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28 May, 16:09

2.088 g sample of a compound containing only carbon, hydrogen, and oxygen is burned in an excess of dioxygen, producing 4.746 g CO 2 and 1.943 g H 2O. What is the empirical formula of the compound?

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  1. 28 May, 16:14
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    The empirical formula is C3H6O

    Explanation:

    Step 1: Data given

    Mass of the sample = 2.088 grams

    The mass contains carbon, hydrogen, and oxygen

    Mass of CO2 produced = 4.746 grams

    Mass of H2O produced = 1.943 grams

    Molar mass of CO2 = 44.01 g/mol

    Molar mass of H2O = 18.02 g/mol

    Atomic mass of C = 12.01 g/mol

    Atomic mass of H = 1.01 g/mol

    Atomic mass of O = 16.0 g/mol

    Step 2: Calculate moles CO2

    Moles CO2 = mass CO2 / molar mass CO2

    Moles CO2 = 4.746 grams / 44.01 g/mol

    Moles CO2 = 0.1078 moles

    Step 3: Calculate moles C

    For 1 mol CO2 we have 1 mol C

    For 0.1078 moles CO2 we'll have 0.1078 moles C

    Step 4: Calculate mass C

    Mass C: moles C * atomic mass C

    Mass C: 0.1078 moles * 12.01 g/mol

    Mass C = 1.295 grams

    Step 5: Calculate moles H2O

    Moles H2O = 1.943 grams / 18.02 g/mol

    Moles H2O = 0.1078 moles

    Step 6: Calculate moles H

    For 1 mol H2O we'll have 2 moles H

    For 0.1023 moles H2O we'll have 2*0.1078 = 0.2156 moles H

    Step 7: Calculate mass H

    Mass H = 0.2046 moles * 1.01 g/mol

    Mass H = 0.218 grams

    Step 8: Calculate mass O

    Mass O = 2.088 grams - 1.295 grams - 0.218 grams

    Mass O = 0.575 grams

    Step 9: Calculate moles O

    Moles O = 0.575 grams / 16.0 g/mol

    Moles O = 0.0359 moles

    Step 10: Calculate the mol ratio

    We divide by the smallest amount of moles

    C: 0.1078 moles / 0.0359 moles = 3

    H: 0.2156 moles / 0.0359 moles = 6

    O: 0.0359 moles / 0.0359 moles = 1

    The empirical formula is C3H6O
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