Question

Calculate the theoretical yield of water, if 5.00g of an unknown organic substance was subjected to combustion. In a separate prior experiment the same unknown organic substance was subjected to elemental analysis and the percent composition was determined to be 39.06% carbon, 8.77% hydrogen, and 52.17% oxygen. The molar mass of the unknown is 93 g/mol.

Answer 1

We know that the compound is 39.06% carbon, which means:

93 g/mol * 0.3906 = 36.3258 grams of Carbon / mol

We divide it by 12 grams of carbon per mol, we have 3 atoms of carbon in the molecule.

We do the exact same procedure for hydrogen and oxygen, and we get 3 atoms of oxygen and 8 atoms of hydrogen, getting a molecule of glycerol, C₃H₈O₃.

Combustion of glycerol has the following stoichiometrical relation:

C₃H₈O₃ + 7/2 O₂ -> 3CO₂ + 4H₂O

If we are given 5 g of glycerol, then we have:

5 g of C₃H₈O₃ * (1 mol / 93 g) = 0.05376 moles

We convert them to moles of water:

0.05376 moles of C₃H₈O₃ * (4 moles of H₂O / 1 mol of C₃H₈O₃) = 0.215 moles of H₂O

Converting them to grams:

0.215 grams of H₂O * (18 g / mol) = 3.87 grams of H₂O <- Theoretical yield