Empirical Formula of Combustion Reaction

Question

A sample of a substance (containing only C, H, and N) is burned in oxygen.
8.395 g of CO2, 1.504 g of H2O and 2.146 g of NO are the sole products of combustion.
What is the empirical formula of the compound?

 

Answer 1

From moles of CO2, H2O and NO, you can determine the moles of C, H and N in the original staring material.  From that, you can deduce the empirical formula.

 

Moles of C:  8.395 g CO2 x 1 mole CO2/44.01 g x 1 mole C/mole CO2 = 0.19075 moles C

Moles of H:  1.504 moles H2O x 1 mole H2O/18.02 g x 2 moles H/mole H2O = 0.1669 moles H

Moles of N:  2.146 g NO x 1 mole NO/30.01 g x 1 mole N/mole NO = 0.0715 moles N

Divide all by 0.0715 in an attempt to get all to be whole numbers.

C:  0.1908/0.0715 = 2.668

H:  0.1669/0.0715 = 2.334

N: 0.0715/0.0715 = 1.000

Now we can multiply all by a factor of 3 to get all to be whole numbers.

C:  2.668 x 3 = 8.00

H:  2.334 x 3 = 7.00

N:  1.00 x 3 = 3.00

Empirical formula = C8H7N3