Question

a 0.490 g sample of a compound is heated through the successive evolution of the following gases, all at 1.00atm pressure: 280 ml of h2o vapor at 182ºC, 112ml of ammonia vapor at 273ºC, 0.0225g ofwater at 400ºC and 0.200g of SO3 at 700ºC, At the end of the heating, 0.090 g of FeO remains. Deduce the empirical formula for the compound

Answer 1

Molar volume of all gases at 273K = 22.414L/mol

At 182C, volume of H2O vapor = 0.28L
Molar volume at 273K = 22.414L/mol
At 182C=182+273=455K molar volume = 22.414/273*455 = 37.36 L/mol
moles of H2O vapor = 0.28/37.36 = 0.0075 mol
No of H atoms in H2O vapor = 0.0075*2 = 0.015
No of O atoms in H2O vapor = 0.0075mol

At 273C=273+273=546K, volume of ammonia vapor = 0.112L
Molar volume at 546K = 22.414*546/273=44.828L
moles of NH3 vapor = 0.112/44.828 = 0.0025mol
No of H atoms in NH3 vapor = 0.0025*3 = 0.0075
No of N atoms in NH3 vapor = 0.0025mol

0.0225g ofwater at 400ºC
Molar mass of water = 18gm/mol
moles of H2O = 0.0225/18 = 0.00125mol
No of H atoms in H2O vapor = 0.00125*2 = 0.0025
No of O atoms in H2O vapor = 0.00125mol

0.200g of SO3 at 700ºC
Molar mass of SO3 = 80gm/mol
moles of SO3 = 0.2/80 = 0.0025mol
No of O atoms in SO3 vapor = 0.0025*3 = 0.0075
No of S atoms in SO3 vapor = 0.0025mol

0.090g of FeO remains.
Molar mass of FeO = 56+32=88
moles of FeO = 0.09/88 = 0.001mol
No of Fe atoms in FeO = 0.001mol
No of O atoms in FeO = 0.001mol

Total no of moles of atoms:
Fe = 0.001
O = 0.001+0.0075+0.00125+0.0075=0.01725
H = 0.0025+0.0075+0.015=0.025
N = 0.0025
S = 0.0025

Dividing by smallest no of atoms 0.001
Fe = 0.001/.001 = 1
O = 0.01725/.001 =17.25
H = 0.025/.001=25
N = 0.0025/.001=2.5
S = 0.0025/.001=2.5

Making Integers by multiplying with 4
Fe = 1*4 = 4
O = 17.25*4 = 69
H = 25*4 =100
N = 2.5*4 =10
S = 2.5*4=10

Empirical formula: Fe(4)O(69)H(100)N(10)S(10)