Articles & Tutorials

chemistry: nice!
Some basics on cross-multiplying

Problem:
A vitamin tablet contains 328 mg of Vitamin C (C6H8O6). How many Vitamin C molecules are contained in the tablet?

  1. First we'll assume the 328 mg are pure Vitamin C.
    1. You have 328 mg of stuff. But you need to know the # of molecules of that stuff you have. Um, what time is it? Yes, it's RULE OF THUMB time!!! Nice!

      1. RULE OF THUMB:  what units do we have?  And what units do we need? 

    2. We get from miliigrams (units we have) to molecules (units we need) by multiplying by known relationships, so that we can cancel out the terms we don’t want.  [Note the repeating theme.... ]
    3. To go from mass to molecules involves Avagodro's number. Check it:

      4. 328 mg C6H8O6
      X
      1 g
      X
      1 mole C6H8O6
      X
      6.022 x 1023 thingys
      =
      thingys of C6H8O6
       
      1,000 mg
      176 g
      1 mole C6H8O6

      1. See how things cancel? The only unit left after cancelling is "thingys" (molecules). Something key: to use Avogodro's # we need grams/mole. But we have miligrams. So, a conversion step is needed.

      [ahem]... kapeesh?

      :)

       

Problem:
How many magnesium ions are present in 9.1 mg of magnesium oxide?
How many iodide ions are present in 9.1 mg of magnesium oxide?

Here’s how I’d solve these kinds of problems.  Some of these steps are borrowed from what you already know, which is good.  It reinforces what you’ve learned. 

IMPORTANT: DO NOT try and memorize these steps. I explain them in text, which makes this all seem really long. I'm reasoning this all out in writing, so the important thing is to "think" about these steps instead of just "doing" them.

  1. First, if you could view just one ‘molecule’ of this stuff under a microscope, how many atoms of each would there be?
    1. To answer this, you have to know the charge of each ion.  Why?  Because to put ions together in a ‘molecule’ ["molecular unit"], the charges have to balance out to zero. 
    2. So, Magnesium is most stable as 2+.  What about Oxygen?  It’s a 2-. 
    3. So these two ions are equally charged, but in the opposite direction.  Magnesium goes up the number line by 2 charge units, and Oxygen goes back down by 2 charge units.  So it’s a one for one relationship to get the 'molecule' to a zero/neutral charge.
    4. And so you’d write it like this: MgO

  2. Next step: we know we have 9.1 mg of this stuff: MgO.  So how do we get to the ‘number’ of individual ions?  Think Avagodro’s number!  That dude’s ‘number’ gives you an important relationship; grams per mole, or grams/mole.. 
    1. Since we know the milligrams of MgO, and we can know the grams/mole from the periodic table, we can get to moles.  And once we find out how many moles we have of either Mg2+ or O2- we can then know how many actual ionic particles there are.  It will look like this:

      grams MgO
      X
      moles MgO
      X
      6.022 x 1023 particles
      =
      grams MgO
       
      # of grams
      moles MgO

    1. Above is a general form to get from grams to moles to particles.  Don’t memorize it.  Trust me.  It’s WAY better to understand how each step makes you able to cancel out terms, and by virtue of doing that each step gets you to the units you want; particles, in this case.
    2. RULE OF THUMB:  what units do we have?  And what units do we need? 
      1. We get from grams (units we have) to ions (units we need) by multiplying by known relationships, so that we can cancel out the terms we don’t want.  Nice! 
  1. So…. We have 9.1 mg of MgO.  Let’s work backwards a bit. To know how many particles of Mg, we need to know how many moles of Mg. 
    1. Since the formula MgO tells us there’s one Mg for every O, we know that whatever quantity of moles there are for Mg there must be for O. 
      1. As an example, in H20: if I said there were 10 moles of water, you’d know from the formula that there are 10 moles of O, and 20 moles of H.  Cuz there’s two H’s for every O. 
    1. So, to go from milligrams of MgO to moles of MgO, let’s get the Molecular weight of MgO.  From the periodic table, Mg is 24.3 grams/mole, and O is 15.9 grams/mole.   Add ‘em together, and you get 40.2 grams/mole.  (they have the same units, so they can just be added together). 
    2. OK.  So we have 9.1 mg of MgO, and we need to get to grams of MgO before we can multiply by grams/mole, so we set up our thingy-ding equation like this:

      9.1 mg MgO
      X
      1 g
      X
      1 mole MgO
      =
      2.26 x 10-4 moles MgO
       
      1,000 mg
      40.2 g MgO

    3. In the above, milgrams cancel out, and grams cancel out, leaving just moles of MgO. Nice!
      1. Since there's one mole of Mg in every mole of MgO, we now know exactly how many moles we have of both Mg and O. Nice!
    4. Now, with the # of moles, we figure out # of ions of Mg. Check it:
      2.26 x 10-4 moles Mg
      X
      6.02 x 1023 ions of Mg
      =
      the answer!
       
      1 mole

    5. Hey, I can't do the entire problem for you, right?!
    6. So how to you answer the 2nd question: how many ions of O in MgO? (You could run through all the above steps again, right? Or..... you could simply use the fact that for every mole of Mg there is a mole of O. So, however many ions there are of Mg, there's an equal number of ions of O.

      1. As in the example of H20: the relationship is 2 H's for every O. So for however many O's there are, there must be twice as many H's.  
    7. Kapeesh?