Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: KungKemi on March 04, 2017, 11:08:54 PM
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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 AxBy. 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):
1/2xA2 (g) + 1/2yB2 (g) :rarrow: AxBy (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 AxBy
Let ni = initial number of moles
Let nf = final number of moles
Let ρi = initial density of mixture
Let ρf = final density of mixture
Let m = mass of mixture
Let T = temperature
Let P = pressure
Let Vi = initial volume of container
Let Vf = final volume of container
ρf = m/Vf
ρi = m/Vi
ρf = 1.5ρi = 1.5m/Vi
1.5m/Vi = m/Vf
1.5Vf = Vi
ni = z + z = 2z
nf = c + 2z - a - b
a mol A2 × 1 mol AxBy/0.5x mol A2 = c mol AxBy
a/x = 1/2c
b mol B2 × 1 mol AxBy/0.5y mol B2 = c mol AxBy
b/y = 1/2c
a/x = b/y
PVi = niRT
1.5PVf = 2z × 8.31 × T
PVf/T = 11.08z
PVf = nfRT
PVf = (c + 2z - a - b) × 8.31 × T
PVf/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, AxBy 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
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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?
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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)A2Bx + (1-2/x)A2 for x>≥2
gives always a final volume equal to 1V
reaction
A2 + B2 = (2/x)ABx + (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
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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)A2Bx + (1-2/x)A2 for x>2
gives always a final volume equal to 1V
reaction
A2 + B2 = (2/x)ABx + (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
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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 Vi = 1.5Vf would still hold up. And, since the empirical formula is simply a ratio (like Vi = 1.5Vf) then it would be assumed that the following would not affect the results.
Is this feasible intuition?
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Hmm, okay, I see. So A3B is not a feasible solution?
A and B are interchangeable. NH3 and H3N both will work.
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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
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I guess that value of density may be sufficient for compound indentification.
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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 Vi = 1.5Vf would still hold up. And, since the empirical formula is simply a ratio (like Vi = 1.5Vf) 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.
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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 Vi = 1.5Vf would still hold up. And, since the empirical formula is simply a ratio (like Vi = 1.5Vf) 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