Question

A 5.274 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 11.99grams of CO₂ and 4.909 grams of H₂O are produced.

In a separate experiment, the molar mass is found to be 116.2 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

let in compound number of moles of C, H and O be x, y and z respectively

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

= 11.99/44

= 0.2725

Number of moles of H2O = mass of H2O / molar mass H2O

= 4.909/18

= 0.2727

Since 1 mol of CO2 has 1 mol of C

Number of moles of C in CO2= 0.2725

so, x = 0.2725

Since 1 mol of H2O has 2 mol of H

Number of moles of H = 2*0.2727 = 0.5454

Molar mass of O = 16 g/mol

mass O = total mass - mass of C and H

= 5.274 - 0.2725*12 - 0.5454*1

= 1.459

number of mol of O = mass of O / molar mass of O

= 1.459/16.0

= 9.116*10^-2

so, z = 9.116*10^-2

Divide by smallest to get simplest whole number ratio:

C: 0.2725/9.116*10^-2 = 3

H: 0.5454/9.116*10^-2 = 6

O: 9.116*10^-2/9.116*10^-2 = 1

So empirical formula is:C3H6O

Molar mass of C3H6O,

MM = 3*MM(C) + 6*MM(H) + 1*MM(O)

= 3*12.01 + 6*1.008 + 1*16.0

= 58.078 g/mol

Now we have:

Molar mass = 116.2 g/mol

Empirical formula mass = 58.078 g/mol

Multiplying factor = molar mass / empirical formula mass

= 116.2/58.078

= 2

So molecular formula is:C6H12O2

empirical formula is:C₃H₆O

molecular formula is:C₆H₁₂O₂