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An organic compound contains carbon, hyd...

An organic compound contains carbon, hydrogen and oxygen. IF the percentage of ``C: the percentage of `H=6:1`, calculate the simplest formula of the compound, given that one molecule of the compound contains half as much oxygen as would be required to burn all the corbon and hydrogen atoms in it to `CO_2` and `H_2O`

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Percentage of C: percentage of `H=6:1`
Pecentage of molar ratio
`=(6)/(12):(1)/(1)-=(1)/(2):1[{:(at.wt. of C=12),(at.wt.of H=1):}]`
`-=1:2`
molar ratio `C:H::1:2`
General formula `C_xH_(2x)O_y+O_2rarrxCO_2+xH_2O`
Oxygen atom in product `=3x`
According to question.
`(3x)/(2)=y`,`3x=2y`
`y=(3x)/(2)`
Formula is `C_xH_(2x)O_((3x)/(2))`, simplifying we get `C_2H_4O_3`.
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An organic compound contains carbon, hydrogen and oxygen. If the ratio percentage of C and H is 6:1 calculate the simplest formula of the compound, given that one molecule of the compound contains half as much oxygen as would be required to burn all the carbon and hydrogen atoms in it to CO_2 and H_2O

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Knowledge Check

  • An organic compound contains C,H and O . If C (%):H^(%) = 6:1 , what is the simplest formula of the compound, given that one mole of the compound contains half as much oxygen as would be required to burn all the C and H atoms in it to CO_(2) and H_(2)O ?

    A
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    B
    `C_(2)H_(2)O_(3)`
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    D
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  • An organic compound contains carbon, hydrogen and oxygen. Its elemental analysis gave C, 38.71% and H, 9.67%. The empirical formula of the compound would be

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    D
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  • An organic compound contains carbon, hydrogen and oxygen. Its elemental analysis gave C,38.71% and H,9.67% . The empirical formula of the compound would be :

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