Combustion analysis of 0.213 g of a compound containing C, H, and O produces 0.468 g of CO2 and 0.255 g of H2O. Mass spectral analysis shows that the compound has a molar mass around 120 g mol-1. What is the compound's:

Question

Empirical Formula:(-------) Molecular Formula:(--------)

Answer 1

g C = 0.468 x (12.01/44.01) = 0.1277 g C.
g H = 0.255 x (2/18.015) = 0.0283 g H.
g O = 0.213 - 0.1277 - 0.2831 =0.057 g O.

Now convert those to moles.
0.1277 g C/atomic mass C = ??
0.0283 g H/atomic mass H = ??
0.057/atomic mass O = ??

Now find the ratio of these to each other in small whole numbers. The easy way to do that is to divide the smallest number by itself which assures you of getting 1.000 for that value, then divide the other numbers by that same small number. Finally, round to whole numbers and you should have the empirical formula.

To find the molecular formula, you want to know how many units of the empirical formula are linked together to make the molecule. To determine this, just divide 120 (the molar mass) by the empirical formula mass, and round to a whole number. For example, just suppose the empirical formula turns out to be C3H8O, the empirical formula mass is 60 (3*C+8*H+1*O), Then 120/60 = 2 so you know the molecular formula is (C3H8O)₂ or you can write it as C6H16O2.