Stoichiometry Question

[Lunasly]

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Hey guys,

Lets say that you are doing a combustion analysis experiment and you take an unknown hydrocarbon and oxidize it into water and carbon dioxide by using excess oxygen and various drying agents (to separate the water from the carbon dioxide). If you start with 1.00 grams of the liquid hydrocarbon and you oxidize it, then how is it that the carbon dioxide when collected can be 3.045g at STP. That is, why the hell does only one of the two compounds produced by this hydrocarbon weight more the original compound itself?

Thanks,
Lunasly.

 

[ingramw1202]

The key here is excess oxygen. The grams of carbon dioxide is not as important as understanding the moles of carbon dioxide. So let's say you combust one mole of an unknown hydrocarbon. In the first chamber the mass of the drying agent changes by x grams. And it's given in the problem that the first chamber absorbs water and the second absorbs carbon dioxide. So the change in mass in chamber one comes from water. Chamber two has a mass change by x number of grams and this is strictly due to carbon dioxide. Now, the only thing that was combusted was the hydrocarbon and excess oxygen. So, when the mass of the first chamber changes, the change is due to formation of water, from which all hydrogen can from the hydrocarbon. So by calculating how many moles of water you have, you can calculate the number of moles of hydrogen you have. Two moles of hydrogen per mole of water. Then in chamber two, the mass changes due to carbon dioxide, In which all of the carbon came from the hydrocarbon. You can use the mass change to calculate moles of co2 that were created. Then you know that for every one mole of co2 you have one mole of c. This gives you the fundamentals for writing a combustion reaction. And the moles of carbon and hydrogen you started with are the same as the moles of the two calculated by the above method.

 

[Lunasly]

Okay, so if I start with 1.00g of a liquid hydrocarbon and the final mass of the carbon dioxide produced by the combustion of the hydrocarbon by excess oxygen is 3.045g, then I know that the difference in that mass 2.045g can be used to determine the moles of CO2 created. RIght?

How did you come to the conclusion that that for every mole of CO2, that one mole of C was present? Can the same conclusion be made by saying that for every 1 mole of CO2, there are two moles of O2?

Also, if you have the BR Chem Book, do you know how to figure out #24 from chapter 1?

Thanks,
Lunasly.

 

[tttgo]

Then you know that for every one mole of co2 you have one mole of c.

I think he means that you can use the mass of co2 to find out the mass of C in the unknown. The co2 tells you 1 mol c/co2 in the product, but not necessarily 1 mol C in the unknown. You can do the same to find grams of H in the unknown. The grams of C and h you found for the unknown are relative, so it doesn't tell you how many mols of C and H are in your 1g unknown. Using mass % you can find the empirical formula. If you had the molar mass of the unknown, you could figure out the molecular formula too. Look at Passage3 for TBR gchem.

 

[ingramw1202]

You got it bud. Thanks for clarifying that

 

[Lunasly]

As far as I know, the mass of carbon in the unknown hydrocarbon is going to be the same mass as the carbon in CO2 (I assume that because carbon is found in only one compound one each side of the reaction [the hydrocarbon and CO2 being the two compounds] that this is the reason). Now say that I am at STP like I described before and the drying agent (NaOH) that originally captured all of the CO2 was heated, thus releasing the CO2. Lets say that this released CO2 occupies 1.55L of the container that is it in (once again at STP). I can take this 1.55L and divide it by the molar volume of CO2 with is 22.41L at STP to be able to get moles of CO2. I can then take these moles of CO2 and multiply it the molecular mass of carbon dioxide (12g). I can then divide that number by the mass of the original hydrocarbon liquid sample which is 1g as I described in my first post. I can then multiply this entire answer by 100% to get the mass percent of carbon in the unknown hydrocarbon sample.

This is essentially the question I am getting stuck on. I understand that dividing the volume that CO2 occupies in the container by the molar volume at STP will give me moles of CO2, however how does multiplying the moles of CO2 by the molecular mass of carbon (12g) and then dividing it by the original mass of the hydrocarbon compound give me mass percent of carbon of the unknown compound.

