An Unknown Compound

  1. 1. The problem statement, all variables and given/known data

    A compound containing element A and oxygen has a mole ratio of A:O = 2:3. If 8.0 grams of the oxide contains 2.4 grams of oxygen...

    a) what is the atomic weight of A?
    b) what is the weight of one mole of the oxide?
    c) what is the theoretical weight of the oxide formed when 28 g of A is heated in excess oxygen?
    d) what is the % yield if 38 grams was produced from 28 grams of A?

    2. Relevant equations


    3. The attempt at a solution

    I didn't have any because I barely understood what to do :( can someone please guide me on how to solve this? It'll be of great help ^^
  2. jcsd
  3. symbolipoint

    symbolipoint 3,070
    Homework Helper
    Gold Member

    Start with the oxygen. How many moles of oxygen is equivalent to that 2.4 grams of oxygen from the compound?
  4. so I'll convert the oxygen grams to moles right? :)
  5. Borek

    Staff: Mentor

    Yes. Then try to find out how many moles of the oxide you have (look at the formula).
    Last edited by a moderator: Aug 13, 2013
  6. 8.0 - 5.6 = 2.4 gO

    2.4 gO x 1 mole O / 16 gO = 0.15 moles O

    A:O = 2:3
    A:0.15 = 2:3

    [tex]\frac{A}{0.15}[/tex] x [tex]\frac{2}{3}[/tex]

    3A = (0.15)(2)
    3A = 0.30
    A = 0.1 moles Oxygen

    AW = g/mole
    AW = 5.6 gO / 0.1 moles O
    AW = 56 g/mole

    = (56)(2) + (16)(3)
    = 160 g

    Is this correct? I'm sure with my a & b but with c and d, I don't really get how that happened :( I asked for help in solving this, and this was the solution he gave.
  7. Borek

    Staff: Mentor

    No - this is not "moles of Oxygen". This is number of moles of A in the sample.

  8. oops! I'm very sorry for that :)) the one in my paper was 0.1 moles of A :D

    haha xD

    thank you! um, but the part that I'm really confused about is this one ...

    4A + 3O2 -> 2 A2O3

    if we take the balancing off..
    A + O2 -> A2O3

    Where did we get A + O2 ? Is this common knowledge? because all I understood was A2O3 since we had a mole ratio of 2:3 right? :D
  9. Borek

    Staff: Mentor

    Not sure what your problem is. There are many ways of producing oxides, but direct reaction between the element and oxygen is the simplest one (even if technically not always possible). Even if it is not possible, you were told in the question that 38 grams of A2O3 were produced from 28 grams of A - no matter what the real reaction was, molar ratio of the A and A2O3 will be always the same.

    Then A2O3 was given, O[ub]2[/sub] is a common knowledge, A is just the simplest way of approaching the problem.
    Last edited by a moderator: Aug 13, 2013
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