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    Combustion Analysis of Toluene

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    Combustion analysis of toluene, a compound containing carbon and hydrogen only, gives 5.86 mg carbon dioxide and 1.37 mg of water. What is its empirical formula?

    Menthol is composed of carbon, hydrogen and oxygen. A 1.005 g sample is combusted, producing 0.2829 g carbon dioxide and 0.1159 g of water. If the molar mass of menthol is 156 g/mol, what are the empirical and molecular formulas of menthol?

    Please provide detailed explanation.

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    https://brainmass.com/chemistry/physical-chemistry/combustion-analysis-toluene-516097

    Solution Preview

    SOLUTION:
    Steps:
    1. Solve for the mass of carbon from the given the mass of carbon dioxide.

    mass of CO2 = 5.86 mg = 0.00586 g

    mass of C = 0.00586 g CO2 x (12 g C/44 g CO2)

    mass of C = 1.598 x 10-3 g

    Note: 1 mole CO2 contains 1 mole C, hence, the mass of C relative to the mass of CO2 is (12 g C/44 g CO2).

    2. Solve for the mass of hydrogen from the given the mass of water.

    mass of H2O = 1.37 mg = 0.00137 g

    mass of H = 0.00137 g H2O x (2 g H/18 g H2O)

    mass of H = 1.522 x 10-4 g

    Note: 1 mole H2O contains 2 moles H, hence, the mass of H relative to the mass of H2O is (2 g H/18 g H2O).

    3. Determine the empirical formula of the compound.

    Mnemonics in finding the empirical formula:
    *Mass to Mole*
    *Divide by small*
    *Multiply 'till whole*

    Elements in the ...

    Solution Summary

    The expert examines a combustion analysis of toluene. The molar mass of menthol is analyzed.

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