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 = $52.93 g *\frac{1 mol}{12gC} =4.41 mol C$

Mass of H = 5.92 g

=> Moles of H=$5.92g*\frac{1mol}{1.01g}=5.86molH$

Mass of N = 41.15 g

=>Moles of N=$41.15g*\frac{1 mol}{14 g}=2.94mol N$

Simplest mole ratio of the atoms in the compound:

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

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

Multiple n = $\frac{204.2}{68.04}=3$

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