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Homework Help: Titration Mass Problem

  1. Mar 27, 2010 #1
    1. The problem statement, all variables and given/known data
    A titration revealed I have .23M acetic acid (HC2H3O2) in a 10.0mL soln.

    I was not asked the weigh the sample at any time but our question is:
    "The manufacturer of this vinegar claims it is 5% by weight. Ue your results and a density of 1.0g/mL to investigate this claim"

    I'm at a loss to determine any results based weight with no weighing involved...I thought the molarity answered any questions about concentration.

    2. Relevant equations
    Acetic acid 60.05g/mol
    H2O 18.02g/mole
    No samples weighed at any time

    titration began with 10.0mL vinegar and 10.0mL H2O

    3. The attempt at a solution
    My only guess is acetic acid is 60.05 g/mol at .23M gives me 13.81 grams times .05 gives me .69 g/mL which would be 5% of the molecular weight...?

    I'm I getting close?

    Thanks as always!

  2. jcsd
  3. Mar 27, 2010 #2
    If you have 5% by weight, how much do you have in 1 litre? Which is how many moles?
  4. Mar 27, 2010 #3
    See that's confusing because I know the molarity is .23M. So I have .23 moles/liter which is 13.81 g/L acetic acid. That's a fact. This should relate exactly to 5% in manufacturers claim is accurate......
  5. Mar 28, 2010 #4
    Perhaps that's the point - sometimes claims can be misleading. Or perhaps you made a mistake in your calculation. If there was more acid than claimed I'd look at the ingredients to see if there was any other source but as it's less..?
  6. Mar 28, 2010 #5


    User Avatar

    Staff: Mentor

    You are given volume (10 mL), you are given density... Can't you simply calculate mass of the sample?

  7. Mar 28, 2010 #6
    We're not given the density just molarity. All I can figure since it's vinegar they have told us the soln is HC2H3O2 + H2O with 100% H2O being 1.0g g/mL.

    Here's my final answer:
    .24M= (.24 moles/L HC2H3O2)(60.05 g/mole)= 14.41g/L= .014g/mL = 1.4% by weight.....

    For 5% the eqn would be:
    .83M= (.83moles/L)(60.05g/mole)=50.0g/L= .05g/mL= 5% by weight......right?


  8. Mar 29, 2010 #7


    User Avatar

    Staff: Mentor

    You were specificially told to use 1.0 g/mL density, aren't you?

    And you did exactly what you were expected to do - you used density to calculate mass of the solution (assuming 1 mL is 1 g is exactly the same), yet you claim you don' know how :grumpy:

  9. Mar 29, 2010 #8
    actually...you have done it...:biggrin:
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