Combustion of 47.6mg cumene (with only Hydrogen and Carbon) produces "some" CO2 and 42.8mg H2O: Determine...?

Question

I'm supposed to find the empirical and molecular formula if the molar mass of cumene is between 115 and 125 g/mol. Seriously, what the h**l is "some" CO2? I calculated it like the only by products were the 42.8mg of H2O and the remaining weight was CO2...and I got H2C97 as the empirical forumula (haha, WAY wrong!).

What is going on here? That's wayyyy too much water it seems like?

Answer 1

Moles H2O = 0.0428 g / 18.02 g/mol = 0.00237

Moles H = 2 x 0.00237 = 0.00474

Mass H = 1.008 g/mol x 0.00474 mol = 0.00478 g

Mass C = 0.0476 g -0.00478 =  0.0428 g

Moles C = 0.0428 / 12.011 g/mol = 0.00356

the ratio between C and H is 0.00356 : 0.00474

we divide by the smallest number

0.00356 / 0.00474=  0.75 => C

0.00474 / 0.00474 = 1 => H

to get whole numbers we multiply by 12

0.75 x 12 = 9 => C

1 x 12 = 12 => H

the formula is C9H12