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?

Question

Jordyn King

Chemistry

16 March, 07:44

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?

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Answers (1)

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Zoe Mcconnell · Answer 1

C₈H₇N

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