Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Determining an Emperical Formula using Combustion Analysis data

  1. Jun 8, 2012 #1
    the problem im working with is:
    Naphthalene is a carbon-hydrogen compound that finds use as mothballs. A sample of naphthalene is subjected to combustion analysis, producing 1.100g of CO2 and .1802g of H2O. Based on these data, calculate the empirical formula of naphthalene.

    So when i set up the demensional analysis and get the moles of C and H i get for C .02499/
    .02499 and for H i get .02000/.02499. For H the book multiplies the fraction for H by 5 because .02000/.02499 = .800320128 which is not close to a whole number.

    My question is: How do they determine what the fraction is. Which is in this case is 4/5ths.

    The Emperical Formula for napthalene is C5H4
     
  2. jcsd
  3. Jun 8, 2012 #2

    Borek

    User Avatar

    Staff: Mentor

    0.8=8/10=4/5.

    It is not always that obvious, but here it is pretty simple.
     
  4. Jun 8, 2012 #3
    Ya but how do i know when to multiply is by some fraction? Like .99999 is pretty obvious but what about .4154 or .5900 etc etc. What is the rule of thumb?
     
  5. Jun 8, 2012 #4
    At a simple level, just multiply by 2, 3, 4, 5 etc until you get reasonably close to whole numbers
     
  6. Jun 8, 2012 #5
    Ok thanks Sjb,
    Now i have a different problem that i have been working:
    Ethyl Alcohol, the alcohol present in alcoholic beverages, is a carbon-hydrogen-oxygen compound. Combustion analysis of a 1.000g sample of ethyl alcohol produces 1.913g of CO2 and 1.174g of H2O. Based on these data, calculate the empirical formula of ethyl alcohol.

    So i set the dimensional analysis and solve for grams of Carbon(.5220) and grams of Hydrogen(.1316). Then i add C and H up and get .6536g mass. Next i solve for O by sutracting .6536 from 1.000 and come up with a .3464g of O. Then I convert H, C, and O to moles and come up with .04346 mol of C, .130 mol of H, and .02165 mol of O.

    At this point either the book is wrong or i am wrong. I used the moles of C (.04346) as my divisor to calculate the ratio. I came up with an empirical formula of CH3O5 but the book uses the moles of O (.02165) as the divisor and comes up with a empirical formula of C2H6O.

    Can someone check my calculations because i have done this problem twice. Thanks
     
  7. Jun 9, 2012 #6
    What is .04346 / .02165 ?
     
  8. Jun 9, 2012 #7

    Borek

    User Avatar

    Staff: Mentor

    Numbers of moles you calculated so far are OK. Answer sjb's question.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Determining an Emperical Formula using Combustion Analysis data
  1. Emperical formula (Replies: 2)

Loading...