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Titration Mass Problem

  • Thread starter Whalstib
  • Start date
  • #1
119
0

Homework Statement


Hi,
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.





Homework 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



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!

Warren
 

Answers and Replies

  • #2
445
5
If you have 5% by weight, how much do you have in 1 litre? Which is how many moles?
 
  • #3
119
0
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......
 
  • #4
445
5
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..?
 
  • #5
Borek
Mentor
28,328
2,717
"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...
You are given volume (10 mL), you are given density... Can't you simply calculate mass of the sample?

--
methods
 
  • #6
119
0
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?

Thanks,

Warren
 
  • #7
Borek
Mentor
28,328
2,717
We're not given the density just molarity.
Ue your results and a density of 1.0g/mL to investigate this claim"
You were specificially told to use 1.0 g/mL density, aren't you?

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?
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:

--
 
  • #8
1
0
actually...you have done it...:biggrin:
 

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