Answer:
The empirical formula is C4H5ClO2
Step-by-step explanation:
Step 1: Data given
Mass of the sample = 40.10 grams
Mass of CO2 produced = 58.57 grams
Mass of H2O produced = 14.98 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
In experiment 2, mass = 75.00 grams and 22.06 grams is Cl
Step 2: Calculate moles CO2
Moles CO2 = mass CO2 / molar mass CO2
Moles CO2 = 58.57 grams / 44.01 g/mol
Moles CO2 = 1.33 moles
Step 3: Calculate moles C
For 1 mol CO2 we have 1 mol CO2
For 1.33 moles CO2 we have 1.33 moles C
Step 4: Calculate mass C
Mass C = 1.33 grams * 12.01 g/mol
Mass C = 15.97 grams
Step 5: Calculate moles H2O
Moles H2O= 14.98 grams /18.02 g/mol
Moles H2O = 0.831 moles
Step 6: Calculate moles H
For 1mol H2O we have 2 moles H
For 0.831 moles H2O we have 2*0.831 = 1.662 moles H
Step7: Calculate mass H
Mass H = 1.662 moles * 1.01 g/mol
Mass H = 1.68 grams
Step 8: Calculate mass %
%C = (15.97 grams / 40.10) * 100 %
%C = 39.8 %
%H = (1.68 / 40.10 ) *100%
%H = 4.2 %
%Cl = (22.06 / 75.00 ) * 100%
%Cl = 29.4 %
%O = 100 % - 39.8% - 4.2 % - 29.4 %
%O = 26.6 %
Step 9: Calculate moles in compound
We assume the compound has a mass of 100 grams
Mass C = 39.8 grams
MAss H = 4.2 grams
MAss Cl = 29.4 grams
Mass O = 26.6 grams
Moles C = 39.8 grams / 12.01 g/mol
Moles C = 3.314 moles
Moles H = 4.2 moles / 1.01 g/mol
Moles H = 4.158 moles
Moles Cl =29.4 grams / 35.45 g/mol
Moles Cl = 0.829 moles
Moles O = 26.6 grams / 16.0 g/mol
Moles O = 1.663 moles
Step 10: calculate the mol ratio
We divide by the smallest amount of moles
C: 3.314 moles / 0.829 moles = 4
H: 4.158 moles / 0.829 moles = 5
Cl: 0.829 moles /0.829 moles = 1
O: 1.663 moles / 0.829 moles = 2
This means For each Cl atom we have 4 C atoms, 5 H atoms and 2 O atoms
The empirical formula is C4H5ClO2