Mass of precipitate.

  • Thread starter m0286
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  • #1
m0286
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Hello
Im taking first year chem in university, and I am having trouble with one review question.. I don't know where to begin, I am 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
 

Answers and Replies

  • #2
whozum
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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.
 
  • #3
m0286
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still not gettin it.

Wow i guess i shouldn't be in university if i can't do this simple problem. So would it be just, 0.973g x 0.473g x (mass of 369 mL of water) and that's 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!
 
  • #4
johnw188
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m0286 said:
Hello
Im taking first year chem in university, and I am having trouble with one review question.. I don't know where to begin, I am 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
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.

169.824501g

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.
 
  • #5
m0286
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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!
 
  • #6
HallsofIvy
Science Advisor
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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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