Chemical Forums

Chemistry Forums for Students => High School Chemistry Forum => Topic started by: aduke on July 25, 2005, 04:00:14 PM

Title: Combustion of unkown compound
Post by: aduke on July 25, 2005, 04:00:14 PM

Here's the problem:

A compound contains only C, H, N, and O. Combustion of 0.157 g of the compound produces 0.213 g CO2 and 0.0310 g H2O.  In another experiment, it is found that 0.103 g of the compound produces 0.0230 g NH3.  What is the empirical formula of the compound?  Hint: Cobustion involves reacting with excess O2.  Assume that all the carbon ends up in CO2 and all the hydrogen ends up in H2O.  Also assume that all the nitrogen ends up in the NH3 in the second experiment.
From Zumdahl Chemistry 5e, Chapter 3, #121

I started by writing the chemical equation:

CxHxNxOx + O2 --> CO2 + H2O + NH3
With x = unknown.

Is this correct?  Can I add NH3 to the equation?  It has to go somewhere.

Second, I used a proportion to find the amound of NH3 produced if 0.157 g of compound combusted.  I got 0.0351g NH3.  I then found the amount of O2 that was used in combustion by subtracting reactant masses from product masses and got 0.122 g O2 combusted.  Next, I found how many grams of each element was used up in each product by using molar masses.  I found that there was 0.0581g C, 0.00970g H, 0.0289g N, and 0.061g O used in the compound.  I then used % composition in the compound and found 37.0%C, 6.18%H, 18.4% N, and 38.9% O by mass.  I then divided those numbers by the molar mass of each element to get 6.13 mol H, 3.08 mol C, 1.31 mol N, and 2.43 mol O.  Even allowing for err in rounding, these numbers are nowhere near integers nor do they have a small GCF to multiply by to get integers.  Book answer is C7H5N3O6.  This is an AP Chem summer assignment so it is not mission critical that I arrive at the correct answer, but help would be greatly appreciated.  

Title: Re:Combustion of unkown compound
Post by: Donaldson Tan on July 25, 2005, 04:49:58 PM

Quote from: aduke on July 25, 2005, 04:00:14 PM

A compound contains only C, H, N, and O. Combustion of 0.157 g of the compound produces 0.213 g CO2 and 0.0310 g H2O.  In another experiment, it is found that 0.103 g of the compound produces 0.0230 g NH3.  What is the empirical formula of the compound?  Hint: Cobustion involves reacting with excess O2.  Assume that all the carbon ends up in CO2 and all the hydrogen ends up in H2O.  Also assume that all the nitrogen ends up in the NH3 in the second experiment.
From Zumdahl Chemistry 5e, Chapter 3, #121

CwHxNyOz + oxygen => CO2 + H2O + NH3

Quote

Combustion of 0.157 g of the compound produces 0.213 g CO2 and 0.0310 g H2O.

mass of C in CO2 = 12/44 * .213 = 0.0581g
=> %mass composition of carbon in compound = 0.0581/0.157 = 37%
mass of H in H2O = 2/18 * 0.310 = 0.0344g
=> %mass composition of hydrogen in compound = 0.0344/0.157 = 21.9%

Quote

In another experiment, it is found that 0.103 g of the compound produces 0.0230 g NH3.

mass of nitrogen in NH3 = 14/17 * 0.0230 = 0.0189g
=> %mass composition of nitrogen in compound = 0.0189/0.103 = 18.4%

%mass compositoin of oxygen in compound = 100% - 37%  - 21.9%  - 18.4% = 22.7%

Now given the %mass composition of the compound, solve using the standard emperical formula method.

Title: Re:Combustion of unkown compound
Post by: sdekivit on July 25, 2005, 05:13:42 PM

this is wrong, since combustion of the compound results in NO2, not NH3.

--> experiment 2 must be used separately from experiment 1 as done in the calculation

Title: Re:Combustion of unkown compound
Post by: aduke on July 25, 2005, 08:55:24 PM

Thanks to both of you who replied.

I got the same results as you did geodome, except you calculation of % H was off by a factor of 10 due to a copying error on your part.  You used 0.310 g instead of 0.0310 g.  That said, i got 3.08 mol C, 8.18 mol H, 1.31 mol N, and 2.28 mol O.  Nowhere near the book answers.  Either the book answer of C7H5N3O6 is wrong or we are missing something.

I then attempted the problem using NO2 instead of NH3 like sdekivit said, using 0.0230 g NO2 instead of NH3.  I got the same percentage C as I expected, but obviously different H, N, and O amounts.  I got 3.08 mol C, 2.19 mol H, 0.485 mol N, and 3.38 mol O.  I don't know what is wrong and am beginning to believe there is a book problem.

