percent comp.

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musiclink213

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I have an unknown sample of KHP, and i need to find the percent comp. I did a titration with NaOH, but i don't know how to find the percent comp. help please!
 
I know this might sound like a 'dumb' question, but does your textbook have a page devoted to the calculations? Or the lab manual?

It should be pretty straightforward to convert the NaOh amount back to the orginial starting material on a one-to-one basis (I think).

Then take the starting material, plus the 1:1 ration you just established, and the calculate the % composition based on that. You definitely need to recheck your textbook because the calculations should be in there. Or try one of the Schaum's texts for GenChem. They're great.
 
nah, i don't have the textbook. i took gen chem last year, but this is quant. the lecture i took last semester, the lab this semester, and so i conveniantly forgot everything. and we don't exactly have a lab manual. he tried to save us some money, so he typed up the labs, sent it off to a print shop and they printed out copies for the 6 of us in there and colated them. so there is no equation sheet.

what i have is the grams of unknown sample, (1.2854), the amount of 1M NaOH it took to titrate it (~29) and that's it.

can i get onto the gen chem textbook website without an ID or anything? the one that i used to use had us type in the book number.
 
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Uh-oh, a question concerning a homework assignment...

Beware the wrath of Doctorcynical...
 
Originally posted by klonoff
google.com is amazing. Even if your professor was paper-stingy, many others post theirs online, there are many chemistry help sites, any equation you need should be available online.

I typed in KPD titration calculation and this was the first page the popped up. Looks like it has everything you need.

http://www.ic.sunysb.edu/Class/che133/lectures/calc011.html

thanks for your help. i too scoured the depths of google, typing in different things, and i did manage to find what i needed, actually i found basically the exact same experiment online.