Okay so I did this lab last week and since then have forgoten something so stupid that I will be hitting myself over the head when I finally remember how to do it.
I need to find the concentration (or percent composition?) of vinegar in terms of % HC2H3O2 by weight. I calculated the molarity of vinegar to be 0.9552 and the molarity of NaOH that I titrated with was 0.1940.

What in the world am I doing? How do I solve this? If you could show me that'd be great, because I have another one with NH4 that's just like this.

Thanks For the Help!!

Once you find the concentration of CH3CO2H, you can convert that to a weight by multiplying the concentration by the molecular weight:



To get a percentage you divide the number of grams of acetic acid by the weight the sample (and multiply by 100). You can assume a 1 liter sample size since you calculated the concentration in moles per liter. If you don't know the weight of that sample and you don't know the density, you can simply use the density of water as an estimate if the concentration of acetic acid isn't very large (it isn't). Water weighs approximately 1000 g / Liter.

Sorry I really don't get what your saying in the second bit.
I did the first one and got 0.0159 as the concentration, but the percent composition has me baffled.
:-(

The molecular weight of acetic acid is 60.05 g/mol. If the concentration is 0.0159 moles/liter (make sure of your units and make sure you're correct -- I didn't check you), then the mass of acetic acid in one liter is:



So, there is 1.134 grams of acetic acid per liter of solution. If we assume that 1 liter of solution weighs 1000 g then

Quote:

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Originally Posted by Perito

The molecular weight of acetic acid is 60.05 g/mol. If the concentration is 0.0159 moles/liter (make sure of your units and make sure you're correct -- I didn't check you), then the mass of acetic acid in one liter is:



So, there is 1.134 grams of acetic acid per liter of solution. If we assume that 1 liter of solution weighs 1000 g then

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Uh ho... you're a little tired ol'boy? Well, putting in 1.034g/L, you'll obtain 0.1034%

Thanks. I was afraid that I might make a typo.

It's OK Perito! :)

Okay, I think I've got it now.
Thanks!