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0.1152 g of a compound containing carbon, hydrogen, nitrogen and oxygen are burned in...

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0.1152 g of a compound containing carbon, hydrogen, nitrogen and oxygen are burned in excess oxygen. The gases produced are treated further to convert nitrogen-containing products into N2. The resulting mixture of CO2, H2O and N2 and excess O2 is passed through a CaCl2 drying tube, which gains 0.09912 g. The gas stream is bubbled through water where the CO2 forms H2CO3. Titration of this solution to the second endpoint with 0.3283 M NaOH requires 28.81 mL. The excess O2 is removed by reaction with copper metal and the N2 is collected in a 225.0 mL measuring bulb where it exerts a pressure of 65.12 mmHg at 25 ˚C. In a separate experiment the molar mass of this compound is found to be approximately 150 g/mol.

Find the molecular formula of the compound.

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I have answered the previous part of the question by;


It will give you the amount of `CO_2,H_2O` and `N_2` amounts as follows.

Amount of `CO_2 =0.00473mols`

Amount of `H_2O = 0.0055mols`

Amount of `N_2 = 0.000788mols`


C,H,N in the compound will be subjected to following reactions.

`C+O_2 rarr CO_2`

`2H+1/2O_2 rarr H_2O`

`2N rarr N_2`


`C:CO_2 = 1:1`

`H:H_2O = 2:1`

`N:N_2 = 2:1`


Amount of C in the compound `= 0.00473mols`

Amount of H in the compound`= 2xx0.0055 = 0.011mols`

Amount of N in the compound

`= 2xx0.000788`

`= 0.001576mols`

mass of C in the compound

`= 0.00473xx12`

`= 0.05676g`

mass of H in the compound

`= 0.011xx1`

`= 0.011g`


mass of N in the compound

`= 0.001576xx14`

`= 0.02206g`


It is given that the mass of the compound is 0.1152g.


Mass of O in the compound 

`= 0.1152-0.02206-0.011-0.05676`

`= 0.02538g`


Moles of O in the compound `= 0.02538/16 = 0.001586`


Mole ratio

`C:H:O:N = 0.00473:0.011:0.001586:0.001576`

`C:H:O:N = 0.00473/0.001576:0.011/0.001576:0.001586/0.001576:0.001576/0.001576`

`C:H:O:N = 3:7:1:1`

So empirical formula is `C_3H_7NO`

It is given that the molecular formulae is 150g/mol.

`n(12xx3+1xx7+14xx1+16) = 150`

`n = 150/73`


So the molecular formulae of the compound is `C_6H_14N_2O_2`


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