OT: Anyone Good At Chemistry?

[Code001]

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 . . .

See, that's the thing. The teacher told me the reaction isn't a 1:1. With a 1:1, wouldn't both moles of NaOH and moles of HC2H3O2 present be the same answers? Thanks for all the help, Olaf!

 

[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.

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

I'm 20. I took Chemistry in highschool, but that was 5 years ago. I remember I did well in the class, but I had a great teacher who was real cool. I'm not bad at mathmatics and the sciences that involve heavy math (chemistry, physics, etc.), but I really need to work at it because I don't retain any of that information. If it's history, biology, govt., and especially English, I can do that stuff no problem since I tend to retain it. In highschool, it was more about learning the theory of chemistry rather than learning the actual mathmatical aspects of it. We spent more time learning about the people who created it and during which era vs. calculations. Moles were introduced, but in a very primitive form, and they were at the very end, so I was able to get by without having to remember too much.

Edit: Forgot to mention that I need this class to get into the nursing program that the school offers. At the moment, I'm moving towards working in the ER, but who knows, maybe I'll go ahead and get my Computer Science degree as well and do that instead.

 

[jkdsteve]

Congrats for going for some useful occupation (instead of waste of space corporate shill)...GL

 

[nbarts]

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.

So this is how it looks like when I talk to people about cutting & boosting frequencies, delay times, reverb tails, etc....

 

[noseminer]

oh god i have forgotten everything i know about chemistry, thanks for making me realize it

 

[Code001]

So this is how it looks like when I talk to people about cutting & boosting frequencies, delay times, reverb tails, etc....

Only difference is that boosting/cutting frequencies, setting up delay times, and configuring reverb trails have a use in real life.

/rimshot

 

[jkdsteve]

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 . . .

See, that's the thing. The teacher told me the reaction isn't a 1:1. With a 1:1, wouldn't both moles of NaOH and moles of HC2H3O2 present be the same answers? Thanks for all the help, Olaf!

If you look at his (and my) answers..they are the same, it's the concentration of the solution that's different.

As far as I know it has to be a 1:1 reaction...be different with a more 'powerful' acid such as H2SO4 .. which gives up 2H+ ions right? (from memory)

 

[Jakem]

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 . . .

Exactly. Calculating with "units" is such an easy way to make sure that one's calculations are correct and easy way to understand what one's calculating instead of trying to remember all the formulas.

 

[duesentrieb]

If you look at his (and my) answers..they are the same, it's the concentration of the solution that's different.

As far as I know it has to be a 1:1 reaction...be different with a more 'powerful' acid such as H2SO4 .. which gives up 2H+ ions right? (from memory)

Yup. I was thinking that a confimation would be good, you know.

Yup, something like H2SO4 or H3PO4 would have reacted stochiometrically (sp in english?) diff.