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21 November, 05:56

The combustion of 40.10 g of a compound which contains only C, H, Cl and O yields 58.57 g of CO2 and 14.98 g of H2O. Another sample of the compound with a mass of 75.00 g is found to contain 22.06 g of Cl. What is the empirical formula of the compound?

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  1. 21 November, 06:24
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    The empirical formula is C4H5ClO2

    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
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