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7 July, 03:15

A 5.701 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 12.28 grams of CO2 and 5.029 grams of H2O are produced. In a separate experiment, the molar mass is found to be 102.1 g/mol. Determine the empirical formula and the molecular formula of the organic compound.

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  1. 7 July, 03:22
    0
    The empirical formula is the same as the molecular formula = C5H1002

    Explanation:

    Step 1: Data given

    Mass of the compound = 5.701 grams

    Mass of CO2 = 12.28 grams

    Mass of H2O = 5.029 grams

    Molar mass = 102.1 g/mol

    Molar mass of CO2 = 44.01 g/mol

    Molar mass of H2O = 18.02 g/mol

    Step 2: Calculate moles CO2

    Moles = mass / molar mass

    Moles CO2 = mass CO2 / molar mass CO2

    Moles CO2 = 12.28 grams / 44.01 g/mol

    Moles CO2 = 0.279 moles

    Step 3: Calculate moles C

    In 1 mol CO2 we have 1 mol C

    In 0.279 moles CO2 we have 0.279 moles C

    Step 4: Calculate mass C

    Mass C = 0.279 moles * 12.01 g/mol

    Mass C = 3.35 grams

    Step 5: Calculate moles H2O

    Moles H2O = 5.029 grams / 18.02 g/mol

    Moles H2O = 0.279 moles

    Step 6: Calculate moles H

    In 1 mol H2O we have 2 moles H

    In 0.279 moles we have 2 * 0.279 = 0.558 moles H

    Step 7: Calculate mass H

    Mass H = 0.558 moles * 1.01 g/mol

    Mass H = 0.564 grams

    Step 8: Calculate mass O

    Mass O = mass compound - mass C - mass H

    Mass O = 5.701 grams - 3.35 grams - 0.564 grams

    Mass O = 1.787 grams

    Step 9: Calculate moles O

    Moles O = 1.787 grams / 16.0 g/mol

    Moles O = 0.1117 moles

    Step 10: Calculate the mol ratio

    We divide by the smallest amount of moles

    C: 0.279 moles / 0.1117 moles = 2.50

    H: 0.558 moles / 0.1117 moles = 5.0

    O: 0.1117 moles / 0.1117 moles = 1

    For every 1 O atom we have 2.5 C atoms and 5 H atoms

    Or for every 2 O atoms we have 5 C atoms and 10 H atoms

    The empirical formula is C5H1002

    The molecular mass of this formula is 102.1 g/mol

    The empirical formula is the same as the molecular formula = C5H1002
  2. 7 July, 03:25
    0
    C10H5O2

    Explanation:

    First you want to find the number of mols of CO2 and H2O.

    mols of CO2 = 12.28 g/44.01gmol^-1

    = 0.2790 mols C

    mols of H2O = 5.029 g/18.02gmol^-1

    = 2 (0.2790 mols)

    = 0.5580 mols of H2

    then you want to find the samples of each of the compounds

    Carbon: 0.2790 mols * 12.01 g/mol = 3.351 g

    Hydrogen : 0.5580 mols * 1.01 g/mol = 0.564 g

    subtract those values from total sample to find the amount of oxygen in the compounds

    5.701g - 3.351g - 0.564g = 1.786 g of oxygen in sample

    mols of O: 1.786g/16 gmol^-1 = 0.1116 mols

    Divide all ratios by the smallest ration

    C:H:O

    0.5580/0.1116: 0.2790/0.1116 : 0.1116/0.1116

    (5:2.5:1) 2

    10:5:2

    C10H5O2
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