A non-volatile hydrocarbon has 5.6% hydrogen. Determine the molecular formula of the non-volatile hydrocarbon (Molecular mass of benzene is 78 a.m.u.)

Answer

`c_(14)H_(10)`

Explanation

Step I. Calculation of empitical fromula of the hydrocarbon.
in hydrocarbon, H=5.6% , C=100-5.6= 94.4 %

Empirical formula of hydrocarbon `=C_(7)H_(5)`
`"Empirical formula mass of hydrocarbon"=7xx12+5xx1=89" amu"=89 g`
Step II. Calculation of molecular mass of the hydrocarbon.
According to Raoult's law,
`(P_(A)^(@)-P_(S))/(P_(A)^(@))=X_(B)=n_(B)/n_(A)`
`(P_(A)^(@)-P_(S))/(P_(A)^(@))=1.306xx10^(-2),W_(B)=3g, W_(A)=100g, M_(A)=78" amu"=78" g mol"^(-1)`
`1.306xx10^(-2)=(W_(B)//M_(B))/(W_(A)//M_(A))=((3g)//M_(B))/((100g)//(78" g mol"^(-1)))=(3xx78(g//mol^(-1)))/(100xx(M_(B)))`
`M_(B)=(378(g//mol^(-1)))/(100xx(1.306xx10^(-2)))=179.17("g mol"^(-1)).`
Step III. Calculation of molecular formula of the hydrocarbon.
`n=("Molecular mass of hydrocarbon")/("Empirical formula mass of hydrocarbon")=(179.17)/(89.0)~~2.
`"Molecular formula"=nxx"Empirical formula"=2C_(7)H_(5)=C_(14)H_(10).`