In Sulphur estimation, 0.7 g of an organic compound gives 1 g of BaSO4 by Carius method. What is the % of ‘S’ in compound?

To find the percentage of sulfur ($\text{S}$) in the organic compound using the Carius method, we can follow these step-by-step calculations:

1. Identify the given values:
- Mass of the organic compound = $0.7\text{ g}$
- Mass of barium sulfate (${\text{BaSO}}_4$) formed = $1\text{ g}$

2. Determine the molar masses:
- Atomic mass of Sulfur ($\text{S}$) = $32\text{ g/mol}$
- Atomic mass of Barium ($\text{Ba}$) = $137\text{ g/mol}$
- Atomic mass of Oxygen ($\text{O}$) = $16\text{ g/mol}$
- Molar mass of ${\text{BaSO}}_4$ = $137+32+(4\times 16)=233\text{ g/mol}$

This means that $233\text{ g}$ of ${\text{BaSO}}_4$ contains $32\text{ g}$ of sulfur.

3. Calculate the mass of sulfur in $1\text{ g}$ of ${\text{BaSO}}_4$:
Mass of sulfur = $\frac{32}{233}\times 1\text{ g}\approx 0.1373\text{ g}$

4. Calculate the percentage of sulfur in the organic compound:
The percentage of sulfur is given by the formula:
$\text{\% of S}=\frac{\text{Mass of sulfur}}{\text{Total mass of organic compound}}$
Substituting the values:
$\text{\% of S}=\frac{32}{233}\times \frac{1\text{ g}}{0.7\text{ g}}$
$\text{\% of S}=\frac{3200}{233\times 0.7}$
$\text{\% of S}=\frac{3200}{163.1}\approx 19.6198\text{\%}$

Rounding to two decimal places, we get 19.61%.

Therefore, the percentage of sulfur in the organic compound is 19.61%.