Question

An organic compound contains C, H, N, and O. Combustion of 0.3069 g of the compound produces 0.8301 g CO2 and 0.2975 g H2O. A sample of 0.5322 g of the compound was analyzed for nitrogen. At STP, 30.54 mL of dry N2 (g) was obtained. In a third experiment, the density of the compound as a gas was found to be 3.68 g/L at 143 deg C and 244 torr. Calculate the empirical formula and the molecular formula of the compound.

Answer 1

mass of CO2 :0.8301g mass of water 0.2795g mass of compound =0.3069

moles of CO2 = 0.019 = moles of carbon; mass of carbon = 0.23gm

moles of water =0.016 ; moles of H = 0.031moles mass of H =0.031g

.5322g compound gave 30.54 ml of dry N2 at STP

1 mole of nitrogen gas = 22.4L so we have .03054 / 22.4 moles of N2 =0.001363 moles of N2 in .5322g

moles of N =0.001363x .3069 / .5322g x 2 in .3069g or 0.00156moles of N =0.0219g

mass of C+H +N=0.280g ; mass of O =0.027g moles of O =0.00170

molar ratio of C : H :N :O = 0.019:0.031:0.001560:00170

or after dividing by the smallest molar ratio of C : H :N ; O = 12.08:19.88:1.00 :1

empirical formula is :C12H20NO;empirical formula mass = 12x12+20+16+14 or 194g

compound has 3.68 g/L at 143 degrees celsius and 244 torr

n= PV /RT=244x1 /( 62.363 x (273+143) =0.00940 moles which weigh 3.68g
1 mole weighs 3.68 / .0094 =391.5g which is 2 x the empirical mass so the molecular formula is

C₂₄H₄₀O₂N₂This is the empherical formula of the compound.