Answer:
Empirical C5H3
Molecular
C20H12
Step-by-step 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