Answer to Question #183503 in General Chemistry for joshua

Question

An unknown compound contains only C

C , H

H , and O

O . Combustion of 6.60 g

6.60 g of this compound produced 16.1 g

16.1 g 

CO2 and 6.60 g

6.60 g H

2

O

H2O . What is the empirical formula of the unknown compound?

Answer 1

$\frac{16.1g(CO₂ )×1 mole( CO_2)}{44.01 g (CO_2)} = 0.366mole (CO_2$

$\frac{0.366mole (CO_2) ×1 mole( C)}{1 mole (CO_2 )}= 0.366mole (C)$

$\frac{6.60 g (H₂O) × 1 mole( H_2O)}{18.02 g (H_2O )}=0.366mole (H_2O)$

$\frac{0.366 mole (H_2O) ×2 mole( H)}{1 mole (H2O)} =0 .732mole( H)$

 

0.366 mole (C ) x 12.01 g (C)/1 mole C = 4.40g (C)

 

0.732mole (H) x 1.01 g (H)/1 mole (H) = 0.739 g (H)

 

Total= mass C + mass H + mass O

 

6.10 g = 4.40 g + 0.739 g + mass( O)

 

1.461 g = mass of O

 

1.461 g O x 1 mole O/16.00 g O = 0.091 mole O

 

0.366 mole C/0.09 mole = 4.02 mole C

 

0.732 mole H/0.091 = 8.04 mole H

 

0.091mole O/0.91 mole = 1 mole O

 

$C_4H_8O$