Again, thanks for your *delete me*

Title: Re:Combustion of unkown compound
Post by: sdekivit on July 26, 2005, 02:15:13 AM

this is my answer:

1: we have 0,157 g and let that react with O2 to form 0,213 g CO2 and 0,0310 g H2O

--> this results into the following calculation:

--> 0,213 g CO2 contains:
0,213 / 44,01 = 0,0048398091 mol C
2 * 0,0048398091 = 0,0096796183 mol O

--> 0,0310 g H2O contains:
0,0310 / 18,016 = 0,0017206927 mol O
2 * 0,0017206927 = 0,0034413854 mol H

So the compound of 0,157 g Contains:
0,0048398091 mol C
0,0034413854 mol H

--> so the ratio C : H = 7 : 5

the mass% C and H are 37m% and 2,21% respectively

so 0,103 g compound contains 0,03811 g C and 0,0022763 g H
--> that's 0,0022582341 mol H and 0,003173189 mol C

the compuond contained 0,0230 / 17,034 = 0,0013502407 mol N

so the mass is now 0,0593031721 g so 0,0436968279 g is O
--> that's 0,0027310517 mol O
--> N : O = 1 : 2

now let's complete the whole story:

N : C = 1 : 2,35, so that's 3 : 7

so the book is totally correct, since N : O = 1:2 we get indeed the simplest formula of:

C7H5N3O6

Title: Re:Combustion of unkown compound
Post by: Dude on July 26, 2005, 10:46:11 AM

Looks correct.

Title: Re:Combustion of unkown compound
Post by: Donaldson Tan on July 26, 2005, 11:00:42 AM

corrected %mass composition:

carbon: 37%
hydrogen: 2.2%
nitrogen: 18.4%
oxygen: 42.4%

I also arrive at the same answer: C7H5N3O6

Title: Re:Combustion of unkown compound
Post by: aduke on July 26, 2005, 05:51:23 PM

I finally understand this now.
Thanks to all!

Title: Re:Combustion of unkown compound
Post by: aduke on July 28, 2005, 11:04:26 PM

Just curious.
How does this balance?

C7H5N3O6 + O2 --> CO2 + H2O + NH3

Adding coefficients to the NH3 product to add more N product increases the needed coefficient for the CHNO reactant to balance the H, and increasing the CHNO coefficient to balance the H increases the NH3 coefficient to increase the N supply.

Title: Re:Combustion of unkown compound
Post by: sdekivit on July 29, 2005, 03:42:09 AM

Quote from: aduke on July 28, 2005, 11:04:26 PM

Just curious.
How does this balance?

C7H5N3O6 + O2 --> CO2 + H2O + NH3

Adding coefficients to the NH3 product to add more N product increases the needed coefficient for the CHNO reactant to balance the H, and increasing the CHNO coefficient to balance the H increases the NH3 coefficient to increase the N supply.

combustion will give NO2 instead of NH3.
So we get:

2 C7H5N3O6 + 16,5 O2 --> 14CO2 + 5H2O + 6NO2

so we need to multiply with 2 to get the combustion reaction:

4 C7H5N3O6 + 33 O2 --> 28 CO2 + 10 H2O + 12 NO2

Title: Re:Combustion of unkown compound
Post by: Borek on July 29, 2005, 04:05:42 AM

Quote from: sdekivit on July 29, 2005, 03:42:09 AM

combustion will give NO2 instead of NH3.

Not necesarilly. In most cases of organic compunds combustion nitrogen evolves as N₂.

Title: Re:Combustion of unkown compound
Post by: Borek on July 29, 2005, 04:06:45 AM

Quote from: aduke on July 28, 2005, 11:04:26 PM

How does this balance?

C7H5N3O6 + O2 --> CO2 + H2O + NH3

Not every equation can be balanced.

Title: Re:Combustion of unkown compound
Post by: Mitch on July 29, 2005, 04:39:32 AM

This works but I had to use a negative stoichimetric coefficient, so its not to be trusted at all. But here you go.

C7H5N3O6 + (3)O2 --> (7)CO2 + (-2)H2O + (3)NH3

Title: Re:Combustion of unkown compound
Post by: AWK on July 29, 2005, 04:47:38 AM

During combustion organic compounds form always N₂.
Ammonia is formed in another experiment (Kjeldahl method)

http://www.rosesci.com/Products/Chemical%20Analysis/Kjeldahl%20Chemistry%20-%20Overview.htm

Title: Re:Combustion of unkown compound
Post by: Donaldson Tan on July 29, 2005, 03:55:25 PM

Quote from: aduke on July 25, 2005, 04:00:14 PM

Assume that all the carbon ends up in CO2 and all the hydrogen ends up in H2O.  Also assume that all the nitrogen ends up in the NH3 in the second experiment.

although the question might be theoretically wrong, we were already told to assume that all nitrogen ends up as ammonia.

Title: Re:Combustion of unkown compound
Post by: sdekivit on July 29, 2005, 04:18:03 PM

in another experiment, which is probably another experiment than a combustion as in experiment 1.