Also, how can we make the assumption that the mass of carbon in CO2 is the same as the mass of carbon in the unknown hydrocarbon?

 

[tttgo]

Ok, so you found the moles of Co2. Molar mass is grams/mol. If you multiply them mol cancels out and you're left with grams of Co2. For every 1 mol of that Co2, you have 1 mol of carbon. Now we have to assume grams carbon in product=grams carbon in hydrocarbon. Let's put it this way: Say we have 2H20 becomes 2 H2 and 1O2. We know that 1 mol of H2 (or 2 grams of H) means that 1 mols of water reacted (aka 2 grams of H).

Whatever carbon formed in Co2 came from the hydrocarbon since the hydrocarbon was the limiting reagent. We don't know the mols of C in the hydrocarbon since the ratio in CxHx could be anything. That's why we need the mass percent. Mass percent is just grams (of Carbon)/total mass of compound.

 

[Lunasly]

Ok, so you found the moles of Co2. Molar mass is grams/mol. If you multiply them mol cancels out and you're left with grams of Co2. For every 1 mol of that Co2, you have 1 mol of carbon. Now we have to assume grams carbon in product=grams carbon in hydrocarbon. Let's put it this way: Say we have 2H20 becomes 2 H2 and 1O2. We know that 1 mol of H2 (or 2 grams of H) means that 1 mols of water reacted (aka 2 grams of H).

Whatever carbon formed in Co2 came from the hydrocarbon since the hydrocarbon was the limiting reagent. We don't know the mols of C in the hydrocarbon since the ratio in CxHx could be anything. That's why we need the mass percent. Mass percent is just grams (of Carbon)/total mass of compound.

Hmm, I apologize, but I sort of lost you in the first paragraph. If I am understanding you correctly, you are saying that if you have 2H2O (for example) then that is the equivalent of saying that we have two H2 and one O2. I don't understand the bit after that. Don't we have 4 hydrogen atoms and two oxygen atoms (or two hydrogen molecules and 1 oxygen molecule)? Which means that to form 2H2O we require two hydrogen molecules and one oxygen molecule. Am I correct?

With that being said, if I wanted to find out the mass percent of carbon (which we established is the same in the unknown hydrocarbon and the carbon dioxide), then I would divide the volume of which CO2 occupies in the container (1.55L) by the molar volume of CO2 at STP which is 22.41. The volume that CO2 occupies is in units of litres and the molar volume is in units of litres per mole. Thus litres cancel and I am left with moles of CO2. I multiply the moles of CO2 then by the molar mass of carbon which is in units of grams per mole (12g/mol). This leaves me with just grams of CO2. I then divide this entire number by the entire mass of the hydrocarbon which is 1.00g. Multiply by 100% and I get mass percent of carbon of the unknown compound.

What is confusing me is that when I multiply the moles of CO2 by the molar mass of carbon (12g/mole) to be able to get grams of CO2, why am I multiply by 12? I know the molar mass of carbon is 12g/mol, but if I have moles of CO2, my next move is to say that, well, in 1 mole of CO2, I have 44g of CO2. How does multiplying the moles of CO2 by the molar mass of carbon give me the grams of carbon?

 

[ingramw1202]

Hmm, I apologize, but I sort of lost you in the first paragraph. If I am understanding you correctly, you are saying that if you have 2H2O (for example) then that is the equivalent of saying that we have two H2 and one O2. I don't understand the bit after that. Don't we have 4 hydrogen atoms and two oxygen atoms (or two hydrogen molecules and 1 oxygen molecule)? Which means that to form 2H2O we require two hydrogen molecules and one oxygen molecule. Am I correct?

