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Gas problem.

  1. Oct 19, 2007 #1
    This problem was proposed to us by our Chemistry professor. Consider a sample of a hydrocarbon at 0.959atm and 298K. Upon combusting the entire sample in oxygen, you collect a mixture of gaseous carbon dioxide and water vapor at 1.51atm and 375K. This mixture has a density of 1.391 g/L and occupies a volume four times as large as that of the pure hydrocarbon. Determine the molecular formula of the hydrocarbon.

    I really do not know how to go about the problem. I think there's a relationship between the information that I am not seeing properly.
  2. jcsd
  3. Oct 19, 2007 #2
    Well here's a hint. In the formula for pressure. PV=nRT. n is the number of moles. Does that help you?
  4. Oct 19, 2007 #3
    how can you re-write PV=nRT so that you can find molecular mass?

    density equals what? how do you find moles?
  5. Oct 19, 2007 #4
    I know the ideal gas law, how to find moles etc. Density = mass/volume. But I'm still getting confused with what to do with all the data.
  6. Oct 19, 2007 #5
    Well the moles of the 2 are going to be the same. Think you can do it now?
  7. Oct 20, 2007 #6
    This is what's on my scratch paper so far:

    054 moles at 1.51atm/375k yields 1 liter
    now molar wt for water is 18 g/m and co2 is 44 g/m

    say x is number of moles hydrogen needed
    and y is number of moles carbon needed
    from above we build a formula total moles equal to .054
    x + y= .054
    x=.054 -y
    and we know that number moles times molar wt gives the weight and from above we know total weight is 1.391 g
    so x *18g + y* 44g = 1.391g
    18x + 44y = 1.39
    x= (1.39 - 44y)/18
    x= .077 - 2.44y
    so combine
    .077-2.44y = .054 - y
    .077 - .054 = 2.44 y - y
    .023 = 1.44 y
    y = .016 mole
    x= .038 mole
    water to co2 molar ratio is .038 /.016
    or about 2 to 1
    because water has 2 hydrogens and co2 has only 1 carbon
    the hydrogen to carbon ratio is 4 to 1
    now find the size of the hydrocarbon buy finding the ratio of moles hydrocarbon to moles of co2 /water mixture
    ideal gas law ..... pv=nrt or n=pv/rt
    if the 1 subset is for before reaction
    and the 2 subset is for after
    r=(p1 v1)/(n1 t1) before
    r= (p2 v2)/(n2 t2)
    (p1 v1)/(n1 t1) = (p2 v2)/(n2 t2)
    also some equalities
    p2=(1.51/.959) p1= 1.57 p1
    t2= 1.25 t1
    v2 = 4 v1
    ( p1 v1 t2 )/(p2 v2 t1) = n1/n2
    (p1 v1 1.25 t1)/(1.57 p1 4 v1 t1) =n1/n2
    (1.25/ 4 * 1.57) =
    n1/n2 = .2
    5 n1 = n2
    so you need 5 times more moles in the product then in the reactant.

    Am I doing this right?
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