Question

An unknown compound contains only C, H, and O. Combustion of 5.90 g of this compound produced 14.4 g of CO2 and 5.90 g of H2O. What is the empirical formula of the unknown compound?

Answer 1

moles CO2 = 19.3 g/ 44.009 g/mol=0.438 
moles C in the unknown = 0.438 
mass C in the unknown = 0.438 mol x 12.011 g/mol=5.27 g 

moles H2O = 5.27 g/ 18.02 g/mol=0.292 
moles H in the unknown = 0.292 x 2 = 0.584 
mass H in the unknown = 0.584 mol x 1.008 g/mol=0.589 g 

mass O = 8.20 - ( 0.589 + 5.27)=2.34 g 
moles O = 2.34 g / 15.999 g/mol=0.146 

the ratio between C , H and O is 0.438 : 0.584 : 0.146 

divide by the smallest 
0.438/ 0.146 = 3 => C 
0.584/ 0.146 =4 => H 
0.146/0.146 = 1 => O 

C3H4O