In the sulphur estimation, 0.471 g of an organic compound gave 1.44 g of barium sulfate. The percentage of sulphur in the compound is __________ %. (Nearest integer)
(Atomic Mass of Ba=137 u, S=32 u, O=16 u)
Show Hint
For quantitative estimation problems (Carius, Dumas, Kjeldahl, etc.), it's efficient to remember the direct percentage formula. For sulphur as BaSO\(_4\), the key factor is (32/233). For halogens as AgX, the factor is (Atomic mass of X / Molar mass of AgX). This saves time calculating the mass of the element separately.
Step 1: Understanding the Question
This is a quantitative analysis problem (Carius method for Sulphur). All the sulphur in an organic compound is converted into barium sulfate (BaSO\(_4\)). From the mass of BaSO\(_4\) formed, we need to find the percentage of sulphur in the original compound. Step 2: Key Formula or Approach
The percentage of Sulphur (%S) can be calculated using the formula:
\[ % \text{S} = \frac{\text{Atomic mass of S}}{\text{Molar mass of BaSO}_4} \times \frac{\text{Mass of BaSO}_4 \text{ formed}}{\text{Mass of organic compound taken}} \times 100 \]
Step 3: Detailed Calculation Calculate the molar mass of BaSO\(_4\):
Molar Mass = Mass(Ba) + Mass(S) + 4 \(\times\) Mass(O)
Molar Mass = 137 + 32 + 4(16) = 137 + 32 + 64 = 233 g/mol. Substitute the values into the formula:
Atomic mass of S = 32 u
Molar mass of BaSO\(_4\) = 233 u
Mass of BaSO\(_4\) formed = 1.44 g
Mass of organic compound taken = 0.471 g
\[ % \text{S} = \frac{32}{233} \times \frac{1.44}{0.471} \times 100 \]
Calculate the final percentage:
\[ % \text{S} = 0.1373 \times 3.0573 \times 100 \]
\[ % \text{S} = 41.98 % \]
Step 4: Final Answer
Rounding the result to the nearest integer, the percentage of sulphur is 42%.
