What is the process for finding the percent by mass of acetic acid in vinegar?

[majinknight]

I have another question similar to the last one. Ok The question tells me that the density of vinegar is 1.01g/mL and to calculate the mass of the vinegar sample, and the find the percent by mass of acetic acid in the simple. Now I know density= mass/ volume, and they give me the mass, the volume is 25mL as that is how much vinegar i used. So would i just rearrange it to find the answer? Also what does it mean when it says find the percent by mass of acetic acid in sample? Do they want me to once I found the mass of the vinegar divide the mass of the acetic acid which i found in last question to be 0.948g, by the mass of the vinegar?

 

[Monique]

You realize vinegar = acetic acid?

I have a really hard time understanding your questions, since the writing is not so clear..

 

[majinknight]

Ok what I am asking is do you calculate the mass of the vinegar by rearranging the density formula like so:

mass= density x volume
mass= 1.01g/mL x 25mL
mass= 25.25g

Would that be right for calculating the mass of vinegar?

And then the second part asks to find the percent by mass of acetic acid in the sample so would i take the mass of acetic acid from my last question which was 0.948g and divide by mass of vinegar like this:

percent= (0.948/25.25) x 100
percent= 3.75%

Would that be correct?

 

[Monique]

density of vinegar is 1.01g/mL
calculate the mass of the vinegar sample

density= mass/ volume
they give me the mass
the volume is 25mL

I guess they want you to multiply your volume of 25 ml with the density, so that know exactly what the mass of NH₃COOH is and the rest would be the mass of water.

Also what does it mean when it says find the percent by mass of acetic acid in sample? Do they want me to once I found the mass of the vinegar divide the mass of the acetic acid which i found in last question to be 0.948g, by the mass of the vinegar?

They probably mean %(m/v) which stands for mass per volume, you also have %(v/v): percentage volume per volume.

So you calculate the mass of NH₃COOH and divide that of the volume of vinegar you have (which is a mixture of NH₃COOH and H₂O.

 

[Monique]

Originally posted by majinknight
Would that be right for calculating the mass of vinegar?

Yes :)

percent= (0.948/25.25) x 100
percent= 3.75%

Would that be correct?

In principle, yes. But why do you for the same experiment get two different answers for the mass of NH₃COOH in the same sample?

Did you use glacial NH₃COOH in your experiment?

 

[majinknight]

I don't think so, what we did for the experiment is take a beaker with 25 mL of vinegar (CH3COOH) and add a few drops of phenophalen, then slowly add drop by drop Sodium Hydroxide until the solution turns slightly pink and stays for 30 seconds. That is when we knew we were done. That is all we did then had to answer some questions.

 

[Monique]

The reason I asked: glacial acidic acid has a density of 1.01 g/ml. If you have 25 ml of that and start adding NaOH, I think you can expect a very fierce reaction.

What I am thinking is that you had some glacial acidic acid, you diluted that to 25 ml and then started the titration with NaOH.

 

[majinknight]

Ya there is a good chance because the teacher when we ran out had to go into the back and make some more which i thought was weird since we were told we were using vinegar. I know it isn't really strong and either is the base as i sort of spilt some of the acedic acid on my hand and it did nothing and spilt a lot of the base on my arm and it didnt burn or anything either.

 

[radagast]

You need to be somewhat careful, though I doubt this is a factor in this case. There are cases where adding two volumes, x & y, doesn't yeild a combined volume of x + y.

If you add 1 liter of anhydrous ethanol to 1 liter of water, the combined volume, once thoroughly mixed, will be slightly less than 2 liters, assuming constant temperature.