0.316 g of an organic compound, after heating with fuming nitric acid and barium nitrate crystals in a sealed tube gave 0.466 g of the precipitate of barium sulphate. Determine the percentage of sulphur in the compound. (Atomic masses : Ba = 137, S = 32, O = 16, C = 12, H = 1).

0.316 g of an organic compound, after heating with fuming nitric acid

1.	0.316 g of an organic compound, after heating with fuming nitric acid and barium nitrate crystals in a sealed tube gave 0.466 g of the precipitate of barium sulphate. Determine the percentage of sulphur in the compound. (Atomic masses : Ba = 137, S = 32, O = 16, C = 12, H = 1).
Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> of the substance taken = 0.316 g <br/> Mass of `BaSO_(4)` <a href="https://interviewquestions.tuteehub.com/tag/formed-464209" style="font-weight:bold;" target="_blank" title="Click to know more about FORMED">FORMED</a> = 0.466 g <br/> Molecular mass of `BaSO_(4) = <a href="https://interviewquestions.tuteehub.com/tag/137-272694" style="font-weight:bold;" target="_blank" title="Click to know more about 137">137</a> + <a href="https://interviewquestions.tuteehub.com/tag/32-307188" style="font-weight:bold;" target="_blank" title="Click to know more about 32">32</a> + <a href="https://interviewquestions.tuteehub.com/tag/64-330663" style="font-weight:bold;" target="_blank" title="Click to know more about 64">64</a> = 233` <br/> Then, mass of S in 0.466 g of `BaSO_(4)` <br/> `=(0.466xx32)/(233)g` <br/> Percentage of S in compound <br/> `=(0.466xx32xx100)/(233xx0.316)` <br/> = 20.25</body></html>