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Molecular mass of gaseous compound question

  1. Nov 14, 2012 #1
    1. The problem statement, all variables and given/known data
    "A gaseous binary compound has a vapor density that is 1.94 times that of oxygen at the same temperature and pressure. When 1.39g of the gas is burned in an excess of oxygen, 1.21 g water is formed, removing all the hydrogen originally present.

    (a) Estimate the molecular mass of the gaseous compound.
    (b)How many hydrogen atoms are there in a molecule of the compound?
    (c) What is the maximum possible value of the atomic mass of the second element in the compound?
    (d) Are other values possible for the atomic mass of the second element? Use a table of atomic masses to identify the element that best fits the data.
    (e) What is the molecular formula of the compound? "

    p.47 question 29 out of "Principles of Modern Chemistry" (Authored by: Oxtoby Gillis Campion).
    3. The attempt at a solution
    My attempt at this solution I do not believe is worth writing.

    This is not work assigned from a college course, I am simply trying to study chemistry independently.

    With this question, I have not been able to find an effective approach to yield a correct solution. Help provided will be greatly appreciated.
     
  2. jcsd
  3. Nov 15, 2012 #2

    AGNuke

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    Gold Member

    (a)Use the relative Vapor density given. The answer is in front of your eyes.
     
  4. Nov 21, 2012 #3
    I used the "1.94 times" information given from the problem to find the molecular mass of the compound after finding the vapor density formula on wikipedia.
    (32g/mol O2/2.016g/mol H2) = 15.873
    15.873*1.94 = 30.8 .
    30.8 *2.016 = 62.1 g/mol (molar mass of the gaseous binary compound)

    So I have completed a), but I am now unable to find out the number of H atoms.
    How do I go about this?

    I have found the number of g H in the gaseous binary compound : .0677 g H , but from there I do not know where to go.
     
  5. Nov 21, 2012 #4

    Borek

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    Staff: Mentor

    How many moles of the compound? How many moles of hydrogen per one mole of the compound?
     
  6. Nov 21, 2012 #5
    (62.1g/mol / 1.39g [binary gaseous compound] ) = 44.6 mol binary gaseous compound

    1.21g H2O * (1.008g/18.016g) * (1 mol H/ 1.008g H) = 0.0672 mol H
     
  7. Nov 21, 2012 #6

    Borek

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    Staff: Mentor

    You are twice wrong. You approach from the right side, but you miss details or get it reversed.
     
  8. Nov 21, 2012 #7
    Borek, correct, I have since reworked my calculations.

    a) 62.1g/mol
    b)
    1.21g H2O * (1 mol H2O / 18.016 g H2O) * (2 mol H/ 1 mol H2O) = .134 mol H
    1.39g [compound] * (1 mol [compound] / 62.1 g) = .0224 mol [compound]

    .134 mol H/ .0224 mol [compound] = 5.98 , approx. 6
    thus 6 hdron atoms in a molecule of this gaseous binary compound

    c) 6 * 1.008 g H = 6.048
    62.1-6.048=56.052

    d) 56.052/2 = 28.026
    Silicon fits or Nitrogen

    e) thus with 6 Hydrogen atoms and 2 Silicon atoms the molecular formula is
    Si2H6


    I have checked with the appendix of the book and the answers are correct.


    Thanks to Borek and AGNuke. Very much appreciated here.
     
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