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0.2004 g of glucose gave on combustion 0.2940 g of CO_(2) and 0.1202 g of H_(2)O. Find the percentage composition.

Answer» <html><body><p></p>Solution :Weight of <a href="https://interviewquestions.tuteehub.com/tag/organic-1138713" style="font-weight:bold;" target="_blank" title="Click to know more about ORGANIC">ORGANIC</a> compound = 0.2004 g <br/> Weight of carbon dioxide = 0.2940 g <br/> Weight of water = 0.1202 g <br/> Percentage of carbon <br/> 44 g of <a href="https://interviewquestions.tuteehub.com/tag/co2-409421" style="font-weight:bold;" target="_blank" title="Click to know more about CO2">CO2</a> contains, carbon = 12 g <br/> `"0.2940 g of "CO_(2)" contains, carbon"=(12xx0.2940)/(44)xx(100)/(0.2004)=40.01` <br/> Percentage of <a href="https://interviewquestions.tuteehub.com/tag/hydrogen-22331" style="font-weight:bold;" target="_blank" title="Click to know more about HYDROGEN">HYDROGEN</a> <br/> <a href="https://interviewquestions.tuteehub.com/tag/18-278913" style="font-weight:bold;" target="_blank" title="Click to know more about 18">18</a> g of `H_(2)O` contains, hydrogen = 2 g <br/> 0.1202 of `H_(2)O` contains, hydrogen `=(2xx0.1202)/(18)` <br/> Percentage of hydrogen `=(2xx0.1202)/(18)xx(100)/(0.2004)=6.66` <br/> Percentage of <a href="https://interviewquestions.tuteehub.com/tag/oxygen-1144542" style="font-weight:bold;" target="_blank" title="Click to know more about OXYGEN">OXYGEN</a> is therefore by difference, <br/> `=[100-(40.01+6.66)]=53.33`</body></html>


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