Question

A 7.364 gram sample of an organic compound containing only C, H, and O is analyzed by combustion analysis and 9.343 g CO₂ and 2.550 g H₂O are produced.

In a separate experiment, the molar mass is found to be 104.1 g/mol. Determine the empirical formula and the molecular formula of the organic compound.

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

= 9.343/44

= 0.2123

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

= 2.55/18

= 0.1417

Since 1 mol of CO2 has 1 mol of C

Number of moles of C in CO2= 0.2123

so, x = 0.2123

Since 1 mol of H2O has 2 mol of H

Number of moles of H = 2*0.1417 = 0.2833

Molar mass of O = 16 g/mol

mass O = total mass - mass of C and H

= 7.364 - 0.2123*12 - 0.2833*1

= 4.533

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

= 4.533/16.0

= 0.2833

so, z = 0.2833

Divide by smallest:

C: 0.2123/0.2123 = 1

H: 0.2833/0.2123 = 4/3

O: 0.2833/0.2123 = 4/3

multiply by 3 to get simplest whole number ratio:

C: 1*3 = 3

H: 4/3* 3 = 4

O: 4/3* 3 = 4

So empirical formula is:C3H4O4

Molar mass of C3H4O4,

MM = 3*MM(C) + 4*MM(H) + 4*MM(O)

= 3*12.01 + 4*1.008 + 4*16.0

= 104.062 g/mol

Now we have:

Molar mass = 104.1 g/mol

Empirical formula mass = 104.062 g/mol

Multiplying factor = molar mass / empirical formula mass

= 104.1/104.062

= 1

So molecular formula is:C3H4O4

Answer: C3H4O4