A compound contains only C, H, and N. Combustion of 42.0 mg of the compound produces?

Question

40.2 mg CO2 and 49.3 mg H2O. What is the empirical formula of the compound?

Answer 1

Moles CO2 = 0.0402 g/ 44.0098 g/mol =0.000913

mass C = 0.000913 x 12.011 = 0.0110 g

moles water = 0.0493 g / 18.02 g/mol = 0.00274

moles H = 2 x 0.00274 =0.00548

mass H = 0.00548 x 1.008 =0.00552 g

mass N = 0.0420 - ( 0.0110 + 0.00552)=0.0255 g

moles N = 0.0255 / 14.0067 =0.00182

C 0.000913  H 0.00548  N 0.00182

we divide by the smallest number to get the empirical formula

CH6N2