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Complete combustion of 0.858 g of compou...

Complete combustion of `0.858 g` of compound `X` given `2.64g` of `CO_2 and `1.26gg` of `H_2O.` The lowest molecular mass `X` can have:

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Complete combustion of 1.80g of an oxygen containing compound " (C_(x)H_(y)O_(z)) gave 2.64g of CO_(2) and 1.08g of H_(2)O .The percentage of oxygen in the organic compound is:

44g of a sample on complete combustion given 88g CO_2 and 36g of H_2 O. The molecular formula of the compound may be :

Knowledge Check

  • Complete combustion of 0.858 g of compound X gives 2.63 g of CO_(2) and 1.28 g of H_(2) O. The lowest molecular mass X can have

    A
    (a)43 g
    B
    (b)86 g
    C
    (c )129 g
    D
    (d)172 g

  • Complete combustion of 1.18 g of organic compound gives 2.64 g of CO_(2) & 1.26 g " of " H_(2)O . Find emprical formula of compound ?

    A
    `C_(3)H_(7)O`
    B
    `C_(2)H_(4)`
    C
    `C_(3)H_(6)O`
    D
    `C_(3)H_(7)`

  • Complete combustion of a sample of hydrocarbon Q gives 0.66 g of CO_2 and 0.36 g of H_2O .The emptical formula of the compound is

    A
    `CH_(2)`
    B
    `C_(3)H_(4)`
    C
    `C_(3)H_(8)`
    D
    `C_(4)H_(8)`

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    Complete combustion of a hydrocarbon gives 0.66 g of CO_(2) and 0.36g of H_(2)O . Find the empirical formula of the compound.

    On complete combustion, 0.246g of an organic compound gave 0.198g of CO_(2) and 0.1014g of H_(2)O . The ratio of carbon and hydrogen atoms in the compound is:

    On complete combustion, 0.246 g of an organic compound gave 0.198 g of CO_2 and 0.1014 g of H_2 Otha ratio of carbon and hydrogen atoms in the compound is

    44g of a sample on complete combusion given 88 gm CO_2 and 36 gm of H_2O . The molecular formula of the compound may be :-

    44g of a sample on complete combusion given 88gm CO_(2) and 36gm of H_(2)O . The molecular formula of the complound may be

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