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.

A compound of C, H and N contains the three element in the ratio of 9

1.	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» Solution :Step I. Empirical formula of the compound `{:("Element","PERCENTAGE","Atomic mass","Gram ATOMS (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):}` Empirical formula of compound `= C_(3)H_(4)N` Step II. Molecular formula of the compound Empirical formula mass `= 3 xx 12 + 4 xx 1 + 14 = 54 U` Molecular mass = 108 AMU (GIVEN) `nn=("Molecular mass")/("Empirical formula mass")=((108u))/((54u))=2` `:.` Molecular formula `= 2 xx C_(3)H_(4)N=C_(6)H_(8)N_(2)`.