0.2500 g of a known compound was subjected to combustion. 0.1500 g of water and 0.3664 g of carbon dioxide were produced. Determine the empirical formula of this compound?. Note the known compound contains C, H and O. (Molar mass: N=14 g/mol; C=12 g/mol; O = 16 g/mol)

Answer 1

To determine the empirical formula, we need to find the ratio of the number of atoms of each element in the compound.

First, we need to find the number of moles of water and carbon dioxide produced:

moles of water = (0.1500 g) / (18.015 g/mol) = 0.008324 mol
moles of carbon dioxide = (0.3664 g) / (44.01 g/mol) = 0.008328 mol

The number of carbon atoms in the known compound can be found from the moles of carbon dioxide produced (since each molecule of CO2 contains one carbon atom):

number of carbon atoms = 0.008328 mol

The number of hydrogen atoms in the known compound can be found from the moles of water produced (since each molecule of water contains two hydrogen atoms):

number of hydrogen atoms = 2 x 0.008324 mol = 0.016648 mol

The number of oxygen atoms in the known compound can be found by subtracting the number of carbon and hydrogen atoms from the total number of atoms produced:

number of oxygen atoms = (0.008328 mol x 2) - 0.016648 mol = 0.000008 mol

Next, we need to convert the number of atoms to whole numbers by dividing by the smallest number of atoms (in this case, the number of oxygen atoms):

ratio of carbon atoms = 0.008328 mol / 0.000008 mol = 1035
ratio of hydrogen atoms = 0.016648 mol / 0.000008 mol = 2081

Finally, we can write the empirical formula by dividing the ratios by their greatest common factor:

empirical formula = C1H2O