Question

Extra Credit: Calculate the empirical formulas for the compounds given the following information: 49.32% C, 43.84% O, and 6.85% H by mass a 0.4647 g sample of a compound that only contains C, H, and O is burned to give 0.8635 g CO_z and 0.1767 g H_2 O.

Answer 1

(A): Let's consider 100 g of the compound.

Mass of carbon = 49.32 g

molecular mass of carbon = 12.01 g/mol

Hence moles of carbon = mass / molecular mass = 49.32 g / 12.01 g/mol = 4.11 mol carbon

Mass of oxygen = 43.84 g

molecular mass of oxygen = 16.0 g/mol

Hence moles of oxygen = mass / molecular mass = 43.84 g / 16.0 g/mol = 2.74 mol oxygen

Mass of oxygen = 43.84 g

molecular mass of oxygen = 16.0 g/mol

Hence moles of oxygen = mass / molecular mass = 43.84 g / 16.0 g/mol = 2.74 mol oxygen

Mass of hydrogen = 6.85 g

molecular mass of hydrogen = 1.0 g/mol

Hence moles of hydrogen = mass / molecular mass = 6.85 g / 1.0 g/mol = 6.85 mol hydrogen

Ratio of moles of C, O and H are

moles of C : moles of O : moles of H = 4.11 mol C : 2.74 mol O : 6.85 mol H

Dividing the above ratio by 2.74

moles of C : moles of O : moles of H = 1.5 mol C : 1 mol O : 2.5 mol H = 3 mol C : 2 mol O : 5 mol H

Hence empirical formulae is C₃O₂H₅ (answer)

(B): mass of CO2 formed = 0.8635 g

molecular mass of CO2 =44.01 g/mol

Hence moles of CO2 formed = mass/mlecular mass = 0.8635 g / 44.01 g/mol = 0.01962 mol

Hence moles of C-atom in the sample = 0.01962 mol C

Hence mass of C-atoms in the sample = 0.01962 mol x 12.01 g/mol = 0.2356 g

mass of H2O formed = 0.1767 g

molecular mass of H2O = 18.0 g/mol

Hence moles of H2O formed = mass/mlecular mass = 0.1767 g / 18.0 g/mol = 0.009817 mol

Hence moles of H-atom in the sample = 2 x 0.009817 mol = 0.01963 mol H

mass of H-atoms in the sample = 0.01963 mol x 1.0 g/mol = 0.01963 g

Hence mass of O-atoms in the sample = 0.4647g - 0.2356g - 0.01963g = 0.2095 g

moles of oxygen in the sample = mass / molecular mass = 0.2095 g / 16.0 g/mol = 0.01309 mol

Ratio of moles of C, O and H are

moles of C : moles of O : moles of H = 0.01962 mol C : 0.01309 mol O : 0.01963 mol H

Dividing the above ratio by 0.01309

moles of C : moles of O : moles of H = 1.5 mol C : 1 mol O : 1.5 mol H = 3 mol C : 2 mol O : 3 mol H

Hence empirical formulae is C₃O₂H₃ (answer)