[gogglespinoy]

So there is .1235g of the unknown being neutralized with 15.55 mL of 0.1087M NaOH.
so i did 15.55(.1087)=.00169
then i did .1235/.00169 and got 73.07 for the molar mass of the diprotic acid.

the second part of the question is to find the empirical formula for the diprotic acid. the acid (.3469g) is burned with pure oxygen producing .6268g of CO2 and .2138g of H20
so i did (.6268*12)/44 to find the grams of carbon and got .171g. then .171/12 for .01425moles of Carbon
then to find grams of H i did (.2138*2)/18 and got .0238g. .0238/1=.0238 moles of hydrogen
to find grams of Oxygen i added .171 and .0238 and got .1948. then did .3469-.1948=.152g of Oxygen
then .152/16g=.0095moles of oxygen

then after that i did divided carbon and hydrogen by the moles of oxygen and my empirical formula is C3H5O2.

the third part is the molecular formula.
the mm of C3H5O2 is 73 so it must be the X-Factor (or the way my teacher says it) would be one and the molecular formula is C3H5O2.

I missed about a week of my chemistry class and had to figure this out on my own. just checking my answers. if i'm doing anything wrong please tell me

[Borek]

So there is .1235g of the unknown being neutralized with 15.55 mL of 0.1087M NaOH.
so i did 15.55(.1087)=.00169
then i did .1235/.00169 and got 73.07 for the molar mass of the diprotic acid.

I can't see "diprotic" part, what you did is OK for monoprotic acid.

[gogglespinoy]

so then how would i do it if it was diprotic? thats what confused me

[Borek]

What is stoichiometric ratio of the diprotic acid and NaOH? Do they react 1:1?

[gogglespinoy]

gahhh its 2!!! thanks.