Question

A 12 g sample of a smelly compound was tested by combustion analysis. The products were 21.41 g carbon dioxide, 14.59 g water, and 17.51 g N2O5. Further analysis showed that oxygen was NOT present in the molecule. What is the empirical formula of the compound?​

Answer 1

mass of CO2 = 21.41 g

moles of CO2 = 21.41 / 44.01 = 0.4865

moles of C = 0.4865

mass of water = 14.59 g

moles of water = 14.59 / 18.02 = 0.8097

moles of H = 1.6194

moles of N2O5 = 17.51 / 108 = 0.162 mol

moles of N = 0.324

ratio :       C          H        N

           0.4865     1.6194     0.324

              3/2          5               1

Emperical formula = C₃H₁₀N₂