Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produced 39.01 g CO2 and
10.65 g H2O. The molar mass of the unknown compound is 272.38 g/mol.
Find the molecular formula of the unknown compound.

The empirical and molecular formula for the given organic compound is and

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

Mass of

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

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

So, in 39.01 g of carbon dioxide, of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 10.65 g of water, of hydrogen will be contained.

Mass of oxygen in the compound = (13.42) - (10.64 + 1.18) = 1.6 g

To formulate the empirical formula, we need to follow some steps:

Moles of Carbon =

Moles of Hydrogen =

Moles of Oxygen =

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.1 moles.

For Carbon =

For Hydrogen =

For Oxygen =

The ratio of C : H : O = 9 : 12 : 1

The empirical formula for the given compound is

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

We are given:

Mass of molecular formula = 272.38 g/mol

Mass of empirical formula = 136 g/mol

Putting values in above equation, we get:

Multiplying this valency by the subscript of every element of empirical formula, we get:

Hence, the empirical and molecular formula for the given organic compound is and

the answer is learned