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  • In the estimation of Sulphur by Carius Method, 0.468g of organic compound gave 668mg <img alt="\mathrm{BaSO _{4}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BBaSO%20_%7B4

In the estimation of Sulphur by the Carius Method, 0.468g of organic compound gave 668mg \mathrm{BaSO _{4}}. Calculate % S (Rounding off up to 1 decimal place).

 

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Given-

  • Mass of organic compound = 0.468 g
  • Mass of barium sulphate (BaSO4) formed = 668 mg = 0.668 g

To find the percentage of Sulphur (% S) in the compound

Solution- In the Carius method, sulphur in the compound is converted to BaSO4.

  • Molar mass of BaSO4 = 137 (Ba) + 32 (S) + 64 (O4) = 233 g/mol
  • Molar mass of Sulphur (S) = 32 g/mol

So, mass of S in 0.668 g BaSO4 = $$
\frac {32}{233} \times 0.668 = 0.0917 g
$$ $$
S \%= \frac {\text{Mass of S}}{\text{Mass of compound}} \times 100= \frac {0.0917}{0.468}\times 100= 19.6 \%
$$

Therefore, the percentage of sulfur (S) in the given organic compound( based on the Carius method) is 19.6%.

Posted by

Saniya Khatri

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