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Percent acetic acid by weight

  • Thread starter Integral0
  • Start date
47
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The molarity and percent by weight of the acetic acid in the vinegar. Use a value of 1.005 g/mL for the density of the vinegar.

Acetic acid molarity in vinegar = .208 mol / L

How do you calculate the Percent acetic acid by weight (using the acetic acid molarity)?

Thanks for the help.
 
Assume you've got one liter of solution, to simplify things. How many moles of acetic acid is that? How many grams of acetic acid is that? How many grams of solution do you have? Should be easy to figure out from there.
 
47
0
RE

Well, I have .208 moles of solution and 13.3 g of Acetic acid . . . I have 78 grams of Vinegar so 13.3 / 78 = 17.0% ahh I see

thanks for the help

P.S. How do I become a chem super freak like you? :wink:
 
hey how did u get 13.3 g of acetic acid from 0.208 moles provided that the molar mass is 60.5 . since, moles = mass/molar mass
therefore , mass = 60.5 * 0.208 = 12.58 grams ?
i have an assignment due on this , in the next 3 hrs. plz help someone ???
 

Borek

Mentor
28,036
2,546
Molar mass is 60.05, which makes 13.3 even more difficult to understand.

Check how to convert percentage to molarity - you can solve final formula for percentage.
 

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