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Formula from systematic name

  1. Mar 8, 2017 #1
    1. The problem statement, all variables and given/known data
    Sodium chlorate(I)
    Sodium chlorate(V)
    Potassium nitride(III)
    Phosphorus(III) chloride
    Magnesium iodate(I)

    2. Relevant equations
    None

    3. The attempt at a solution
    I know the number in the bracket refers to what's before it, i.e. in sodium chlorate(V), the (V) means that the chlorine has an oxidation number of +5, but how does this help me find the fomula of the copound.

    For example with sodium chlorate(V):
    I know its made from sodium (Na+) ions, and chlorate(ClO3-) ions.
     
  2. jcsd
  3. Mar 8, 2017 #2

    Borek

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    Staff: Mentor

    Molecule must be neutral, assume charge on oxygen to be -2.
     
  4. Mar 8, 2017 #3
    I'm not familiar with the use of this style for negative oxidation numbers, e.g. potassium nitride(III), if this is really K3N and not a typo for potassium nitrate(III) = potassium nitrite = KNO2.
     
  5. Mar 8, 2017 #4
    For sodium chlorate(I)
    Na would be +1
    Cl would be +1
    and O would be -2
    1 + 1 - 3(2) = -4 (what would I do with this?)

    Or

    Na + 1 -6 = 0
    Na = 5 (?)

    For sodium chlorate(V)
    Na would be +1
    Cl would be +5
    and O would be -2
    1 + 5 - 3(-2) = 0
    NaClO3
     
  6. Mar 8, 2017 #5
    .
     
  7. Mar 8, 2017 #6
    How many oxygens are there in sodium chlorate(I)?
    Note: in this nomenclature you must not get hung up on the notion that e.g. "chlorate" means "ClO3". All oxysalts are "ates" with different oxidation number of the central atom. This is quite different from the "ate", "ite", "per...ate" system.
     
  8. Mar 8, 2017 #7
    I'm not sure, I have to find the formula just from the name. I know sodium chlorate(v) has 3.
     
  9. Mar 8, 2017 #8

    Borek

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    Staff: Mentor

    And assuming one Na and one Cl, how many oxygens are needed for the molecule to be neutral?
     
  10. Mar 8, 2017 #9
    While I was away, amd therefore didn't see your reply, I was thinking and I think i got it:
    1 + 1 -2O = 0
    -2O = -2
    O = 1
    NaClO?
     
  11. Mar 8, 2017 #10

    Borek

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    Staff: Mentor

    Yep.

    Try with chlorate(III) and chlorate(V).
     
  12. Mar 8, 2017 #11
    Sodium chlorate(III):
    1+3-2O=0
    -2O=-4
    O=2
    NaClO2

    Sodium chlorate(V):
    1+5-2O=0
    -2O=-6
    O=3
    NaClO3
     
  13. Mar 8, 2017 #12

    Borek

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    Staff: Mentor

    At least with these you should have no problems now :wink:
     
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