Answer:
The empirical formula is C6H5Cl
Explanation:
Step 1: Data given
Mass of the sample = 1.50 grams
Mass of CO2 = 3.52 grams
Molar mass CO2 = 44.01 g/mol
In a separate reaction, a 1.00-g sample reacts with Ag+, producing 1.27 g of AgCl.
Mass of AgCl = 1.00 grams
Molar mass AgCl = 143.32 g/mol
Molar mass Ag = 107.87 g/mol
Step 2: Calculate moles CO2
Moles CO2 = mass CO2 / molar mass CO2
Moles CO2 = 3.52 grams / 44.01 g/mol
Moles CO2 = 0.0800 moles
Step 3: Calculate moles C
For 1 mol CO2 we have 1 mol CO2
For 0.0800 moles CO2 we have 0.0800 moles C
Step 4: Calculate mass C
Mass C = 0.0800 moles C * 12.01 g/mol
Mass C = 0.961 grams
Step 5: Calculate moles AgCl
Moles AgCl = 1.27 grams / 143.32 g/mol
Moles AgCl = 0.00886 moles
Step 6: Calculate moles Ag+
For 1 mol AgCl we have 1 mol Cl
For 0.00886 moles AgCl we have 0.00886 moles Cl
Step 7: Calculate mass Cl
Mass Cl = 0.00886 moles *35.45 g/mol
Mass Cl = 0.314 grams
Step 8: Calculate mass %
% C = (0.961 grams / 1.50 grams) * 100 %
% C = 64.1 %
% Cl = (0.314 grams / 1.00 grams) *100%
% Cl = 31.4 %
Step 9: Calculate mass % of H
Mass % H = 100 % - 64.1 % - 31.4 %
Mass % H = 4.5 %
Step 10: Calculate the number of moles in the compound
moles C = 64.1 grams / 12.01 g/mol
Moles C = 5.34 moles
Moles Cl = 31.4 grams / 35.45 g/mol
Moles Cl = 0.886 moles
Moles H = 4.5 grams / 1.01 g/mol
Moles H = 4.46 moles
Step 11: Calculate mol ratio
We divide by the smallest amount of moles
C: 5.34 moles / 0.886 moles = 6
Cl: 0.886 moles / 0.886 moles = 1
H: 4.46 moles / 0.886 moles = 5
This means for 1 Cl atom we have 6 C atoms and 5 H atoms
The empirical formula is C6H5Cl
