In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of barium sulphate. Find out the percentage of sulphur in the given compound.
Solution and Explanation
Total mass of organic compound = 0.468 g [Given]
Mass of barium sulphate formed = 0.668 g [Given]
1 mol of BaSO4 = 233 g of BaSO4 = 32 g of sulphur
Thus, 0.668 g of BaSO4 contains \(\frac{32\times 0.668}{233}g\) of sulphur = 0.0917 g of sulphur
Therefore, the percentage of sulphur = \(\frac{0.0197}{0.468}\times 100 =19.59\)%
Hence, the percentage of sulphur in the given compound is 19.59 %.
Top Questions on Organic Chemistry - Some Basic Principles and Techniques
- Which of the following compounds has the highest boiling point?
- BITSAT - 2025
- Chemistry
- Organic Chemistry - Some Basic Principles and Techniques
- The compound which does not react with ammonical silver nitrate solution is:
- UPCATET - 2025
- Chemistry
- Organic Chemistry - Some Basic Principles and Techniques
- The amine which is most basic in gas phase is:
- UPCATET - 2025
- Chemistry
- Organic Chemistry - Some Basic Principles and Techniques
- Which of the following amines will give most easily \(\mathrm{N}_2\) gas on treatment with \(\mathrm{NaNO}_2 + \mathrm{HCl}\) at 275 K:
- UPCATET - 2025
- Chemistry
- Organic Chemistry - Some Basic Principles and Techniques
- Electrophile used in Kolbe’s reaction is:
- UPCATET - 2025
- Chemistry
- Organic Chemistry - Some Basic Principles and Techniques
Questions Asked in CBSE Class XI exam
- We have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?
- CBSE Class XI
- Motion in a straight line
- If U = {a, b, c, d, e, f, g, h}, find the complements of the following sets:
(i) A = {a, b, c}
(ii) B = {d, e, f, g}
(iii) C = {a, c, e, g}
(iv) D = {f, g, h, a}- CBSE Class XI
- Complement of a Set
- What are the oxidation numbers of the underlined elements in each of the following and how do you rationalise your results?
(a) KI3 (b) H2S4O6 (c) Fe3O4 (d) CH3CH2OH (e) CH3COOH- CBSE Class XI
- Oxidation Number
- Using s, p, d notations, describe the orbital with the following quantum numbers.
- n=1, l=0
- n = 3, l=1
- n = 4, l =2
- n=4, l=3
- CBSE Class XI
- Developments Leading to the Bohr’s Model of Atom
- Write the following as intervals:
(i) {\(x: x ∈ R, -4 < x ≤ 6\)}
(ii) {\(x: x ∈ R, -12 < x < -10\)}
(iii) {\(x: x ∈ R, 0 ≤ x < 7\)}
(iv) {\(x: x ∈ R, 3 ≤ x ≤ 4\)}- CBSE Class XI
- Sets and their Representations
Concepts Used:
Organic Chemistry – Some Basic Principles and Techniques - Reaction Mechanism
SN1 Reaction Mechanism:
SN1 reaction mechanism takes place by following three steps –
- Formation of carbocation
- Attack of nucleophile
- Deprotonation
SN2 Reaction Mechanism:
The SN2 reaction mechanism involves the nucleophilic substitution reaction of the leaving group (which generally consists of halide groups or other electron-withdrawing groups) with a nucleophile in a given organic compound.
Electrophilic Aromatic Substitution Reaction Mechanism:
The mechanism of an electrophilic aromatic substitution reaction contains three main components which are:
- A new sigma bond from C=C is formed during the reaction in the arene nucleophile.
- Due to the breaking of the C-H sigma bond, a proton is removed.
- The C=C bond is reformed and it restores the aromaticity of the compound.
Electrophilic Substitution Reaction Mechanism:
The electrophilic substitution reaction mechanism is composed of three steps, which will be discussed more below.
- Electrophile Generation
- Carbocation Formation
- Proton Removal