Question

A 8.048 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 12.00 grams of CO2 and 3.684 grams of H2O are produced.

In a separate experiment, the molar mass is found to be 118.1 g/mol. Determine the empirical formula and the molecular formula of the organic compound.

Answer 1

Atomic weights:
H: 1
C: 12
O: 16
I'm calling the mystery compound CHO?

Combustion analysis adds O2, which is where the extra mass on the consumption side comes from.
8.048g CHO? + 7.636g O2 -> 12.00g CO2 + 3.684g H2O
(12.00 + 3.684 - 8.048 = 7.636)

1mol H2O is 18g: 2g H + 16g O
3.684g H2O is produced
3.684 * (2/18) = 0.4122 g H
3.684 * (16/18) = 3.272g O

1mol CO2 is 44g: 12g C + 32g O
12.00g CO2 is produced
12.00 * (12/44) = 3.275g C
12.00 * (32/44) = 8.725g O

1mol O2 is 32g: 32g O
7.636g O2 is consumed (all from O)

The O provided by CHO? is 3.272 + 8.725 - 7.636 = 4.361g. All the C and H produced only came from the CHO? so that's easy to determine.

8.048g CHO? contains:
0.4122g H
3.275g C
4.361g O
But what portion of the 118.1g/mol do each of those represent?
(0.4122/8.048) * 118.1 = 6.05g H/mol CHO? = 6 H atoms (1g/atom)
(3.275/8.048) * 118.1 = 48g C = 4 C atoms (12g/atom)
(4.361/8.048) * 118.1 = 64g O = 4 O atoms (16g/atom)

So the molecular formula is C4H6O4. The empirical formula is C2H3O2. According to http://www.nauticus.org/chemistry/chememperical.html.