Question

A 20.49 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 20.03 grams of CO₂ and 4.100 grams of H₂O are produced.

In a separate experiment, the molar mass is found to be 90.04 g/mol. Determine the empirical formula and the molecular formula of the organic compound.

Answer 1

Answer – Given, mass of compound = 20.49 g , mass of CO₂ = 20.03 g ,

mass of H₂O = 4.100 g

Now calculating the moles of CO₂ and H₂O

Moles of CO₂ = 20.03 g / 44.0 g.mol^-1 = 0.455 mole

Moles of H₂O = 4.100 g / 18.015 g.mol^-1 = 0.228 moles

Moles of C -

1 moles of CO₂ = 1 moles of C

So, 0.455 moles of CO₂ = ?

= 0.455 moles of C

Moles of H -

1 moles of H₂O = 2 moles of H

So, 0.228 moles of H₂O = ?

= 0.455 moles of H

Mass of C = 0.455 moles * 12.011 g/mol

                  = 5.47 g of C

Mass of H = 0.455 moles * 1.0079 g/mol

                  = 0.459 g of H

We know, total mass of compound = mass of C + mass of H + mass of O

20.49 g = 5.47 g + 0.459 g + mass of O

Mass of O = 20.49 g – 5.47 g – 0.459 g

                  = 14.56 g of O

Moles of O = 14.56 g / 15.998 g.mol^-1

                   = 0.910 moles of O

Now we need to divided each mole by this mole of H or C due to the moles of H or C atom is smallest

C = 0.455 /0.455 = 1

H = 0.455 / 0.455 = 1

O = 0.910 /0.455 = 2

So empirical formula is CHO₂

Now we need to calculate the molecular formula

We know formula

Molecular formula = n* empirical formula

n = molecular mass / empirical formula mass

      = 90.04 /45.02

      = 2

So, molecular formula = 2* CHO₂

                                     = C₂H₂O₄