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A compound of C, H and N contains the three element in the ratio of 9 : 1 : 3.5. Calculate the molecular formula given that the molecular mass of the compound is 108 amu Hint : Percentgae of C=(9)/(13.5)xx100=66.7,H=(1)/(13.5)xx100=7.4,N=(3.5)/(13.5)xx100=25.9.

Answer» <html><body><p><br/></p>Solution :Step I. Empirical formula of the compound <br/> `{:("Element","<a href="https://interviewquestions.tuteehub.com/tag/percentage-13406" style="font-weight:bold;" target="_blank" title="Click to know more about PERCENTAGE">PERCENTAGE</a>","Atomic mass","Gram <a href="https://interviewquestions.tuteehub.com/tag/atoms-887421" style="font-weight:bold;" target="_blank" title="Click to know more about ATOMS">ATOMS</a> (Moles)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("C",66.7,12,(66.7)/(12)=5.56,(5.6)/(1.85)=3.0,3),("H",7.4,1,(7.4)/(1)=7.40,(7.40)/(1.85)=4.0,4),("N",25.9,14,(25.9)/(14)=1.85,(1.85)/(1.85)=1.0,1):}` <br/> Empirical formula of compound `= C_(3)H_(4)N` <br/> Step II. Molecular formula of the compound <br/> Empirical formula mass `= 3 xx 12 + 4 xx 1 + 14 = 54 <a href="https://interviewquestions.tuteehub.com/tag/u-1435036" style="font-weight:bold;" target="_blank" title="Click to know more about U">U</a>` <br/> Molecular mass = 108 <a href="https://interviewquestions.tuteehub.com/tag/amu-25369" style="font-weight:bold;" target="_blank" title="Click to know more about AMU">AMU</a> (<a href="https://interviewquestions.tuteehub.com/tag/given-473447" style="font-weight:bold;" target="_blank" title="Click to know more about GIVEN">GIVEN</a>) <br/> `nn=("Molecular mass")/("Empirical formula mass")=((108u))/((54u))=2` <br/> `:.` Molecular formula `= 2 xx C_(3)H_(4)N=C_(6)H_(8)N_(2)`.</body></html>


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