Question

A 5.797 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 9.656 grams of CO₂ and 3.163 grams of H₂O are produced.

In a separate experiment, the molar mass is found to be 132.1 g/mol. Determine the empirical formula and the molecular formula of the organic compound.

**** Do the problem in Conversion factor style!

Answer 1

Calculate mol of C:

mol of CO2 = mass/MW = 9.656/44 = 0.21945

mol of H2O = mass/MW = 3.163/18 = 0.1757

total mol of:

C --> 0.21945

H --> 2*0.1757 = 0.3514

C + O2 --> CO2

2H + 1/2O2 --> H2O

Mol of O used from combustion --> 0.21945 + 0.3514 = 0.57085

Total amount of mol = 0.21945*2 + 0.1757*2 - 0.57085 = 0.21945 mol of O

then

C:O ratio = 1

C:H ratio = 0.3514/0.21945 = 1.601275

let it be

C1O1H2

MW of empirical formula = CH2O = 12+2+16 = 30

x times = MW of compund / MW empirical = 132.1/30 = 4.4033

nearest --> 4 --> 9

C4O4H8