Determine the empirical and molecular formula?

Question

An organic compound contains C, H, N, and O. Combustion of 0.1023 g of the compound in excess oxygen yielded 0.2766 g of CO2 and 0.0991 g of H2O. A sample of 0.4831 g of of the compound was analyzed for nitrogen by the Dumas method. At STP, 27.6 mL of dry N2 was obtained. In a third experiment the density of the compound as a gas was found to be 4.02 g/L at 127°C and 256 torr. What are the empirical formula and the molecular formula of the compound?

I got C24H42O3N for the empirical formula. but that is totally wrong. could someone please help me
please and thank you

Answer 1

Molar mass of CO2: 44.01 g/mol
Molar mass of H2O: 18.015 g/mol
In 0.1023 g sample, 0.2766g CO2 <==> 0.006285 mole CO2 <==> 0.006285 mole C atom <==> 0.07548g carbon
and 0.0991g H2O <==> 0.005501 mole H2O <==> 0.011002 mole H atom <==> 0.01108g hydrogen
In 0.4831 g sample, moles of N2 = PV/RT = 1*0.0276/(0.08206*273) = 0.001232
In 0.4831 g sample, 0.002464 mole N atom <==> In 0.1023 g sample, 0.0005218 mole N atom <==> In 0.1023 g sample, 0.00731g nitrogen.
Thus, in 0.1023 g sample, there are 0.006285 mole C, 0.011002 mole H, 0.0005218 mole N, and (0.1023-0.07548-0.01108-0.00731)g oxygen = 0.00843g oxygen = 0.000527 O atom.
From the third test, 1 mole of this organic compound would occupy a volume of:
V = RT/P = 0.08206*400*(760/256) = 97.45L ==> 392g/mol
With the mole ratio of C:H:N:O approximately 6285 :11002 :522 :527, the empirical formula must be C12H21NO, and the molecular formula must be C24H42N2O2.