Question

A combustion analysis of 5.214 g of a compound yields 5.34 g CO2​, 1.09 g H2​O, and 1.70 g N2​. If the molar mass of the compound is 129.1 g/mol, what is the chemical formula of the compound?

Answer 1

mass of CO2 = 5.34 g

moles of CO2 = 5.34 / 44 = 0.1214

moles of C = 0.1214

moles of H2O = 1.09 / 18.02 = 0.0605 mol

moles of H = 0.121 mol

moles of N2 = 1.70 / 28 = 0.0607

moles of N = 0.1214

mass of O = 1.9366

moles of O = 0.121

ratio :    C         H           N            O

         0.1214     0.121    0.1214      0.121

              1              1            1            1

Emperical formula = CHNO

mass of emperical formula = 43 g/mol

n = 129 / 43 = 3

chemical formula = C₃H₃N₃O₃