Solve Titration Mass Problem: Unweighed Sample, 5% Acetic Acid

[Whalstib]

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

 

[sjb-2812]

If you have 5% by weight, how much do you have in 1 litre? Which is how many moles?

 

[Whalstib]

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...

 

[sjb-2812]

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..?

 

[Borek]

"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?

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methods

 

[Whalstib]

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

 

[Borek]

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:

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[clp991636]

actually...you have done it...