[KungKemi]

Hey,

So basically I've set myself some chemistry questions to do over the holidays which are derived from a text at my school. As such, I don't have access to the answers, and I was just wondering if my answer to the following chemical marathon seems valid:

"A sample consisting of 22.7 g of nongaseous, unstable compound X is placed inside a metal cylinder with a radius of 8.0 cm, and a piston is carefully placed on the surface of the compound so that, for all practical purposes, the distance between the bottom of the cylinder and the piston is zero. (A hole in the piston allows trapped air to escape as the piston is placed on the compound; then this hole is plugged so that nothing inside the cylinder can escape.) The piston-and-cylinder apparatus is carefully placed in 10.00 L of water at 298 K. The barometric pressure is 778 torr.

When the compound spontaneously decomposes, the piston moves up, the temperature of the water reaches a maximum of 302.52 K, and then it gradually decreases at the water loses heat to the surrounding air. The distance between the piston and the bottom of the cylinder, at the maximum temperature, is 59.8 cm. Chemical analysis shows that the cylinder contains 0.30 mol carbon dioxide, 0.25 mol liquid water, 0.025 mol oxygen gas, and an undetermined amount of a gaseous element A.

It is known that the enthalpy change for the decomposition of X, according to the reaction stated above, is -1893 kJ/mol X. The standard enthalpies of formation for gaseous carbon dioxide and liquid water are -393.5 kJ/mol and -286 kJ/mol, respectively. The heat capacity for water is 4.184 J/K·g. The conversion factor between L·atm and J can be determined from the two values for the gas constant R, namely, 0.08206 L·atm/mol·K and 8.3145 J/mol·K. The vapor pressure of water at 302.5 K is 31 torr. Assume that the heat capacity of the piston-and-cylinder apparatus is negligible and that the piston has negligible mass.

Given the preceding information, determine the formula for X."

Okay, so basically I determined the following...

Maximum temperature in vessel:

Volume = 12.024 L
Temperature = 302.520 K
Internal pressure = external pressure = 103.725 kPa

Determines moles of each gas:

n_total = PV/RT = (103.725×12.024)/(8.31×302.52) = 0.496 mol
Gaseous moles = 0.300 mol CO₂ + 0.025 mol O₂ + x-mol H₂O + y-mol A

From questions...the vapour pressure of H₂O at local temperature is 31 torr...

n_{gaseous H2O} = PV/RT = (4.133×12.024)/(8.31×302.52) = 0.020 mol H₂O

y-mol A = 0.496 mol - 0.020 mol - 0.025 mol - 0.300 mol = 0.151 mol A

If A = 4.194 g mass in the 22.7 g sample, then molar mass A = 27.775 gmol^-1

Elemental analysis from the acquired information:

22.7 g sample = 3.603 g C + 0.504 g H + 14.399 g O + 4.194 g A

Moles of each element:

n_C = 3.603/12.011 = 0.300 mol C
n_H = 0.504/1.0079 = 0.500 mol H
n_O = 14.399/15.999 = 0.900 mol O
n_A = 4.194/27.775 = 0.151 mol C

Divide by lowest common mole (0.151):

X = (0.300 mol C + 0.500 mol H + 0.900 mol O + 0.151 mol A)/0.151 = (2.0 mol C + 3.3 mol H + 6.0 mol O + 1.0 mol A)

Multiply by 3...

X = 6.0 mol C + 10.0 mol H + 18.0 mol O + 3.0 mol A = C₆H₁₀O₁₈A₃.

I'm a little sceptical about this answer, and I'm uncertain on whether the small amount of water vapour would even exist, however, that was the only thing that I could see that last bit of information hinting about in the question.

If anyone can prove or disprove my working that would be appreciated!
KungKemi

[KungKemi]

Wait, if one assumes that A is the diatomic form of Nitrogen (N₂), since the molecular formula is (27.775 gmol^-1≈28.014 gmol^-1 N₂), then that implies that the molecular formula is C₆H₁₀O₁₈N₆.

As such, the molecular/empirical formula of X must be C₃H₅O₉N₃.

I think, at least.

Does that seem right?
KungKemi

[AWK]

From other data you can calculate the molecular mass of your compounds.
Your formula fits to nitroglycerin.

[KungKemi]

From other data you can calculate the molecular mass of your compounds.
Your formula fits to nitroglycerin.

Yes, okay. I think that the answer seems pretty valid, because if one were to find the energy required for the 10.0 kg of water to increase a total of 4.52 K in temperature as it does in the question...then one would find the following:

q = mCΔT = 10000 × 4.184 × 4.52 = 189.117 kJ

And if one were to assume that the molecular formula is nitroglycerin, then the potential energy stored if it has a heat of combustion of -1893 kJ/mol is:

22.7 g × 1 mol/227.0845 g × 1893 kJ/1 mol = 189.23 kJ ≈ 189.117 kJ

So...from that I think that nitroglycerin seems a fairly valid candidate for the molecular formula of X.

KungKemi

[KungKemi]

Finally, I think I'm pretty confident about the initial answer, however, the question goes on to ask:

"Given the preceding information, determine...

2. The pressure-volume work (in kJ) for the decomposition of the 22.7-g sample of X.
3. The molar change in internal energy for the decomposition of X and the approximate standard enthalpy of formation for X."

For question 2, I just said that Energy (J) = volume (L) × pressure (kPa) = 12.024 × 103.725 = 1247.20 J or 1.25 kJ.

For question 3 (i), I said that molar change in internal energy is found as ΔU = ΔH - ΔPV, where ΔU is molar change in energy, and ΔH is enthalpy.

So... ΔU = -1893 - 1.25 = -1894.25 kJ/mol

and for question 3 (ii), I said that the reaction (according to the product mole ratios) was:

C₃H₅O₉N₃ 2.5H₂O + 3CO₂ + 0.25O₂ + 1.5N₂

, so... °ΔH_f of C₃H₅O₉N₃ from the following is -2.5 kJ/mol.

Does this look correct?

KungKemi

[mjc123]

Looks OK to me; just a couple of points:

(i) Pressure-volume work is P_extΔV, not Δ(PV). It works out the same in this case, but it doesn't always, so take care.

(ii) You are only given ΔH to the nearest kJ/mol, so it is not justified to give ΔU to 2 decimal places.

[KungKemi]

Looks OK to me; just a couple of points:

(i) Pressure-volume work is P_extΔV, not Δ(PV). It works out the same in this case, but it doesn't always, so take care.

(ii) You are only given ΔH to the nearest kJ/mol, so it is not justified to give ΔU to 2 decimal places.

I'll keep that in mind for next time. Thank you for your assistance!

KungKemi