Grade 12th passPhysical Chemistry

combustion of 6.51 mg of a compound gave 20.47 mg of carbon dioxide and 8.3 mg of water. The molecular weight was found to be 84 gm/mole.
Calculate molecular formula of the compound
(Atomic Weight of C=12.01, H=1.008, O=16.0 gm/mole)

Devendra

8 Years agoGrade 12th pass

1 Answer

Arun

8 Years ago

moles CO2 = mass / molar mass = 20.47 mg/44.01 g/mol = 0.4651 mmol
moles C = moles CO2 = 0.4651 mmol
mass C = 0.4651 mmol x 12.01 g/mol = 5.586 mg
% mass C = 5.586 mg / 6.51 mg x 100/1 = 85.8 %

moles H2O = 8.36 mg / 18.016 g/mol = 0.4640 mmol
moles H = 2 x moles H2O = 0.9281 mmol
% mass H = 100% - 85.8 % = 14.2 %

% composition
85.8% C, 14.2 % H

ratio moles C : moles H
= 0.4651 : 0.9281

divide each number in the ratio by the smallest number
0.4651 / 0.4651 : 0.9281 / 0.9281
= 1 : 2

empirical formula
CH2

to determine the molecular formula divide the molar mass by the mass of the empirical formula
= 84 g/mol / 14 g/mol
= 6

there are 6 empirical units in the molecular formula

molecular formula = 6 x (CH2)
= C6C12