1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Magnesium oxide smoke lost during the reaction?

  1. Jan 6, 2009 #1
    1. The problem statement, all variables and given/known data
    Would the Mg : O ratio be affected if some magnesium oxide smoke escaped during the reaction of Mg with oxygen gas?

    Please explain with a sample calculation. Thanks.

    2. Relevant equations
    I know that the empirical formula is MgO.

    3. The attempt at a solution
    I don't understand how to do this question., but I think that the ratio will remain unchanged?
  2. jcsd
  3. Jan 6, 2009 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I assume you are calculating the ratio by weighing the magnesium ribbon before combustion and then weighing the ash after combustion. If some ash mass is lost, what will this do to your calculation?

    You need to set up a thought experiment and then remove a little of the mass of ash and recalculate. What happens to the ratio?
  4. Jan 6, 2009 #3
    I have already done an experiment to find out the empirical formula for Magnesium Oxide, but this is a follow-up question..

    If some ash mass is lost, then I guess the numbers to calculate the ratio would be smaller?
    But since MgO is already reduced, then I guess the ratio will remain unchanged?

    Also, how can I show it with a sample calculation? Thanks.
  5. Jan 6, 2009 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    When you weighed the MgO, the balance didn't know it was MgO. All the balance tells you is how much mass is in the pan. YOU assign meaning to the reading.
  6. Jan 6, 2009 #5
    Sorry, I still don't understand.
  7. Jan 7, 2009 #6


    User Avatar

    Staff: Mentor

    When doing calculations for th eesxperiment, you are making two assumptions - even if you are not aware of the fact that you are doing them.

    First, you assume that the product that you have weighted is pure MgO.

    Second, you assume that all Mg was converted to the oxide and none was lost.

    Do you see now?

    As for the calculations:

    If there were some MgO lost in the form of "smoke", final mass of the product will be smaller. Subtract 10% from the mass you have found and repeat the calculations identical to those you have already performed. What ratio did you get this time? Is it identical to the one you got previous time?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook