Cumene is a compound containing only carbon and hydrogen that is used in the production of acetone and phenol in the chemical industry Combustion of 47.6 mg cumene produces some CO2 and 42.8 mg water?

Answer 1

The molar mass of H2O = 2 + 16 = 18 g/mol.

The ratio of H in H2O is therefore 2/18 = 1/9.

Thus, amount of H in 42.8 mg H2O = 42.8 * 1/9 = 4.76 mg.

This must then be the amount of H in the 47.6 mg cumene.

Subtracting, 47.6 - 4.76 = 42.84 mg,

is therefore the amount of C in the cumene.

Summarising: 47.6 mg cumene = 42.84 mg C + 4.76 mg H

Converting mass of C and H to millimoles, we have :

C = 42.84 /12 = 3.57 mmol and H = 4.76/1 = 4.76 mmol.

The molar ratio C:H is therefore 3.57:4.76.

Dividing each by the smallest (that is, 3.57) gives :

C:H = 1:1.33

Now, multiplying each by 3 brings us to the integers :

C:H = 3:4

So the empirical formula for cumene is C3H4.

This has a molar mass of 3*12 + 4*1 = 40 g/mol.

Given that the molar mass is between 115 g/mol

and 125 g/mol, if we multiply 40 by 3, this gives

120 g/mol, smack bang in the middle.

So, molecular formula of cumene is C9H12.