Question

In a Carius determination, 0.234 g of an organic substance gave 0.334 g of barium sulphate. Calculate the percentage of sulphur in the given compound. [Ba = 137, S = 32, O = 16]

Answer 1

Molecular mass of barium sulphate, having the chemical formula as $BaS{O_4}$ is 233 g/mol. 233 g of $BaS{O_4}$ contains 32 g of sulphur. You need to calculate the amount of sulphur contained in 0.334 g of barium sulphate. Then, find out the percentage of sulphur in the given organic substance.

Complete step by step solution:
We are given that 0.234 g of an organic substance gives 0.334 g of barium sulphate ($BaS{O_4}$).
Thus, given weight of barium sulphate = 0.334 g.
Now, molecular mass of $BaS{O_4}$ = ${\text{Atomic weight of Ba + Atomic weight of S + 4(Atomic weight of O)}}$
We are given in the question,
At. wt. Of Ba = 137, At. wt. of S = 32, At. wt. of O = 16
Therefore, molecular mass of $BaS{O_4}$= $137 + 32 + 4(16) = 233g$
1 mole of $BaS{O_4}$ i.e, 233 g of $BaS{O_4}$ contains 32 g of sulphur.
So, by unitary method, 0.334 g (given weight) of $BaS{O_4}$ would contain $\dfrac{{32}}{{233}} \times 0.334 = 0.0458g$
Now, calculating the percentage of sulphur in the given organic substance, having weight 0.234 g.
% of Sulphur = $\dfrac{{{\text{Weight of sulphur in 0}}{\text{.334 g of }}BaS{O_4}}}{{{\text{Weight of the organic substance}}}} \times 100$
$\therefore$ % of Sulphur = $\dfrac{{0.0458}}{{0.234}} \times 100 = 19.6\%$

Hence, 19.6 % of sulphur is present in the given organic substance having weight 0.234 g.

Note: So many times, the information regarding the percentage of a particular element present in the compound is required. We can check the purity of a given sample by analysing the data of percentage composition of the elements present in the compound.
If we are asked to calculate the mass percentage of the sulphur in barium sulphate. Then, it can be calculated by using the formula given below:
Mass % of an element = $\dfrac{{{\text{mass of that element in the compound}}}}{{{\text{molar mass of the compound}}}} \times 100$
Mass % of sulphur in $BaS{O_4}$ = $\dfrac{{{\text{Mass of sulphur in BaS}}{{\text{O}}_{\text{4}}}}}{{{\text{Molar mass of BaS}}{{\text{O}}_{\text{4}}}}} \times 100 = \dfrac{{32}}{{233}} \times 100 = 13.73\%$