Quantitative analysis

H2 O has a MW of 18 & CO2 has a MW of 44

So 3.384 g of CO2 is 0.0761 gmol of CO2 & 0.693 g H2O is 0.0385 gmol

Now there is 1 atoms of Carbon in CO2 and 2 atoms of H in H2O

so there is 0.0761 mol of C and 0.077 mol of H

the ratio of C to H in the initial sample is equal (within sample error) so the mol formulae is (CH)n. This empirical mol formulae has a mol weight of 78

CH=12+1=13 & n*13=78 so n=78/13=6

So the compound is C6H6

This is benzene.

Rap