewis structures.
The actual structure of nitromethane is a resonance hybrid of the two canonical forms I and II. Resonance energy: The difference in the energy between the most stable contributing structure for a compound and its resonance hybrid is called as resonance energy or resonance stabilisation energy.
Resonance Effect: The polarity produced in the molecule by the interaction of two π-bonds or between a π-bond and a lone pair of electrons present on an adjacent atom. There are two types of resonance or mesomeric effects designated as R or M effect.
Positive Resonance Effect (+R effect):
Those atoms which lose electrons towards a carbon atom are said to have a +M effect or +R effect. For example:
—Cl, —Br, —I, —NH2, —NR2, —OH, —OCH3
Negative Resonance Effect (-R effect): Those atoms or groups which draw electrons away from a carbon atom are said to have a -M effect or -R effect.
Electromeric Effect (E Effect):
The electrom’eric effect refers to the polarity produced in a multiple bonded compound when it is attacked by a reagent when a double or a triple bond is exposed to an attack by an electrophile E+ (a reagent) the two π electrons which from the π bond are completely transferred to one atom or the other. The electromeric effect is represented as:
The curved arrow shows the displacement of the electron pair. The atom A has lost its share in the electron pair and B has gained this share. Therefore A acquires a positive charge and B a negative charge.
Hyperconjugation or No Bond Resonance: When the alkyl group is attached to an unsaturated system such as —CH=CH2 group the order of inductive effect gets reversed. The behaviour can be explained by hyperconjugation effect.
Such structures are arrived at by shifting the bonding electrons from an adjacent C —H bond to the electron deficient carbon. In this way, the positive charge originally on carbon is dispersed to the hydrogen.
This way of electron release by assuming no bond character in the adjacent C—H bond is called No-Bond Resonance or Hyperconjugation.
• Orbital Concept of Hyperconjugation
It involues delocalisation of o electrons of C—H bond of an alkyl group which is attached directly to an atom of unsaturated system or to an atom with an unshared p-orbital.
Let us consider CH3CH2 (ethyl cation) in which the positively charged carbon atom has an empty p-orbital. One of the C—H bonds of the methyl group can align in the plane of this empty p-orbital and electron constituting the C—H bond in plane with this p-orbital can then be delocalised into the empty p-orbital as in Fig.
In general, greater the number of alkyl groups attached to a positively charged carbon atom, the greater is the hyperconjugation.
• Qualitative Analysis of Organic Compounds
Detection of Carbon and hydrogen: Put Copper Oxide Test:
The organic substance is mixed intimately with about three times its weight of dry copper oxide. The mixture is then placed in a hard glass test-tube fitted with a bent delivery tube. The other end of which is dipping into lime water in another test tube.
The mixture is heated strongly and the following reactions take place.
Thus if carbon is present it is oxidised to carbon dioxide which turns lime water milky. If hydrogen is also present, it will be oxidised to water droplets on the cooler wall of the test tube.
Detection of other elements
Lassaigne’s test: Nitrogen, sulphur, halogens and phosphorus present in an organic compound are detected by ‘Lassaigne’s test’. Covalent compounds are converted into ionic form by fusing the compound with sodium metal. Following reaction occurs:
Test for Nitrogen:
(i) The substance is heated strongly with sodium metal.
Na + C + N ———–> NaCN
(ii) The water extract of the fused mass is boiled with ferrous sulphate solution
(iii) To the cooled solution is then added a little ferric choride solution and excess of concentrated hydrochloric acid.
The formation of prussian blue or green colouration confirms the presence of nitrogen.
Test for sulphur:
Sodium Test: Sulphur, if present, in the given organic compound, upon fusion with sodium reacts to form sodium sulphide
Thus the sodium extract obtained from the fused mass is tested as:
(i) Add freshly prepared sodium nitroprusside solution. A deep violet colouration indicates sulphur.
(ii) Acidify the portion of the extract with acetic acid and then add lead acetate solution. A black precipitate of lead sulphide confirms the presence of sulphur.
If nitrogen and sulphur both are present in an organic compound, sodium thiocyanate is formed. It gives blood red colour and no prussian blue since there are no free cyanide ions.
Test for Halogens
Sodium Test: Upon fusion with sodium, the halogens in the organic compound are converted to sodium halides.
Cl + Na ——-> NaCl
Br + Na ———> NaBr
I + Na ———>NaI
Acidify a portion of ‘sodium extract’ with dilute nitric acid and add to it silver nitrate solution. White ppt. soluble in ammonia indicates Chlorine.
Yellow ppt. sparingly soluble in ammonia indicates Bromine.
Yellow ppt. insoluble in ammonia indicates Iodine.
When nitrogen or sulphur is also present in the compound, the sodium extract before testing for halogens is boiled with strong nitric acid to decompose the cyanide and the sulphite formed during the sodium fusion. If not removed, these radicals will form a white and black precipitate respectively on the addition of silver nitrate.
