A 0.600 g sample of a carboxylic acid is burned in oxygen, producing 1.07 g of CO2 and 0.440 g of H2O. Determine the empirical formula of the carboxylic acid.

Answer 1

In the combustion, all of the C and H that were in the carboxylic acid ends up in the CO₂ and H₂O respectively. So we first find the masses of the C & the H in these molecules.

(12 g C/44 g CO₂) 1.07 g CO₂ = 0.2918 g C

(2.02 g H/18.0 g H₂O) 0.440 g H₂O = 0.0489 g H

Subtracting these from the original mass of the acid we get the mass of O in the sample. 0.600 g CHO - (0.292 g C + 0.0489 g O) = 0.259 g O

Now, calculate the number of moles of each atom and get the smallest whole number ratio (the empirical formula).

(1 n C/12.0 g C) 0.292 g C = 0.0243 n C

(1 n H/1.01 g H) 0.0489 g H = 0.0484 n H

(1 n O/16.0 g O) 0.259 n O = 0.0162 n O

Dividing by the smallest value, 0.0162 gives the ratio 1.50 C: 2.99 (or 3) H: 1.00 O

We must multiply thru by 2 to get the lowest whole number ratio of 3:6:2 for a formula of C₃H₆O₂