A Combustion of unokwn compound that contains only N, C, H and O?

Question

A compound contains only carbon, nitrogen, hydrogen and oxygen. Combustion of 0.157g of that compound produced 0.213g of CO2 and 0.0310g of H2O. In another experiment, 0.103g of that compound produced 0.0230g of NH3. What is the empirical formula of the compound?

The answer should be C7H5N3O6 (the answer below is wrong)

Answer 1

This is one of those trick questions designed by an individual with a theoretical background. C7H5N3O6 is the emperical formula for one of the isomers of TNT. C6H2(NO2)3CH3 is the chemical formula.

Without the balance equation, the answer cannot be determined.

Unknown + O2 --> CO2 + H2O + ???
Do the other products include NO2? CO?

Assuming 0.157g of unknown produces 0.213g of CO2 and 0.0310g of H2O, this means:
0.213 g/ 44.01 g/mol = 0.004840 moles CO2 produced and
0.0310 g/ 18.02 g/mol = 0.001720 moles H2O
0.00484 : (0.00172/2) = 1.407 which is where the 7C:5H ratio came from.

0.103 g unknown --> 0.0230 g of NH3
0.0230 g/ 17.03 g/mol = 0.00135 mole of NH3
This would have been 0.157/0/103 * 0.00135 moles of NH3 if the samples had been the same size. This means 0.00206 moles of NH3.

0.00484 moles of C = 7C
0.00206 moles of N = 3N (the ratio is 2.349 is close to 2.333)

This gives us C7H5N3Ox

We do not know the MW of the unknown, nor the number of moles used in either experiment.

BTW, Detonation of TNT is:
2 C7H5N3O6 --> 3 N2 + 5 H2O + 7 CO + 7 C

Burning TNT in excess air is probably:
4 C7H5N306 + 33 O2 --> 28 CO2 + 10 H2O + 12 NO2

Answer 2

Lancenigo di Villorba (TV), Italy

THE ANSWER IS "C1.00 H7.10 O0.28 N1.00".

You report several composition's data about an unknown organic compound.

w, x COEFFICIENTs
By means of "Total Combustion Experiments"

CwHxOyNz + (w + x / 4) O2(g) --->
---> (w - y / 2) CO2(g) + (x / 2) H2O + (z / 2) N2 (g)

I rescue the w, x coefficients analizing the data about CO2, H2O's recovered amount.
0.213 g of CO2 are equal to 0.213 / 44.0 = 4.8E-3 moles of CO2
0.031 g of H2O are equal to 0.031 / 18.0 = 1.7E-3 moles of H2O
Hence

w° = 4.8 E-3
x° = 1.7E-3 * 2 = 3.4E-3

y, z COEFFICIENTs
I sum the mass related to carbon's atoms and hydrogen's ones
4.8E-3 * 12.0 + 1.7E-3 * 2 * 1.0 = 6.1E-2 g
hence I determine the remaining mass in the unknown compound
0.157- 0.061 = 9.6E-2 g
The latter mass owing to oxygen and nitrogen's atoms, the remaining species to test.
On the other hand, you reported that 0.023 g of NH3 lift up from 0.103 g of the compound.
0.023 g of NH3 as 0.023 / 17 = 1.35E-3 moles of NH3
so I retrieve a mass like 1.35E-3 * 14.0 = 1.9E-2 g in the compound. The latter is the mass of the nitrogen's atoms in the compound, so I join to the oxygen atom's mass
9.6E-2 - 1.9E-2 = 7.7E-2 g
that is 7.7E-2 / 16-0 = 4.8E-3 "atom-grams" of Oxygen.
Hence I assume

y° = 1.35E-3
z° = 4.8E-3

Back to the "w, x, y and z coefficients", I scale-up the former determinations toward CONVENTIONAL EXPRESSION CARBON-BASED

w = 1 ; x = x° / w° = 0.708 ; y = y° / w° = 0.28 ; z = z° / w° = 1

I hope this helps you.

Answer 3

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