Test for phosphorous
The compound is heated with an oxidising agent (sodium peroxide). The phosphorous present in the compound is oxidised to phosphate.
The solution is boiled with HN03 and treated with ammonium molybdate. A yellow coloured ppt. indicates the presence of phosphorous.
• Quantitative Analysis
Estimation of Carbon and Hydrogen
Both carbon and hydrogen are estimated together in one operation. A known weight of an organic compound is burnt in the presence of excess of oxygen and copper (II) oxide. Carbon and hydrogen are oxidised to carbon dioxide and water respectively.
The weight of carbon dioxide and water thus formed are determined and the amounts of carbon and hydrogen in the original substance calculated.
Estimation of Nitrogen
(i) Dumas method: This method is based on fact that the nitrogenous compounds when heated with copper oxide in an atmosphere of carbon dioxide yield free nitrogen,
The traces of oxides of nitrogen, which may be formed in some cases, are reduced to elemental nitrogen by passing over heated copper spiral.
Apparatus can be shown as:
(ii) Kjeldahl’s Methods: Kjeldahl’s method is based on the fact that when an organic compound containing nitrogen is heated with con. H2S04 the nitrogen in it is converted to ammonium sulphate. The resultant liquid is then treated with excess of alkali and the liberated ammonia gas absorbed in excess of standard acid. The amount of ammonia is determined by finding the amount of acid neutralised by back filtration with some std. alkali.
• All organic compounds contain carbon and hydrogen as essential constituents.
• In a homologous series two successive members differ in their molecular formula by -CH2 unit.
• Aliphatic Compounds are open chain compounds contain straight or branched chain of carbon atoms.
• Alicyclic Compounds: Compounds containing closed ring of carbon compounds.
• Aromatic Compounds: Benzene and its derivatives are called aromatic compounds.
• Functional group: A functional group is an atom or group of atoms bonded together in a unique fashion and which determines the physical and chemical properties of the compounds.
• Homolytic Bond Fission: It leads to the formation of free radicals.
• Crystallisation is used to purify organic solids by dissolving them in suitable solvents.
• Simple distillation is used to purify liquids with non-volatile impurities.
• Steam distillation is used to purify organic compounds which give sufficient vapours at the boiling of water and are insoluble in water.
• Chromatography is used to purify and separate the constituents from a sample.
• Lassaigne’s test is used to detect carbon, nitrogen, sulphur and halogen in organic compound.
• Dumas or Kjeldahl’s method: Nitrogen is estimated by this method.
• Halogens: Halogens are estimated by Carius method.
• Sulphur and phosphorous: Sulphur and phosphorous are estimated by oxidising them to sulphuric and phosphoric acid respectively.
• Oxygen: The percentage of oxygen is usually determined by difference between the total percentage (100) and the sum of the percentages of all other elements present.
CBSE Class 11 Chemistry Chapter-12 Important Questions
1 Marks Questions
1.How many σ and π bonds are present in each of the following molecules?
(a) HC≡CC≡CCH3 (b) CH2=C=CHCH3.
Ans.(a) σ C = C : 4 (b) σ C = C : 3
σ C – H : 6 σ C – H : 6
π C = C : 3 π C = C : 2
2.Why are electrons easily available to the attacking reagents in π – bonds?
Ans.The electron charge cloud of the π – bond is located above and below the plane of bonding atoms. This results in the electrons being easily available to the attacking reagents.
3.Write the bond line formula for
4.How are organic compounds classified?
Ans.(i) Acyclic or open chain compounds
(ii) Alicyclic or closed chain or ring compounds.
(iii) Aromatic compounds.
5.Define homologous series?
Ans.A group or a series of organic compounds each containing a characteristic functional group forms a homologous series and the members of the series are called homologous.
6.Write an example of non – benzenoid compound.
7.What is the cause of geometrical isomerism in alkenes?
Ans. Alkene have a π – bond and the restricted rotation around the π – bond gives rise to geometrical isomerism.
8.Name the chain isomers of C5H12 which has a tertiary hydrogen atom.
Ans. 2 – Methyl butane (CH3)2 CH – CH2 – CH3
9.Define heterolytic cleavage.
Ans. In heterolytic cleavage the bond breaks in such a fashion that the shared pair of electrons remains with one of the fragments.
Ans.A species having a carbon atom possessing sextet of electrons and a positive charge is called carbocation.
11.What are the nucleophiles?
Ans. The electron rich species are called mucleopiles. A nucleophile has affection for a positively charge centre.
eg OH–, I–, CN–, : NH3, NO2–.
12.How can the mixture of kerosene oil and water be separated?
Ans.The mixture of kerosene oil and water can be separated by using a separating funnel.
