Question

An organic compound contains only carbon, hydrogen, and chlorine.

Combustion of a 1.50-g sample produces 3.52 g of CO2. In a separate reaction, a 1.00-g
sample reacts with Ag+, producing 1.27 g of AgCl. What is the compound’s empirical
formula?

Answer 1

The empirical formula is C6H5Cl

Step-by-step 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