Answer:
The empirical formula is C3H10N2O2
Explanation:
Step 1: Data given
Mass of the sample in experiment 1 = 1.38 grams
Mass of CO2 produced = 1.72 grams
Mass of H2O produced = 1.18 grams
Molar mass CO2 = 44.01 g/mol
Molar mass H2O = 18.02 g/mol
Atomic mass C = 12.01 g/mol
Atomic mass O = 16.0 g/mol
Atomic mass H = 1.01 g/mol
MAss of the sample in experiment 2 = 22.34 grams
Mass of O = 6.75 grams
Step 2: Calculate moles CO2
Moles CO2 = mass CO2 / molar mass CO2
Moles CO2 = 1.72 grams / 44.01 g/mol
Moles CO2 = 0.0391 moles
Step 3: Calculate moles C
For 1 mol CO2 we have 1 mol C
For 0.0391 moles CO2 we have 0.0391 moles C
Step 4: Calculate mass C
Mass C = 0.0391 moles * 12.01 g/mol
Mass C = 0.470 grams
Step 5: Calculate moles H2O
Moles H2O = 1.18 grams / 18.02 g/mol
Moles H2O = 0.0655 moles
Step 6: Calculate moles H
For 1 mol H2O we have 2 moles H
For 0.0655 moles H we have 2*0.0655 = 0.131 moles H
Step 7: Calculate mass H
Mass H = 0.131 moles * 1.01 g/mol
Mass H = 0.132 grams
Step 8: Calculate the mass %
% C = (0.470 grams / 1.38 grams) * 100 %
% C = 34.06 %
%H = (0.132/1.38)*100 %
%H = 9.57 %
%O= (6.75 grams / 22.34 grams)*100 %
%O = 30.21 %
%N = 100 % - 34.06 % - 9.57 % - 30.21 %
%N = 26.16 %
Step 8: Calculate moles in compound
Let's assume 100 grams sample then we have:
Mass C = 34.06 grams
Mass H = 9.57 grams
Mass O = 30.21 grams
Mass N = 26.16 grams
Moles C= 34.06 grams / 12.01 g/mol
Moles C= 2.836 moles
Moles H = 9.57 grams / 1.01 g/mol
Moles H = 9.475 moles
Moles O = 30.21 grams / 16.0 g/mol
Moles O = 1.888 moles
Moles N = 26.16 grams / 14.0 g/mol
Moles N = 1.869 moles
Step 9: Calculate mol ratio
We have to divide by the smallest amount of moles
C: 2.836 moles / 1.869 moles = 1.5
H: 9.475 moles /1.869 moles = 5
O: 1.888 moles /1.869 moles = 1
N: 1.869 moles / 1.869 moles = 1
This means for each mol O we have 1.5 moles C, 5 moles H and 1 mol N
OR
For every 2 O atoms we have 3 C atoms, 10 H atoms and 2 N atoms
The empirical formula is C3H10N2O2
