[trinhn812]

Carminic acid,a naturally occuring red pigment extracted from the cochineal insect, contains only C,H,and O. It was commonly used as a dye in the first half of the nineteenth century. It is 53.66% C and 4.09% H by mass. A titration of the 0.3602 g sample pf the carminic acid requires 18.02 Ml of 0.0406 M NaOH to neutralize. Assuming there is only 1 acidic hydrogen per molecule, what is the molecular formula of carminic acid?

So I'm guessing that you find the molecular formula first.
53.66g C/12.01g =4.468 mol / 2.641 = 1.629 *6 =10
4.09gH/1.01g=4.050 mol /2.641= 1.534 *6= 9
42.25g O/16.00g = 2.641 mol /2.641= 1 *6 = 6
Emp Formula: C10H9O6

From here I don't know what to do.

[enahs]

How many mols of NaOH were required to neutralize the acid? You are told it is 1:1, so therefor however many mols that were in the NaOH is how many mols of acid you have. You have the grams of acid. MW is g/mol.

http://chemed.chem.purdue.edu/genchem/probsolv/stoichiometry/molecular2/mf2.0.html

[trinhn812]

Thank you. I think I got it now.