0.3g of an organic compound on combustion liberated 0.18g of water vapour and 0.44 g of carbonioxide. Calculate the percentage composition of the compound

0.3g of an organic compound on combustion liberated 0.18g of water vap

1.	0.3g of an organic compound on combustion liberated 0.18g of water vapour and 0.44 g of carbonioxide. Calculate the percentage composition of the compound
Answer» <html><body><p></p>Solution :Weight of the compound =W = 0.3g <br/> Weight of `CO_(2)= W_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>) = 0.44g` <br/> Weight of `H_(2)O = W_(2) = 0.18g` <br/> % of carbon `=(W_(1) xx 12 xx 100)/(W xx 44) = (0.44 xx 12 xx 100)/(0.3 xx 44) = 40.0%` <br/> % of hydrogen `= (W_(2) xx 2xx 100)/(W xx <a href="https://interviewquestions.tuteehub.com/tag/18-278913" style="font-weight:bold;" target="_blank" title="Click to know more about 18">18</a>) = (0.18 xx 2 xx 100)/(0.3 xx 18) = 6.67%` <br/> After calculating the percentage <a href="https://interviewquestions.tuteehub.com/tag/composition-22493" style="font-weight:bold;" target="_blank" title="Click to know more about COMPOSITION">COMPOSITION</a> of the given elements in the compound, if their sum is not 100, then the difference is assumed as oxygen <br/> `:.` % of oxygen `=[100-(40+6.67)]= 53.33%`</body></html>