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 3.649 mg sample of benzo[a]pyrene burns to give 12.73 mg of CO2. Determine its empirical and molecular formulas. (Omit states-of-matter from your answer.) Empirical formula

Empirical C5H3

Molecular

C20H12

Explanation:

Like all other hydrocarbons, benzopyrene would burn in oxygen or air to form carbon iv oxide and water only.

From the mass of carbon iv oxide produced, we can get the mass of carbon and hence the mass of hydrogen.

It can be seen from the question that 12.73mg of carbon iv oxide was produced. Let us convert this to grammes, we simply divide by 1000 = 0.01273g

Now we need to calculate the number of moles of carbon iv oxide produced. To do this, we simply divide the mass by the molar mass of carbon iv oxide. The molar mass of carbon iv oxide is 44g/mol

The number of moles is thus 0.01273/44 = 0.00028392moles

As we can see that there is only one atom of carbon in a molecule of carbon iv oxide. Hence, the number of moles of carbon iv oxide present is equal to 0.00028392moles

From here we can get the mass of carbon which is equal to the number of moles of carbon multiplied by the atomic mass of carbon. The atomic mass of carbon is 12 a. m. u

The mass of carbon in the benzopyrene is thus 0.00028392 * 12 = 0.003472g

The mass of hydrogen in the compound is 0.003649g - 0.003472g = 0.000177g

We can now deduce the empirical formula by dividing the masses by the atomic masses.

H = 0.000177/1 = 0.000177

C = 0.003572/12 = 0.000289333333

We now divide by the smallest which is that of hydrogen.

H = 0.000177/0.000177 = 1

C = 0.000289333333/0.000177 = apprx 1.63

Multiply both by 3 to yield C5H3

The molecular formula is as follows:

[ (12 * 5) + (3 * 1) ]n = 252.3

63n = 252.3

n = 4

The molecular formula is thus C20H12