1.

0.45 g of an organic compound gave 0.44 g of CO_(2) and 0.09 g of H_(2)O. The molecular mass of the compound is 90 amu. Calculate the molecular formula.

Answer» <html><body><p><br/></p>Solution :Step I. Percentage of elements <br/> Percentage of carbon `= (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> ("Mass of "CO_(2))/("Mass of compound")xx100` <br/> `=(12)/(44)xx(0.44)/(0.45)xx100=26.67` <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> `=(2)/(18)xx("Mass of" H_(2)O)/("Mass of compound")xx100` <br/> `=(2)/(18)xx(0.09)/(0.45)xx100=2.22` <br/>Percentage of oxygeb `= 100 - (26.67 + 2.22) = 100-28.89 = 71.11` <br/> Step II. Empirical formula of the compound <br/> `{:("<a href="https://interviewquestions.tuteehub.com/tag/element-969236" style="font-weight:bold;" target="_blank" title="Click to know more about ELEMENT">ELEMENT</a>","Percentage","Atomic mass","Gram atoms (Moles)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("C",26.67,12,(26.67)/(12)=2.22,(2.22)/(2.22)=1.0,1),("H",2.22,1,(2.22)/(1)=2.22,(2.22)/(2.22)=1.0,1),("O",71.11,16,(71.11)/(16)=4.44,(4.44)/(2.22)=2.0,2):}` <br/> Empirical formula of the compound `= CHO_(2)`<br/> Step <a href="https://interviewquestions.tuteehub.com/tag/iii-497983" style="font-weight:bold;" target="_blank" title="Click to know more about III">III</a>. Molecular formula of the compound <br/> Empirical formula mass `= 12+1 +2 xx 16 = 45 u` <br/> Molecular mass = 90 <a href="https://interviewquestions.tuteehub.com/tag/amu-25369" style="font-weight:bold;" target="_blank" title="Click to know more about AMU">AMU</a> (Given) <br/> `n=("Molecular mass")/("Empirical formula mass")=((90u))/((45u))=2` <br/>Molecular formula `= n xx` Empirical formula `= 2 xx CHO_(2)=C_(2)H_(2)O_(4)`.</body></html>


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