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
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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.
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