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[Impending Doom]

So I've been working on this chemistry assignment, and I seem to have run into a roadblock.

Spoiler (click to show/hide)

1. Use the following information to identify diatomic molecule A and compound B, then answer the following questions.

An empty glass container has a mass of 658.572 grams. It has a mass of 660.356 g after it has been filled with nitrogen gas at a pressure of 790 torr and a temperature of 15.0 degrees Celsius. When the container is evacuated and refilled with the diatomic molecule A at a pressure of 745 torr and a temperature of 26 degrees Celsius, it has a mass of 667.812 g.

So first, you need three things (either temperature, pressure, moles of substance, or volume) to plug into the Ideal Gas Equation and get anything meaningful. For diatomic molecule A (henceforth referred to as dmA) we only have two - temperature and pressure. So instead we look at the info for nitrogen gas. Here we have temperature, pressure, and because we know the molar mass of N₂ (28.02^g/_mol) we can subtract the mass of the empty tank from that of the full tank and get the mass of our gas (1.784 g) and use that to determine how many moles of substance we have (I got 0.064 moles). To plug these values into the Ideal Gas Equation, some of them need to be converted into the appropriate units:

15 degrees C + 273 = 288 degrees K

790 torr / 760 = 1.039 atmospheres

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Solving for volume using the Ideal Gas Equation:

      nRT
  V = ---
       P

Plug in our values + the gas constant (0.08208):

      (0.064)(0.08208)(288)
  V = --------------------- = 1.456
             (1.039)

So the container has a volume of 1.456 liters. now that we have that, we can figure out how many moles of dmA we have.

22 degrees C + 273 = 299 degrees K

745 torr / 760 = 0.980 atmospheres

Code: [Select]

Solving for moles using the Ideal Gas Equation:

      PV
  n = --
      RT

Plug in values + gas constant:

      (0.980)(1.456)
  n = -------------- = 0.058
      (0.08208)(299)
So now we know that there are 0.058 moles of dmA in the container. We also know that there are 9.24 grams of the substance (again, full tank - empty tank) Now by dividing grams / moles, we get that there are 159 grams in one mole of dmA.

There are only a few elements that can form diatomic molecules: Hydrogen, Nitrogen, Oxygen, Fluorine, Chlorine, Bromine, and Iodine. of these, the closest in molar mass to our number is Bromine. So:

dmA = Br₂

Compound B is a gaseous organic compound that consists of 79.89% carbon and 20.11% hydrogen by mass. Exactly 18.384 g of the compound is placed in a stainless steel vesel (11.47 L) with 4.63 mol of oxygen gas at 22.0 degrees Celsius. The total pressure is 83d1 torr.

From what I can tell, the whole bottom part about putting the compound in a container with oxygen is pretty much meaningless. What is important is that we know how many grams of the compound we have, and its percent composition by mass.

79.89% of 18.384 = 14.687 g

20.11% of 18.384 = 3.697 g

carbon = 12.01 ^g/_mol

hydrogen = 1.008 ^g/_mol

So we do grams  / moles, and find that there 1.223 moles of carbon and 3.668 moles of hydrogen. this gives us a 1 to 3 ratio of carbon to hydrogen, for a formula of CH₃.

A and B react quantitatively in a 1:1 molar ratio to form one mole of the single product, gas X. Write a balanced equation to determine the formula of gas X.

Okay, this is where I run into trouble. When CH₃ and Br₂ react, the only logical thing they could form is CH₃Br. The problem being, you would need a 2:1 molar ratio of CH₃ to Br₂ to get a balanced reaction. If you tried to do it like the problem say, you would get 0.5 moles of the product, and still have 0.5 moles of Br₂ left over.

Where am I going wrong?

[Virex]

The last part, where you put oxygen and the organic compound in a container is actually very important. If you were to put CH₃* and oxygen in a container, the first thing you have on your hand is a load of shrapnel, as CH₃* is a radical and a pretty nasty one at that. You actually can't have isolated CH₃* at moderate temperatures, you have to heat ethane (c₂H₆) up to fairly high temperatures to even get a bit of it to fracture into radicals (and you're going to get it mixed with other fun stuff such as H* and C₂H₅*, not to mention rubbish such as atomic carbon)


Now what you've actually got on your hands is an organic compound with the empirical formula C_nH_3n. Now, you can use the density you get when mixed with oxygen to obtain the molar weight and from that you can find the right compound.


