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Grade 12th passPhysical Chemistry

Combustion of 6.51 mg of an organic compound gave 20.46 mg of carbon dioxide and 8.36 mg of water. The molecular weight was found to be 84. Calculate the: a) mass of C, b) %C, c) mass of H; d) %H

Profile image of sophia
5 Years agoGrade 12th pass
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1 Answer

Profile image of Govind Sharma
5 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

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