A compound with a molar mass of 180 g is made of carbon, hydrogen and oxygen. When it is burned, the products are 44.0 g of #CO_2#, and 8 g of #H_2O#. Based on this, what are the empirical formula and the molecular formula of the compound?

To solve this problem, we look at the amount of products. 44 g of #CO_2# is one mole of this compound, and so, contains one mole of carbon atoms and two moles of oxygen atoms.

Eight grams of water is #8 -: 18 g/"mol" = 0.44# mol of water, and so, contains 0.88 mole of hydrogen atoms (plus 0.44 mol of oxygen atoms).

The combustion of the compound would have added an unknown amount of oxygen to the products, but did not add any carbon or hydrogen. So, the original compound must have consisted of 1 mol of carbon for each 0.88 mol of hydrogen.

The combined mass of 1.0 mol of C and 0.88 mol of H would be #12+0.88 = 12.88 g#

This means, the remaining 7.12 g of the compound (it was 20 g in mass, remember?) must have been oxygen, as this is the only other element found in the products.

#7.12 "g O" = 7.12 -: 16g/"mol" = 0.445# mol

So, we now know the compound contained 1.0 mol of C, 0.88 mol of H and 0.445 mol of O.

Dividing each mole quantity by the smallest will help us get a whole number ratio.

#1.0 -: 0.445 = 1.25# mol C

#088 -: 0.445 = 2.0# mol H

#0445 -: 0.445 = 1.0# mol O (of course)

Now, multiply each value by 4, and we have it!

In terms of a simple, whole number ratio, this is 9 mol C to each 8 mol H and 4 mol O.

The empirical formula is #C_9H_8O_4#

To find the molecular formula, we test this empirical to find the mass of one mole of it:

#(9xx12) + (8xx1)+(4xx16) = 180 g#

Since this matches the given molar mass, we can conclude that the empirical formula is also the molecular formula.
(If this did not turn out so nicely, we would still find that the molar mass to be a whole number multiple of our "test result", meaning the molecular formula is an identical multiple of the empirical formula.)