Combustion of 29.26 gg of a compound containing only carbon, hydrogen, and oxygen produces 33.86 gCO2gCO2 and 13.86 gH2OgH2O. Part A What is the empirical formula of the compound

C2H4O3

Explanation:

Empirical formula is defined as the simplest whole number ratio of atoms present in a molecule.

To solve this question we need to find the moles of carbon = Moles of CO2, the moles of hydrogen (Using moles of H2O) and the moles of oxygen (Finding the mass of the mass of each atom) as follows:

Moles Carbon -Molar mass CO2: 44.01g/mol-:

33.86g CO2 * (1mol/44.01g) = 0.769 moles CO2 = Moles C * (12g/mol) =

9.23g C

Moles Hydrogen -Molar mass H2O: 18.01g/mol-

13.86g H2O * (1mol/18.01g) = 0.770 moles H2O * (2mol H / 1mol H2O) = 1.54 moles H * (1g/mol) = 1.54g H

Moles Oxygen:

Mass: 29.26g - 9.23g C - 1.54g H = 18.49g O * (1mol/16g) = 1.156 moles O

Dividing each number of moles in the moles of C (Lowe number of moles):

C = 0.769 moles C / 0.769 moles C  = 1

H = 1.54 moles H / 0.769 moles C = 2

O = 1.156 moles O / 0.769 moles C = 1.5

As the number must be a whole number each ratio twice:

C = 2

H = 4

O = 3

Empirical formula is: