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Chemistry - measuring the ammount of oxygen i need to to make MgO

  1. Dec 19, 2011 #1
    i have done the maths based on an online conversion calculator that said 1 liter of air is equal to 0.001m^3 in volume. my chemistry teacher tells me that my maths is wrong but wont tell me how...

    here is what i wrote:

    Failing the above, a larger crucible would hold more oxygen and the magnesium would be able to burn for longer before the lid would have to be lifted. This wouldn’t prevent the error given by lifting the crucible lid but it would minimise it to best practice. It would be possible to measure how large the crucible would need to be by working out the molecular weight of oxygen in a given area of air (air is roughly 21% oxygen).

    To react the magnesium with enough oxygen the same amount of each element is needed to give the 1:1 ratio meaning if the weight of magnesium given was 24.31 grams we would need 16 grams of oxygen to complete the reaction.
    24.31 / 16 = 1.519375. This gives the difference between the weights required. 1.519375 grams of magnesium is necessary if you had 1g of oxygen to react with.
    The magnesium weight at the beginning of the experiment was 0.26 grams.
    24.31 / 0.26 = 93.5. The amount of magnesium given would fit into 1 mole 93.5 times.
    16 / 93.5 = 0.171122994. This is the minimal amount of oxygen required in grams for this reaction to have worked completely without additional oxygen having to be added by lifting the crucible lid.

    Having this number would mean that you could work out how much air would need to be in contact with the magnesium, for there to be enough oxygen available to react with the magnesium without completely using it up and having to lift the lid.
    For example if we had 21 grams of air inside our crucible, in theory 1 gram of oxygen would be available for this reaction and it would have been enough.

    If it is said that air is 100% and 21% of it is needed (the oxygen).
    100/21 = 4.761904762. This is the amount of times 21 fits into the 100.
    Now multiplying the amount of oxygen required by this number should give us the total mass of the air needed for our reaction to be successful without having to add additional air.
    0.171122994 x 4.761904762 = 0.8148714 grams of air would be required.
    The density of air is approximately 1.19g/cm3

    The crucibles were larger than 1cm3 on the inside so there should have been enough oxygen overall inside to react with the magnesium.

    please can someone help me figure out where i went wrong?
     
  2. jcsd
  3. Dec 19, 2011 #2

    chemisttree

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    Found it.

    Redo this whole mess with units and you will avoid this mistake in the future.

    Check your work! You state that the reaction stoichiometry is 24.31 g Mg per 16 g O2. Clearly you need less oxygen by mass than magnesium to achieve stoichiometry. The bolded answer reads, "For every 0.17 g Mg, I need 0.8 g air (0.17 g O2) for complete oxidation. 21 percent of that is O2 or 0.2*.8=0.17 g O2." Clearly these two statements are not compatible.

    Teacher wins again!
     
  4. Dec 19, 2011 #3
    i went that the ammount of magnesium i have to react fits into a single mole (24.31 grams) 93.5 times. so i need the ammount of oxygen that fits into a single mole 93.5 times. so 16:00 is a single mole of oxygen. 16/93.5 gives me the ammount of oxygen that i needed.

    the oxygen is 21% of the air. 100% would be the total air. so 100/21 gives me a number which i can multiply the oxygen number by to get the total mass of the air required.

    you may be right with how ive done the maths wrong but the way you explained it confuses me.
     
  5. Dec 20, 2011 #4

    Borek

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    Staff: Mentor

    First of all - I have no idea what the question is. You are doing strange things moving is circles around something, but I have no idea around what.

    There is a reason we expect you not ignore the template - if you expect help, you need to explain what is the question you are solving. You failed.

    Please note that ignoring the template can get your post deleted. The only reason it didn't happen this time is that you already get an answer.
     
  6. Dec 20, 2011 #5
    sorry the reason why i didn't use the template is because i am writing up an experiment report and this was part of my evaluation, there wasn't a direct question asked of me.

    i am trying to find out the volume of air that i would need to be able to completely react 0.26 grams of magnesium into magnesium oxide.

    for the experiment, we used a very small air tight crucible to heat the magnesium in, i want to find the size of the crucible we would have needed to use without having to lift the lid in order to let more oxygen inside. does that make it a little more clear?
     
  7. Dec 20, 2011 #6

    Borek

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    Staff: Mentor

    Now it starts to make sense. Note that is easily fits the template.

    So, assuming you have 0.26g Mg, you calculated you need 0.17g of oxygen - that's correct. However, oxygen makes 21% of the atmosphere by volume, not by mass - so the rest of your calculations is incorrect.

    Use ideal gas equation to calculate volume of the oxygen. Don't ask me what was the gas temperature - as I have no idea. Magnesium can burn well over 2000 °C.
     
  8. Dec 20, 2011 #7
    yeah i was thinking about using the ideal gas laws as we are learning that in physics class at the moment. the issue is that we dont have a temp. not having that value will affect everything else and if i give the results just assuming STP i dont think it would be worthwhile including it in my evaluation as its going to be way off the mark.
     
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