Problem is the third question. Assuming we're really dealing with  C_nH_3n, a reaction with Br₂ should either result in the breaking of a carbon-carbon bond, yielding two moles of an organic bromide, or you get Br₂ -> 2 Br*, Br*+ C_nH_3n ->  C_nH_3n-1Br + H*, meaning you're going to get H₂ and HBr as side products. From that I presume that you've made an error in calculating the molar ratios in your compound and you're actually dealing with an alkene, in which case you're getting addition of bromide across a double bond (which is actually a pretty awesome process if you look closer at it, see this), which would result in a single molecule.

Edit: wat? Huh? I can't find any errors in your calculation? That's odd. You may want to double check the figures with your teacher because you can't get ethane (the only compound with the formula C_nH_3n that isn't something nasty) to react with bromine and yield only 1 compound, you'd get pentavalent carbon atoms and stuff like that...

[Miggy]

I wish I could say something helpful, but Virex pretty much covered everything.

[Eoganachta]

I wish I could say something helpful, but Virex pretty much covered everything.

I saw this and thought I might be able to help but some of the material is stuff I haven't covered yet. Sorry.

[Another]

One more nitpick: since Br₂ boils (max pressure of 760 torr) at 59C and has vapor pressure of 76 torr at 3C I highly doubt that it could provide 745 torr at 26C in fully gaseous state. An ideal gas approximation for Br at room temperatures looks unwarranted.

Given the above, maybe CH₃CH₃+Br₂->2CH₃Br, while not the reaction that would happen in nature, is what is asked here? The reality had already been severely violated in the first question, can it get much worse? The pressure in the second question also supports that there is about 0.6 moles of substance with 1.2 moles of carbon in it.

Edit:oops, "one mole of a single product" went past me. Formally the molar ratio and not absolute quantities of A and B were mentioned. It could technically be "[0.5 moles of] A and [0.5 moles of] B react quantitatively in a 1:1 molar ratio to form one mole of the single product, gas X" but that is not the only problem with this assignment.

[Virex]

I don't think they treat virial equations at their level though, so they probably use the ideal gas law to keeps things simple.

One more nitpick: since Br₂ boils (max pressure of 760 torr) at 59C and has vapor pressure of 76 torr at 3C I highly doubt that it could provide 745 torr at 26C in fully gaseous state. An ideal gas approximation for Br at room temperatures looks unwarranted.

Given the above, maybe CH₃CH₃+Br₂->2CH₃Br, while not the reaction that would happen in nature, is what is asked here? The reality had already been severely violated in the first question, can it get much worse? The pressure in the second question also supports that there is about 0.6 moles of substance with 1.2 moles of carbon in it.

Given the right catalyst, the second reaction could potentially happen, but that would violate the description of the last question: "A and B react quantitatively in a 1:1 molar ratio to form one mole of the single product, gas X. Write a balanced equation to determine the formula of gas X." So we're looking for a reaction that has two inputs and 1 output, and Br₂+C₂H₆ -> 2 CH₃Br yields two molecules.


I suggest that we just pull these gasses through a GCMS and be done with it, who's with me?

[Impending Doom]

I have no idea what a GCMS is, so... go right ahead, I guess?

[Virex]

GCMS = Gas Chromatograph with Mass Spectrometer. Basically a one-stop wonder for analyzing anything you can vaporise at moderate temperatures. (Well, typically you'll want to take an IR or NMR spectrum for cross reference, but if you've got a good idea what you may be dealing with a GCMS will usually give you the answer. Plus, it can handle mixtures)

Edit:oops, "one mole of a single product" went past me. Formally the molar ratio and not absolute quantities of A and B were mentioned. It could technically be "[0.5 moles of] A and [0.5 moles of] B react quantitatively in a 1:1 molar ratio to form one mole of the single product, gas X" but that is not the only problem with this assignment.

Seen that way it could work. It's a weird reaction for sure, but if you've got a catalyst that can weaken carbon-carbon single bonds, then you could get the bromine to preferentially attack the carbon-carbon bond.  (Wouldn't platinum be good for that? Car catalysts use platinum/palladium to burn unburnt fuel, but palladium is only really useful for oxidizing carbon monoxide and adsorbing oxygen so I assume the platinum binds the alkanes)

[Miggy]

I'd say that when the assignments seem to be geared towards learning how to use the ideal gas law (or whatever you call it, PV=nRT) and work around with pressures, breaking carbon-carbon bonds with platinum catalysts is pretty far-fetched.

That said, here's where our instructors would, instead of using methyl and bromine, use imaginary substances A and B, and their reaction to form C.

[Virex]

Meh, it doesn't really matter for his situation anyway, he just has to figure out the right compound, not make it on an industrial scale (In which case I'd look at using methyl alcohol as a stock instead of ethane anyway...)