Ethylene glycol used in automobile antifreeze and in the production of polyester. the name glycol stems from the sweet taste of this poisonous compound. combustion of 6.38 g of this compound gives 9.06 g of co2 and 5.58 g of h2o. the compound only contains c, h, and o. if the molecular mass of this compound is 124 amu, what is the empirical and molecular formula?

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

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

So, in 9.06 g of carbon dioxide, of carbon will be contained.

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

So, in 5.58 g of water, of hydrogen will be contained.

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 0.206 moles.

For Carbon =

For Hydrogen  =

For Oxygen  =

The ratio of C : H : O = 1 : 3 : 1

Hence, 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 = 124 amu = 124 g/mol

Mass of empirical formula = 31 g/mol

Putting values in above equation, we get:

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

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

the answer is c. nacl