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    Carius Method - Practice Questions & MCQ

    Edited By admin | Updated on Sep 18, 2023 18:35 AM | #JEE Main

    Quick Facts

    • Carius Method (Halogen and Sulphur) is considered one of the most asked concept.

    • 13 Questions around this concept.

    Solve by difficulty

    QuestionMEDIUM

    In Carius method of estimation of halogen, 0.172 g of an organic compound showed presence of 0.8 g of bromine. Which of these is the correct structure of the compound ?

     

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    In Carius tube, an organic compound ' X ' is treated with sodium peroxide to form a mineral acid ' Y '.The solution of $\mathrm{BaCl}_2$ is added to ' Y ' to form a precipitate ' Z '.' Z ' is used for the quantitative estimation of an extra element. ' X ' could be

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    Which method is used for the quantitative estimation of halogen in an organic compound.

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    In sulphur estimation, 0.157g of an organic compound gave 0.4813g of $\mathrm{BaSO}_4$ . The % of sulphur in the organic compound is

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    Match List-I with List-II

    List-I

    (Purification technique)

    List-II

    (Mixture of organic compounds)
    (A) Distillation (simple) (I) Diesel + Petrol
    (B) Fractional distillation (II) Aniline + Water
    (C) Distillation under reduced pressure (III) Chloroform + Aniline
    (D) Steam distillation (IV) Glycerol + Spent-lye

    Choose the correct answer from the options given below :

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    In the expression of percentage oxygen calculation $\% \mathrm{O}=\frac{32 * \mathrm{~m}_1 * 100}{88 * \mathrm{~m}}, \mathrm{~m}_1, \mathrm{~m}$ ,    respectively are

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    Which of the following is not correct about the detection of Oxygen using Carius Method?

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    Which compound is used in the quantitative estimation of carbon monoxide

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    Concepts Covered - 2

    Carius Method (Halogen and Sulphur)

    Halogens
    A known mass of an organic compound is heated with fuming nitric acid in the presence of silver nitrate contained in a hard glass tube known as Carius tube, in a furnace. Carbon and hydrogen present in the compound are oxidised to carbon dioxide and water. The halogen present forms the corresponding silver halide (AgX). It is filtered, washed, dried and weighed.

    Let the mass of organic compound taken = m g
    Mass of AgX formed = m1 g
    1 mol of AgX contains 1 mol of X
    \text {Mass of halogen in } \mathrm{m}_{1} \mathrm{g} \text { of } \mathrm{AgX} \: =\: \frac{\text { atomic mass of } \mathrm{X} \times m_{1} \mathrm{g}}{\text { molecular mass of } \mathrm{AgX}}

    \begin{aligned} \text { Percentage of halogen } = \frac{\text { atomic mass of } \mathrm{X} \times m_{1} \times 100}{\text { molecular mass of } \operatorname{Ag} \mathrm{X} \times m} \end{aligned}

    Sulphur
    A known mass of an organic compound is heated in a Carius tube with sodium peroxide or fuming nitric acid. Sulphur present in the compound is oxidised to sulphuric acid. It is precipitated as barium sulphate by adding excess of barium chloride solution in water. The precipitate is filtered, washed, dried and weighed. The percentage of sulphur can be calculated from the mass of barium sulphate.

    Let the mass of organic compound taken = m g
    and the mass of barium sulphate formed = m1 g
    1 mol of BaSO4 = 233 g BaSO4 = 32 g sulphur

    \begin{array}{l}{\mathrm{m}_{1} \mathrm{g} \mathrm{BaSO}_{4} \text { contains } \frac{32 \times m_{1}}{233} \mathrm{g} \text { sulphur }} \\\\ {\text {Percentage of sulphur }=\frac{32 \times m_{1} \times 100}{233 \times m}}\end{array}

    Carius Method (oxygen)

    The percentage of oxygen in an organic compound is usually found by difference between the total percentage composition (100) and the sum of the percentages of all other elements. However, oxygen can also be estimated directly as follows:
    A definite mass of an organic compound is decomposed by heating in a stream of nitrogen gas. The mixture of gaseous products containing oxygen is passed over red-hot coke when all the oxygen is converted to carbon monoxide. This mixture is passed through warm iodine pentoxide (I2O5) when carbon monoxide is oxidised to carbon dioxide producing iodine.

    On making the amount of CO produced in equation (A) equal to the amount of CO used in equation (B) by multiplying the equations (A) and (B) by 5 and 2 respectively; we find that each mole of oxygen liberated from the compound will produce two moles of carbon dioxide.

    \begin{aligned} &\mathrm{2 C + O _{2} \stackrel{1373 K }{\longrightarrow} 2 CO\quad\quad\quad\ \ (A)\ \quad\ \times 5 }\\ & \mathrm{I _{2} O _{5}+5 CO \longrightarrow I _{2}+5 CO _{2} (B)\quad\quad\times 2} \end{aligned}

    ___________________________________________

    \mathrm{10 C + 5 O_2 + 2I_2O_5 \longrightarrow 2I_2 + 10 CO_2}

    Thus 88 g carbon dioxide is obtained if 32 g oxygen is liberated.
    Let the mass of organic compound taken be \mathrm{m\ g }
    Mass of carbon dioxide produced be \mathrm{m_1\ g }
    \mathrm{\therefore m_1 \ g } carbon dioxide is obtained from \mathrm{\frac{32 \times m_{1}}{88} \mathrm{g}\: \mathrm{O}_{2}}


    \mathrm{\therefore \text { Percentage of oxygen }=\frac{32 \times m_{1} \times 100}{88 \times m} \%}

    Study it with Videos

    Carius Method (Halogen and Sulphur)
    Carius Method (oxygen)

    Study other Related Concepts

    Carius Method Current Topic

    Purification of Organic Compounds
    Qualitative Analysis of Organic Compounds
    Lassaigne’s Test - Test for Nitrogen, Sulphur, Halogens
    Duma's Method
    Kjeldahl's Method
    Carius Method

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    Books

    Reference Books

    Carius Method (Halogen and Sulphur)

    Chemistry Part II Textbook for Class XI

    Page No. : 367

    Line : 7

    Carius Method (oxygen)

    Chemistry Part II Textbook for Class XI

    Page No. : 368

    Line : 47

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