Question

0.53 g of substance gave 1.158 g of ${\text{BaS}}{{\text{O}}_4}$ by Carius method. Calculate the percentage of sulphur.
${\text{A}}{\text{.}}$ 42.10%
${\text{B}}{\text{.}}$ 23.3%
${\text{C}}{\text{.}}$ 32.03%
${\text{D}}{\text{.}}$ 30%

Answer 1

Hint- Here, we will proceed by defining the term known as quantitative analysis. Then, we will discuss the Carius method used for the estimation of percentage of sulphur present in a chemical compound.

Complete answer:
Formula Used- Percentage of S = \[\dfrac{{\left( {{\text{Atomic mass of S}}} \right)\left( {{\text{Mass of BaS}}{{\text{O}}_4}} \right)}}{{\left( {{\text{Molecular mass of BaS}}{{\text{O}}_4}} \right)\left( {{\text{Mass of the substance}}} \right)}} \times 100\].
An analysis method used to determine the number of elements or molecules produced during a chemical reaction is known as quantitative analysis. Organic compounds are composed of carbon, hydrogen, oxygen, nitrogen, phosphorus, sulphur and halogens. There are various methods which are used for the measurement of percentage composition of elements in an organic compound.
Carius method is used for the quantitative determination of halogens, sulphur and phosphorus in any chemical substance.
Estimation of Sulphur using Carius method is given as under:
A known mass of the substance is heated with concentrated nitric acid (${\text{HN}}{{\text{O}}_3}$) in the presence of barium chloride (${\text{BaC}}{{\text{l}}_2}$) solution in Carius tube. Sulphur is oxidised to sulphuric acid (${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$) and precipitated as ${\text{BaS}}{{\text{O}}_4}$. It is then dried and weighed. Then, the percentage of the sulphur present in the substance can be determined using the formula below given.
Percentage of S = \[\dfrac{{\left( {{\text{Atomic mass of S}}} \right)\left( {{\text{Mass of BaS}}{{\text{O}}_4}} \right)}}{{\left( {{\text{Molecular mass of BaS}}{{\text{O}}_4}} \right)\left( {{\text{Mass of the substance}}} \right)}} \times 100{\text{ }} \to {\text{(1)}}\]
Given, Mass of the substance = 0.53 g
Mass of ${\text{BaS}}{{\text{O}}_4}$ obtained = 1.158 g
We know that,
Atomic mass of sulphur atom (S) = 32 g
Molecular mass of ${\text{BaS}}{{\text{O}}_4}$ = Atomic mass of Ba + Atomic mass of S + 4(Atomic mass of O)
$ \Rightarrow $ Molecular mass of ${\text{BaS}}{{\text{O}}_4}$ = 137 + 32 + 4(16) = 233 g
By substituting all the above values in the formula given by equation (1), we get
$ \Rightarrow $Percentage of S = \[\dfrac{{\left( {{\text{32}}} \right)\left( {1.158} \right)}}{{\left( {233} \right)\left( {{\text{0}}{\text{.53}}} \right)}} \times 100 = \dfrac{{37.056}}{{123.49}} \times 100 = 30\]%

Therefore, the required percentage of the sulphur present in the given mass of the substance is 30%.

Hence, option D is correct.

Note- Carius method is also used for the percentage of halogens in the given mass of the chemical substance. For this, the compound is heated with Concentrated ${\text{HN}}{{\text{O}}_3}$ in the presence of ${\text{AgN}}{{\text{O}}_3}$ in a hard glass tube called Carius tube. C and H are oxidised to ${\text{C}}{{\text{O}}_2}$ and ${{\text{H}}_2}{\text{O}}$. The halogen forms the corresponding AgX. It is filtered, dried and weighed.