A 4.917 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 11.18 grams of CO2 and 4.577 grams of H2O are produced.

Question

In a separate experiment, the molar mass is found to be 116.2 g/mol. 

Determine the empirical formula and the molecular formula of the organic compound.

Answer 1

Step 1: Calculate the number of grams of each element in the products by multiplying the grams of the product species by the ratio of the atomic mass of an element to the molecular mass of the whole product molecules.

e.g. atomic mass C / mol. mass CO2 x 11.18 g CO2 = 12.011 / 44.009 x 11.18 = 3.051 g C. Repeat calculation for H and O.

Results: C 3.015g H 0.5086 g O 4.917g - (3.015g + 0.5086g) = 1.393 g

Step 2: Now divide each mass by the element's atomic mass.

C 3.015 / 12.011 = 0.2510 H 0.5086/1.008 = 0.5046 O 1.393 / 15.999 = 0.0871

Step 3: Divide the results above by the smallest ratio (0.0.0871) and round to whole numbers.

C3 H6 O1 empirical formula mol. mass = 58.08

Step 4: Divide 116.2 by 58.08 = 2 and multiply the empirical formula subscripts by 2.

Molecular formula = C6H12O2