One of the components of natural crude oil and coal deposits is benzo[a]pyrene, a compound with a molecular mass of about 252.3 amu, containing only carbon and hydrogen. A 3.320 mg sample of benzo[a]pyrene burns to give 11.58 mg of CO2. Determine its empirical and molecular formulas. (Omit states-of-matter from your answer.)

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

Explanation:

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

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

We are given:

Mass of     (Conversion factor:  1 g = 1000 mg)

Mass of sample = 3.320 mg = 0.00332 g  

For calculating the mass of carbon:

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

So, in 0.01158 g of carbon dioxide, of carbon will be contained.

Mass of hydrogen in the compound = (0.00332) - (0.00316) = 0.00016 g

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

Moles of Carbon =

Moles of Hydrogen =

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00016 moles.

For Carbon =

For Hydrogen =

The ratio of C : H = 2 : 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 = 252.3 g/mol

Mass of empirical formula = 25 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 c. nacl