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The ratio of mass percent of C and H of an organic compound (CXHYOZ) is 6 : 1. If one molecule of the above compound
(CXHYOZ) contains half as much oxygen as required to burn one molecule of compound CXHY completely to CO2 and
H2O. The empirical formula of compound CXHYOZ is :

  • Option 1)

    C2H4O3
     

  • Option 2)

    C3H6O3
     

  • Option 3)

    C2H4O
     

  • Option 4)

    C3H4O2

 

Answers (2)

best_answer

As we learned

Empirical formula -

An Empirical formula represents the simplest whole number ratio of various atoms present in a compound.

- wherein

for glucose, empirical formula is CH2O

 

 

 

 

Mass ratio of C & H is 6 :

So mole ratio of C & H is 1 : 2

\therefore   X : Y = 1 : 2

To burn one molecule of C_{X}H_{Y}

CH_{2}+\frac{3}{2}O_{2}\rightarrow CO_{2}+H_{2}O

\frac{3}{2} molecule of O_{2} is required i.e. 3 atoms of O are required so X:Y:Z= 1:2:\frac{3}{2}\Rightarrow X:Y:Z= 2:3:4

So empirical formula is C_{2}H_{4}O_{3}


Option 1)

C2H4O3
 

Option 2)

C3H6O3
 

Option 3)

C2H4O
 

Option 4)

C3H4O2

Posted by

prateek

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