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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

a) Calculate the number of moles of
i. H2O
ii. CO2
iii. N2

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Here we have to understand the test clearly.

First of all the compound consist of C,H,N and O.

After burning with excess `O_2` we will get `CO_2,H_2O` and` N_2` .

When the gaseous mixture is passed through `CaCl_2` it will absorb all the moisture in the mixture or in other words `H_2O` .

When the mixture with `H_2CO_3` reacted with NaOH it will react at two stages.

`CO_2+H_2O rarr H_2CO_3`

`H_2CO_3+NaOH rarr NaHCO_3+H_2O`

`NaHCO_3+NaOH rarr Na_2CO_3+H_2O`

At the second stage of the reaction all the `H_2CO_3` will be reacted with NaOH.

`H_2CO_3+2NaOH rarr Na_2CO_3+2H_2O`

So the elements in the compound will be converted to the following during burning.

`C rarr CO_2`

`H rarr H_2O`

`N rarr N_2`


All the C in the compound is consumed for the reaction with NaOH.

All the H in the compound was removed as `H_2O` by `CaCl_2` .

All the N in the compound was collected as `N_2` gas.


Determination of `CO_2`

Amount of NaOH used `= 0.3283xx28.81/1000 = 0.00946`

`H_2CO_3:NaOH = 1:2`


So amount of `H_2CO_3` reacted  `= 0.00473 mol`

Amount of `CO_3^(2-)` is same as amount of `CO_2` in the compound.

`CO_2` moles present in the medium is 0.00473moles.


Determination of H_2O

Amount of `H_2O` collected = 0.09912g

Amount of `H_2O` collected = 0.09912/18 mol

Amount of `H_2O` collected = 0.0055 mol


`H_2O` moles present in the medium is 0.0055moles.

Determination of `N_2`

Using `PV = nRT`

`P = 65.12xx0.001316 = 0.0857atm`

`V = 225ml = 0.225L`

`R = 0.0821(Latm)/(molK)`

`T = 298K`


`n = (0.0857xx0.225)/(0.0821xx298)`

`n = 0.000788 mol`


`N_2` moles present in the medium is 0.000788moles.


`N_2` act as a ideal gas

No wastage of material during the process


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