Question

Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produced 39.61 g CO2 and 9.01 g H2O. The molar mass of equilin is 268.34 g/mol. Find the molecular formula of the unknown compound. Express your answer as a chemical formula.

Answer 1

1) Determine the grams of each element present in the original compound. Carbon is always in CO2 in the ratio (12 g / 44 g), hydrogen is always in H2O in the ratio (2 g / 18 g), etc.

For C: 39.61 g x (12 g / 44 g) = 10.8 g
For H: 9.01 g x (2 g / 18 g) = 1 g
For O: 13.42 g - 10.8 g - 1 g = 1.62 g

(To find O, subtract each element's mass from the total mass)


2) Convert grams of each elment to the number of moles. You do this by dividing the grams by the atomic weight of the element.

For C: 10.8 g / (12 g/mol) = 0.9 mol
For H: 1 g / (1 g/mol) = 1 mol
For O: 1.62 g / (16g/mol) = 0.1 mol


3) Divide each molar amount by the lowest value, seeking to modify the molar amounts into small, whole numbers to get the empirical formula.

For C: 0.9 mol / 0.1 mol = 9
For H: 1 mol / 0.1 mol = 10
For O: 0.1 mol / 0.1 mol = 1

So, the empirical formula is C9H10O

4) Find the molecular formula from the empirical formula and the given molar mass of equilin.

The molar mass of this compound is:
(9 x 12g/mol) + (10 x 1g/mol) + (1 x 16g/mol) =134 g/mol

The molar mass of equilin is 268.34g/mol, which is approximately double 134g/mol, so the empirical formula must be doubled.

The molecular formula for equilin is thus C18H20O2