How do do I work this problem?

Question

Valproic acid, a medication used to treat epilepsy and bipolar mania, contains only C, H, and O. Combustion of 1.000 g of valproic acid produces 2.441 g of CO₂ and 1.000 g of H₂O. Determine the molar mass of the compound if it is between 80.0 and 150.0 g/mol. Provide an answer to four significant figures.

Answer 1

The idea here is to find the MOLES of C, H and O. To find moles of C and H, we use the info provided, as follow:

For C: 2.441 g CO2 x 1 mol CO2/44.01g = 0.05546 mols CO2 x 1 mol C/mol CO2 = 0.05546 moles C

For H: 1.000 g H2O x 1 mol H20/18.02 g = 0.05549 mol H2O x 2 mol H/mol H2O = 0.11099 mols H

To find grams and moles of O, we proceed as follows:

0.05546 mol C x 12 g/mol = 0.6655 g C

0.11099 mol H x 1 g/ mol = 0.11099 g H

g of O = 1.000 g - 0.6655g - 0.11099g = 0.2235 g O x 1 mol O/16 g = 0.01397 mol O

SUMMARY

0.05546 moles C

0.11099 moles H

0.01397 moles O

Divide all by 0.01397 to get…

moles C = 4

moles H = 8

moles O = 1

EMPIRICAL FORMUL = C4H8O

molar mass = 72.11 g/mole

Since the molar mass of the valproic acid has to be between 80 and 150, we must make the formula twice this

C8H16O2 molar mass = 144.2 g/mole