OT: Anyone Good At Chemistry?

[Code001]

I need some help with my titration (acid/base) school work. Basically, we added NaOH to HC2H3O2 to form a salt and water. Here's the data on test 1:

Volume of vinegar solution = 5.0mL
Concentration of NaOH = 0.200 M
Initial buret reading = 0.00mL
Final buret reading = 21.60mL

I need to find:
Moles of NaOH added
Moles of acetic acid present in sample
Molarity of acetic acid in sample
Average molarity
Mass of acetic acid in sample
Weight percent of acetic acid in sample (assuming 5 mL weighs 5g)
Average weight percent

Now, I'm not looking for people to just give me the answers, but someone who's willing to help teach me (fat chance, right?). My teacher is a rasta-man and can barely speak English, so it's hard for people to understand him. Anyway, any help would be greatly appreciated.

 

[Copperhead]

Oh geez. My son's a Dean's List chem major at U of A, unfortunately he left this morning driving my wife's mom cross country.
Sorry

 

[Gainfreak]

I need some help with my titration (acid/base) school work. Basically, we added NaOH to HC2H3O2 to form a salt and water. Here's the data on test 1:

Volume of vinegar solution = 5.0mL
Concentration of NaOH = 0.200 M
Initial buret reading = 0.00mL
Final buret reading = 21.60mL

I need to find:
Moles of NaOH added
Moles of acetic acid present in sample
Molarity of acetic acid in sample
Average molarity
Mass of acetic acid in sample
Weight percent of acetic acid in sample (assuming 5 mL weighs 5g)
Average weight percent

Now, I'm not looking for people to just give me the answers, but someone who's willing to help teach me (fat chance, right?). My teacher is a rasta-man and can barely speak English, so it's hard for people to understand him. Anyway, any help would be greatly appreciated.

talk to Olaf!!

 

[Code001]

Oh geez. My son's a Dean's List chem major at U of A, unfortunately he left this morning driving my wife's mom cross country.
Sorry

Hah, just my luck.

talk to Olaf!!

Olaf! Wo bist du?

 

[glpg80]

computer engineering and electrical engineering here.. i had (did have =p) the highest average in physics back in high school out of the entire school 98.8.

but its been way too long, and ive slept many times since then sorry man.

i do know what you mean about the teachers and speaking. i had a russian college trig II class-teacher that was the same way. luckly by the end i got used to it and was a teachers pet also lol. but it does make a hard class even harder

-matt

 

[Capulin Overdrive]

it was Chinese grad students at my college? the foreign students were achievers, but God Damn, you think maybe the uni. could got them to learn english a little better if they're gonna be teaching hungover white trash rednecks the hardest classes?

 

[jkdsteve]

acid + base = salt + water

This stuff should be easy to find on the net.... NaOH + acetic acid is going to form Sodium Acetate + water right?

Sodium is gonna displace the active Hydrogen and I think Sodium is Na+ in solution right so it's a 1 to 1 reaction with the Acetyl group which has one -ve charge

or (expressed as ions) Na+OH- and H+C2H3OO- makes it easy to see the reaction.

Write out the chemical reaction...

NaOH + HC2H3O2 = NaC2H302 + H20

You know the molar concentration of the NaOH solution...molar concentration is M per litre....assume the thing is titrated until all the acid is used up....stuff's gotta be easy but it's been >20 years for me...so don't quote me on teh rest (do your own research man!

21.6ml of NaOH is added right? (this is the buret reading)

0.2M concentration so # of moles NaOH 0.2 * 21.6 * 10^-3 (for ml) ?

I could be totally wrong but I think this is also teh # of moles of acetic acid (1:1 reaction)

Molarity of acetic acid is then 0.2*21.6*10^-3/5x10^-3 = 0.2*21.6/5

Not sure what average molarity is but am guessing it's the sum of # of moles of acid+# of moles of base / total liquid volume (21.6ml+5ml)

There must be an easy way to get mass of acetic acid from # of moles...need to know what one mole weighs...can't think of that right now....lookup molar mass..since we have calculated # of moles.

Weight percent I'm not familiar with at all or just can't remember...it might be the molar concentration of the solution expressed as a % somehow...

I might have done more damage than good...and sorry about all the edits, memory of this shit is foggy and I looked up Molar concentration

 

[Code001]

acid + base = salt + water

This stuff should be easy to find on the net.... NaOH + acetic acid is going to form Sodium Acetate + water right?

Sodium is gonna displace the active Hydrogen and I think Sodium is Na+ in solution right so it's a 1 to 1 reaction with the Acetyl group which has one -ve charge

or (expressed as ions) Na+OH- and H+C2H3OO- makes it easy to see the reaction.

Write out the chemical reaction...

NaOH + HC2H3O2 = NaC2H302 + H20

You know the molar concentration of the NaOH solution...molar concentration is M per litre....assume the thing is titrated until all the acid is used up....stuff's gotta be easy but it's been >20 years for me...so don't quote me on teh rest (do your own research man!

Yeah, all of that has been given to us already on the front sheet. The only things we need to work out are the things I gave.

21.6ml of NaOH is added right? (this is the buret reading)

0.2M concentration so # of moles NaOH 0.2 * 21.6 * 10^-3 (for ml) ?

Not 100% sure, since I really don't know how to calculate moles, but the mL needs to be converted to L and then setup. I was gonna learn how to do moles when I woke up tomorrow.

