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0.1152 g of a compound containing carbon, hydrogen, nitrogen and oxygen are burned in...
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
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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
Posted by jeew-m on June 6, 2013 at 8:00 PM (Answer #1)
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