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, 0.4808 g sample containing only carbon, hydrogen, and nitrogen is subjected to combustion analysis, resulting in the formation of 0.6859 g CO2, 0.6973 g H2O, and 0.4646 g NO. What is the empirical formula of the sample

Answer 1

C₂H₅N₂

Step-by-step explanation:

In the combustion analysis, all NO comes from the nitrogen of the sample, that means:

Moles NO = Moles N in the sample

In the same way:

Moles CO2 = Moles C in the sample

Moles H2O = 1/2 moles H in the sample

To solve this problem we must convert the mass of each gas to moles in order to find the moles of each atom. The empirical formula is the simplest whole number ratio of atoms presents in the molecule:

Moles NO = N -Molar mass: 30g/mol

0.4646g * (1mol / 30g) = 0.01549 moles N

Moles CO2 = C -Molar mass: 44.01g/mol-

0.6859g * (1mol / 44.01g) = 0.01559 moles C

Moles H2O = 1/2moles H -Molar mass: 18.02g/mol-

0.6973g * (1mol / 18.02g) = 0.03870 moles H

The ratio of atoms is -Dividing in the low number of moles = moles N-:

C = 0.01559 / 0.01549 = 1

N = 0.01549 / 0.01549 = 1

H = 0.03870 / 0.01549 = 2.5

Twice this ratio of atoms -Because empirical formula must be given only with whole numbers:

C = 2

N = 2

H = 5

The empirical formula is:

C₂H₅N₂