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Homework Help: Help with finding molecular formula at a specific P, V, T

  1. Oct 5, 2008 #1
    1. The problem statement, all variables and given/known data
    A gaseous compound is 30.4% nitrogen and 69.6% oxygen by mass. A 5.25 g sample of the gas occupies a volume of 1.00 L and and exerts a pressure of 1.26 atm at -4.0C. Which of the following is its molecular formula?

    30.4% Nitrogen
    69.6% Oxygen
    M(sample)= 5.25g
    P= 1.26 atm
    V= 1.00 L
    T= -4.0C = 269K

    2. Relevant equations
    PV= nRT

    3. The attempt at a solution
    1. I found the mass of each gas in the compound using the given % by mass.
    I got:
    mass of N2= .304*5.25= 1.596g
    mass of O2= 5.25 - 1.596= 3.654g
    2. I found the mol of each gas in the compound using the above masses convert to mol.
    I got:
    mol of N2 = 0.0570 mol
    mol of O2 = 0.114 mol

    That's all I have tried, but I stuck here, I didn't know what to do next, even not sure what I have tried was right. Can anyone help, please!?
  2. jcsd
  3. Oct 5, 2008 #2
    You don't need to use N2 or O2 because they are together in the compound (this rule only applies when they are used as separate elements), so you should only use N and O.

    Before you can find the molecular formula, you need to find the empirical formula. In order to do this, you must find the moles of N and O. You can find this by dividing the percent of the element that was given to you in the problem by the mass of the element found on the periodic table. Also if the question gives you a mass you would use that in place of the percent, or vice versa. After you find the moles of both elements you must divide both of these amounts by the smallest amount of moles. For example, if you got .0234 moles of N and .567 moles of O, you would divide both amounts by .0234. With this answer you can somewhat round sloppily. Ex: if one of your answers was 5.91 you could consider it 6. If it is not close like this, try to find a common fraction. Then multiply both of your answers so that they are both whole numbers. If there are no decimals, then there is no need to multiply anything, and you will have your empirical formula answer.

    In order to find the molecular formula, you need to find the molar mass of the compound. To do this you would use the equation PV= nRT to find a g/mol ratio. I have not worked with this since last year, so this part might not be right.

    You would first alter the equation to solve for moles (or n). Then plug the values into the equation. It should look something like this:

    since R = (.0821L)(atm)/(mol)(K) you would flip this equation so that everything cancels itself out and moles are on top.
    Everything is now canceled out except for moles.
    Do the math to get your amount of moles.

    Once you have your moles, take the 5.25g and divide that by the amount of moles. This will give you your molar mass.

    Next, find the periodic table grams of your N?O?. Take your molar mass/PT g. You will then multiply this number by the subscripts of you empirical formula. Then you will have a molecular formula!

    Let me know what you get.

    If you need anything else just let me know!
  4. Oct 5, 2008 #3
    Hi, I found the right answer: N2O4.

    Here is my works:
    1. I found the mass of each gas in the compound using the given % by mass.
    mass of N= .304*5.25= 1.596g N
    mass of O= 5.25 - 1.596= 3.654g O

    2. Mass-Mole conversion:
    1.596g N* 1mol of N/ 14.01g of N = 0.1139 mol N
    3.654g O* 1mol of O/ 16g of O = 0.2284 mol of O

    3. Writing the Emperical Formula:
    0.1139/0.1139= 1
    0.1139/0.2284= 2
    => NO2

    Emp. Formulas mass= 46.01g

    4. Find the Molar Mass:
    M= m*RT/PV
    => M= {5.25g*0.0821(L*atm/K*mol)*269K}/(1.26atm*1L)
    M= 92.02g

    5. Find the Multiplier (x)
    x= Molar mass/Emp.Formula mass= 92.02g/46.01g
    x= 2

    Molecular Formula:
    (NO2)*2 => N2O4

    Wink! That's awesome! Thank you for being so helpful to me. I'll find you if I need more helps! hihihi.
    Best wishes!
  5. Oct 5, 2008 #4
    Your welcome! Glad I could help. :]
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