0.30 g of an organic compound containing C, H and O on combustion yields 0.44 g CO_(2) and 0.18 g H_(2)O. If one mole of compound weighs 60, then molecular formulaof the compound is

0.30 g of an organic compound containing C, H and O on combustion yiel

1.	0.30 g of an organic compound containing C, H and O on combustion yields 0.44 g CO_(2) and 0.18 g H_(2)O. If one mole of compound weighs 60, then molecular formulaof the compound is
Answer» <html><body><p>`C_(3)H_(8)O`<br/>`C_(2)H_(4)O_(2)`<br/>`CH_(2)O`<br/>`C_(4)H_(<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>)O`</p>Solution :`%<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a> = (12)/(44) <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> (0.44)/(0.30) xx 100 = 40` <br/> `%H = (2)/(18) xx (0.8)/(0.30) xx 100 = 6.66` <br/> `%O = 100 - (40.0 + 6.66) = 53.34` <br/> Now `C:H:O = (40)/(12) : (6.66)/(1.0) : (53.34)/(16)` <br/> = 3.33: 6.66 : 3.33 <br/> `:. E.F. = CH_(2)O` <br/> E.F. wt. = 12+2+16 = 30 <br/> But Mol. wt. = <a href="https://interviewquestions.tuteehub.com/tag/60-328817" style="font-weight:bold;" target="_blank" title="Click to know more about 60">60</a> <br/> `:. M.F. = CH_(2)O xx (60)/(30) = C_(2)H_(4)O_(2)`</body></html>