1.

A compound contains `42.3913 % K,15.2173 % Fe,19.5652 % C and 22.8260 % N`. The molecular mass of the compound is 368 u. Find the molecular formula of the compound.

Answer» Step I : Determination of empirical formula of the compound
`{("Element","Percentage","Atomic mass","Gram atoms (Moles)","Atomic ratio (Molar ratio)","Simplest whole. no ratio"),("K",42.3913,39,(42.3913)/(39)=1.087,(1.087)/(0.2717)=4,4),("Fe",15.2173,56,(15.2173)/(56)=0.2717,(0.2717)/(0.2717)=1,1),("C",19.5652,12,(19.5652)/(12)=1.630,(1.630)/(0.2717)=6,6),("N",22.8260,14,(22.8260)/(14)=1.630,(1.630)/(0.2717)=6,6):}`
The empirical formula of the compound `= K_(4)FeC_(6)N_(6)`
Step II. Determination of molecular formula of the compound
Empirical formula mass `= 4xx39 +56 + 6 xx 12 + 6 xx 14`
`= 156 + 56 + 72 + 84 = 368` u
Molecular mass = 368 u (Given) `:. n=("Molecular mass")/("Empirical formula mass")=(368)/(368)=1`
Molecular formula `= n xx` Empirical formula `= 1 xx (K_(4) FeC_(6)N_(6)) or K_(4)[Fe(CN)_(6)]`
The name of the compounbd is Potassium ferrocyanide.


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