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anonymous

  • 4 years ago

Combustion of an unknown hydrocarbon produces 0.954 mole CO2 and 1.27 mole H2O. what is the emperical formula of the unknown?

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  1. Xishem
    • 4 years ago
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    This is very similar to the question about the tetraphosphorous decaoxide, except there are a lot fewer steps to go through. Given the information that hydrocarbons consist of ONLY hydrogen and carbon, the equation for this reaction, ignoring coefficients, will look something like this...\[C_xH_y+O_2 \rightarrow CO_2+H_2O\]This means that all atoms of carbon from the hydrocarbon go to the carbon dioxide. Similarly, all the atoms of hydrogen from hydrocarbon go to the water. Try this: convert the moles of carbon dioxide into moles of carbon, and convert the moles of water into moles of hydrogen. Then find the ratio of moles, similar to what we did in the previous question. If you have trouble, indicate where you had trouble, and I'll help you out a bit.

  2. anonymous
    • 4 years ago
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    ok (0.954)(44.01g CO2) (12.01g C)

  3. anonymous
    • 4 years ago
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    is that the right conversion set up to gt moles of carbon?

  4. Xishem
    • 4 years ago
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    Be careful. You don't want to convert to grams of carbon. Just moles. Hint: There is 1 mol of carbon atoms per mol of compound.

  5. anonymous
    • 4 years ago
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    (0.954 mol)(44.01 g CO2/1mol CO2)?

  6. Xishem
    • 4 years ago
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    You don't need to use a molar mass. It's already in moles. Let's try doing it intuitively before I just give you a conversion factor. Then maybe you'll understand the conversion factor more. If you have 1 carbon dioxide molecule, how many carbon atoms do you have?

  7. anonymous
    • 4 years ago
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    6.022 x 10^23

  8. anonymous
    • 4 years ago
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    no?

  9. Xishem
    • 4 years ago
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    If you were responding to my old question, then yes, you are correct. If you have 6.022*10^23 molecules of CO_2, then you'll have 6.022*10^23 molecules of CO_2. Therefore...\[\frac{6.022*10^{23}\ molecules\ CO_2}{6.022*10^{23}\ atoms\ C}=\frac{1mol\ CO_2}{1mol\ C}=1\]Any value that is equal to one when you evaluate it can be used as a conversion factor. Therefore, we can use this relationship as a conversion factor in our problem.

  10. anonymous
    • 4 years ago
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    oh i think i see now. 0.954 x 6.022 E^23 1.27 x 6.022 E^23 i take the lower of these two and divide the higher by it.

  11. Xishem
    • 4 years ago
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    Not quite! That finds the ratio of moles of carbon dioxide to water. What you want to find is the ratio of atoms of carbon to atoms of hydrogen, because that is what you need to find the formula for the unknown hydrocarbon. First, you need to convert moles of carbon dioxide to moles of carbon, and moles of water to moles of hydrogen.

  12. anonymous
    • 4 years ago
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    somehow i got C2H8

  13. Xishem
    • 4 years ago
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    Without even doing any calculations, I can tell you that that formula isn't correct. Why? Because the atoms aren't expressed in the lowest possible whole-number ratios.

  14. anonymous
    • 4 years ago
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    ok but im confused here. i was taught moles-grams-atoms. i have never seen a conversion between moles and moles

  15. Xishem
    • 4 years ago
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    You are converting between moles of CO2 and moles of C within the same molecule. You seemed to understand the conversion intuitively. You are basically just saying there are x number of atoms of C in each mol of CO2.

  16. anonymous
    • 4 years ago
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    can you just spill it out lol. its 2:10 am here and my brain is in fatigued

  17. anonymous
    • 4 years ago
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    so how exactly do you do this if you dont mind.

  18. Xishem
    • 4 years ago
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    \[0.954mol(CO_2)*\frac{1mol(C)}{1mol(CO_2)}=0.954mol\ C\]\[1.27mol\ H_2O*\frac{2mol\ H}{1mol\ H_2O}=2.54mol\ H\]There is something inherently wrong with this problem, in that to get a number close enough to consider rounding, you have have to get up to about 22,000 atoms of hydrogen. Either whoever wrote the problem didn't do their calculations very well, or the problem wasn't reported exactly. Anyway, I'm going to assume that 5.3259 is close enough to round to 5 (even though it's nowhere close), and assume that the writer of the problem is an idiot. I think the answer they are looking for is... \[C_2H_5\]This answer is no way reflective of what you should get if you worked the problem correctly as it stands, though.

  19. anonymous
    • 4 years ago
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    the writer has C3H8 as the final answer

  20. Xishem
    • 4 years ago
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    Oh! I completely neglected to see that. I'm an idiot, not the writer :P The factor you have to multiply by is 3. So... \[0.954mol\ C*3=2.862 \approx 3\]\[2.54mol\ H*3=7.987 \approx 8\] Even still, that 2.862 is pretty far to consider rounding. Anyway, using that, you get...\[C_3H_8\]

  21. anonymous
    • 4 years ago
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    ok thanks once again

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