# 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

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

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...

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..?

"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

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.

.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

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?

actually...you have done it... 