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Grade: 12th pass

                        

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)

2 years ago

Answers : (1)

Arun
24742 Points
							
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
2 years ago
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