13.Lasaigne’s test is not shown by diazonium salts. Why?
Ans.Diazonium salts usually leave N2 on heating much before they have a chance to react with the fused sodium metal. Therefore, diazonium salts do not show positive lassaigne’s test for nitrogen.
14.In which C – C bond of CH3CH2CH2Br, the inductive effect is expected to be the least?
Ans .Magnitude of inductive effect diminishes as the number of intervening bonds increases. Hence the effect is least in C3 – H bond.
15.Can you use potassium in place of sodium for fusing an organic compound in Lassaigne’s test?
Ans . No, because potassium is more reactive than sodium.
16.Give the reason for the fusion of an organic compound with sodium metal for testing nitrogen, sulphur and halogens.
Ans. The element present in the compound are converted from covalent form into ionic form by fusing the compound with sodium metal.
17.Write the chemical composition of the compound formed when ferric chloride is added containing both N and S.
2 Marks Questions
1.Write the expanded form of the following condensed formulas into their complete structural formulas.
2.How does hybridization affect the electronegativity ?
Ans. The greater the s – character of the hybrid orbital’s, the grater is the electro negativity.
3.Why is sp hybrid orbital more electronegative than sp2 or sp3 hybridized orbitals?
Ans. The greater the s – character of the hybrid orbital’s, the greater is the electro negativity. Thus, a carbon atom having an sp hybrid orbital with 50% s – character is more electro negative than that possessing sp2 or sp3 hybridized orbital’s.
eg: hydroxyl group (- OH)
aldehyde group (- CHO)
carboxylic acid group (-COOH) etc.
4.Give two examples of aliphatic compounds.
5.Write an example of alicyclic compound.
6.For each of the following compounds write a condensed formula and also their bondline formula.
(a) HOCH2 CH2 CH2CH (CH3) CH (CH3) CH3
(a) HO (CH2)5CH CH3 CH (CH3)2
(b) HOCH (CN)2.
Bond line formula.
7.Write the structural formula of
(a) p – Nitro aniline (b) 2,3 – Dibromo-1-phenylpentane.
8.Derive the structure of 3 – Nitrocyclohexene.
Ans.Six membered ring containing a carbon – carbon double bond is implied by cyclohexene, which is numbered. The prefix 3 – nitro means that a nitro group is parent on C – 3. Thus complete structured formula of the compound is derived. Double bond is suffixed functional group whereas NO2 is prefixed functional group; therefore double bond gets preference over – NO2 group:
9.Give the IUPAC of the following –
Ans.(a) 2,5 – dimethyl heptanes (b) 2,2 – dichloro ethanol.
10.Draw the two geometrical isomers of, but – 2 – en – 1, 4 dioic acid. Which of the will have higher dipole movement?
11.How many structural isomers and geometrical isomers are possible for a cyclohexane derivative having the molecular formula C9H16?
Ans.Five structural isomers
12.Alkynes does not exhibit geometrical isomers. Give reason.
Ans.Because of linear geometry.
13.Which of the following shows geometrical isomerism?
(a) CH Cl = CH Cl (b) CH2 = C Cl2 (c) C Cl2 = CH Cl.
Ans.Only compound (a) will show geometrical isomers.
(a) CH Cl = CH Cl
14.What is a functional group?
Ans. It may be defined as an atom or group of atoms joined in a specific manner which is responsible for the characteristic chemical properties of the organic compounds.
15.How many isomers are possible for monosubstituted and disubstituted benzene?
Ans.There is one, monosubstituted benzene as
There are three disubstituted benzenes.
16.Identify electrophilic centre in the following:
Ans. The shared carbon atoms are electrophilic centres as they will have partial positive charge due to polarity of the bond. CH3 HC = O, H3 CC = N, H3 C – I.
17.For the following bond cleavages, use curved arouse to the electron flow and classify each as photolysis or heterolysis. Identify the reaction intermediates products as free radical carbocation or carban ion.
(a) CH3 O – O CH3 →CH3
18.Write resonance structures of CH2 = CH – CHO. Indicate relative stability of the contributing structure.
19.Write the resonance structures of
(a) CH3 NO2 (b) CH3 COO–
20.Explain why is (CH3)3 C+ more stable than CH3CH2+ and CH3+ is the least stable cation.
Ans.Hyper conjugation interaction in (CH3)3C+ is greater than in CH3CH2+ as (CH3)3C+ has nine C-H bonds. In CH3+, The C-H bond the nodal plane of the vacant 2p orbital and hence can not overlap with it. Thus, CH3+ locus hyper conjugate stability.
21.Show how hyper conjugation occurs in propene molecule.
22.Draw the orbital diagram showing hyperconjugation in ethyl cations
23.Name the common techniques used for purification of organic compounds.
