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Mass of precipitate.

  1. Sep 18, 2005 #1
    Hello
    Im taking first year chem in university, and Im having trouble with one review question.. I dont know where to begin, im hoping just for some guidance or a formula...

    0.973g of silver nitrate and 0.473 g of potassium bromate are added to 369mL of water. Solid silver bromate is formed, dried and weighed. What is the mass, in g, of the precipitated silver bromate? Assume the silver brmate is completely insoluble.
    THANKS
     
  2. jcsd
  3. Sep 18, 2005 #2
    Find the mass of the whole solution including the water, and then noting that potassium nitrate will dissolve in the water, when dried out the silver bromate will have the mass off the entire solution - the water and potassium bromate.
     
  4. Sep 22, 2005 #3
    still not gettin it.

    Wow i guess i shouldnt be in university if i cant do this simple problem. So would it be just, 0.973g x 0.473g x (mass of 369 mL of water) and thats my answer? I found somewhere on the internet that that mass of water would just be 369g is that right??? which would give you a final answer of 169.824501g. Or am i going about this wrong still.
    THANKS!
     
  5. Sep 22, 2005 #4
    Its been a while since I've taken chem, but I have a feeling that whozum is making it more complex than it needs to be. If I was to solve this problem I'd take these steps:

    1. Work out the moles of each substance
    2. Find the limiting reagent
    3. Work out the moles of product formed by the reaction
    4. Calculate the mass of the product

    If you still need more help I'll go pull out my text and look up some numbers/formulas.

    One other note: Always remember significant figures.

    This answer is actually wrong, as you were only given 3 significant figures worth of data. The actual answer to the calculation would be 170g.
     
  6. Sep 22, 2005 #5
    Thanks soo much you were a big help :smile: I understand it now. One of the nicest helps I have had on this site, it was kind of you to offer to lookit up. Thanks a ton YOUR GREAT!
     
  7. Sep 23, 2005 #6

    HallsofIvy

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    Am I missing something here? You say "Solid silver bromate is formed, dried and weighed." And then ask " What is the mass, in g, of the precipitated silver bromate?"

    Since you already know the weight (in dynes?) just divide by the "gravitational factor" 981 m/s2 to get the mass in grams.

    Of course, normally, chemicals are "weighed" on a balance scale, using standard weights that are marked as mass, in grams, so perhaps that is not what you want. What is the "weight" and, most importantly, what units is it in?
     
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