Question

Combustion analysis of a compound yielded 235.50 g CO2, 72.33 g H2O, 61.57 g NO2, and 85.72 g SO2. (a) What is the empirical formula of the compound? (Assume it contains no oxygen.) chemPad Help C2H2NO Your answer appears to substitute one element for another. (b) If the molar mass of the compound is 200.34 g/mol, what is the molecular formula of the compound?

Answer 1

Number of moles of CO₂, n = mass/molar mass

                                        = 235.50 g / 44(g/mol)

                                        = 5.35 moles

So number of moles of C is 5.35 mol

Number of moles of H₂O is , n' = mass/molar mass

                                             = 72.33 g / 18(g/mol)

                                            = 4.02 moles

So the number of moles of H = 2x4.02 = 8.04 mol

Number of moles of NO₂ is , n'' = mass/molar mass

                                              = 61.57 g / 46(g/mol)

                                               = 1.34 mol

So number of moles of N = 1.34 mol

Number of moles of SO₂ is , n"' = mass/molar mass

                                               = 85.72 g / 64(g/mol)

                                               = 1.34 moles

So the number of moles of S is 1.34 mol

The ratio of number of moles of C : H : N : S = 5.35 : 8.04 : 1.34 : 1.34

                                                           = 4 : 6 : 1 : 1.34

So the emperical formula is C₄H₆N₁S₁

Emperical formula mass = (4xAt.mass of C) + (6xAt.mass of H) + At.mass of N +At.mass of S

                                    = (4x12) + (6x1) +14+ 32

                                    = 100

Given Molar mass is 200.34

So n = Molar mass / Emperical formula mass

       = 200.34 / 100

       = 2

So Molecular formula is = 2(C₄H₆N₁S₁ )

                                   = C₈H12N₂S₂