Question

Carbon, hydrogen, and oxygen are not the only elements that can be characterized by combustion analysis. If a compound also contains sulfur or nitrogen, then it will form CO​2​​ (from the carbon), H2​​O (from the hydrogen), N2​​ (from any nitrogen), and SO​2​​ (from any sulfur). The amount of oxygen in the original sample is determined from subtracting the masses of the other elements from the total, as in the combustion analysis described earlier. If a 0.500 g sample of a compound yields 0.814 g CO​2​​, 0.204 g H​2​​O, 0.0288 g N2​​, and 0.132 g SO​2​​ when burned, then what is its empirical formula? Give your answer in the form C#H#N#O#S#, where the number is the subscript.

Answer 1

From grams calculate moles

moles C = 0.814 g/44 g/mol = 0.0185 mol

mass C = 0.222 g

moles H = 2 x 0.204 g/18 = 0.0227 mol

mass H = 0.023 g

moles N = 2 x 0.0288 g/28 g/mol = 0.002 mol

mass N = 0.029 g

moles S = 0.132 g/64.066 g/mol = 0.00206 mol

mass S = 0.066 g

mass O in sample = 0.5 - (0.222 + 0.023 + 0.029 + 0.066) = 0.16 g

moles O = 0.16 g/16 g/mol = 0.01 mol

divide each moles by 0.002

C = 0.0185/0.002 = 9

H = 0.0227/0.002 = 11

N = 0.002/0.002 = 1

S = 0.00206/0.002 = 1

empirical formula = C9H11NSO5

O = 0.01/0.002 = 5