9.9g of amide with molecular formula C_(4)H_(5)N_(x)O_(y) on heating with alkali liberated 1.7g of ammonia. If the percentage of oxygen is 33.33% then the ratio of 'N' and 'O' atoms in the compound is:

9.9g of amide with molecular formula C_(4)H_(5)N_(x)O_(y) on heating w

1.	9.9g of amide with molecular formula C_(4)H_(5)N_(x)O_(y) on heating with alkali liberated 1.7g of ammonia. If the percentage of oxygen is 33.33% then the ratio of 'N' and 'O' atoms in the compound is:
Answer» <html><body><p>`1:1`<br/>`1:2`<br/>`2:3`<br/>`3:2`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Amide `C_(4)H_(5)N_(x)O_(y)` will give x mol `NH_(3)`. <br/> `:.` <a href="https://interviewquestions.tuteehub.com/tag/molecular-562994" style="font-weight:bold;" target="_blank" title="Click to know more about MOLECULAR">MOLECULAR</a> <a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> of amide `=(9.9)/(1.7) xx 17x = 99x` <br/> % <a href="https://interviewquestions.tuteehub.com/tag/nitrogen-1118291" style="font-weight:bold;" target="_blank" title="Click to know more about NITROGEN">NITROGEN</a> in the amide `= (14x)/(99x) xx 100 = <a href="https://interviewquestions.tuteehub.com/tag/14-272882" style="font-weight:bold;" target="_blank" title="Click to know more about 14">14</a>.14` <br/> Ratio of number of atoms of 'N' and 'O' <br/> `= (14.14)/(14): (33.33)/(16) = 1:2`</body></html>