[KungKemi]

I was doing this question and I was wondering if my process and conclusion were valid...

Consider an equimolar mixture (equal number of moles) of two diatomic gasses (A₂ and B₂) in a container fitted with a piston. The gasses react to form one product (which is also a gas) with the formula A_xB_y. The density of the sample after the reaction is complete (and the temperature returns to its original state) is 1.50 times greater than the density of the reactant mixture.

a) Specify the formula of the product, and explain if more than one answer is possible based on the given data.

b) Can you determine the molecular formula of the product with the information given or only the empirical formula?

This was my process for (a):

1/2xA_{2 (g)} + 1/2yB_{2 (g)}  A_xB_{y (g)}

Let z = initial moles of A₂
Let z = initial moles of B₂
Let a = used moles of A₂
Let b = used moles of B₂
Let c = produced moles of A_xB_y
Let n_i = initial number of moles
Let n_f = final number of moles
Let ρ_i = initial density of mixture
Let ρ_f = final density of mixture
Let m = mass of mixture
Let  T = temperature
Let  P = pressure
Let V_i = initial volume of container
Let V_f = final volume of container

ρ_f = m/V_f
ρ_i = m/V_i
ρ_f = 1.5ρ_i = 1.5m/V_i
1.5m/V_i = m/V_f
1.5V_f = V_i

n_i = z + z = 2z
n_f = c + 2z - a - b

a mol A₂ × 1 mol A_xB_y/0.5x mol A₂ = c mol A_xB_y
a/x = 1/2c
b mol B₂ × 1 mol A_xB_y/0.5y mol B₂ = c mol A_xB_y
b/y = 1/2c

a/x = b/y

     PV_i = n_iRT
1.5PV_f = 2z × 8.31 × T
  PV_f/T = 11.08z

   PV_f = n_fRT
   PV_f = (c + 2z - a - b) × 8.31 × T
PV_f/T = 8.31(c + 2z - a - b)

11.08z = 8.31(c + 2z - a - b)
   4/3z = c + 2z - a - b
  -2/3z = c - a - b

Now, let B₂ be the limiting reagent in the following reaction, that is b = z, and y > x...

Let z = 1 mole, and x = 1...

Find x:y...

-2/3z = c - a - b
-2/3z = c - a - z
-1/3z = c - a
  1/3 = c - a
  1/3 = 2a/1 - a
  1/3 = a

  a/x = b/y
  1/3 = 1/y
     y = 3

Therefore, the ratio of x:y is 1:3 respectively. Likewise, if one were to assume that A₂ was the limiting reagent, then the ratio of x:y would be found to be 3:1 respectively.

As such, A_xB_y can be either AB₃, or A₃B.

(b) Not enough information is given in order to determine the molecular formula of the compound. As such, only the empirical formula can be calculated.

If anyone could let me know if this looks like the correct working and logic, that would be greatly appreciated!

Thank you,
KungKemi

[Borek]

I have not checked thoroughly, perhaps I will get back to the problem once I drink my morning coffee, but... Would the final volume be identical for AB₃ and A₂B₆?

[AWK]

Lets take 2 moles (volumes) of mixture A₂ and B₂ (1:1). New volume for density 1.5 times higher will be 2/1.5=4/3V of 2V
AB gives 2 volumes
reaction
A₂ + B₂ = (2/x)A₂B_x + (1-2/x)A₂ for x>≥2
gives always a final volume equal to 1V
reaction
A₂ + B₂ = (2/x)AB_x + (1-1/x)A₂ for x≥2
gives the final volume equal [(1+x)/x]V which is (3/2, 4/3, 5/4...)V . for x=2, 3, 4 ... respectively.
Hence condition for density gives unique solution for AB₃
correction

[KungKemi]

Lets take 2 moles (volumes) of mixture A₂ and B₂ (1:1). New volume for density 1.5 times higher will be 2/1.5=4/3V of 2V
AB gives 2 volumes
reaction
A₂ + B₂ = (2/x)A₂B_x + (1-2/x)A₂ for x>2
gives always a final volume equal to 1V
reaction
A₂ + B₂ = (2/x)AB_x + (1-1/x)A₂ for x≥2
gives the final volume equal [(1+x)/x]V which is 3/2, 4/3, 5/4 . for x=2, 3, 4 ... respectively.
Hence condition for density gives unique solution for AB₃

Hmm, okay, I see. So A₃B is not a feasible solution? I mean, as soon as I saw AB₃ I thought of ammonia (where A₂ is nitrogen gas and B₂ is hydrogen gas) or something homologous to the following. But, similarly, depending on perspective, couldn't A₂ instead be hydrogen gas, and B₂ be nitrogen gas? If this is the case, then the formula becomes A₃B. Of course, this isn't the typical notation for writing such a formula, but does that correlate to this question as well?

Also, I'm assuming that the molecular formula is unobtainable using the given data, or is that not the case?

Thank you for the help AWK,
KungKemi

[KungKemi]

I have not checked thoroughly, perhaps I will get back to the problem once I drink my morning coffee, but... Would the final volume be identical for AB₃ and A₂B₆?

Also, the volumes would not be equivalent, however, the relationship V_i = 1.5V_f would still hold up. And, since the empirical formula is simply a ratio (like V_i = 1.5V_f) then it would be assumed that the following would not affect the results.

Is this feasible intuition?

[AWK]

Hmm, okay, I see. So A₃B is not a feasible solution?

A and B are interchangeable. NH₃ and H₃N both will work.

[KungKemi]

Hmm, okay, I see. So A₃B is not a feasible solution?

A and B are interchangeable. HN₃ also will work far over boiling point.

Okay, excellent. I'm also certain that the molecular formula is unquantifiable using the given information (as pointed out by your previous post), since if the formula is A₂B₆, then the following conditions are still met indifferently.

Thank you for the help,
KungKemi

[AWK]

I guess that value of density may be sufficient for compound indentification.

[Borek]

I have not checked thoroughly, perhaps I will get back to the problem once I drink my morning coffee, but... Would the final volume be identical for AB₃ and A₂B₆?

Also, the volumes would not be equivalent, however, the relationship V_i = 1.5V_f would still hold up. And, since the empirical formula is simply a ratio (like V_i = 1.5V_f) then it would be assumed that the following would not affect the results.

Is this feasible intuition?

Try to estimate final volume for both cases (AB₃ and A₂B₆), quite easy if you just follow the stoichiometry.

[KungKemi]

I have not checked thoroughly, perhaps I will get back to the problem once I drink my morning coffee, but... Would the final volume be identical for AB₃ and A₂B₆?

Also, the volumes would not be equivalent, however, the relationship V_i = 1.5V_f would still hold up. And, since the empirical formula is simply a ratio (like V_i = 1.5V_f) then it would be assumed that the following would not affect the results.

Is this feasible intuition?

Try to estimate final volume for both cases (AB₃ and A₂B₆), quite easy if you just follow the stoichiometry.

Oh, yes, I didn't do the stoichiometry, but I see what you mean, Borek. The answer can only be AB₃ or A₃B.

Thanks,
KungKemi