Question

A 4.626 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 9.968 grams of CO₂ and 4.082 grams of H₂O are produced.

In a separate experiment, the molar mass is found to be 102.1 g/mol. Determine the empirical formula and the molecular formula of the organic compound.

• Enter the elements in the order C, H, O
  

  empirical formula =

  ____

  molecular formula =

  _____

Answer 1

Molar mass of CO2 = 44.01 gm/mol

Number of moles of CO2 = Mass/Molar mass = 9.968/44.01 = 0.2265 moles

Number of moles of Carbon present in sample = Number of moles of CO2 formed = 0.2265 moles

Molar mass of H2O = 18.01 gm/mol

Number of moles of water = Mass/Molar mass = 4.082/18.01 = 0.2266 moles

1 mole of H2O requires two moles of hydrogen in the sample, hence moles of Hydrogen in sample will be 0.2266 * 2 = 0.4533 moles

Total mass of sample = Mass of C + Mass of O + Mass of H

4.626 = 0.2265 * 12 .01 + Mass of O + 0.4533 * 1

Mass of O = 1.454

Number of moles of O = Mass/Molar mass = 1.454/16 = 0.090875

Hence the ratio will be C:H:O is 0.2266:0.4533:0.090875, which is 2.5:5:1 or 5:10:2

Hence the empirical formula will be C5H10O2, which is having the mass of 102.1 g/mol, hence the empirical formula must be same as molecular formula

Hence the molecular formula will be C5H10O2