Question

A compound containg only C,H, and N only. A chemist azalyzes it by doing the following experiments. Complete combustion of 84.0 g of the compound converted all its carbon in to 84.4 g CO2(g). A 156.5 g sample of the compound was analyzed for nitrogen, giving 85.44 mL N2(g) at 740 torr and 25 C. Calculate the emperical formula of the compound

Answer 1

At first we calculate the % carbon in the sample = %C = [(84.4*12/44)/84]*100 = 27.402%

now using ideal gas law we calculate no. of moles of N2
PV = nRT
n = PV/RT need T in K, V in L, and P in atm R= 0.0821 L atm mol-1 K-1

thus P = 740/760 atm , T= 273.15 + 25 = 298.15 C and V = 85.44/1000 L

thus n = (740/760)*(0.08544)/(0.0821*298.15) = 0.00339 moles

N2 molecular weight = 28
thus N2 present= 28 * 0.00339 = 0.09516 gm
% N2 = (0.09516/156.5) *100 = 0.060 %

as these contain only C H and N thus % H = 100 - 0.06 - 27.402 = 72.538 %

Atomic ratios = C = 27.402/12 = 2.2835
N = 0.06 / 14 = 0.004285
H = 72.538 / 1 = 72.538

Thus Dividing by 0.004285=

C = 533
N= 1
H = 16928

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