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Homework Help: Magnesium Oxide lab

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  1. Sep 4, 2017 #1
    So in my experiment we combusted Magnesium ribbon to form Magnesium oxide.
    These were the reactions:

    Mg(s) + N2(g) + O2(g) → MgO(s) + Mg3N2(s)


    MgO(s) + Mg3N2(s) + H2O(l) → MgO(s) + Mg(OH)2(s) + NH3(g)


    MgO(s) + Mg(OH)2(s) → MgxOy(s) + H2O(g)
    -----------------------------------------------------------------------------
    Questions:
    In each case decide whether the situation described below would lead to a calculated ratio of too much oxygen or too little oxygen. (The calculated ratio contains the mistake)
    a. putting in more water than is need
    b. Having some Mg(OH)2 left in the crucible instead of converting it to MgO
    c. Having some Mg3N2 in the crucible as product, instead of pure MgO
    d. Forgetting to weigh the crucible cover along with the crucible and contents.
    e. Letting a lot of the dense white smoke escape during heating.
    ------------------------------------------------------------------------------
    My answers:
    Short version:
    a. too much
    b. too much
    c. too little
    d. too little
    e. too little

    Long version:
    a. If you did not dry out the excess water you would calculate the ratio of MgO to contain too much oxygen. Water contains oxygen, this oxygen would be calculated into the composition of MgO. If you don’t heat the crucible then the water cannot be released as water vapor, thus the ratio of MgO would be too high in oxygen.



    b. If you didn’t convert the Mg(OH)2 left in the crucible to MgO the calculated ratio of MgO would contain too much oxygen, because you are subtracting the MgO mass by the mass of the starting magnesium to find oxygen. There are also 2 atoms of Oxygen in Mg(OH)2, compared to MgO’s one, leading to an excess of oxygen.



    c. If you left some Mg3N2 in the crucible as product you would calculate a ratio of MgO to contain too little oxygen. When we weigh the final contents we wont know that the contents have more magnesium, we assume it’s pure MgO. Mg3N2 does not contain any oxygen, so there would be too little oxygen, but there is magnesium present so there is more magnesium than. oxygen.



    d. The cover would significantly reduce the weight of the crucible and its contents. We would calculate he mass of MgO to be significantly less. This would lead us to calculate a negative number for the amount of magnesium oxide. There would be too less of oxygen.



    e. If you let the dense white smoke escape from the crucible you would calculate the ratio of MgO to contain too little oxygen. The smoke contain magnesium and oxygen, when we weigh the crucible and its contents and subtract this from the mass of the crucible and ribbon we would calculate a number that we assume is pure MgO, resulting in too little oxygen.

    Thanks!
     
  2. jcsd
  3. Sep 5, 2017 #2

    Borek

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    Staff: Mentor

    First of all - these:

    are not reaction equations. These are sums of more than one reaction and they can be balanced in many ways.

    Then, you have not described the experiment, so it is not clear what you are trying to calculate. If we don't know what you are calculating, how can we know what kind of error you are making?
     
  4. Sep 5, 2017 #3
    We took some magnesium ribbon and we heated it so that it would react with the oxygen in the air to form magnesium oxide. There was also nitrogen in the air so that turned into some magnesium nitride. To convert the magnesium nitride to magnesium oxide we added water so it turned into magnesium hydroxide, we heated this and were left with magnesium oxide.
     
  5. Sep 5, 2017 #4

    Borek

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    Staff: Mentor

    And you calculated ratio of what to what? And from what data?
     
  6. Sep 5, 2017 #5
    When calculating the empirical formula i got 1.09 moles of magnesium to O. So MgO
     
  7. Sep 5, 2017 #6

    Borek

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    Staff: Mentor

    That doesn't answer my question.

    Ratio requires knowing two numbers. What numbers where these and where did you got them from?
     
  8. Sep 5, 2017 #7
     
  9. Sep 5, 2017 #8
    I got the number from converting the magnesium grams with the grams of oxygen reacted. Converting these to moles and dividing both by the smaller number.
     
  10. Sep 5, 2017 #9

    Borek

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    Staff: Mentor

    And where did you got masses of magnesium and oxygen from?

    This all matters, you assume some things to be obvious and known, but in fact they are not. I can guess what you did - but it will be still a guess, not necessarily correct.
     
  11. Sep 5, 2017 #10
    I got these numbers from the scale when conducting the experiment. So we took a crucible and put some magnesium in it and weighed it. After heating the magnesium it turned into magnesium oxide i then weighed this and found the change in mass. So now I had the mass of starting magnesium and magnesium oxide. I subtracted these 2 numbers and that was the amount of oxygen that reacted with the magesium during the heating. Mg + O2--> 2MgO
     
  12. Sep 7, 2017 #11

    Borek

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    Staff: Mentor

    Note: hard to say where you got the magnesium mass from (you have mentioned weighing crucible and magnesium together, but not crucible or magnesium alone), but that's not that important, let's just assume you got the magnesium mass.

    a: "adding too much water" is not equivalent to "leaving the sample wet". You can add too much water, but still dry the sample correctly.

    b:

    what about hydrogens in Mg(OH)2? do they matter, or not?
     
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