A 17.11 gram sample of an organic compound containing only C, H, and O is analyzed by combustion analysis and 21.71 g CO2 and 5.926 g H2O are produced. In a separate experiment, the molar mass is found to be 104.1 g/mol. Determine the empirical formula and the molecular formula of the organic compound. Enter the elements in the order C, H, O empirical formula

 The empirical formula and the molecular formula of the organic compound is and respectively.

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of = 21.71 g

Mass of = 5.926 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 21.71 g of carbon dioxide, = of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 5.926 g of water, = of hydrogen will be contained.

Mass of oxygen in the compound = (17.11) - (5.921+0.658) = 10.53  g

Mass of C = 5.921 g

Mass of H = 0.658 g

Mass of O = 10.53 g

Step 1 : convert given masses into moles.

Moles of C =

Moles of H=

Mass of O=

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =

For H =

For O=

The ratio of C : H: O =  1: 1: 1

Hence the empirical formula is .

empirical mass of CHO = 12(1) + 1(1) + 1 (16) = 29

Molecular mass = 104.1 g/mol

Thus molecular formula =