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26 August, 16:28

1. Ethylene glycol, commonly used in antifreezes, contains only carbon, hydrogen, and oxygen. When a sample of it is combusted in air, a 23.46 g sample produces 20.42 g of water and 33.27 g of carbon dioxide. Determine its empirical formula. If it has a molar mass of 62.0 g, what is its molecular formula

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  1. 26 August, 16:57
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    Empirical formula = CH3O

    Molecular formula = C2H6O2

    Explanation:

    Step 1: Data given

    Mass of the sample = 23.46 grams

    Mass of H2O = 20.42 grams

    Molar mass of H2O = 18.02 g/mol

    Mass of CO2 = 33.27 grams

    Molar mass of CO2 = 44.01 G:mol

    Atomic mass of C = 12.01 g/mol

    Atomic mass of O = 16.0 g/mol

    Atomic mass of H = 1.01 g/mol

    Molar mass of the compound = 62.0 g/mol

    Step 2: Calculate moles of H2O

    Moles H2O = 20.42 grams / 18.02 g/mol

    Moles H2O = 1.133 moles

    Step 3: Calculate moles H

    For 1 mol H2O we have 2 moles H

    For 1.133 moles H2O we have 2 * 1.133 = 2.266 moles H

    Step 4: Calculate mass H

    Mass H = 2.266 moles * 1.01 g/mol

    Mass H = 2.29 grams

    Step 5: Calculate moles CO2

    Moles CO2 = 33.27 grams / 44.01 g/mol

    Moles CO2 = 0.7560 moles

    Step 6: Calculate moles C

    For 1 mol CO2 we have 1 mol C

    For 0.7560 moles CO2 we have 0.7560 moles C

    Step 7: Calculate mass C

    Mass C = 0.7560 moles * 12.01 g/mol

    Mass C = 9.08 grams

    Step 8: Calculate mass O

    Mass O = 23.46 grams - 9.08 grams - 2.29 grams

    Mass O = 12.09 grams

    Step 9: Calculate moles O

    Moles O = 12.09 grams / 16.0 g/moles

    Moles O = 0.7556

    Step 10: Calculate mol ratio

    We divide by the smallest amount of moles

    C: 0.7560 moles / 0.7556 moles = 1

    H: 2.266 moles / 0.7556 moles = 3

    O; 0.7556 / 0.7560 moles = 1

    This means for 1 mol C we have 3 moles H and 1 mol O

    The empirical formula is CH3O

    Step 11: Calculate the molecular formula

    The molar mass of the empirical formula is 31 g/mol

    Step 11: Calculate molecular formula

    We have to multiply the empirical formula by n

    n = 62.0 g/mol / 31g/mol = 2

    Molecular formula = 2 * (CH3O)

    Molecular formula = C2H6O2
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