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"Combustion of 0.1023 g of the compound in excess oxygen yielded 0.2766 g of CO2 and 0.0991 g of H2O" tells you that .1023 g of the material contains [12/44]*.2766 = .0754 g Carbon, which means the compound has a mass fraction of .0754/.1023 = .7374 Carbon,
and [2/18]*.0991 g = .011011 g Hydrogen and a hydrogen mass fraction of .011011/.1023 = .1076
"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."
tells you the Nitrogen content, via PV = nRT and the Nitrogen mass fraction.
The Oxygen mass fraction will be whatever is left over [Omassfrac = 1-Cmassfrac -Hmassfrac -Nmassfrac]
From the mass fractions you can then get the atomic fractions [dividing by atomic weights] which will give you the empirical formula.
"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"
tells you the molecular weight of the compound [assuming it acts as an ideal gas], so now you can find the actual molecular formula
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