With that being said, if I wanted to find out the mass percent of carbon (which we established is the same in the unknown hydrocarbon and the carbon dioxide), then I would divide the volume of which CO2 occupies in the container (1.55L) by the molar volume of CO2 at STP which is 22.41. The volume that CO2 occupies is in units of litres and the molar volume is in units of litres per mole. Thus litres cancel and I am left with moles of CO2. I multiply the moles of CO2 then by the molar mass of carbon which is in units of grams per mole (12g/mol). This leaves me with just grams of CO2. I then divide this entire number by the entire mass of the hydrocarbon which is 1.00g. Multiply by 100% and I get mass percent of carbon of the unknown compound.

What is confusing me is that when I multiply the moles of CO2 by the molar mass of carbon (12g/mole) to be able to get grams of CO2, why am I multiply by 12? I know the molar mass of carbon is 12g/mol, but if I have moles of CO2, my next move is to say that, well, in 1 mole of CO2, I have 44g of CO2. How does multiplying the moles of CO2 by the molar mass of carbon give me the grams of carbon?

Okay so before what I was saying is this in a nutshell: Because we are using EXCESS Oxygen, and the only source of hydrogen and carbon comes from the hydrocarbon itself, we can use the change in mass of the chambers as a direct representation of the Carbon and hydrogen present in the hydrocarbon. So if in chamber one we are given that the mass changes by lets say 3 grams, and we know that chamber one absorbs water, then that means that the mass of chamber one changed by 3 grams of water. Water (H2O) can not form unless there is hydrogen present, correct? SO that means that all of the hydrogen in the water that formed came directly from the hydrocarbon.So if 3 grams of water formed, then (3 grams water)/ (MW water) gives you the number of moles of water that formed. Well, we know that in every 1 mole of water, there are 2 moles of hydrogen and one mole of oxygen. Does this make sense? Its like saying for every one water molecule you have two hydrogen atoms and one oxygen atom. So this means that the number of moles of hydrogen in the sample can be calculated as follows:
(grams water absorbed)/(mw water)= moles water formed
(moles water formed) x (2 moles hydrogen/ 1 mole of water) = moles of hydrogen.

Then, for chamber two, CO2 is formed correct? And the only source of carbon comes from the hydrocarbon. SO this means that the change in mass of chamber two = amount of CO2 formed in grams. Then, you can take the number of grams of CO2 and divide by the MW of CO2 to get number of moles of CO2 that formed. Well, for every one mole of CO2, there is one mole of Carbon present. This is similar to saying for every molecule of carbon dioxide, there is always one molecule of carbon. So, the number of moles of CO2 formed = the number of moles of carbon present in the sample. This is like saying the following:

C(s) +O2(g)--> CO2(g) .

This is a balanced equation, so the stoichiometric coefficient is equal to the ratio combination (it takes 1 mole of C and 1 mole of O2 to make 1 mole of CO2).

So, a combustion reaction has the general formula of CxHy + O2--> xCO2 + (y/2)H2O.
This is true because for every mole of CO2 produced, one mole of carbon is used. So the moles of CO2 produced =the moles C in the sample because they are in a 1:1 ratio. Oxygen is in excess, so the coefficient of it does not matter as long as you have plenty of it and you understand that you need oxygen to make CO2 and H2O from CxHy. Now, we know that for every mole of H2O produced, 2 hydrogen atoms are needed. SO, if we started out with 4 hydrogen atoms, and all of the hydrogen is consumed to produce water in the presence of excess oxygen, then two water molecules are created because for every two hydrogens, one water molecule can be produced.

To sum up:

We know the mass of water formed and the mass of CO2 formed. Chamber A will absorb only H2O, and Chamber B will absorb only CO2. So the change in mass of chamber A is equal to the mass of water absorbed, and the change in mass of CO2 is equal to the mass of carbon dioxide absorbed. Now, we also can calculate the number of moles of each compound absorbed by dividing by molecular weight. So:

moles H2O formed= mass H2O formed/ MW H2O
moles CO2 formed = mass CO2 formed/ MW CO2

