Cumene is a compound containing only C and H that is used in the production of Phenol and acetone in industry. combustion of 47.6 mg cumene produce some CO2 and 42.80 mg water. the molar mass of cumene is between 115-125 g/mol. Determine the empirical and molecular formula of cumene?

Answer 1

42.80 mg H2O. Convert to mg H (not H2).
47.6 mg cumene - mg H = mg C
moles H = mg H/atomic mass H.
moles C = mg C/atomic mass C.

Now fine the ratio. The easy way to do that is divide the smaller moles by itself thereby assuring you of getting 1.00 for the smallest value. Then divide the other number by the same small number. I did this and obtained 1:1.33 and those aren't whole numbers. So you multiply by 2, 3, 4, 5, etc until you come up with values that can be rounded to whole numbers. (Hint: Thae multiplier is 3). That will give you the empirical formula.
To fine the molecular formula,
115/empirical formula mass = n. Round to a whole number and the molecular formula will be
(empirical formula)_n
According to google the molecular formula is C9H12.

Answer 2

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