Question

Given,0.23 g of an organic compound gave 0.332 g of $BaS{{O}_{4}}$ . Determine the percentage of sulphur in the compound.
A. 32.61 %
B. 8.5 %
C. 19.77 %
D. None of these

Answer 1

Hint: The molecular weight of the barium sulphate is 233. The molecular formula of barium sulphate is $BaS{{O}_{4}}$ . From the chemical formula we can easily see that there is one sulphur atom present in barium sulphate.

Complete answer:
- In the question it is asked to find the percentage of the sulphur in 0.332 gm of barium sulphate.
- The given data is 0.23 g of organic compound giving 0.332 g of $BaS{{O}_{4}}$ .
- The molecular weight of the $BaS{{O}_{4}}$ (barium sulphate is 233) .
- We know the molecular weight of the sulphur is 32 g.
- From the chemical formula of barium sulphate we can easily say that barium sulphate contains one mole of sulphur.
- Means 233 g of barium sulphate contains 32 g of sulphur.
- We have to find the sulphur amount in 233 g of barium sulphate.
- Therefore the percentage of sulphur in barium sulphate calculation will be as follows.
The percentage of sulphur in 0.332 g of sulphur $=\dfrac{0.332\times 32}{233\times 0.23}\times 100=19.77$ %
- Therefore the percentage of sulphur in barium sulphate which is obtained from 0.23 g of organic compound is 19.77%.

Therefore the correct option is C.

Note:
We are supposed to check how many moles of sulphur is present in the barium sulphate before going to do calculation. If we are not going to check then we will get a wrong solution, which is not correct here.