Ans.(i) Sublimation (ii) Crystallization (iii) Distillation (iv) Differential extraction and (v) Chromatography.
24.Will C Cl4 give white precipitate of Ag Cl on heating it with Ag NO3?
Ans. CCl4 does not give white precipitate with silver nitrate solution.
CCl4 + Ag NO3 →No reaction.
Carbon tetrachloride contains chlorine but it is bonded to carbon by a covalent bond. Therefore it is not in ionic form. Hence, it does not combine with Ag NO3 solution.
25.Without using column chromatography, how will you separate a mixture of camphor and benzoic acid?
Ans. Sublimation can not be used since both camphor and benzoic acid sublime on heating. Therefore a chemical method using NaHCO3 solution is used when benzoic acid dissolves leaving camphor behind. The filtrate is cooled and then acidified with dil HCl, to get benzoic acid.
26.A liquid (1.0g) has three components. Which technique will you employ to separate them?
27.Name two methods which can be safely used to purify aniline.
Ans.(i) vacuum distillation method
(ii) steam distillation method.
28.What is the basic principle of chromatography?
Ans.The method of chromatography is based on the difference in the rates at which the components of a mixture are adsorbed on a suitable adsorbent.
29.How will you separate a mixture of two organic compounds which have different solubility’s in the same solvent?
Ans. By fractional crystallization.
3 Marks Questions
1.What is the shape of the following molecules:
(a) H2 C=O (b) CH3F (c) HC≡N.
Ans.(a) sp2 hybridized carbon, trigocal planar
(b) sp3 hybridized carbon, tetrahedral
(c) sp hybridized carbon, linear.
2.Giving justification, categories the following molecules or ions as nucleophle or electrophile: HS–, BF3, C2H5O–, (CH3)3N:, Cl–, CH3C+ = O,
Ans.Nucleophiles : HS–, C2H5O–, (CH3)3 N:, H2N–: (have unshared pair of electrons which can be donated and shared with an electrophile)
Electrophile : BF3, Cl+, CH3C+ = O+ NO2[have only six electrons which can be accept electron from a nucleophile].
3. Using curved – arrow notation, show the formation of reactive intermediates when the following covalent bond undergo heterolysis cleavage.
(a) CH3 – SCH3, (b) CH3 – CN, (c) CH3 – Cu.
4.Benzyl carbonation is more stable than ethyl carbonation. Justify.
Ans. In ethyl carbocation, there is only hyper conjugation of the three α – hydrogen atoms and as a result, the following contributing structures are feasible.
But benzyl carbocation is more stable due to the presence of resonance and the following
resonating structures are possible
5.Which of the following pairs of structures do not constitute resonance structures?
(c) CH3CH=CHCH3 and CH3CH2CH = CH2.
(b) (CH3)2 CO
6.Write resonance structures of
(a)CH3COO– (b) C6H5NH2.
7.Draw the resonance structures for the following compounds
(a) C6H5OH (b) C6H5 –
8. 0.395 g of an organic compound by Carius method for the estimation of sulphur gave 0.582 g of BaSO4. Calculate the percentage of sculpture in the compound.
Ans. Mass of BaSO4 = 0.582g
BaSO4 = S
233g of BaSO4 contain sulphur = 32g
0.582g of BaSO4 contains sulphur
Percentage of sulphur =
9. 0.40g of an organic compound gave 0.3g of Ag Br by Carious method. Find the percentage of bromine in the compound.
Mass of the compound = 0.40g
Now 188g of Ag Br will contain Br = 80g
Therefore, 0.3g of Ag Br will contain Br =
The percentage of Br in the organic compound
10. 0.12g of organic compound containing phosphorus gave 0.22g of Mg2P2O7 by the usual analysis. Calculate the percentage of phosphorus in the compound.
Ans. Here the mass of the compound taken = 0.12g
Mass of Mg2P2O7 formed = 0.22g of atoms of P
Now 1 mole of Mg2P2O7 = (2×24+2×31+1687)
= 222g of Mg2P2O7
i.e; 222g of Mg2P2O7 contain phosphorus = 62g.
∴ 0.22g of Mg2P2O7 will contain phosphorus.
But this is the amount of phosphorus present in 0.12g of organic compound
Hence, percentage of phosphorus
11. Ammonia produced when 0.75g of a substance was kjeldahlized, neutralized 30cm3 of 0.25 N H2SO4. Calculate the percentage of nitrogen in the compound.
Ans.Mass of organic compound = 0.75g
Volume of H2SO4 used us = 30cm3
Normality of H2SO4 = 0.25N
30cm3 of H2SO4 of normality 0.25N ≡ 30ml of NH3 solution of normality 0.25N
But 1000cm2 of NH3 of normality 1 contains 14g of nitrogen
∴ 30cm3 of 0.25N NH3 contains nitrogen
% of nitrogen =