
In Sulphur estimation, 0.157 grams of an organic compound gave 0.4813 grams of Barium Sulphate. What is the percentage of Sulphur in the compound?
Answer
161.1k+ views
Hint: In the Carius method of estimation of sulphur, barium Sulphate is precipitated and then the obtained ppt. is filtered, washed, and dried. After that, the weight of the ppt is determined.
Complete Step by Step Solution:
Let’s first understand the method of estimation of Sulphur.
1) In the Carius method, an organic compound whose mass is known undergoes heating with sulphur in presence of nitric acid (in excess) is a Carius tube which is sealed.
2) In the tube, conversion of sulphur of the organic compound to sulphuric acid takes place.
3) Then, the treatment of sulphuric acid with barium chloride (in excess).
4) In the reaction, formation of barium sulphate occurs and it gets separated from the solution as precipitated.
5) Then, the precipitate is filtered, then washed, then dried and weighed.
6) The weighted percentage helps in estimation of Sulphate in the compound.
Now, come to the question. Here, we have to find out the percentage of element sulphur of the given compound.
We know that sulphur possesses an atomic mass of 32u. And now we have to find out the molar mass of \[{\rm{BaS}}{{\rm{O}}_{\rm{4}}}\] . Oxygen possesses an atomic mass of 16 u and barium possesses atomic mass of 137u.
So, molar mass of \[{\rm{BaS}}{{\rm{O}}_{\rm{4}}} = 137 + 32 + 4 \times 16 = 233\,\,{\rm{u}}\]
Now, we have to find out the weight of sulphur in \[{\rm{BaS}}{{\rm{O}}_{\rm{4}}}\]. Given that, sulphur estimation gives 0.4813 gram of barium Sulphate.
Weight of sulphur in \[{\rm{BaS}}{{\rm{O}}_{\rm{4}}} = \dfrac{{32}}{{233}} \times 0.4813 = 0.066\,{\rm{g}}\]
Now, we have to calculate % of sulphur in the organic compound. Given, weight of the compound is 0.157 g and weight of Sulphur in \[{\rm{BaS}}{{\rm{O}}_{\rm{4}}}\]is 0.066 g.
% composition=\[\dfrac{{0.066}}{{0.157}} \times 100 = 42.03\]
Hence, the percentage of Sulphur in the compound is 42.03.
Note: \[{\rm{BaS}}{{\rm{O}}_{\rm{4}}}\] is a white coloured solid (crystalline forms) which has no odour and has the property of insolubility in water. It is the chief commercial source of the element barium and many materials are prepared from it.
Complete Step by Step Solution:
Let’s first understand the method of estimation of Sulphur.
1) In the Carius method, an organic compound whose mass is known undergoes heating with sulphur in presence of nitric acid (in excess) is a Carius tube which is sealed.
2) In the tube, conversion of sulphur of the organic compound to sulphuric acid takes place.
3) Then, the treatment of sulphuric acid with barium chloride (in excess).
4) In the reaction, formation of barium sulphate occurs and it gets separated from the solution as precipitated.
5) Then, the precipitate is filtered, then washed, then dried and weighed.
6) The weighted percentage helps in estimation of Sulphate in the compound.
Now, come to the question. Here, we have to find out the percentage of element sulphur of the given compound.
We know that sulphur possesses an atomic mass of 32u. And now we have to find out the molar mass of \[{\rm{BaS}}{{\rm{O}}_{\rm{4}}}\] . Oxygen possesses an atomic mass of 16 u and barium possesses atomic mass of 137u.
So, molar mass of \[{\rm{BaS}}{{\rm{O}}_{\rm{4}}} = 137 + 32 + 4 \times 16 = 233\,\,{\rm{u}}\]
Now, we have to find out the weight of sulphur in \[{\rm{BaS}}{{\rm{O}}_{\rm{4}}}\]. Given that, sulphur estimation gives 0.4813 gram of barium Sulphate.
Weight of sulphur in \[{\rm{BaS}}{{\rm{O}}_{\rm{4}}} = \dfrac{{32}}{{233}} \times 0.4813 = 0.066\,{\rm{g}}\]
Now, we have to calculate % of sulphur in the organic compound. Given, weight of the compound is 0.157 g and weight of Sulphur in \[{\rm{BaS}}{{\rm{O}}_{\rm{4}}}\]is 0.066 g.
% composition=\[\dfrac{{0.066}}{{0.157}} \times 100 = 42.03\]
Hence, the percentage of Sulphur in the compound is 42.03.
Note: \[{\rm{BaS}}{{\rm{O}}_{\rm{4}}}\] is a white coloured solid (crystalline forms) which has no odour and has the property of insolubility in water. It is the chief commercial source of the element barium and many materials are prepared from it.
Recently Updated Pages
Two pi and half sigma bonds are present in A N2 + B class 11 chemistry JEE_Main

Which of the following is most stable A Sn2+ B Ge2+ class 11 chemistry JEE_Main

The enolic form of acetone contains a 10sigma bonds class 11 chemistry JEE_Main

The specific heat of metal is 067 Jg Its equivalent class 11 chemistry JEE_Main

The increasing order of a specific charge to mass ratio class 11 chemistry JEE_Main

Which one of the following is used for making shoe class 11 chemistry JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts of Chemistry

NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction

JEE Advanced 2025 Notes