I could be totally wrong but I think this is also teh # of moles of acetic acid (1:1 reaction)

Teacher said this is wrong, even though we asked him before and he said it was right. I don't really understand it (yes, he said one thing and then said the other all within about 30 minutes of class).

Molarity of acetic acid is then 0.2*21.6*10^-3/5x10^-3 = 0.2*21.6/5

Not sure what average molarity is but am guessing it's the sum of # of moles of acid+# of moles of base / total liquid volume (21.6ml+5ml)

There must be an easy way to get mass of acetic acid from # of moles...need to know what one mole weighs...can't think of that right now....lookup molar mass..since we have calculated # of moles.

Weight percent I'm not familiar with at all or just can't remember...it might be the molar concentration of the solution expressed as a % somehow...

I might have done more damage than good...and sorry about all the edits, memory of this shit is foggy and I looked up Molar concentration

Rest of the stuff, I have no clue how to do. Gotta look it up on the net, but I was just hoping for someone to help me in case I had questions on certain aspects. Thanks for helping out so far!

 

[jkdsteve]

That's a shame ... the top stuff is the stuff I remembered

Pretty sure my example for moles is correct from M and was about 95% sure that 1 mole of NaOH + 1 mole of acid -> 1 mole of salt + 1 mole of water :-/

I tried to put enough down to give you a nod in the general direction.... maybe too much but you should definately do it yourself if only to check it

 

[Code001]

That's a shame ... the top stuff is the stuff I remembered

Pretty sure my example for moles is correct from M and was about 95% sure that 1 mole of NaOH + 1 mole of acid -> 1 mole of salt + 1 mole of water :-/

I tried to put enough down to give you a nod in the general direction.... maybe too much but you should definately do it yourself if only to check it

I definitely want to do it myself. I mean, it's useless for me to just get answers and write them down when I'm gonna have a test on this stuff next week. Again, thanks for the help, and I'll check it out more in depth tomorrow when I wake up. I'm dead tired at the moment since I only got about 3 hours of sleep.

 

[duesentrieb]

Will reply soon, Tom.

 

[duesentrieb]

CH3COOH + NaOH --> CH3COONa + H2O
so a 1:1 reaction

21.6 ml of a 0.2 mol solution: 0.0216l*0.2mol/l = 0.00432 Mol NaOH
which equals 0.00432 mol CH3COOH

5 ml CH3COOH --> 0.004321mol/0.005l = 0.864 mol/l conc. of CH3COOH

molar weight CH3COOH: 12+3+12+32+1 = 60 g/mol

60g/mol * 0.864 mol/l * 0.005l = 0.2592 g (mass of CH3COOH)

percentage: 0.2592g/5g = 0.0518 = 5.2%

and so on . . .

 

[duesentrieb]

one tip:

if you calculate, always calculate with the dimensions too and see what happens - f.i.

g/mol * mol/l = g/l

and so on . . .

 

[mysticaxe]

one tip:

if you calculate, always calculate with the dimensions too and see what happens - f.i.

g/mol * mol/l = g/l

and so on . . .

a big +1 on that. Follow the calculations the way Olaf set them up. The way I always did it was draw the equation and then under each compound put all of the information - moles, volume, grams, whatever. Then you can use that information to set up the mass/mole balance. This case has a 1:1:1:1 stoichiometry, so you won't have annoying 2's and 3's to forget.

 

[iplayloudly]

chemistry = hardest 2 classes I ever took. Seriously. I studied harder for those 2 classes than any other class all through my 4 years of college, and I still only passed with a C in both.

 

[jkdsteve]

That's a shame ... the top stuff is the stuff I remembered

Pretty sure my example for moles is correct from M and was about 95% sure that 1 mole of NaOH + 1 mole of acid -> 1 mole of salt + 1 mole of water :-/

I tried to put enough down to give you a nod in the general direction.... maybe too much but you should definately do it yourself if only to check it

I definitely want to do it myself. I mean, it's useless for me to just get answers and write them down when I'm gonna have a test on this stuff next week. Again, thanks for the help, and I'll check it out more in depth tomorrow when I wake up. I'm dead tired at the moment since I only got about 3 hours of sleep.

Hey man..this set me wondering how old you are? I think we first learn this stuff at 14 in the UK (Chemistry O Level..it might even have been sooner..pre o level primer stuff)...figured you were a little older than that? ..course if you didn't do any Chemistry and are just ploughing into that in College...my condolences, it's going to be a rough ride

 

[duesentrieb]

life = chemistry + physics

 

[tweed]

CH3COOH + NaOH --> CH3COONa + H2O
so a 1:1 reaction

21.6 ml of a 0.2 mol solution: 0.0216l*0.2mol/l = 0.00432 Mol NaOH
which equals 0.00432 mol CH3COOH

5 ml CH3COOH --> 0.004321mol/0.005l = 0.864 mol/l conc. of CH3COOH

molar weight CH3COOH: 12+3+12+32+1 = 60 g/mol

60g/mol * 0.864 mol/l * 0.005l = 0.2592 g (mass of CH3COOH)

percentage: 0.2592g/5g = 0.0518 = 5.2%

and so on . . .

So....how many 12ax7's is that exactly?

 

[duesentrieb]

Pi*r^2

 

[tweed]

Pi*r^2

3 it is!