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14 March, 08:02

Combustion analysis of a hydrocarbon produced 33.01 g g CO2 C O 2 and 10.15 g g H2O H 2 O. Part A Calculate the empirical formula of the hydrocarbon.

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  1. 14 March, 09:31
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    The empirical formula is C2H3

    Explanation:

    Step 1: Data given

    Mass of CO2 = 33.01 grams

    Molar mass of CO2 = 44.01 g/mol

    Mass of H2O = 10.15 grams

    Molar mass of H2O = 18.02 g/mol

    Step 2: Calculate moles CO2

    Moles CO2 = mass CO2 / molar mass CO2

    Moles CO2 = 33.01 grams / 44.01 g/mol

    Moles CO2 = 0.7501 moles

    Step 3: Calculate moles C

    In 1 mol CO2 we have 1 mol C

    In 0.7501 moles CO2 we have 0.7501 moles C

    Step 4: Calculate moles H2O

    Moles H2O = 10.15 grams / 18.02 g/mol

    Moles H2O = 0.5633 moles

    Step 5 Calculate moles H

    In 1 mol H2O we have 2 moles H

    In 0.5633 moles H2O we have 2*0.5633 moles = 1.1266 moles

    Step 6: Calculate the mol ratio

    We divide by the smallest amount of moles

    C: 0.7501 moles / 0.7501 moles = 1

    H: 1.1266 moles / 0.7501 moles = 1.5

    For each C atom we have 1.5 H atoms OR for each 2 C atoms we have 3 H atoms

    The empirical formula is C2H3
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