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30 April, 13:41

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?

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Answers (1)
  1. 30 April, 13:48
    0
    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
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