Unknown compound combustion

Question

An unknown compound contains only C, H, and O. Combustion of 6.10 g of this compound produced 14.4 g of CO₂ and 3.92 g of H₂O. What is the empirical formula of the unknown compound?

Answer 1

14.4 g of CO₂ x 1 mole CO2/44.01 g CO2 = .327 mole CO2 x 1 mole C/1 mole CO2 = .327 mole C

 

3.92 g of H₂O x 1 mole H2O/18.02 g H2O = .218 mole H2O x 2 mole H/1 mole H2O = .435 mole H

 

.327 mole C x 12.01 g C/1 mole C = 3.93 g C

 

.435 mole H x 1.01 g H/1 mole H = .439 g H

 

Total= mass C + mass H + mass O

 

6.10 g = 3.93 g C + .439 g H + mass O

 

1.73 g = mass of O

 

1.73 g O x 1 mole O/16.00 g O = .108 mole O

 

.327 mole C/.108 mole = 3.03 mole C

 

.435 mole H/.108 = 4.03 mole H

 

.108 mole O/.108 mole = 1 mole O

 

C3H4O

 

You better check my math...