The combustion of 15.5 g of the primary amine released 5.2 liters of nitrogen. Determine the molecular formula of the amine.

1. Let’s compose the equation of primary amine combustion:

CnH2n + 1NH2 + (1.5n + 0.75) O2 → nCO2 ↑ + (n + 1.5) H2O + 0.5N2 ↑;

2.Calculate the chemical amount of released nitrogen:

n (N2) = V (N2): Vm = 5.6: 22.4 = 0.25 mol;

3. Determine the amount of amine combusted:

n (CnH2n + 1NH2) = n (N2): 0.5 = 0.25: 0.5 = 0.5 mol;

4. find the molar mass of the amine:

M (CnH2n + 1NH2) = m (CnH2n + 1NH2): n (CnH2n + 1NH2) = 15.5: 0.5 = 31 g / mol;

5. calculate and equate the relative molecular weight to the found molar to establish the amine formula:

Mr (CnH2n + 1NH2) = M (CnH2n + 1NH2);

14n + 1 + 14 + 2 = 31;

14n + 17 = 31;

14n = 14;

n = 1.

Answer: methylamine CH3NH2.