Question

The most common experimental technique to perform elemental analysis is combustion analysis, where a sample is burned in a large excess of oxygen and the combustion products are trapped in a variety of ways. A 99.99% pure, 0.4986 g sample containing only carbon, hydrogen, and nitrogen is subjected to combustion analysis, resulting in the formation of1.423 g CO2, 0.3377 g H2O, and 0.1606 g NO. What is the empirical formula of the sample?

Answer 1

C₈H₇N

Step-by-step explanation:

This problem is solved by calculating the moles of atoms of C, H, N present in the sample and then find their proportions in whole numbers.

MW CO2 = 44 g/mol ⇒ mol CO2 = 4.23 g/ 44 g/mol = 0.096

MW H2O = 18 g/mol ⇒ mol H2O = 0.3377 g/ 18 g/mol =0.019

MW NO = 30 g/mol ⇒ mol NO = 0.1606 g/ 30 g/mol = 0.054

In terms of mol of atoms of C, H , and N :

0.096: 0.038 : 0.054 (H is 0.38 because there are 2 H per molecule H₂O)

Dividing by the smallest 0.054

7.8 :7 : 1 rounding 8:7:1

empirical formula = C₈H₇N