We also know that the only source of carbon and hydrogen present in the chamber with which H2O and CO2 can be created came directly from the sample of unknown hydrocarbon with a formula of CxHy. We also know that for every mole of water formed, 2 moles of hydrogen are needed, hydrogen that came from the hydrocarbon sample. And for every mole of CO2 formed, one mole of carbon is needed, carbon that came directly from the hydrocarbon sample. So:

number of moles H2O formed= (0.5) x moles of hydrogen in original sample
number of moles of CO2 formed = number of moles of carbon in original sample

HOWEVER: It should be understood that the empirical formula only can be derived from combustion data because CxHy can be in a variety of different ratios. For example, CH4 has the same ratio as C2H8 and so forth. In order to determine the empirical formula, you must calculate the mass percent. By your question, I am assuming that you know the volume that the CO2 occupies. And we know that at STP, 1 mole of gas will occupy 22.4 L. So:

volume CO2 occupies x (1 mole CO2/ 22.4 L)= moles of CO2.

This is an alternative method for determining the number of moles of CO2 in sample. In the MCAT review classes I teach we go over the kaplan question that gives the change in mass of a chamber due to CO2 formation. Also note that this volumetric method only works for gases behaving ideally. At high pressures, the volume that the gas occupies will be slightly more than the volume calculated due to the volume of the individual atoms.

So now we know how many moles of carbon we started out with (calculations mentioned above) and how many moles of hydrogen we started out with (also mentioned above). To determine the empirical formula you simply divided the moles of each by the smallest number.

If X<Y, then we say X/X and Y/X
If Y<X, then we say X/Y and Y/Y

We know that 1 carbon can bond four hydrogens in sigma bonds. In an alkene, two carbons can each bind two hydrogens. In an alkyne, two carbons can each bond 1 hydrogen. So any hydrocarbon may have any multiple of the following formulas: CH4, C2H4 (or CH2), and C2H2 (or CH).
If moles of carbon<moles of hydrogen:
C(x/x)H(y/x)=C1H(y/x)
If moles of carbon are equal to moles hydrogen, then we know that the empirical formula must be CH.
Moles C can not be higher than moles of hydrogen because for every 1 carbon, the smallest empirical formula possible is C1H1.

In regards to mass percent:

By taking the original mass of CO2 and multiplying by (12/44) we can find the mass carbon contained in our original sample. Then, dividing this number by original mass of sample tells you mass percent of carbon in the sample. But the way I explained above is exactly the same but simplified. Here is why with some example numbers:

Mass carbon dioxide obtained: 7 grams
Mass percent of carbon in CO2: 7 grams CO2 x (12/44)= 1.909 g C
Moles carbon present in sample: 1.909g C x (1 mol/ 12 g carbon) =0.159 moles carbon

Now the way I explained above with the ratios:

Mass carbon dioxide obtained: still 7 grams
Moles carbon dioxide obtained: 7 grams CO2 x (1 mol CO2/44 g CO2) =0.159 moles carbon dioxide
In 1 molecule of CO2, there is 1 atom of carbon, so for every mole of CO2 formed, one mole of Carbon is needed. (disregarding oxygen)
0.159 moles CO2 x (1 mole C/ 1 mole CO2)= 0.159 moles carbon

Lets try with water:

Mass water obtained: 3.5 g
Mass percent of Hydrogen in water recovered: 3.5 g x (2/18) =0.389 g Hydrogen
Moles hydrogen= 0.389 g H x (1 mol/1 g H) = 0.389 mol Hydrogen

Now the way I explained it above:

Mass of water obtained: 3.5 g
moles of water obtained: 3.5 g x (1 mole water/ 18 g water)= 0.194 moles water
For every one molecule of water produced, 2 hydrogen atoms are required, so 2 moles of Hydrogen can yield 1 mole of H2O.
Moles of Hydrogen: 0.194 moles water x (2 mole H/ 1 mole H2O) =0.389 moles Hydrogen.

As you can see, the mass percent is the same thing that I have explained in my reasoning, however, it just do it in a different way and choose to work with moles instead of grams and realize the molar ratios of an atom in its molecule.
I hope this cleared up stuff. Sorry for any confusion and delay on the response. Feel free to ask me any more questions.
Cheers