Question

A 15.44 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 21.89 grams of CO₂ and 13.44 grams of H₂O are produced.

In a separate experiment, the molar mass is found to be 62.07 g/mol. Determine the empirical formula and the molecular formula of the organic compound.

Empirical Formula?

Molecular Formula?

Answer 1

Empirical formula is the least coefficient formula, that is

CxHyOz

we must find x,y,z via gravimetry.

Typically we do this relating to moles of C,H,O

mol of CO2 = mass of CO2/MW of CO2 = (21.89)/(44) = 0.4975 mol of CO2

1 mol of CO2 = 1 mol of C ---> 0.4975mol of CO2 = 0.4975mol of C

mol of H2O = mass of H2O/MW of H2O= (13.44)/(18) = 0.7466 mol of H2O

1 mol of H2O= 2 mol of H ---> 0.00981 mol of H2O= 0.7466*2 = 1.4932 mol of H

Now... mol of Oxygen:

Mass of O = Mass of sample - Mass of C - Mass of H

Mass of C = mol of C * MW of C = 0.4975 *12 = 5.97 g

Mass of H = mol of H * MW of H = 1.4932*1= 1.4932 g

Mass of O = 15.44- 5.97 -1.4932= 7.9768 g of O

mol of O = mass of O / MW of O = (7.9768)/(16) = 0.49855 mol of O

Ratios:

C:O = 0.4975 /0.49855 = 0.99 = 1

H:O =1.4932 /0.49855 = 2.99 = 3

H:C = 1.4932 /0.4975 = 3.01 = 3

then

C1H3O1

CH3O

empirical = CH3O

molar mass = 12 + 3*1 + 16 = 31 g/mol

now... molar mass s actually 62

ratio = 62/31 = 2x

then

C2H6O2 is the actual molecular formula