Question

Triethylenemelamine is used in the plastics industry and as an anticancer drug. Its analysis is 52.93% carbon, 5.92 % hydrogen, 41.15% nitrogen. The molecular mass of triethylenemelamine is 204.2 g/mol. What is its empirical formula? What is its molecular formula?

Answer 1

Assuming total mass to be 100 g

Mass of C = 52.93 g

=> Moles of C =
`52.93 g *(1 mol)/(12gC) =4.41 mol C`

Mass of H = 5.92 g

=> Moles of H=
`5.92g*(1mol)/(1.01g)=5.86molH`

Mass of N = 41.15 g

=>Moles of N=
`41.15g*(1 mol)/(14 g)=2.94mol N`

Simplest mole ratio of the atoms in the compound:

`C_{(4.41mol)/(2.94mol) } H_{(5.86mol)/(2.94mol) }N_{(2.94mol)/(2.94mol) } =C_(1.5)H_(2)N_(1) or(C_(1.5)H_(2)N_(1))_(2)=>C_(3)H_(4)N_(2)`

Empirical formula mass=
`(3*12)+(4*1.01)+(2*14)=68.04g`

Multiple n =
`(204.2)/(68.04)=3`

Molecular formula=
`(Empirical formula)_(n) =(C_(3)H_(4)N_(2))_(3)=C_(9)H_(12)N_(6)`