Question

Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produced 39.61 g CO2 and 9.01 g H2O. The molar mass of the unknown compound is 268.34 g/mol . Find the molecular formula of the unknown compound.

Answer 1

Sol :-

Number of moles of CO2 = Mass of CO2 in g / Gram molar mass of CO2

= 39.61 g / 44 g/mol

= 0.900 mol

So, Number of moles of C in 0.900 mol of CO2 = 0.900 mol of C

Mass of C = Moles of C x Gram atomic mass of C

= 0.900 mol x 12 g/mol

= 10.8 g

Also,

Number of moles of H2O = 9.01 g / 18 g/mol

= 0.500 mol

Number of moles of H in 0.500 mol of H2O = 2 x 0.500 mol = 1.00 mol

Mass of H = 1.00 mol x 1 g/mol = 1 g

Now, sum of mass of C and H = 10.8 g + 1 g = 11.8 g

Mass of O in sample = 13.42 g - 11.8 g = 1.62 g

Moles of O = 1.62 g / 16 g/mol

= 0.10125 mol

Now,

Molar ratio of C : H : O = 0.900 : 1.00 : 0.10125

= 8.9 : 9.9 : 1

Simplest whole number ratio of C : H : O = 9 : 10 : 1

So,

Empirical formula of the compound = C₉H₁₀O

Empirical mass = 9 x 12 + 10 x 1 + 1 x 16 = 134 g/mol

Given molecular mass = 268.34 g/mol

Now,

Simplest whole number (n) = Molecular mass / Empirical mass

= 268.34 g/mol / 134 g/mol

= 2

Therefore,

Molecular formula = n ( Empirical formula)

= 2 (C₉H₁₀O)

= C₁₈H₂₀O₂

Hence, Molecular formula of the compound = C₁₈H₂₀O₂